Anomalies in (2+1)D Fermionic Topological Phases and (3+1)D Path Integral State Sums for Fermionic SPTs

Abstract

Given a (2+1)D fermionic topological order and a symmetry fractionalization class for a global symmetry group G, we show how to construct a (3+1)D topologically invariant path integral for a fermionic G symmetry-protected topological state (G-FSPT), in terms of an exact combinatorial state sum. This provides a general way to compute anomalies in (2+1)D fermionic symmetry-enriched topological states of matter. Equivalently, our construction provides an exact (3+1)D combinatorial state sum for a path integral of any FSPT that admits a symmetry-preserving gapped boundary, which includes the (3+1)D topological insulators and superconductors in class AII, AIII, DIII, and CII that arise in the free fermion classification. Our construction proceeds by using the fermionic topological order (characterized by a super-modular tensor category) and symmetry fractionalization data to define a (3+1)D path integral for a bosonic theory that hosts a non-trivial emergent fermionic particle, and then condensing the fermion by summing over closed 3-form $\mathbb Z_2$ background gauge fields. This procedure involves a number of non-trivial higher-form anomalies associated with Fermi statistics and fractional quantum numbers that need to be appropriately canceled off with a Grassmann integral that depends on a generalized spin structure. We show how our construction reproduces the $\mathbb Z_{16}$ anomaly indicator for time-reversal symmetric topological superconductors with ${\textbf{T}}^2 = (-1)^F$. Mathematically, with some standard technical assumptions, this implies that our construction gives a combinatorial state sum on a triangulated 4-manifold that can distinguish all $\mathbb Z_{16}$ $\textrm{Pin}^+$ smooth bordism classes. As such, it contains the topological information encoded in the eta invariant of the pin$^+$ Dirac operator, thus giving an example of a state sum TQFT that can distinguish exotic smooth structure.

Appendices

Topological Preliminaries

Here we will provide some mathematical background necessary to understand various technical details and geometric content of this paper, and particularly for derivations in Appendix B, E, G, J. We start by reviewing chains/cochains on a triangulation and Poincaré duality. Then we discuss how branching structures connect geometric to algebraic constructions by inducing geometrical structures like $\cup _i$ products, Stiefel-Whitney classes, (s)pin structures.

Throughout this appendix we will refer to $M^d$ as a d-manifold equipped with a triangulation, as opposed to the notation $(M^d,T)$ we used in the main text.

1.1 Triangulations and cellulations of manifolds

Let $M^d$ be a triangulated d-manifold. We denote a k-simplex of M as $\langle r_0 \cdots r_k \rangle$, with $r_i$ labeling vertices. We will often abuse language in discussing these k-simplices, as it could be possible that multiple $r_i$ are identified on a single k-simplex. Most precisely, we will be working in the setting of PL cellulations of d-dimensional manifolds, which means that for every k-simplex $\langle r_0 \cdots r_k \rangle$, $\textrm{Link}(\langle r_0 \cdots r_k \rangle )$ is homeomorphic to a $(d-k-1)$-sphere. Here $\textrm{Link}(\langle r_0 \cdots r_k \rangle )$ is the collection of $(d-k-1)$-simplices $\langle y_0 \cdots y_{d-k-1} \rangle$ such that the vertices $\{r_0 \cdots r_k,y_0 \cdots y_{d-k-1}\}$ form a d-simplex. This collection of simplices can be visualized as ‘linking’ with the original k-simplex (see Fig. 12 for an example). This is the setting for which Pachner’s theorem about equivalent PL cellulations of manifolds applies. A reader interested in PL-manifold theory might find [92] helpful, although we will not make much use of it here.

Fig. 12

$\textrm{Link}(s)$ for a 1-simplex s in $d=3$. In general $\textrm{Link}(s_{d-k})$ of a $(d-k)$-simplex is a $(k-1)$-sphere

We note that often a triangulation is defined such that all subsimplices consist of distinct vertices and for which any two k-simplices can only share at most one $(k-1)$-simplex. This is too restrictive for our purposes and we instead use PL-cellulations so we can deal with more compact cellulations. In any case, passing to a barycentric subdivision (see Sect. E 2 for a definition) of a PL-cellulation will give a triangulation in the standard sense and one can pass to the barycentric subdivision via Pachner moves (again see Sect. E 2). Since our state sum is invariant under Pachner moves, this abuse of language will not be an issue for us.

Often we will refer to k-simplices as $\langle 0 \cdots k \rangle$ for ease.

1.1.1 Chains and cochains

To set notation, here we briefly review the concepts of $\mathbb Z_2$-valued chains and cochains.

For $0 \le k \le d$, the set of ‘k-chains’ on M is denoted $C_k(M,\mathbb Z_2)$. Each $C_k(M,\mathbb Z_2)$ is a $\mathbb Z_2$ vector space with one basis element per k-simplex, spanned by basis vectors $\{| r_0 \cdots r_k \rangle \}$. So chains $c \in C_k(M,\mathbb Z_2)$ are in bijection with subsets of the k-simplices of the triangulation. We write

$$\begin{aligned} c = \sum _{\langle 0 \cdots k \rangle \in M} c_{\langle 0 \cdots k \rangle } | 0 \cdots k \rangle , \;\;\; c_{\langle 0 \cdots k \rangle } \in \{0,1\}. \end{aligned}$$

(A1)

The sum of two chains $c+d$ returns a sum over k-simplices that are part of exactly one (not both) of c or d.

The ‘boundary’ operator $\partial$ from $C_k(M,\mathbb Z_2) \rightarrow C_{k-1}(M,\mathbb Z_2)$ is a linear operator determined by its action on the basis elements $| 0 \cdots k \rangle$, which is given as

$$\begin{aligned} \partial | 0 \cdots k \rangle = \sum _{i=0}^k | 0 \cdots \hat{i} \cdots k \rangle \end{aligned}$$

(A2)

where $\hat{i}$ means skipping over i in the vertex collection. This corresponds exactly to the $(k-1)$-simplices on the boundary of $\langle 0 \cdots k \rangle$. So $\partial c$ consists of the $(k-1)$-simplices that are contained in an odd number of k-simplices of c, and that set can be identified with the ‘mod 2’ boundary of the k-simplices of c. We say that c is a ‘closed’ chain, or a ‘cycle’, if $\partial c \equiv 0$ is the zero vector. The set of closed k-chains is denoted $Z_k(M,\mathbb Z_2)$. We say that c is a ‘boundary’ if $c = \partial b$ for some $b \in C_{k+1}(M,\mathbb Z_2)$, and denote $B_k(M,\mathbb Z_2)$ as the set of boundaries. Every boundary is closed, so that $\partial \partial c = 0$ for any chain.

Now, we describe cochains on M, denoted by $C^k(M,\mathbb Z_2)$. These are all linear functionals from $C_k(M,\mathbb Z_2) \rightarrow \mathbb Z_2$. In particular, we can write a cochain $\alpha \in C^k(M,\mathbb Z_2)$ in terms of the dual basis vectors $\langle 0 \cdots k |$, so

$$\begin{aligned} \alpha = \sum _{\langle 0 \cdots k \rangle \in M} \alpha _{\langle 0 \cdots k \rangle } \langle 0 \cdots k |, \;\;\; \alpha _{\langle 0 \cdots k \rangle } \in \{0,1\} \end{aligned}$$

(A3)

and equivalently $\alpha$ can be thought of as a collection of k-simplices.

The ‘coboundary’ operation $\delta$, which is a linear function $\delta : C^k(M,\mathbb Z_2) \rightarrow C^{k+1} (M,\mathbb Z_2)$, is defined as the adjoint of $\partial$. Specifically,

$$\begin{aligned} \delta \alpha = \sum _{\langle \tilde{r}_0 \cdots \tilde{r}_{k+1} \rangle \in M} \alpha (\partial | \tilde{r}_0 \cdots \tilde{r}_{k+1} \rangle ) \langle \tilde{r}_0 \cdots \tilde{r}_{k+1} | \end{aligned}$$

(A4)

which can be interpreted as the set of $(k+1)$-simplices whose boundaries give an odd number of k-simplices of $\alpha$. We say $\alpha$ is ‘closed’, or a ‘cocycle’, if $\delta \alpha = 0$ and that $\alpha$ is a ‘coboundary’ if $\alpha = \delta \beta$ for some $\beta \in C^{k-1}(M,\mathbb Z_2)$. The set of closed and coboundary k-cochains are denoted $Z^k(M,\mathbb Z_2)$ and $B^k(M,\mathbb Z_2)$ respectively.

The (mod 2) ‘homology’ groups of M are $H_k(M,\mathbb Z_2) := Z_k(M,\mathbb Z_2) / B_k(M,\mathbb Z_2)$. The (mod 2) ‘cohomology’ groups are $H^k(M,\mathbb Z_2) := Z^k(M,\mathbb Z_2) / B^k(M,\mathbb Z_2)$.

1.1.2 Poincaré Duality and the dual cellulation

We denote by $M^\vee$ the dual cellulation of M. The dual of a triangulation will in general not be a triangulation, but a cell complex (see, e.g., [93]). Every k-simplex $\langle r_0 \cdots r_k \rangle$ is dual to a $(d-k)$-cell $P_{\langle r_0 \cdots r_k \rangle }$, as shown for the $d=2$ case in Fig. 13. A precise statement of this construction is not important for our purposes, but can be done by gluing together pieces from the barycentric subdivision (definition in Appendix E 2 a ii) of each d-simplex.

Fig. 13

Dual cellulation (blue) of a triangulation (red) in $d=2$

We can also consider chains of the dual cellulation as formal linear combinations of dual k-cells, which give the elements of a vector space $C_k(M^\vee ,\mathbb Z_2)$ . An element $c \in C_k(M^\vee ,\mathbb Z_2)$ can be written as

$$\begin{aligned} c = \sum _{P_{0 \cdots {d-k}}} c_{P_{0 \cdots {d-k}}} | P_{0 \cdots {d-k}} \rangle \end{aligned}$$

(A5)

There is also a boundary operator $\partial$ acting on $M^\vee$ which gives the boundaries of the dual cells. The boundary of a dual k-cell $P_{0 \cdots {d-k}}$ can be expressed in terms of the cochain $\langle 0 \cdots d-k | \in C^{d-k}(M,\mathbb Z_2)$ of the original triangulation. In particular we have

$$\begin{aligned} \partial | P_{0 \cdots {d-k}} \rangle = \sum _{\langle \tilde{r}_0 \cdots \tilde{r}_{d-k+1} \rangle } \big ((\delta \langle 0 \cdots {d-k} |)| \tilde{r}_0 \cdots \tilde{r}_{d-k+1} \rangle \big ) | P_{\tilde{r}_0 \cdots \tilde{r}_{d-k+1}} \rangle . \end{aligned}$$

(A6)

That is, the boundary of a dual k-cell consists of all the $(k-1)$-cells whose dual $(d-k+1)$-simplices’ boundaries contain the k-cell’s corresponding dual $(d-k)$ simplex. Using the boundary operator, we can define closed dual chains $Z_k(M^\vee ,\mathbb Z_2)$ and boundary dual chains $B_k(M^\vee ,\mathbb Z_2)$ as usual.

k-cochains on $M^\vee$, denoted by $C^k(M^\vee ,\mathbb Z_2)$, are linear functionals $C_k(M^\vee ,\mathbb Z_2) \rightarrow \mathbb Z_2$. There’s an analogous formula for coboundaries,

$$\begin{aligned} \delta \langle P_{0 \cdots {d-k}} | = \sum _{\langle \tilde{r}_0 \cdots \tilde{r}_{d-k+1} \rangle } \big (\langle 0 \cdots {d-k} | \partial | \tilde{r}_0 \cdots \tilde{r}_{d-k+1} \rangle \big ) \langle P_{\tilde{r}_0 \cdots \tilde{r}_{d-k+1}} | \end{aligned}$$

(A7)

Note that this means we can interchangeably refer to chains on M and cochains on $M^\vee$ and vice-versa, since the boundary, coboundary operators on $C_k(M,\mathbb Z_2),C^k(M,\mathbb Z_2)$ give the same action as the coboundary, boundary operators on $C^{d-k}(M^\vee ,\mathbb Z_2),C_{d-k} (M^\vee ,\mathbb Z_2)$. This is the chain-level statement of Poincaré duality. And also we will have that $Z_k(M,\mathbb Z_2) \cong Z^{d-k}(M^\vee ,\mathbb Z_2)$ and $Z^k(M,\mathbb Z_2) \cong Z_{d-k}(M^\vee ,\mathbb Z_2)$, and the same statements for the boundaries.

See Fig. 14 for examples of closed chains and cochains on a two-dimensional triangulation and its dual.

Fig. 14

(Left) A closed cochain in $C^1(M,\mathbb Z_2)$ (red) dual to a closed cochain in $C_1(M^\vee ,\mathbb Z_2)$ (green). (Right) A closed chain in $C_1(M,\mathbb Z_2)$ (yellow) dual to a closed cochain on $C^1(M^\vee ,\mathbb Z_2)$ (blue). Black lines are 1-simplices of M, and green/blue lines are 1-cells of $M^{\vee }$

1.2 Branching structures, $\cup _i$ products, and induced geometrical structures

A branching structure on a triangulation is a local ordering of vertices, which can be specified by an arrow on each 1-simplex $\langle ij \rangle$, such there are no closed loops on any 2-simplices. One can show this defines a total ordering of vertices on every k-simplex $\langle r_0 \cdots r_k \rangle$.

Below we will review how the branching structure gives definitions of cochain-level product operations and underlying geometric structures associated with them. In particular, the branching structure canonically defines a frame of vector fields in the triangulation. These vector fields can be used to give a geometric account of the cochain-level formulas of the $\cup$ and $\cup _i$ products. Next, via so-called obstruction theory, we review how the frame of vector fields define the Stiefel-Whitney classes $w_1$ and $w_2$ and how they are related to the theory of (s)pin structures. Therefore the branching structure gives a bridge between various algebraic formulas and the concrete geometry that underlies the path integral.

For a discussion of spin structures similar to one used here, the reader may find the description in textbook [94] useful and related. We also note that [59] discusses many of these ideas and was an instrumental resource in formulating [49], which much of this subsection is summarizing.

1.2.1 $\cup _i$ products

The $\cup$ product is a function

$$\begin{aligned} - \cup -: C^k(M,\mathbb Z_2) \times C^\ell (M,\mathbb Z_2) \rightarrow C^{k + \ell }(M,\mathbb Z_2) \end{aligned}$$

whose explicit formula for $\alpha \in C^k(M,\mathbb Z_2)$ and $\beta \in C^\ell (M,\mathbb Z_2)$ is

$$\begin{aligned} (\alpha \cup \beta )(0 \cdots {k+\ell }) = \alpha (0 \cdots k) \beta (k \cdots k+\ell ) \end{aligned}$$

(A8)

Note that the branching structure is needed in order to give the vertices $r_0 \rightarrow \cdots \rightarrow r_{k+\ell }$ an ordering to unambiguously define the formula. There is the important Leibniz rule

$$\begin{aligned} \delta (\alpha \cup \beta ) = (\delta \alpha ) \cup \beta + \alpha \cup (\delta \beta ) \end{aligned}$$

(A9)

that is true on the cochain level.

The $\cup$ product can be interpreted geometrically in the Poincaré dual picture. It is well known that, for closed $\alpha ,\beta$, the $\cup$ product induces a cohomology operation $H^k(M,\mathbb Z_2) \times H^\ell (M,\mathbb Z_2) \rightarrow H^{k+\ell }(M,\mathbb Z_2)$ that can be interpreted as the intersection product of the $(d-k)$-manifold $\alpha ^\vee$ and the $(d-\ell )$-manifold $\beta ^\vee$ dual to $\alpha ,\beta$. In particular, the dual of $\alpha \cup \beta$ is homologous to the intersection $\alpha ^\vee \cap \beta ^\vee$. On the cochain level, this means that for any cohomologous pairs of closed $\alpha , \alpha + \delta A$ and $\beta , \beta + \delta B$, we will have $(\alpha + \delta A) \cup (\beta + \delta B) = \alpha \cup \beta + \delta (A \cup \delta B + \alpha \cup B + A \cup \beta )$, which means that the cup product induces a product on cohomology classes. Even though the cochain level formulas all depend on a branching structure, on the cohomology level everything is independent of branching structure.

While this intersection property is well-known at the level of homology and cohomology, there is a way to directly visualize it on the cochain level. In particular, given cochains $\alpha \in C^k(M,\mathbb Z_2)$ and $\beta \in C^\ell (M,\mathbb Z_2)$ and their dual chains $\alpha ^\vee \in C_{d-k}(M^\vee ,\mathbb Z_2)$ and $\beta ^\vee \in C_{d-\ell }(M^\vee ,\mathbb Z_2)$, the dual chain $(\alpha \cup \beta )^\vee$ can be thought of as the intersection of $\alpha ^\vee$ with a shifted version $\beta ^\vee _{\text {shifted}}$ of $\beta ^\vee$, where the shifting vector is determined by the branching structure.¹⁹ We illustrate this in Fig. 15 for the simple case of a 2-manifold with $\alpha ,\beta \in C^1(M,\mathbb Z_2)$. The vector field used to define the shifting is discussed in the subsequent subsection.

Fig. 15

The cup product $(\alpha \cup \beta ) = \alpha (01)\beta (12)$ is interpreted as shifting the cells dual to $\beta$ by a vector field (red arrows). The only potential intersection between dual 1-cells is between $\alpha (01)$ and $\beta (12)$

The $\cup _i$ products for $i > 0$ are product operations

$$\begin{aligned} - \cup _i -: C^k(M,\mathbb Z_2) \times C^\ell (M,\mathbb Z_2) \rightarrow C^{k + \ell - i}(M,\mathbb Z_2) \end{aligned}$$

whose explicit formula can be written

$$\begin{aligned}&(\alpha \cup _i \beta )(0 \cdots k+\ell -i)\nonumber \\&\quad = \sum _{0 \le j_0< \cdots < j_i \le k+\ell -i} \alpha (0 \rightarrow j_0, j_1 \rightarrow j_2, \dots ) \beta (j_0 \rightarrow j_1, j_2 \rightarrow j_3, \dots ) \end{aligned}$$

(A10)

where we use $\{0,1,2,\dots \}$ to refer to the vertices of the $(k+\ell -i)$ simplex, the notation $m \rightarrow n$ refers to all vertices $\{m, m+1, \dots , n-1, n\}$, and we restrict our attention to those combinations $\{j_0,\dots ,j_i\}$ such that the sets $\{0 \rightarrow j_0, j_1 \rightarrow j_2,\dots \}$ and $\{j_0 \rightarrow j_1, j_2 \rightarrow j_3,\dots \}$ consist of exactly $(k+1)$ and $(\ell + 1)$ elements respectively. The $\cup _i$ products also have a generalized Leibniz rule

$$\begin{aligned} \delta (\alpha \cup _i \beta ) = (\delta \alpha ) \cup _i \beta + \alpha \cup _i (\delta \beta ) + \alpha \cup _{i-1} \beta + \beta \cup _{i-1} \alpha \end{aligned}$$

(A11)

that inductively relate $\cup _i$ products to $\cup _{i-1}$ products. From this formula one can check examples that $\alpha \cup _i \beta$ for $i>0$ are generally not even closed for closed $\alpha ,\beta$. Therefore there cannot be a cohomology level interpretation of $\alpha \cup _i \beta$, for general $\alpha ,\beta$. Nevertheless, the operations $\alpha \mapsto \alpha \cup _i \alpha$ are indeed cohomology operations that map cohomology classes to cohomology classes, and define the so-called ‘Steenrod operations’ ${{\textrm{Sq}}}$. These are written as

$$\begin{aligned} \begin{aligned}&{{\textrm{Sq}}}^{k-i}: Z^k(M^d,\mathbb Z_2) \rightarrow Z^{2k-i}(M^d,\mathbb Z_2), {\text { with}} \\&{{\textrm{Sq}}}^{k-i} \alpha := \alpha \cup _i \alpha \end{aligned} \end{aligned}$$

(A12)

because the generalized Leibniz rule above gives $(\alpha + \delta A) \cup _i (\alpha + \delta A) = \alpha \cup _i \alpha + \delta (\alpha \cup _i A+A\cup _i\alpha +A\cup _{i-1}A+A\cup _i\delta A)$. Additionally, changing the branching structure leaves the cohomology class of the Steenrod operations invariant.

The Steenrod operations similarly have an interpretation [57] in terms of Poincaré dual submanifolds (see also [95]). Concretely, we can thicken $\alpha ^\vee$ along some $(k-p)$ generic vector fields and shift it along a $(k-p+1)^{th}$ vector field to produce some $\alpha ^\vee _{\text {thickened,shifted}}$. Then Sq$^p(\alpha )$ is Poincaré dual to the intersection $\alpha ^\vee \cap \alpha ^\vee _{\text {thickened,shifted}}$. Abstractly, this means that Sq$^p(\alpha )$ can be interpreted in terms of pushforwards of Stiefel-Whitney classes of the normal bundles of the embedded $\alpha ^\vee \subset M$; this translates to the concrete statement above via ‘obstuction-theoretic’ definitions of Stiefel-Whitney classes.

While globally the above statement has been known since the 1950’s at the level of cohomology, there is a way to see this locally simplex-by-simplex, directly on the cochain-level. In particular, the cells of $\alpha \cup _i \beta$ can be interpreted in terms of some canonical frame of vector fields defined in the neighborhood of the dual 1-skeleton. In particular, $\alpha \cup _i \beta$ is given by first thickening $\beta ^\vee$ in i directions and shifting it along an $(i+1)^{th}$ direction to produce $\beta ^\vee _{\text {thickened,shifted}}$. Then, $\alpha \cup _i \beta$ on a $(k+\ell -i)$ simplex is obtained by counting the number of intersections of $\alpha ^{\vee } \cap \beta ^\vee _{\text {thickened,shifted}}$. A brief description of the frame of vector fields used to define the shifting is given next.

1.2.2 A frame of vector fields from a branching structure

We now briefly review the explicit vector fields used to define the $\cup$ and $\cup _i$ products, and which we later use to define the first and second Stiefel-Whitney classes, spin structures, and induced spin structures on loops. We refer the reader to [49] for a more thorough treatment.

Specifically, we use the branching structure to define d vector fields, ${\tilde{v}}_1,\cdots , {\tilde{v}}_d$, which are defined in the neighborhood of the dual 1-skeleton.

First, we recall that a d-simplex $\Delta _d$ can be defined as the subset $\Delta _d \subset \mathbb R^{d+1}$ defined in coordinates,

$$\begin{aligned} \Delta _d = \{(x_0 \cdots x_{d}) | x_0 + \cdots x_{d} = 1, {\text { and each }} x_i \ge 0\} \end{aligned}$$

Next, we define a collection of vectors $\vec {b}_1,\dots ,\vec {b}_d \in \mathbb R^{d+1}$,

$$\begin{aligned} \vec {b}_j = \left( 1^j, \frac{1}{2^j}, \frac{1}{3^j},\dots , \frac{1}{(d+1)^j}\right) . \end{aligned}$$

(A13)

The fact that $\vec {b}_0 = (1 \cdots 1)$ and $\vec {b}_j$ together form a Vandermonde matrix leads to some nice algebra that give the formulas for $\cup _i$.

At the center C of each d simplex, we define ${\tilde{v}}_j$ such that

$$\begin{aligned} {\tilde{v}}_j (C) = \vec {b}_j - \vec {b}_j \cdot \vec {b}_0 {\text { for }} j=1,\dots ,d . \end{aligned}$$

(A14)

Thus ${\tilde{v}}_j$ in the center corresponds to $\vec {b}_j$ with the direction along $\vec {b}_0$ projected out. Next, we define ${\tilde{v}}_j$ along the dual 1-skeleton by gradually projecting away the direction normal to the $(d-1)$ simplex as it is approached along the dual 1-skeleton. The precise manner in which this projection gradually occurs is arbitrary and unimportant for our purposes. Once ${\tilde{v}}_j$ is defined along the dual 1-skeleton, it can then be smoothly extended to a neighborhood of the dual 1-skeleton. The Vandermonde structure of the ${\tilde{v}}_j$ will show that any $(d-1)$ of these ${\tilde{v}}_j$ are independent after projecting away that coordinate normal to the $(d-1)$ simplex being approached.

We note that one can continue and use the vector defined at the center of each $(d-1)$-simplex to extend the vector field to be defined along the dual 1-skeleton of the $(d-1)$-simplex. If we inductively continue this process until we approach the 1-simplices, the fact that the coordinates of each $\vec {b}_i$ are decreasing leads to the fact that ${\tilde{v}}_i$ degenerate opposite to the branching structure. This inductive procedure is expected to allow us to define a frame everywhere in the triangulation, although the precise technical construction would involve detailed constructions in PL-manifold theory beyond the scope of this paper.

The vector fields ${\tilde{v}}_j$ along the dual 1-skeleton can be used to give a geometric interpretation of higher cup products $\alpha \cup _i \beta$ when $\beta$ is a $(d-1)$-cochain. Namely, we define $\beta ^\vee _{\text {thickened,shifted}}$ for a collection of 1-cells $\beta ^\vee$ by thickening $\beta ^\vee$ along i directions ${\tilde{v}}_1,\cdots , {\tilde{v}}_i$ and shifting it along ${\tilde{v}}_{i+1}$.

When $\beta$ is an $\ell$-cochain, we can define the thickening and shifting in principle by using ${\tilde{v}}_i$ extended to the dual $\ell$-skeleton, although a prescription with $\ell >1$ for such ${\tilde{v}}_i$ has not yet been given rigorously in the scope of PL-manifold theory. (See [96, 97] for earlier related but more rigorous constructions along these lines.) Ref. [49] gave a prescription to define $\beta ^\vee _{\text {thickened,shifted}}$ by utilizing instead the constant vector field defined by $\vec {b}_j$ for the thickening and shifting directions, which gives a geometric interpretation of higher cup products evaluated on an individual d-simplex.

1.2.3 ± assignments of simplices, $w_1$, and vector fields on the dual 1-skeleton

Given a triangulation, we can assign an arbitrary orientation ± to each d-simplex. As we will explain, this choice, together with the branching structure, determines a closed chain $w_1 \in Z_{d-1}(M,\mathbb Z_2)$ consisting of the $(d-1)$-simplices for which the labeling of the neighboring d-simplices is inconsistent. This chain $w_1$ defines a 1-cochain on the dual cellulation, which with some abuse of notation we also refer to as $w_1$, and which gives a representative of the first Stiefel-Whitney class.

Let us define a collection of vector fields ${v}_i$, for $i = 1,\cdots , d$, such that

$$\begin{aligned} {v}_i = {\tilde{v}}_i, \;\; i = 1,\cdots , d-1 , \end{aligned}$$

(A15)

where ${\tilde{v}}_i$ were defined in the previous subsection. We remarked there that these ${v}_i = {\tilde{v}}_i$ with $i \le d-1$ are all linearly independent near the dual 1-skeleton even after projecting away the components normal to the face. The dth vector field ${v}_d$ is chosen as follows. Away from the center of the d-simplices, ${v}_d$ is chosen to be parallel to the dual 1-skeleton, with the direction chosen so that the determinant of the frame in the local coordinates is positive on a $+$ simplex and negative on a − simplex. Near the centers of the d-simplices, ${v}_d$ is chosen to smoothly interpolate from its values away from the center. See Fig. 16 for a depiction of these vector fields in $d = 2$ and 3. In particular, note that on a ± simplex along the dual 1-skeleton edge $P_{\langle 0 \cdots \hat{i} \cdots d \rangle }$, ${v}_d$ points away from the center of the d-simplex if $\pm (-1)^i = +1$ and points towards the center if $\pm (-1)^i = -1$.

Fig. 16

Illustration of vector field frames with respect to ± assignments of simplices. Note that for each $d=2,3$ $v_d$ will typically be parallel to the dual 1-skeleton except near the barycenter. (Top) Vector fields in $d=2$. (Bottom) $v_1$ is the ‘thickening’ direction that determines the framing of curves on the dual skeleton. For each $i=0,\dots ,3$, $\hat{i}$ refers to the 1-cell opposite to vertex i. For ease of drawing, the $v_2$ drawn here is not the one described in the main text, but is instead related to $v_1,v_3$ by the right-hand rule. The $v_2$ described in the main text will be almost parallel to $v_3$, but still lie on the same side of the thickened sheet as the one drawn

Fig. 17

The dual of $w_1$ is the collection of all $(d-1)$-simplices for which the assignments of vectors $v_d$ along the dual 1-skeleton are inconsistent crossing it. In particular, this is equivalent to the places where the frame of vectors $\{v_1 \dots v_d\}$ become degenerate. (Top) In $d = 2$. (Bottom) In $d = 3$

The direction of ${v}_d$ along $P_{\langle 0 \cdots \hat{i} \cdots d \rangle }$ determines the induced orientation of the boundary $(d-1)$-simplex from the perspective of the d-simplex: the induced orientation is, say, positive (negative) if ${v}_d$ points towards (away from) the center of the d-simplex, i.e. away from (towards) the $(d-1)$-simplex ${\langle 0 \cdots \hat{i} \cdots d \rangle }$.

If the manifold is orientable, then there is always a choice of ± assignments for the d-simplices such that the induced orientations on each $(d-1)$-simplex would be opposites as induced from both its d-simplex neighbors. However if the induced orientations are the same due to an inconsistent ± assignment, which necessarily occurs for non-orientable manifolds, this implies that the frame of vector fields must become degenerate crossing the $(d-1)$-simplex. Since the first Stiefel-Whitney class $w_1$ is the obstruction to defining a frame of independent vectors, we can identify such $(d-1)$-simplices with the dual of $w_1$. This leads us to define the chain $w_1 \in Z_{d-1}(M,\mathbb Z_2)$ from the branching structure and ± assignments as all such $(d-1)$-simplices with the same, ‘inconsistent’, induced orientations. In addition, we can label the $(d-1)$-simplices of $w_1$ as ± depending on their shared induced orientations, positive or negative, from the neighboring d-simplices. In fact, this assignment can show that the dual of $w_1$ is orientable with a consistent labeling of ± given by the shared induced orientation. See Fig. 17 for an illustration.

For an orientable manifold, if we require that the chain representative $w_1 = 0$, then we recover the more familiar situation where the branching structure directly determines the ± assignments of the d-simplices, up to an overall global choice.

We can package the ± assignment all into a single equation as follows. Consider the canonical dual of $w_1$, which we name $w_1^{{\text {canonical}}}$, which is formed by the procedure above if all simplices are chosen as $+$. Then $w_1^{{\text {canonical}}}$ on a $(d-1)$-simplex $\langle r_0 \cdots r_{d-1} \rangle$ can be found in terms of the two d-simplices $\{r_0 \cdots r_{d-1} A\},\{r_0 \cdots r_{d-1} B\}$ that it is a part of, where A and B denote two additional vertices. In particular, let $\ell (A), \ell (B)$ be the coordinate in $\{0,\cdots ,d\}$ in which A, B respectively appear in their d-simplices. Then since the induced orientation on $\hat{i}$ on a $+$ simplex is $-(-1)^i$, we have

$$\begin{aligned} w_1^{{\text {canonical}}}(\langle r_0 \cdots r_{d-1} \rangle ) = -(-1)^{\ell (A) + \ell (B)} \end{aligned}$$

(A16)

Flipping the assignment to − on some region changes matching$\leftrightarrow$mismatching on the boundary of said region. Thus we can obtain a ± assignment consistent with a global orientation by choosing the − region to satisfy

$$\begin{aligned} \partial (- {\text { region}}) = w_1^{{\text {canonical}}} \end{aligned}$$

(A17)

which only has a solution if and only if $w_1$ is a coboundary, i.e. if $w_1$ is trivial. If a solution exists, then there are exactly $2^{\# {\text { of connected components of }}M}=|Z_0(M,\mathbb Z_2)|$ solutions.

1.2.4 Obstruction Theory, $w_2$, and spin structures on a triangulated manifold

Now we are in a position to define the second Stiefel-Whitney class $w_2$ and spin structures. To do this, we need the definition of $w_2(M)$ in obstruction theory. Given $M^d$, we pick any collection of $(d-1)$ ‘generic’ vector fields. Generic means that these fields will be linearly independent everywhere except for a closed codimension-2 submanifold for which the zeros in the determinants defining the linear dependency are linear zeroes in local coordinates. This codimension-2 submanifold is Poincaré dual to $w_2(M)$. Another way to define $w_2(M)$ is to first trivialize the tangent bundle TM on the 1-skeleton (which is possible if and only if M is orientable). On every 2-cell, c, this trivialization can be extended with some number n(c) generic singular points where the the vector fields are not independent. Then, $w_2(M)$ is defined as the cocycle whose values on 2-cells are $w_2(c)=(-1)^{n(c)}$, so is the (mod 2) obstruction to extending the trivialization to the entire 2-skeleton.

A spin structure $\xi$ can be thought of as a ‘fix’ of the vector fields so that all singularities become eliminated (mod 2). This can be implemented by choosing some codimension-1 submanifold whose boundary is $w_2$ and adding a $360^\circ$ ‘twist’ to the vector fields along this submanifold. Equivalently, it is some collection of edges such that twisting the vector fields on those edges allows the trivialization to extend (mod 2) along the 2-skeleton. These are both equivalent to specifying an $\xi \in C^1(M,\mathbb Z_2)$ with $\delta \xi = w_2$.

In our case of a triangulated manifold, the vector fields ${v}_1,\cdots , {v}_d$ defined in the previous sections give a trivialization of TM along the dual 1-skeleton. We can therefore use these vector fields to define a representative of $w_2$ and choice of $\xi$. See Figs. 18 and 19 for an illustration of the vector fields associated to a triangulation and how a spin structure acts to ‘correct’ odd-index singularities in the cases $d = 2$ and $d = 3$.

Given a branched triangulation, there are known explicit chain-level formulas for duals of general Stiefel-Whitney classes [97]. We will see in Appendix B that the Grassmann integral can be used to give an alternate formula for $w_2$ in terms of the winding of the vector fields ${v}_i$ around a $(d-2)$-simplex.

Finally, we state a nice canonical formula for $w_2$ that we use in Sect. J 2 a and that was recently discovered in [88], which expresses the dual of $w_2$ on a branched triangulation in terms of higher cup products. In particular, given a $(d-2)$-simplex $\langle r_0 \cdots r_k \rangle$, $\textrm{Link}(r_0 \cdots r_k)$ comprises of a collection of k $(d-1)$-simplices that form a circle $\Delta _{d-1}^1 \rightarrow \Delta _{d-1}^2 \rightarrow \cdots \rightarrow \Delta _{d-1}^{k} \rightarrow \Delta _{d-1}^{k+1} \equiv \Delta _{d-1}^1$. The formula for $w_2$ is

$$\begin{aligned} (-1)^{w_2(\Delta _{d-2})} = - \prod _{i=1}^k (-1)^{\int _M \varvec{\Delta }_{d-1}^i \cup _{d-2} \varvec{\Delta }_{d-1}^{i+1}} \end{aligned}$$

(A18)

where each $\varvec{\Delta }_{d-1}^i$ is the cochain that’s an indicator on $\Delta _{d-1}^i$. This formula turns out to be equivalent to the winding definition reviewed in Appendix B, which is shown in Appendix B 1 a.

Fig. 18

(Left) Vector fields on the dual 1-skeleton; one is drawn and the other is given by the right-hand-rule. Continuing the vector fields into the 2-skeleton leads to $(d-2)$-simplices (in this case vertices, starred) with odd-index singularities that are dual to $w_2$. (Right) A spin structure $\xi$ is dual to a collection of edges (thick blue ones here) whose boundary is dual to $w_2$. The vector fields get ‘twisted’ by $360^\circ$ at every crossing with $\xi$

Fig. 19

In $d=3$, $w_2$ is dual to a one-dimensional submanifold where a frame of vector fields has an odd-index singularity. A spin structure is dual to a two-dimensional submanifold (blue region) whose boundary is $w_2$ and where the vector fields gain a twist by $360^\circ$ to turn the singularities into even index ones

1.2.5 Induced spin structures on framed curves

Let M be an orientable manifold. Another way to think about spin structures in arbitrary dimensions is how they define induced spin structures on framed loops.

Given a loop embedded in a manifold M, we define a ‘framing’ on it to be a collection of $(d-2)$ vectors of TM independent of the tangent of the loop. Since M is orientable this is equivalent to specifying a trivialization of the normal bundle of the loop in M, since the global orientation will specify the $d^{th}$ vector. We will call the trivialization of the ‘tangent framing’ of the loop because one of the vectors is always tangent to the loop’s direction.

Given a framing of the loop, the induced spin structure on the loop is defined as follows. As we reviewed in the previous section, a spin structure on M is equivalent to a frame of vector fields that become degenerate only at some even-index singularities localized on a codimension-2 submanifold of M. Refer to this framing as the ‘background framing’ of M; in the previous section, this framing was given by the d vector fields ${v}_1,\cdots , {v}_d$. WLOG the curve lies on the 1-skeleton (which for us is the dual 1-skeleton) of M where TM is trivial, so we can refer to these framings as literally a $d \times d$ matrix with positive determinant for every point on the loop. Also WLOG, these framings will be homotopic to maps $[0,1] \rightarrow \textrm{SO}(d)$ since the manifold of positive-determinant matrices $GL^+(d)$ deformation retracts onto $\textrm{SO}(d)$ by the orthogonalization procedure.

Call the background framing $F_{\text {bckd}}: [0,1] \rightarrow \textrm{SO}(d)$ and the tangent framing $F_{\text {tang}}: [0,1] \rightarrow \textrm{SO}(d)$. Note that $F_{\text {bckd}}(0) = F_{\text {bckd}}(1)$ and $F_{\text {tang}}(0) = F_{\text {tang}}(1)$. The relative path of these framings is the function $(F_{\text {bckd}}^{-1} F_{\text {tang}})(t) : [0,1] \rightarrow \textrm{SO}(d)$. Note that if we lift this map to $\tilde{F}: [0,1] \rightarrow \textrm{Spin}(d)$, we will have that $\tilde{F}(0) = \pm \tilde{F}(1)$ since $\textrm{Spin}(d)$ is a double-cover of $\textrm{SO}(d)$.

Now we can define the induced spin structure in terms of this ± sign of $\tilde{F}(0) = \pm \tilde{F}(1)$. If $\tilde{F}(0) = -\tilde{F}(1)$, then the induced spin structure on the loop is the bounding/anti-periodic one. And if $\tilde{F}(0) = +\tilde{F}(1)$, then the induced spin structure is the non-bounding/ periodic one.

In $d = 2$, this induced spin structure can be thought of in terms of the amount of times one of the vectors of the background frame ‘winds’ around the tangent of the loop. In particular, if it winds around an even number of times then the induced spin structure will be the periodic one. If it winds around an odd number of times then the induced spin structure will be the anti-periodic one. In particular, loops that are homologically trivial will always have an induced anti-periodic boundary condition. See Fig. 20.

Fig. 20

Induced spin structures on a curve in $d=2$ can be visualized by the number of rotations the background framing makes with respect to the curve. (Top) An odd number of winds makes antiperiodic boundary conditions. (Bottom) An even number of winds makes periodic boundary conditions

In higher dimensions, a very similar picture of the winding of vector fields can be used. In particular, note that the definition of induced spin structure is invariant under continuous deformations of either of the background or tangent frames. And because $\pi _1(\textrm{SO}(d)) = \mathbb Z_2$, where the two elements can be generated from paths of rotations in a single two-dimensional plane, the relative framing $(F_{\text {bckd}}^{-1} F_{\text {tang}})(t)$ can be deformed to a rotation in a two-dimensional plane. This means we can deform the tangent and background frames to have the functional form:

$$\begin{aligned} (F_{\text {bckd}}^{-1} F_{\text {tang}})(t) = \begin{pmatrix} 1 &{} &{} &{} \\ &{} \ddots &{} &{} \\ &{} &{} 1 &{} \\ &{} &{} &{} F_{{\text {rel }} 2 \times 2}(t) \end{pmatrix} \end{aligned}$$

where $F_{{\text {rel }} 2 \times 2}(t)$ is some $2 \times 2$ matrix that tells us how the remaining two vectors in the framing wind with respect to each other. In particular, the first $(d-2)$ vectors of this deformed frame can be thought of as a ‘shared framing’ which is shared by both the tangent and background frames. And, the induced spin structure can be visualized by measuring the number of $2\pi$ rotations the last two vectors of the background frame make with respect to the loop after projecting away the shared framing. See Fig. 21 for an example.

Fig. 21

A framed curve with ‘shared framing’ (blue sheet spanned by $v_1$). The background frame consists of the vectors $v_1, v_2, v_3$ as labeled in the box; $v_3$ is not shown but is chosen to make the orientation positive. The tangent frame consists of the shared $v_1$, the tangent of the pink curve, and some unshown vector such that orientation is positive. The induced spin structure can be computed by projecting away the shared framing and measuring the winding of one of the background frame’s projected vectors ($v_2$ here) with respect to the curve’s tangent in the projected two-dimensional picture

This point of view of a shared framing will become useful to us when reviewing the geometric interpretation of the Gu–Wen/Gaiotto–Kapustin Grassmann integral $\sigma (f)$ later. In particular, the $(d-2)$ vectors $v_1,\dots ,v_{d-2}$ introduced earlier from the branching structure play the role of this shared framing.

1.2.6 $\det (TM)$, $w_1^2$, and pin$^-$ structures

In the previous sections we discussed spin structures and $w_2$ for orientable manifolds. Here we will generalize the discussion to non-orientable manifolds, $w_1^2$, and spin-like structures on them. The main difficulty in describing a spin structure on a non-orientable manifold is the fact that TM is non-orientable. In other words, it is impossible to trivialize TM on the 1-skeleton, meaning that the constructions above do not directly apply.

However we can consider the orientation bundle $\det (TM)$. The bundle $TM \oplus \det (TM)$ is always orientable since $w_1(TM \oplus \det (TM)) = w_1(TM) + w_1(\det (TM)) = w_1 (\det (TM)) + w_1(\det (TM)) = 0$: the first step is the Whitney sum formula and the second step uses $w_1(TM) = w_1(\det (TM))$. So it is always possible to trivialize $TM \oplus \det (TM)$ along a 1-skeleton.

We will see that defining a spin structure on $TM \oplus \det (TM)$ defines a ${\text {pin}}^-$ structure on M. The reason that $TM \oplus \det (TM)$ is related to the $\textrm{Pin}^-$ groups is that $\textrm{Pin}^-(d)$ can be thought of as a subgroup of $\textrm{Spin}(d+1)$. In particular, $\textrm{Pin}^-(d) \subset \textrm{Spin}(d+1)$ consists of all elements whose projections onto $\textrm{SO}(d+1)$ restrict to O(d) on the first d coordinates. This implies that ${\text {pin}}^-$ structures can be viewed in terms of $TM \oplus \det (TM)$, since this $\det (TM)$ direction plays the role of this $(d+1)^{th}$ direction that parameterizes orientation-reversal. See [39] for a more detailed explanation.

Let us start by describing this trivialization for a 2-manifold. We can visualize this $\det (TM)$ piece as an ‘extra dimension’ sticking out transverse to the surface. So overall we will have that $TM \oplus \det (TM)$ is trivialized by three vector fields ${u},{v}_1,{v}_2$. Given an assignment ± of the 2-simplices, the fields ${v}_1,{v}_2$ are the same inside the 2-simplices as in the orientable case. For example see Fig. 22.

Fig. 22

The vector fields ${u},{v}_1,{v}_2$ to trivialize $TM \oplus \det (TM)$ in $d=2$, on the interior of ± 2-simplices. $\det (TM)$ can be thought of as a direction ‘transverse’ to the manifold and the vector field u locally keeps track of the orientations. The relevant vector fields’ colors match those in the box and illustrate what directions their sections point along the dual 1-skeleton

In general dimensions, the same idea applies, where we have the same vector fields ${v}_1,\dots ,{v}_d$ spanning the TM directions within the d-simplices and the extra u vector field in the $\det (TM)$ direction. These constructions are uniquely defined away from the orientation-reversing wall. But since our manifold is non-orientable we have to modify the vectors along the wall dual to $w_1$ since the vector fields $\{{v}_1, \dots ,{v}_d\}$ by themselves degenerate near $w_1$. Given this extra $\det (TM)$ direction, we can in fact trivialize the bundle if we allow the $\{{v}_1,\dots ,{v}_d\}$ to rotate into the $\det (TM)$ direction with respect to the u vector. In particular we choose a scheme for which only ${v}_d$ and u rotate into each other (recall that ${v}_1,\cdots , {v}_{d-1}$ always remain independent). In fact there are two ways to arrange this and roughly correspond to choosing to rotate them into each other by an angle of either $+180^\circ$ or $-180^\circ$. See Fig. 23.

Fig. 23

Choices of how to rotate u and ${v}_d$ into each other across the dual of $w_1$ with respect to the different possibilities of ± $(d-1)$-simplices on $w_1$. These different choices correspond to a perturbation of the $w_1$ wall. The relevant vector fields’ colors match those in the box and illustrate what directions their sections point along the dual 1-skeleton

Also in Fig. 23, we can see that the particular choice of how ${v}_d$ and u rotate into each other determines a canonical perturbation of the $w_1$ wall as follows and is explained pictorially in the figure. The vector u rotates into being parallel to the dual 1-skeleton towards one of the two d-simplices that share this $(d-1)$-simplex. The perturbing direction is the direction that ${v}_d$ points on the side specified by u. This perturbation can be used to define a ‘self-intersection’ of $w_1$. This self-intersection will be some closed collection of $(d-2)$-simplices that form a cycle in $Z_{d-2}(M,\mathbb Z_2)$ dual to $w_1^2$. See Fig. 24. This particular convention for defining $w_1^2$ is chosen to match the representative of $w_1^2$ used in the Grassmann integral $\sigma (f)$ later on. It also gives the winding matrices in Eq. (B7) a somewhat aesthetic form.

Now we are in a position to define a pin$^-$ structure on a manifold. Geometrically, a pin$^-$ structure can be thought of as a trivialization of $TM \oplus \det (TM)$ on the 1-skeleton of a manifold that extends (mod 2) to a trivialization on the 2-skeleton. This is essentially the same as a spin structure except we replace TM with $TM \oplus \det (TM)$. The obstruction to doing this is $w_2(TM \oplus \det (TM)) = w_2(TM) + w_1(TM)w_1(\det (TM)) = w_2(TM) + w_1(TM)^2 = w_2 + w_1^2$. For us, $w_2$ is the same canonical representative as in Eq. (A18) and $w_1^2$ is the same representative based on the perturbation of the dual of $w_1$. So a spin structure is represented by a cochain $\xi$ dual to a collection of $(d-1)$-simplices along which we twist the background vector fields to fix all singularities to be even-index.

A ${\text {pin}}^+$ structure can also be defined similarly, but is instead a trivialization of $TM \oplus 3 \cdot \det (TM)$ on the 1-skeleton that extends (mod 2) to the 2-skeleton. This is because the group $\textrm{Pin}^+$ embeds into $\textrm{Spin}(d+3)$, such that projections onto $\textrm{SO}(d+3)$ restrict to O(d) in the first d coordinates. We do not further consider this in detail because the discussion and geometric constructions of ${\text {pin}}^-$ structures are more closely related to the definition of Grassmann integral $\sigma (f)$. We do not know if there is a meaningful Grassmann integral that is more closely related to pin$^+$ structures.

Fig. 24

Representative of dual of $w_1^2$ from a perturbation of the orientation-reversing wall, as defined in Fig. 23

1.2.7 Induced pin$^-$ structures on framed curves

Similarly to spin structures, pin$^-$ structures give a way to induce spin/pin structures on embedded loops. The idea is exactly the same in the sense that a framing of $(d-2)$ vectors of TM along an embedded loop induces a particular trivialization of $TM \oplus \det (TM)$ along the loop which is a ‘tangent framing’ of the loop $F_{\text {tang}}$. And similarly the background framing $F_{\text {bckd}}$ is the trivialization of $TM \oplus \det (TM)$ given by the vector fields ${{v}_1,\dots ,{v}_d,{u}}$ along the dual 1-skeleton. In this case, $F_{\text {tang}}$ and $F_{\text {bckd}}$ are instead $(d+1) \times (d+1)$ matrices.

Just as before, the relative framing is a function $(F_{\text {bckd}}^{-1} F_{\text {tang}})(t): [0,1] \rightarrow \textrm{SO}(d+1)$ which lifts to a path $\tilde{F}(t): [0,1] \rightarrow \textrm{Spin}(d+1)$. A subtlety is that in this non-orientable case, we may not have $(F_{\text {bckd}}^{-1} F_{\text {tang}})(0) = (F_{\text {bckd}}^{-1} F_{\text {tang}})(1)$ since curves passing through the orientation-reversing wall will have their local orientation be the opposite at the end of the path with respect to what it was in the beginning.

We illustrate this in $d = 2$ in Fig. 25 for a path that crosses an orientation-reversing loop. Note that in that case

$$\begin{aligned} (F_{\text {bckd}}^{-1} F_{\text {tang}})(0) = \begin{pmatrix} -1 &{} 0 &{} 0 \\ 0 &{} -1 &{} 0 \\ 0 &{} 0 &{} 1 \end{pmatrix} (F_{\text {bckd}}^{-1} F_{\text {tang}})(1) \end{aligned}$$

in the coordinate basis given by $\{{u},{v}_1,{v}_2\}$ for the background frame and $\{x,y,z\}$ for the tangent frame. Indeed, in general for orientation-reversing loops, the relative framing gets changed by $\begin{pmatrix} -1 &{} 0 &{} 0 \\ 0 &{} -1 &{} 0 \\ 0 &{} 0 &{} 1 \end{pmatrix}$ across the loop, while for orientation-preserving loops $(F_{\text {bckd}}^{-1} F_{\text {tang}})(0) = (F_{\text {bckd}}^{-1} F_{\text {tang}})(1)$.

Continuing with the $d = 2$ example, we can consider the lifts to $\textrm{Spin}(3) = \textrm{SU}(2)$ that these paths in $\textrm{SO}(3)$ induce. In particular, the lifting to $\textrm{SU}(2)$ will be (fixing WLOG $\tilde{F}(0) = \mathbb {1}$)

$$\begin{aligned} \tilde{F}(1) = {\left\{ \begin{array}{ll} \pm \mathbb {1} &{}\quad {\text { if orientation-preserving}} \\ \pm i Z &{}\quad {\text { if orientation-reversing}} \end{array}\right. } \end{aligned}$$

(A19)

where $Z = \begin{pmatrix} 1 &{} 0 \\ 0 &{} -1 \end{pmatrix}$. In particular, this lets us define a function on $(d-1)$-cochains $f_L$ dual to a loop L

$$\begin{aligned} \sigma (f_L) = - \begin{pmatrix} 1&0 \end{pmatrix} \tilde{F}(1) \begin{pmatrix} 1 \\ 0 \end{pmatrix} = {\left\{ \begin{array}{ll} \pm 1 &{}\quad {\text {if orientation-preserving}} \\ \pm i &{}\quad {\text {if orientation-reversing}} \end{array}\right. } \end{aligned}$$

(A20)

that encodes this lift. Note the minus sign in front of the matrix element. More precisely, each orientation-preserving loop is endowed with an induced spin structure, which corresponds to $+1$ for $\sigma (f_L)$ if it is anti-periodic and $-1$ if it is periodic. Orientation-reversing loops are endowed with an induced pin$^-$ structure on their Möbius bundle labeled by $\pm i$.

The function $\sigma (f_L)$ defined above generalizes beyond $d = 2$ to higher dimensions after homotoping the first $(d-2)$ vectors of the frame to match, as we discussed previously in the orientable case. Also, even though we technically phrased everything in terms of a given ${\text {pin}}^-$ structure on the manifold where all singularities are even-index, the same procedure also gives a quantity when there are odd-index singularities, although isotopy invariance of moving the curve across these singularities is lost.

In this more general context, the function $\sigma (f_L)$ is an important part of the winding number definition of the Grassmann integral discussed in Sect. 4.2 and Appendix B.

Fig. 25

The tangent and background framings for a loop in $d=2$ that crosses the orientation-reversing wall dual to $w_1$ exactly once. The two blue lines going to the right should be thought of as on top of each other but visually displaced to make it easier to see the vector fields. A neighborhood of the loop in this manifold looks like a Möbius strip; note the depiction is that the two black lines identified via opposite arrows. The vector u is in the $\det (TM)$ direction ‘transverse’ to the 2-manifold M except when it crosses $w_1$. Note that the vector u switches sign relative to the loop’s framing from the beginning to the end. The net rotation in $\textrm{SO}(3)$ between the frames $\{x,y,z\},\{u,v_1,v_2\}$ is the diagonal matrix ${\text {diag}}(-1,-1,1)$ which lifts to $\pm i Z \in \textrm{Spin}(3) = \textrm{SU}(2)$

Explicitly Computing Winding Number Definition of Grassmann Integral, $\sigma (f)$

In this appendix, we fill in the details sketched in Sect. 4.2 and expand on the discussion of Appendix A 2 e, A 2 g for how to explicitly compute the winding definition of the Grassmann integral $\sigma (f)$. We further show how the geometric winding number based definition of Sect. 4.2 is equivalent to the algebraic definition given in Sect. 4.3 based on an integration over Grassmann variables.

As discussed in the main text, for a $(d-1)$-cocycle $f_{d-1}$, we can (using an appropriate trivalent decomposition) decompose its Poincaré dual chain into a set of non-intersecting loops on the dual 1-skeleton. Using Eq. (25), in order to compute $\sigma (f_{d-1})$ we need only explain how to compute $\sigma$ on a single such loop L.

1.1 Orientable manifolds

We begin with the simpler case of M orientable. In Appendix A 2 e, we explained how a frame of d “background” vector fields along a loop defines an induced spin structure on that loop, or alternatively a winding of a tangent framing of the loop with respect to the background framing. In that context, the background vector fields arose from a spin structure, but as we showed in Appendix A 2 c, even in the absence of a spin structure, the branching structure of the triangulation of M together with the ± assignments of d-simplices defines a nonsingular background framing along the dual 1-skeleton consisting of the vector fields ${v}_1,\dots ,{v}_d$. The discussion in Appendix A 2 e then applies directly. Given a $(d-1)$-cocycle $f_L$ dual to a single loop L, the background vector fields, together with the sign assignments $\epsilon (\Delta _d)$, determine a tangent framing $F_{\text {tang}}$ of L. Technically there are two choices for the tangent framing depending on which direction we pick the tangent vector along the curve, but this choice will not change the winding. We can write the background and tangent framings as maps $F_{\text {bckd}},F_{\text {tang}}: [0,1] \rightarrow \textrm{SO}(d)$, lift the relative path $F^{-1}_{\text {bckd}}F_{\text {tang}}$ to a map $\tilde{F} : [0,1] \rightarrow \textrm{Spin}(d)$, and use the sign $\tilde{F}(0) = \pm \tilde{F}(1)$ to define the $\mathbb Z_2$ winding.

The above prescription is still somewhat abstract; we presently explain how to calculate this winding number locally. First, pick a direction for L. Then on each simplex $\Delta _d = \langle 0 \cdots d \rangle$ (here the branching structure is assumed to be $0 \rightarrow 1 \rightarrow \cdots \rightarrow d$), L will traverse between two dual 1-cells, going from from $\hat{i} \rightarrow \hat{j}$, where $\hat{i}$ is the 1-cell dual to the sub-simplex $\langle 0 \cdots \hat{i} \cdots d \rangle$ (which is also opposite to the vertex i). The winding along this leg of the curve depends on i, j as well as the sign assignment $\epsilon (\Delta _d)$ to the d-simplex, and will be

$$\begin{aligned} {\text {wind}}(\hat{i} \rightarrow \hat{j}) = {\left\{ \begin{array}{ll} 0 &{}\quad {\text {if }} i \not \equiv j {\text { (mod 2)}} \\ \epsilon (\Delta _d) \pi &{}\quad {\text {if }} i \equiv j {\text { (mod 2) and }} i < j \\ -\epsilon (\Delta _d) \pi &{}\quad {\text {if }} i \equiv j {\text { (mod 2) and }} i > j \end{array}\right. }. \end{aligned}$$

(B1)

These windings are obtained in two dimensions by examining Fig. 26; the background vector fields are shown on the dual 1-skeleton, and a tangent framing can be drawn in for any particular curve passing through the 2-simplex that is drawn. The winding can then be obtained graphically. In higher dimensions, similar logic can be used to obtain Eq. (B1) after projecting away the $(d-2)$ vectors shared by the background and tangent framings. The projected background vector fields are shown in Fig. 27, and the winding of the tangent vector field relative to this background vector field can be computed by inspection. We emphasize that the tangent framing is determined by the loop L, the d-simplex orientations $\epsilon (\Delta _d)$, and, for $d>2$, the background framing.

Fig. 26

In $d=2$, frame of background vector fields (red, pink) on the dual 1-cells of a 2-simplex. The tangent framing of a fermion loop traveling along a green arrow winds by the given angle relative to this background framing

Fig. 27

Projection to 2D of frame of background vector fields (red and pink) on the dual 1-cells of a d-simplex after projecting out the $(d-2)$ vectors shared by the background and tangent framings. The solid lines are the dual 1-skeleton and the dashed lines are the shift of the dual 1-skeleton along the pink vector field. As described in the main text, the pairs of solid and dashed lines that intersect correspond to the $\cup _{d-2}$ pairing. The tangent framing of a fermion loop traversing the interior of this d-simplex experiences the windings in Eq. (B1). Here, $d'$ ($d''$) is the largest even (odd) number less than or equal to d. The clockwise order of the dual 1-cells and the directions of the vector fields are determined by the triangulation, branching structure, and orientation

Then the total winding is just the product of windings along each segment of the loop L, and $\sigma (f_L)$ is defined from that winding:

$$\begin{aligned} \sigma (f_L) = -\prod _{(\hat{i} \rightarrow \hat{j}) \in L} e^{ \frac{i}{2} \times {\text {wind}}(\hat{i} \rightarrow \hat{j})}. \end{aligned}$$

(B2)

As we saw above, the partial windings are either 0 or $\pi$, and since there must always be an even number of $\pi$ rotations, $\sigma (f_L) = \pm 1$. The minus sign in front is present in order to match Eq. (24) and is needed to correctly reproduce quadratic refinement Eq. (27).

As a technical comment, the order in which the dual 1-cells appear and the relative directions of the vector fields in Fig. 27 is not arbitrary and follows from the fact that the background vector fields, which are completely defined by the triangulation, branching structure, and orientation, were defined to give a geometric meaning to the higher cup product [49]. This choice of background vector fields ensures that the higher cup product properly appears in the quadratic refinement property Eq. (27). The pink vector field in Fig. 27 defines a shift of the dual 1-cells, and in this projected picture, the product $\alpha (\hat{i})\beta (\hat{j})$ appears in the formula for $\alpha \cup _{d-2} \beta$ if and only if the shifted $\hat{j}$ 1-cell intersects the unshifted 1-cell $\hat{i}$. One can check that Fig. 27 correctly reproduces the formula for the $\cup _{d-2}$ product:

$$\begin{aligned} (\alpha \cup _{d-2} \beta )(0 \dots d) = \sum _{\begin{array}{c} i < j {\text { both odd, OR}} \\ i > j {\text { both even}} \end{array}} \alpha (\hat{i}) \beta (\hat{j}) \end{aligned}$$

(B3)

As an aside, we note that the trivalent resolution used throughout was chosen in an ad hoc way to reproduce quadratic refinement formulas and equivalence to the Grassmann integral definitions. However, the cyclic order that appears in Fig. 27 looks quite similar to the trivalent resolution Fig. 5. We hope that the appearance of this ordering in Fig. 27 may lead to a more first-principles explanation for why certain trivalent resolutions work, although we have not figured out a precise connection.

1.1.1 Winding formula for $w_2$ is equivalent to Eq. (A18)

As a cross-check, we show in the case of orientable manifolds that inputting the winding formula into Eq. (26) gives the same answer for $w_2$ as Eq. (A18). More precisely, consider the loop of $(d-1)$-simplices $\Delta _{d-1}^1 \rightarrow \cdots \rightarrow \Delta _{d-1}^k \rightarrow \Delta _{d-1}^1$ comprising the link of 2-simplex $\langle 012 \rangle$. The winding formula expresses $\sigma$ in terms the partial windings ${\text {wind}}(\Delta _{d-1}^\ell \rightarrow \Delta _{d-1}^{\ell +1})$ as in Eq. (B1). On the other hand, Eq. (A18) expresses $w_2$ in terms of the indicator cochains $\varvec{\Delta }_{d-1}^\ell$ on each $(d-1)$-simplex. The equality we want to show is

$$\begin{aligned} \begin{aligned} \sigma (\delta {\lambda _{\langle 012 \rangle }})&= -(-1)^{\frac{1}{2\pi } \sum _{\ell =1}^k {\text {wind}}(\Delta _{d-1}^\ell \rightarrow \Delta _{d-1}^{\ell +1})} \\&= -(-1)^{\sum _{\ell =1}^k \int {\varvec{\Delta }}_{d-1}^{\ell } \cup _{d-2} {\varvec{\Delta }}_{d-1}^{\ell +1}} \end{aligned} \end{aligned}$$

(B4)

where we set $\Delta _{d-1}^{k+1} = \Delta _{d-1}^{1}$. Thus we just need to show that $\sum _{\ell =1}^k \int \varvec{\Delta }_{d-1}^{\ell } \cup _{d-2} \varvec{\Delta }_{d-1}^{\ell +1}$ is an alternate way to express the total winding $\frac{1}{2\pi } \sum _{\ell =1}^k {\text {wind}}(\Delta _{d-1}^\ell \rightarrow \Delta _{d-1}^{\ell +1})$, modulo 2.

To do this, we can redefine the partial windings as follows. In the following pictures, we will say that the black arrows on the bottom are the path along $\textrm{Link}(\langle 012 \rangle )$ and always points to the right. The red arrows on top are the background vector field $v_d$ which we can use to define the relative windings. The windings of $\pm \pi$ fall into four cases of counterclockwise / clockwise and $v_d$ starting in the same direction / opposite direction as the path. We will denote as a counterclockwise / clockwise rotations of the red $v_d$, depending on the relative starting orientatations. Below, we list the various possible rotations and give the cases in which they occur for paths $\hat{i} \rightarrow \hat{j}$ on a ± simplex:

However, note that these we can modify these values of these partial windings in such a way that the total windings stay the same. Any angle $\theta$ for which

will give the same total windings. The first conditions that

enforces that windings that can be deformed to the identity have zero total winding. The conditions

are to ensure that a full $2\pi$ rotation indeed has winding $2\pi$. The above cases that we used to define the winding correspond to $\theta = \pi$. However, we just as well could choose $\theta = 2\pi$ to get alternate expressions in terms of the modified “rotations”:

(B5)

From here, we can directly compare to the formula in terms of higher cup products. For a path $\hat{i} \rightarrow \hat{j}$ inside a d-simplex, the corresponding contribution $\int \varvec{\Delta }_{d-1}^{\ell } \cup _{d-2} \varvec{\Delta }_{d-1}^{\ell +1}$ will be:

$$\begin{aligned} \int \varvec{\Delta }_{d-1}^{\ell } \cup _{d-2} \varvec{\Delta }_{d-1}^{\ell +1} = {\left\{ \begin{array}{ll} 1 &{}\quad {\text {if }} i > j {\text { both even, OR if }} i < j {\text { both odd}} \\ 0 &{}\quad {\text {otherwise}} \end{array}\right. } \end{aligned}$$

using the equation Eq. (B3). Since either of the $\pm 2\pi$ rotations contribute $(-1)$ to the winding expression for $w_2$, comparing the $\cup _{d-2}$ expression above to the table of modified windings Eq. (B5), we can see that there is a $\pm 2\pi$ rotation if and only if $\int \varvec{\Delta }_{d-1}^{\ell } \cup _{d-2} \varvec{\Delta }_{d-1}^{\ell +1} = 1$.

This proves the equivalence of the two formulas for $w_2$.

1.2 Non-orientable manifolds

On a non-orientable manifold, the above procedure needs to be modified because any assignment $\epsilon (\Delta _d)$ of local orientations to d-simplices produces inconsistent induced orientations on the $(d-1)$-simplices. Equivalently, it is impossible to define a nondegenerate frame of vector fields everywhere along the 1-skeleton. The solution, as discussed in Sect. A 2 f, is to instead define a frame of vectors on the bundle $TM \oplus \det (TM)$ instead of TM; the former is always orientable.

Such a “background” framing $F_{{\text {bckd}}}$ on the dual 1-skeleton can be constructed from the branching structure, assignments $\epsilon (\Delta _d)$ of ± signs to d-simplices, and some additional (arbitrary, for present purposes) choices of how the framing behaves across the orientation-reversing wall. The first two pieces of data determine a cochain representative of $w_1$, and the rest of this process determines a perturbation of the orientation-reversing wall and thus a cochain representative of $w_1^2$. This construction is explained in Appendices A 2 c and A 2 f. As discussed in Appendix A 2 g, the background framing and the assignments $\epsilon (\Delta _d)$ determine²⁰ a “tangent" framing $F_{{\text {tang}}}$ of each fermion loop L and use it to define induced (s)pin structures on the curve.

We need to describe how to actually compute the windings and $\sigma$ on a $(d-1)$-cocycle $f_L$ dual to a single loop L. In the orientable case, we computed the number of times the tangent framing of a loop winds around the background framing. In the presence of an orientation-reversing wall, we need something like a “half-winding"; following the naming scheme of [39], we will call this $(\# {\text { of right-half-twists)(L)}}$. The reason for this name is that [39] considers a different scheme for background framings in $d=2$ where the background framing always shares one vector with the tangent of the curve and all rotations are about this axis. In their scheme, the total rotation is always quantized as a multiple of $\pm \pi$. For orientation-reversing loops it will be in total $\pm \pi$ (mod $2\pi$), i.e. a half-twist, because the vectors in the $\det (TM)$ direction switch sides relative to each other around the loop; a similar behavior of the vectors in the $\det (TM)$ direction occurs in our scheme when loops cross the orientation-reversing wall, as in Fig. 25. In fact, the relative frames in our scheme can be deformed to one more like [39] where the relative framing rotates by $\pm \pi$ about the z axis, although this is not trivial to see by inspection. For orientation-preserving loops, the total rotation is always an even multiple of $\pm \pi$, i.e., a full twist. We want $\sigma =\pm 1$ for orientation-preserving loops, i.e., an even number of half-twists, so we demand $\sigma = \pm i$ for orientation-reversing loops, i.e., an odd number of half-twists.

Returning to our scheme, similarly to the orientable case, there is a contribution to the windings that can be computed from how the loop traverses between $\hat{i} \rightarrow \hat{j}$ within simplices, although in the non-orientable case, it is important that this traversal direction matches the direction of the tangent vector. There is an additional contribution to the winding when the loop crosses $w_1$. We thus divide the loop L into N legs $k \in 1,\dots ,N$, where each leg consists of either traveling between the $(d-1)$-simplices $\hat{i} \rightarrow \hat{j}$ within a d-simplex or traveling between two d-simplices along a single dual 1-cell. The loop can only cross $w_1$ in the latter case. Associate to the kth leg of the loop a $4 \times 4$ matrix $W_k$ as follows. If the kth leg is traversing $\hat{i} \rightarrow \hat{j}$ within the d-simplex $\Delta _d$, then

$$\begin{aligned} W_k = \epsilon (\Delta _d) {\left\{ \begin{array}{ll} \mathbb {1}_4 &{} \quad {\text {if }} i \not \equiv j {\text { (mod 2)}} \\ \begin{pmatrix} i X &{} 0 \\ 0 &{} -i X \end{pmatrix} &{}\quad {\text {if }} i \equiv j {\text { (mod 2) and }} i < j \\ \begin{pmatrix} -i X &{} 0 \\ 0 &{} i X \end{pmatrix}&\quad {\text {if }} i \equiv j {\text { (mod 2) and }} i > j \end{array}\right. } \end{aligned}$$

(B6)

If instead the kth leg travels between two d-simplices through a $(d-1)$-simplex $\Delta _{d-1}$ (i.e. along the 1-cell dual to $\Delta _{d-1}$), then if $w_1(\Delta _{d-1}) = 0$, assign ${W_k = \mathbb {1}_4}$. If $w_1(\Delta _{d-1}) = 1$, then both d-simplices neighboring $\Delta _{d-1}$ induce the same orientation ± on $\Delta _{d-1}$; given that induced orientation, define

$$\begin{aligned} W_k = \pm {\left\{ \begin{array}{ll} \begin{pmatrix} 0 &{} -iY \\ iY &{} 0 \end{pmatrix} &{}\quad {\text {if crossing in the perturbing direction}} \\ \begin{pmatrix} 0 &{} iY \\ -iY &{} 0 \end{pmatrix}&\quad {\text {if crossing opposite to the perturbing direction}} \end{array}\right. } \end{aligned}$$

(B7)

Here, $X={\begin{pmatrix} 0 &{} 1 \\ 1 &{} 0 \end{pmatrix}}, Y={\begin{pmatrix} 0 &{} -i \\ i &{} 0 \end{pmatrix}},Z={\begin{pmatrix} 1 &{} 0 \\ 0 &{} -1 \end{pmatrix}}$ are the Pauli matrices. The winding of the loop is related to the product of these winding matrices $W_1 \cdots W_N$ by

$$\begin{aligned} i^{\# \text { of right-half-twists}} = \begin{pmatrix} 1&0&1&0 \end{pmatrix} W_1 \cdots W_N \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}. \end{aligned}$$

(B8)

We can now express $\sigma (f_L)$ as

$$\begin{aligned} \sigma (f_L) = -i^{\# \text { of right-half-twists}}. \end{aligned}$$

(B9)

It will be the case that $\sigma (f_L)$ will always be in $\{\pm 1, \pm i\}$, and will be $\pm i$ if and only if the loop is orientation-reversing (i.e. if $\int _{w_1} f = 1$). For a general cocycle f, we can write

$$\begin{aligned} \sigma (f) = (-1)^{\# \text { of loops}} \prod _{\text {loops}} i^{(\# \text { of right-half-twists)(loop)}} \end{aligned}$$

(B10)

Now, we give a brief description of how the winding matrices are derived, deferring details to [49]. These winding matrices can be derived by inspecting Figs. 23 and 27. First, this block form of zeros and $\textrm{SU}(2)$ elements is related to the two possible configurations of the tangent framing along the segment; these two possibilities correspond to whether the ‘orientation vector’ of the tangent framing in the $\det (TM)$ direction starts out in the same direction as the u of the background framing or starts out opposite. This $2 \times 2$ block structure of 0’s and $\textrm{SU}(2)$ matrices is a way to compactly encode these choices. As for why the matrices are in $\textrm{SU}(2)$, recall that the lift of the relative framing $(F_{\text {bckd}}^{-1} F_{\text {tang}})(t)$ involves three vector fields ${u},{v}_{d-1},{v}_d$ of the background framing and three vectors of the tangent framing, thus forming a path in $\textrm{SO}(3)$ that lifts to a path in $\textrm{SU}(2)$. To derive the relevant winding matrices, one has to consider how these partial windings from the figure lift from elements of $\textrm{SO}(3)$ to elements of $\textrm{SU}(2)$. For example, all winding matrices inside a d-simplex involve no rotation or a $\pm \pi$ rotation about the common $u=x$ axis (see Fig. 25 for axis labels) which lift to $\pm i X$ rotations. This process does not reverse orientation locally, so the block structure should be diagonal. Also, one gets opposite rotations if the $\det (TM)$-direction vectors are in the same direction vs. opposite directions, accounting for the sign difference between the blocks. Whereas, the winding matrices across $w_1$ involve $\pm i Y$ corresponding to $\pm \pi$ rotations about the $v_{d-1}=y$ axis. This process reverses orientation and thus gives the off-diagonal block matrices; one can convince themselves via the figures that these lifts to $\textrm{SU}(2)$ are opposites if the $\det (TM)$ vectors start out in the same vs. opposite directions.

The final winding will always be a rotation about the z-axis and thus have an eigenvector $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$. We assume, without loss of generality, that the loop starts with the vectors aligned in the $\det (TM)$ direction, which is why we apply the winding matrices to $\begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$. After traversing the loop, either the vectors in the $\det (TM)$ direction are aligned or anti-aligned; we sum over both possibilities, only one of which will give a nonzero contribution. This can be accomplished by taking an inner product with $\begin{pmatrix} 1&0&1&0 \end{pmatrix}$ which sums over the eigenvalues of the top-left and bottom-left blocks of the product $W_1 \cdots W_N$.

Again, it turns out that $\sigma (f)$ satisfies the main properties Eqs. (26) and (27), where the representative of $w_1^2$ is the one as specified above and $w_2$ is the ‘canonical’ representative that can be expressed by Eq. (A18).

1.3 Equivalence of winding and Grassmann definitions

Now, we demonstrate that the winding and Grassmann definitions are indeed equivalent to each other, i.e., $\sigma (f) = \sigma ^{\text {gr}}(f)$ for all $\delta f = 0$. We start with the orientable case before proceeding to the non-orientable case.

We claim that it is actually sufficient to show $\sigma (f_L) = \sigma ^{\text {gr}}(f_L)$, where $f_L$ is dual to a single closed loop L, which stems from our matching choice of trivalent resolution and Grassmann variable orderings. Proving this will be the contents of Sects. a, b. Presently we will show $\sigma (f) = \sigma ^{\text {gr}}(f)$ for all f assuming they match on single loops.

First, it is clear that if f is nonzero on only two $(d-1)$-simplices per every d-simplex, then even without the trivalent resolution f decomposes into distinct loops that never meet in a common d-simplex. In this case, $f = f_{\text {loop 1}} + \cdots + f_{\text {loop n}}$ and it is simple to see that $\sigma (f) = \sigma (f_{\text {loop 1}}) \cdots \sigma (f_{\text {loop n}}) = \sigma ^{\text {gr}}(f_{\text {loop 1}}) \cdots \sigma ^{\text {gr}}(f_{\text {loop n}}) = \sigma ^{\text {gr}}(f)$.

Now, consider the case where on some $(d-1)$-simplex ${\Delta _{d}}$, f is nonzero on more than two $(d-1)$-simplices, with $\{\hat{i}_1 \cdots \hat{i}_{2k}\}$ listed in the order of the trivalent resolution and Grassmann variables. Then the trivalent resolution gives f to a decomposition into distinct loops for whom the edges are paired up locally as $\{\hat{i}_1, \hat{i}_2\},\dots ,\{\hat{i}_{2k-1} \hat{i}_{2k}\}$. In addition, the $u({\Delta _{d}}) = \vartheta _{\hat{i}_1}^{f(\hat{i}_1)} \cdots \vartheta _{\hat{i}_{2k}}^{f(\hat{i}_{2k})}$ means that we can split $u({\Delta _{d}})$ into a product of k terms $u({\Delta _{d}}) = (\vartheta _{\hat{i}_1}^{f(\hat{i}_1)} \vartheta _{\hat{i}_2}^{f(\hat{i}_2)}) \cdots (\vartheta _{\hat{i}_{2k-1}}^{f(\hat{i}_{2k-1})} \vartheta _{\hat{i}_{2k}}^{f(\hat{i}_{2k})})$.

Using this trivalent resolution we can organize $f = f_{L_1} + \cdots + f_{L_n}$. The Grassmann formulation gives $\sigma ^{\text {gr}}(f) = \sigma ^{\text {gr}}(f_{L_1}) \cdots \sigma ^{\text {gr}}(f_{L_n})$ because of the freedom to reorganize the pairs of Grassmann variables from each trivalent resolution. And the trivalent resolution order and Eq. (B3) mean that

$$\begin{aligned} \sigma (f) = \sigma (f_{L_1}) \cdots \sigma (f_{L_n}) \prod _{1 \le \ell _1 < \ell _2 \le n} (-1)^{\int (f_{L_{\ell _1}} \cup _{d-2} f_{L_{\ell _2}})} \end{aligned}$$

simplifies to $\sigma (f) = \sigma (f_{L_1}) \cdots \sigma (f_{L_n})$ since each $(-1)^{\int (f_{L_{\ell _1}} \cup _{d-2} f_{L_{\ell _2}})}$ will return 1.

This shows that $\sigma (f) = \sigma ^{\text {gr}}(f)$ always assuming each $\sigma (f_L) = \sigma ^{\text {gr}}(f_L)$. Now we prove that each $\sigma (f_L) = \sigma ^{\text {gr}}(f_L)$.

1.3.1 Orientable case

In the case of an orientable manifold, we now verify that $\sigma (f_L) = \sigma ^{\text {gr}}(f_L)$ for $f_L$ dual to a single closed loop L on the dual 1-skeleton.

The argument is inductive. In the winding definition, every segment of the loop that introduces a nonzero winding involves a $\pm \pi$ rotation of the vector $v_d$ along the dual 1-skeleton relative to the tangent of the loop. We thus choose to induct on the number of times the direction of the vector $v_d$ along the dual 1-skeleton switches direction along the course of the loop, where the tangent to the loop is in the $\rightarrow$ direction. This number of switches is always even. The idea is that each pair of direction switches can be ‘removed’ from the loop independently from each other at the cost of introducing a sign of $\pm 1$ to $\sigma$. In the winding definition, we will get a sign of $+1$ or $-1$ depending on whether the direction switches came from a 0 or $\pm 2\pi$ rotation. In the Grassmann definition, there will be some sign associated to reordering the Grassmann variables in the segment between those direction switches. These signs can be computed for each segment independently; the winding is additive and removing a segment does not affect the rest of the loop, while the reorderings of Grassmann variables on different segments are independent. If the winding and Grassmann definitions give the same sign for each pair of direction switches, and if the base case of no direction switches match, then this implies that both definitions of $\sigma$ agree on the loop.

Before proceeding, we recall from Appendix A 2 c that within a $+$ d-simplex $\Delta _d$ while traversing along the dual 1-cell $\hat{i}$ away from the barycenter of $\Delta _d$ (i.e. near the boundary of $\Delta _d$), $v_d$ points towards the barycenter of $\Delta _d$ if i is odd and away if i is even, and vice-versa for − d-simplex. In parallel, recall from Sect. 4.3 that (away from $w_1$) on a $+$ d-simplex, a Grassmann variable associated to $\hat{i}$ is labeled $\theta$/black if i is even and is labeled $\overline{\theta }$/white if i is odd, and vice-versa on a − d-simplex. When we say that we are considering labelings of $(d-1)$-simplices, sign assignments, and Grassmann variables to the d-simplices which are compatible with a particular set of $v_d$, we mean that the labelings and sign assignments obey these rules.

First, let us verify that the functions equal each other when there are no direction switches of $v_{d}$ along the dual 1-skeleton. For an example path see Fig. 28a. Even though the figure shows some specific $\hat{i} \rightarrow \hat{j}$ and ± assignments on simplices, one can readily verify that the same computation follows through with different data compatible with the directions of $v_d$. The winding definition trivially gives $\sigma (f_L) = -1$, since there is no winding. Now consider the Grassmann integral definition. Labeling the $(d-1)$-simplices along the loop as $e_1, \dots , e_n$, we have

$$\begin{aligned} \sigma ^{\text {gr}}(f_L) = \int \prod _{\Delta _{d-1}} d\theta _{\Delta _{d-1}} d\overline{\theta }_{\Delta _{d-1}} (\theta _{e_1} \overline{\theta }_{e_2})(\theta _{e_2} \overline{\theta }_{e_3}) \cdots (\theta _{e_{n-1}} \overline{\theta }_{e_n}) (\theta _{e_n} \overline{\theta }_{e_1}) = -1 \end{aligned}$$

So we have verified $\sigma = \sigma ^{\text {gr}}$ in the base case.

Now let us consider through the example Fig. 28b a case where there are two direction switches of $v_d$. The winding definition for the specific choices in that figure produces

$$\begin{aligned} \sigma (f_L) = - \begin{pmatrix} 1&0&1&0 \end{pmatrix} \begin{pmatrix} i X &{} 0 \\ 0 &{} -i X \end{pmatrix} \begin{pmatrix} i X &{} 0 \\ 0 &{} -i X \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} = +1 \end{aligned}$$

since the winding matrices for $\hat{0} \rightarrow \hat{2}$ on d-simplex k and $\hat{1} \rightarrow \hat{3}$ on d-simplex $\ell$ are both $\begin{pmatrix} i X &{} 0 \\ 0 &{} -i X \end{pmatrix}$ on $+$-oriented simplices. Similarly, one can check by writing out the Grassmann integral in this case that $\sigma ^{\text {gr}}(f_L) = +1$.

Now we claim that this example is independent of the choices made in Fig. 28b, specifically of the choices of $(d-1)$-simplex labelings $\hat{i}$ and d-simplex orientations ±. First, recall that the winding matrices and the Grassmann variable orderings for $\hat{i} \rightarrow \hat{j}$ both only depend on whether $i = j {\text { (mod 2)}}$, whether $i < j$, and the sign ± of the simplex; and the $i < j$ vs $i > j$ only matter if indeed $i = j {\text { (mod 2)}}$. So, the results for $\hat{0} \rightarrow \hat{2}$ and $\hat{1} \rightarrow \hat{3}$ etc imply the same results for general $i < j$ both even and $i < j$ both odd. Next, note that every switch from $i < j$ both even (resp. odd) to $i > j$ both even (resp. odd) introduces a minus sign in the corresponding winding matrix (see Eq. (B6)), which changes $\sigma$ by a $-1$ and permutes $\vartheta _{\hat{i}}$ and $\vartheta _{\hat{j}}$ in the Grassmann integral, introducing one minus sign in $\sigma ^{\text {gr}}$. So all such cases of $\hat{i} \rightarrow \hat{j}$ will have $\sigma$ and $\sigma ^{\text {gr}}$ match. Since the example we gave consists of all $+$ simplices, we have now verified all cases compatible with the $v_d$ directions when all simplices are $+$.

Now we want to check, for these $v_d$ directions, the compatible cases when some simplices are −. In particular, changing $i < j$ (resp. $i > j$) both even on a $+$ simplex to $i > j$ (resp. $i < j$) both odd on a − simplex, will be compatible with the $v_d$ directions. By inspection, this leaves the winding matrices and Grassmann orderings unaffected. Likewise, changing $i < j$ (resp. $i > j$) both even on a $+$ simplex to $i < j$ (resp. $i > j$) both odd on a − simplex will introduce a $-1$ to both $\sigma$ and $\sigma ^{\text {gr}}$.

The changes to Fig. 28b that we have considered above generate all possible $(d-1)$-simplex labelings and d-simplex orientations that are compatible with these $v_d$, and the winding and Grassmann definitions agree on all of them. This proves that $\sigma (f_L) = \sigma ^{\text {gr}}(f_L)$ for loops which do not intersect an orientation-reversing wall.

Fig. 28

Examples of assignments of Grassmann variables going from d-simplices (labeled ‘simp’) $1 \rightarrow n$ and corresponding $\hat{i} \rightarrow \hat{j}$ assignments for loops traversing n d-simplices (where $\hat{i},\hat{j}$ refer to the $(d-1)$-simplex with vertices ordered as $i,j \in \{0,\cdots ,d\}$ missing.). All d-simplices are $+$ oriented; see the main text for − orientations. We consider traversing the loops from left to right. The red arrows on the bottom represent the directions of $v_d$ along the dual 1-skeleton that correspond to the choices of $\hat{i} \rightarrow \hat{j}$. (a,b) give representative cases where there are no crossings with $w_1$. (c,d,e,f,g,h) give representative cases when there are crossings with $w_1$. Here, $w_1$ is located at the solid red and dashed yellow line. The solid red line represents $w_1$ and the dashed yellow line is the perturbation of $w_1$. The small blue arrow on top of $w_1$ gives a direction of perturbation of $w_1$ in the direction of $\theta \rightarrow \overline{\theta }$, which is used to define $w_1^2$ as described in the main text. The $(d-1)$-simplices representing $w_1$ are labeled as ± as described in Fig. 23. (a) The case when there are no direction switches of $v_d$. (b) The case when there are exactly two direction switches of $v_d$. (c), and (e) are edge cases of (d) and (f) respectively where a direction switch within a d-simplex involves a $(d-1)$-simplex on the orientation-reversing wall. (g) is an edge case of (h) where both crossings of $w_1$ occur on the boundary of the same d-simplex

1.3.2 Non-Orientable case

We will use similar methods to prove that $\sigma (f_L) = \sigma ^{\text {gr}}(f_L)$ for loops that cross $w_1$, and thus that $\sigma = \sigma ^{\text {gr}}$. As before we will prove the statement for the case when the number of direction switches of $v_d$ is two, and this implies the general equivalence by induction. The main difference from the orientable case is that $v_d$ always switches direction when the loop crosses $w_1$; when this occurs, $\sigma$ picks up a factor of $\pm i$. The inductive step now follows from the short calculation that for any of the non-trivial winding matrices

$$\begin{aligned} W_{1,\cdots ,2n} \in \left\{ \pm \begin{pmatrix} i X &{} 0 \\ 0 &{} -iX \end{pmatrix} , \pm \begin{pmatrix} 0 &{} i Y \\ -i Y &{} 0 \end{pmatrix} \right\} \end{aligned}$$

we have that

$$\begin{aligned} \begin{pmatrix} 1&0&1&0 \end{pmatrix} W_1 \cdots W_{2n} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} = \left( \begin{pmatrix} 1&0&1&0 \end{pmatrix} W_1 W_2 \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} \right) \left( \begin{pmatrix} 1&0&1&0 \end{pmatrix} W_3 \cdots W_{2n} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} \right) . \end{aligned}$$

As before, we will pick some representative examples of paths with given arrow switches and locations of $w_1$ along the loop, and it will be easiest to illustrate in the cases where all the simplices are $+$ oriented. The representative examples we will check are drawn in Fig. 28c–h. When considering a loop with two direction switches, we need to consider two classes of cases; either the loop crosses $w_1$ once or twice. In the latter case, since there are only two direction switches, the crossings must be on $(d-1)$-simplices with opposite induced orientation, as can be examined from Fig. 28g, h.

For the cases (c), (d) in Fig. 28, there is exactly one crossing of $w_1$ at a − oriented $(d-1)$ simplex; (c) is an edge case of (d) where the direction switch within a d-simplex involves a simplex on the orientation-reversing wall. First, we compute the winding definition of $\sigma (f_L) = - \begin{pmatrix} 1&0&1&0 \end{pmatrix} \begin{pmatrix} i X &{} 0 \\ 0 &{} -i X \end{pmatrix} \begin{pmatrix} 0 &{} i Y \\ -i Y &{} 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} = +i$. The first matrix comes from the $\hat{0} \rightarrow \hat{2}$ on simplex k and the second matrix comes from crossing $w_1$ at a − simplex in the perturbing direction. Next we will compute $\sigma ^{\text {gr}}(f_L)$. The crossing of $w_1$ is associated to a factor of $-i$. And, both of the cases of (c), (d) can be seen to have a Grassmann integral part $\int \prod _{\Delta _{d-1}} d\theta _{\Delta _{d-1}} d\overline{\theta }_{\Delta _{d-1}} \prod _{\Delta _{d}} u({\Delta _{d}}) = -1$. So in total, $\sigma ^{\text {gr}}(f_L) = +i$. So the definitions match for this example.

The cases (e), (f) in Fig. 28 are the same as (c),(d) but the crossing of $w_1$ occurs at a $+$ oriented $(d-1)$ simplex. Then $\sigma (f_L) = - \begin{pmatrix} 1&0&1&0 \end{pmatrix} \begin{pmatrix} i X &{} 0 \\ 0 &{} -i X \end{pmatrix} \begin{pmatrix} 0 &{} -i Y \\ i Y &{} 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} = -i$. In the Grassmann definition, the crossing with $w_1$ is associated to a factor of $+i$. And again, both the cases of (e), (f) can be seen to have a Grassmann integral part $\int \prod _{\Delta _{d-1}} d\theta _{\Delta _{d-1}} d\overline{\theta }_{\Delta _{d-1}} \prod _{\Delta _{d}} u({\Delta _{d}}) = -1$. So in total, $\sigma ^{\text {gr}}(f_L) = -i$. Again the definitions match for this example.

In the cases (g), (h) in Fig. 28, there are two crossings of $w_1$, one at a $+$ oriented $(d-1)$-simplex and one at a − oriented one; (g) is an edge case of (h) where the crossings both occur on the boundary of the same d-simplex. The winding definition gives $\sigma (f_L) = - \begin{pmatrix} 1&0&1&0 \end{pmatrix} \begin{pmatrix} 0 &{} -i Y \\ i Y &{} 0 \end{pmatrix} \begin{pmatrix} 0 &{} i Y \\ -i Y &{} 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} = +1$. In the Grassmann definition, the crossings of $w_1$ at a $+$ oriented $(d-1)$-simplex and at a − oriented $(d-1)$-simplex give signs of $(+i) \cdot (-i)=1$. The Grassmann part can be shown to be $\int \prod _{\Delta _{d-1}} d\theta _{\Delta _{d-1}} d\overline{\theta }_{\Delta _{d-1}} \prod _{\Delta _{d}} u({\Delta _{d-1}}) = +1$, so $\sigma ^{\text {gr}}(f_L)=+1$. Again, the definitions agree.

Above we have only checked the equivalence explicitly for some representative examples. But these examples are sufficient to show the equivalence of $\sigma =\sigma ^{\text {gr}}$ for the same reasons as the orientable case, that $\sigma (f_L)$ and $\sigma ^{\text {gr}}(f_L)$ get changed by the same sign for any ± assignments of the d simplices and $\hat{i} \rightarrow \hat{j}$ assignments compatible with the $v_d$ directions along the path. The only additional thing we have to worry about are how the perturbing directions of $w_1$ affect $\sigma$ and $\sigma ^{\text {gr}}$. Changing the perturbing direction will change the winding matrix across $w_1$ by a sign, and it will switch the order of two Grassmann variables which adds a sign to $\sigma ^{\text {gr}}(f_L)$. This implies that $\sigma = \sigma ^{\text {gr}}$ in all the representative cases, and therefore for all cases.

Spin, Pin, and $\mathcal {G}_f$ Structures $\xi _{\mathcal {G}}$

The purpose of this appendix is to derive the formula that defines a $\mathcal {G}_f$ structure $\xi _{\mathcal {G}}$ as a trivialization of the 2-cocycle $w_2 + w_1^2 + A_b^*\omega _2$, which is used in the definition of our path integral. Along the way, we provide some detailed explanations of spin and pin structures that may be useful for physicists, and which we have not seen in other treatments of the subject.

1.1 Spin structure review: algebraic definition

In order to define a quantum field theory path integral that describes a fermionic system with $G_f = \mathbb Z_2^f$ internal symmetry, it is well-known that a choice of spin structure $\xi$ is required. Mathematically, a spin structure $\xi$ is a trivialization of the 2nd Stiefel-Whitney class, $\delta \xi = w_2$, and two distinct spin structures on the space-time manifold $M^d$ are related to each other by $H^1(M^d, \mathbb Z_2)$. The need for a spin structure arises for the following reason. A relativistic quantum field theory, of which a TQFT can be thought of as a special case, possesses on flat space an $\textrm{SO}(d)$ symmetry. However fermions, described in relativistic field theory by spinors, transform according to the double cover of $\textrm{SO}(d)$, which is $\textrm{Spin}(d)$. To define the field theory on curved space-time, one needs to consider a $\textrm{SO}(d)$ bundle, which can be thought of as the tangent bundle TM of an orientable manifold M. The transition functions $\phi _{ij} \in \textrm{SO}(d)$ between overlapping patches $U_i$ and $U_j$ must be lifted to $\tilde{\phi }_{ij} \in \textrm{Spin}(d)$. The lift $\tilde{\phi }$ is determined by $\phi _{ij}$ together with a choice of sign $(-1)^{\xi _{ij}} = \pm 1$, for $\xi _{ij} = 0,1$ (mod 2), which we can thus refer to as the spin structure. The combination $\xi _{ij} + \xi _{jk} + \xi _{ki}$ is precisely the definition of the second Stiefel-Whitney class $w_2(M^d)$ in Čech cohomology. We thus find that the spin structure $\xi$ must satisfy $\delta \xi = w_2$.

It is useful to understand this in a bit more detail. First, note that we can define a cohomology class $[\overline{w}_2]$ as the non-trivial element in $\mathcal {H}^2(\textrm{SO}(n), \mathbb Z_2) = H^2(B\textrm{SO}(n), \mathbb Z_2) = \mathbb Z_2$. In particular, the class $[\overline{w}_2]$ specifies a non-trivial $\mathbb Z_2$ extension of $\textrm{SO}(n)$, which is $\textrm{Spin}(n)$. A general $\textrm{SO}(d)$ bundle on $M^d$ can be understood as a map $\phi : M^d \rightarrow B\textrm{SO}(d)$. Therefore, using the pullback $\phi ^*$, one obtains a characteristic class $[w_2(M)] = \phi ^* [\overline{w}_2] \in H^2(M^d, \mathbb Z_2)$.

Now consider our transition functions $\tilde{\phi }_{ij} \in \textrm{Spin}(d)$, which are lifts of $\phi _{ij} \in \textrm{SO}(d)$. We can in general write $\tilde{\phi }_{ij} = (\phi _{ij}, \xi _{ij})$, where $\xi _{ij} = 0,1$ (mod 2) determines the choice of lift. Now consider $\tilde{\phi }_{ij} \tilde{\phi }_{jk} \tilde{\phi }_{ik}^{-1}$. We have

$$\begin{aligned} \tilde{\phi }_{ij} \tilde{\phi }_{jk} = (\phi _{ij} \phi _{jk}, \xi _{ij} + \xi _{jk} + \overline{w}_2(\phi _{ij}, \phi _{jk})) \end{aligned}$$

(C1)

Next, note that $\tilde{\phi }_{ik}^{-1}$, defined as the right inverse such that $\tilde{\phi }_{ik} \tilde{\phi }_{ik}^{-1} = 1$, is given by

$$\begin{aligned} \tilde{\phi }_{ik}^{-1} = (\phi _{ik}^{-1}, \xi _{ik} + \overline{w}_2(\phi _{ik}, \phi _{ik}^{-1})). \end{aligned}$$

(C2)

Therefore, we find

$$\begin{aligned} \tilde{\phi }_{ij} \tilde{\phi }_{jk}\tilde{\phi }_{ik}^{-1} = (1, \xi _{ij} + \xi _{jk} + \xi _{ik} + \overline{w}_2(\phi _{ij}, \phi _{jk})) . \end{aligned}$$

(C3)

In order to have a well-defined $\textrm{Spin}(d)$ bundle, we require $\tilde{\phi }_{ij} \tilde{\phi }_{jk}\tilde{\phi }_{ik}^{-1} = 1 \in \textrm{Spin}(d)$, which leads us to

$$\begin{aligned} \xi _{ij} + \xi _{jk} + \xi _{ik} = \overline{w}_2(\phi _{ij}, \phi _{jk}) \end{aligned}$$

(C4)

In coordinate-free notation, this is precisely the Čech cohomology version of the equation

$$\begin{aligned} \delta \xi = \phi ^* \overline{w}_2 = w_2(M). \end{aligned}$$

(C5)

It is useful to note that if a spin structure $\xi$ is not specified, we can canonically pick the lift $\phi _{ij} \rightarrow \tilde{\phi }_{ij} = (\phi _{ij},0)$. In this case, we can see that $w_2$ effectively acts as a source of fermion parity flux: a fermion traversing the three overlapping regions $U_i$, $U_j$, $U_k$ will pick up a minus sign according to $\overline{w}_2(\phi _{ij},\phi _{jk})$. This sign is effectively canceled by the specification of a spin structure $\xi$ that trivializes $w_2$. An alternate viewpoint of this is also explained in Appendix A 2 d.

When $G_f$ is a non-trivial central extension of $G_b$ by $\mathbb Z_2^f$, characterized by $[\omega _2] \in \mathcal {H}^2(G_b, \mathbb Z_2)$, then we need to consider a generalization which leads us to the notion of a $\mathcal {G}_f$ structure, $\xi _{\mathcal {G}}$. This can be understood as follows.

1.2 $\mathcal {G}_f$ structure derivation for unitary symmetries

In the case where $G_b$ only contains unitary symmetries, then we have the bosonic space-time symmetry group $\mathcal {G}_b = \textrm{SO}(d) \times G_b$. The fermionic space-time symmetry group is $\mathcal {G}_f = (\textrm{Spin}(d) \times G_f)/\mathbb Z_2$, where here the $\mathbb Z_2$ equivalence identifies the $-1$ element of $\textrm{Spin}(d)$ with the fermion parity $\mathbb Z_2^f \subset G_f$.

Here, $\mathcal {G}_f$ can be understood as a $\mathbb Z_2$ extension of $\mathcal {G}_b$, specified by a class $[\overline{w}_2(\mathcal {G}_b)] \in \mathcal {H}^2(\mathcal {G}_b, \mathbb Z_2) = H^2(B\mathcal {G}_b, \mathbb Z_2)$. Then, one can define a lift of a $\mathcal {G}_b$ bundle to the double cover $\mathcal {G}_f$ with a Čech 1-cochain $\xi \in C^1(M^d,\mathbb Z_2)$. Letting $\phi _{ij} \in \mathcal {G}_b$ be the transition functions and $\tilde{\phi }_{ij} = (\phi _{ij}, \xi _{ij})$ be the lift, with $\xi _{ij} \in \mathbb Z_2$, we require $\tilde{\phi }_{ij} \tilde{\phi }_{jk} \tilde{\phi }_{ik}^{-1} = 1 \in \mathcal {G}_f$. If we specify the $\mathcal {G}_b$ bundle in terms of a map $\mathcal {A}_b : M^d \rightarrow B\mathcal {G}_b$, then this is equivalent to

$$\begin{aligned} \delta \xi = \mathcal {A}_b^* \overline{w}_2(\mathcal {G}_b) . \end{aligned}$$

(C6)

Next, note that, via the Künneth decomposition, we have $\mathcal {H}^2(\mathcal {G}_b, \mathbb Z_2) = \mathcal {H}^2(\textrm{SO}(d), \mathbb Z_2) \oplus \mathcal {H}^2(G_b, \mathbb Z_2)$. Therefore,

$$\begin{aligned} \overline{w}_2(\mathcal {G}_b) = \overline{w}_2(\textrm{SO}(d)) + \omega _2. \end{aligned}$$

(C7)

It then follows that we have

$$\begin{aligned} \delta \xi = w_2(M) + A_b^* \omega _2 , \end{aligned}$$

(C8)

where $A_b: M \rightarrow BG_b$ is the projection of $\mathcal {A}_b: M \rightarrow B \mathcal {G}_b = B\textrm{SO}(d) \times BG_b$ onto $BG_b$.

We can go through the above calculation in somewhat more detail as follows. Let us denote an element of $\mathcal {G}_b$ as $(\alpha , g)$, where $\alpha \in \textrm{SO}(d)$ and ${\textbf{g}} \in G_b$. Let $\tilde{\alpha } = (\sigma , \alpha )$ be an element in $\textrm{Spin}(d)$, where $\sigma \in \{0,1\}$ specifies the lift of $\alpha \in \textrm{SO}(d)$ to $\tilde{\alpha } \in \textrm{Spin}(d)$, and $\tilde{{\textbf{g}}} = ({\textbf{g}},\mu ) \in G_f$, where $\mu \in \{0,1\}$ specifies the lift of ${\textbf{g}} \in G_b$ to $\tilde{{\textbf{g}}} \in G_f$. Finally we denote an element of $\mathcal {G}_f$ as $(\sigma , \alpha , {\textbf{g}}, \mu )$. The equivalence relation under $\mathbb Z_2$ implies $(\sigma , \alpha , {\textbf{g}}, \mu ) \sim (\sigma + \mu , \alpha , {\textbf{g}}, 0)$. Let us now consider a transition function $\phi _{ij} \in \mathcal {G}_b$, and its lift $\tilde{\phi }_{ij} \in \mathcal {G}_f$. We thus have

$$\begin{aligned} \tilde{\phi }_{ij} = (\sigma _{ij}, \alpha _{ij}, {\textbf{g}}_{ij}, \mu _{ij}) \sim (\sigma _{ij} + \mu _{ij}, \alpha _{ij}, {\textbf{g}}_{ij},0) = (\xi _{ij}, \alpha _{ij}, {\textbf{g}}_{ij}, 0), \end{aligned}$$

(C9)

where we defined $\xi _{ij} = \sigma _{ij} + \mu _{ij}$.

Now, to have a well-defined $\mathcal {G}_f$ bundle, we require

$$\begin{aligned} \tilde{\phi }_{ij} \tilde{\phi }_{jk} \tilde{\phi }_{ik}^{-1} = 1 \end{aligned}$$

(C10)

First, note that

$$\begin{aligned} \tilde{\phi }_{ij} \tilde{\phi }_{jk} = (\xi _{ij} + \xi _{jk} + \bar{w}_2(\alpha _{ij}, \alpha _{jk}), \alpha _{ij} \alpha _{jk}, {\textbf{g}}_{ij} {\textbf{g}}_{jk}, \omega _2({\textbf{g}}_{ij}, {\textbf{g}}_{jk})). \end{aligned}$$

(C11)

Next, consider

$$\begin{aligned} \tilde{\phi }_{ik} \tilde{\phi }_{ik}^{-1}&= 1\nonumber \\ \tilde{\phi }_{ik} \tilde{\phi }_{ki}&= (\xi _{ik} + \xi _{ki} + \bar{w}_2(\alpha _{ik}, \alpha _{ki}), \alpha _{ik} \alpha _{ki}, {\textbf{g}}_{ik} {\textbf{g}}_{ki}, \omega _2({\textbf{g}}_{ik}, {\textbf{g}}_{ki}))\nonumber \\&= (\bar{w}_2(\alpha _{ik}, \alpha _{ki}), 1, \omega _2({\textbf{g}}_{ik}, {\textbf{g}}_{ki}))\nonumber \\&= (\bar{w}_2(\alpha _{ik}, \alpha _{ki}) + \omega _2({\textbf{g}}_{ik}, {\textbf{g}}_{ki}), 1, 0) \end{aligned}$$

(C12)

Here we use the fact that ${\textbf{g}}_{ik} = {\textbf{g}}_{ki}^{-1}$, $\alpha _{ik} = \alpha _{ki}^{-1}$, and $\xi _{ik} = \xi _{ki}$. Therefore, we see that the right inverse of $\tilde{\phi }_{ik}$ is:

$$\begin{aligned} \tilde{\phi }_{ik}^{-1} = \tilde{\phi }_{ki} \times (\bar{w}_2(\alpha _{ik}, \alpha _{ki}) + \omega _2({\textbf{g}}_{ik}, {\textbf{g}}_{ki}), 1, 0) . \end{aligned}$$

(C13)

Therefore,

$$\begin{aligned} \tilde{\phi }_{ij} \tilde{\phi }_{jk} \tilde{\phi }_{ik}^{-1}&= (\xi _{ij} + \xi _{jk} + \xi _{ki} \nonumber \\&\quad + \bar{w}_2(\alpha _{ij}, \alpha _{jk})+ \bar{w}_2(\alpha _{ik}, \alpha _{ki}), 1, 1, \omega _2({\textbf{g}}_{ij}, {\textbf{g}}_{jk}) + \omega _2({\textbf{g}}_{ik}, {\textbf{g}}_{ki}))\nonumber \\&\quad \times (\bar{w}_2(\alpha _{ik}, \alpha _{ki}) + \omega _2({\textbf{g}}_{ik}, {\textbf{g}}_{ki}), 1, 0)\nonumber \\&= (\xi _{ij} + \xi _{jk} + \xi _{ki} + \bar{w}_2(\alpha _{ij}, \alpha _{jk}) + \omega _2({\textbf{g}}_{ij}, {\textbf{g}}_{jk}), 1,1, 0) , \end{aligned}$$

(C14)

where we have used that $\alpha _{ij} \alpha _{jk} \alpha _{ki} = 1$ and similarly for ${\textbf{g}}_{ij}$.

We see that the $\mathcal {G}_f$ bundle is determined from the $\mathcal {G}_b$ bundle by the choice of lift characterized by $\xi _{ij}$, and to have a well-defined $\mathcal {G}_f$ bundle we need

$$\begin{aligned} \delta \xi [ijk] + \bar{w}_2(\alpha _{ij}, \alpha _{jk}) + \omega _2({\textbf{g}}_{ij}, {\textbf{g}}_{jk}) = 0 . \end{aligned}$$

(C15)

Expressing this in coordinate-free notation then gives Eq. (C8).

As an example, let us consider the case of $G_f = \textrm{U}(1)^f$ (i.e. the group $\textrm{U}(1)$ where the $\pi$ rotation equals fermion parity). In this case $\xi _{\mathcal {G}}$ should be a $\textrm{Spin}^c$ connection, which requires

$$\begin{aligned} \delta \xi _{\textrm{Spin}^c} = w_2 + c {\text { mod 2}} , \end{aligned}$$

(C16)

where c is the first Chern class of the $\textrm{U}(1)$ bundle specified by $A_b$. To check this against our explicit formula above, we need to see that $A_b^*\omega _2$ is the mod 2 reduction of the first Chern class. For $\textrm{U}(1)^f$, we have $\omega _2(a,b) = (-1)^{a + b - [a+b]}$, where the $\textrm{U}(1)$ group elements are $e^{2\pi i a}$, $e^{2\pi i b}$ and $[a] \equiv a {\text { mod }} 1$. We thus have $\omega _2(A_b[01], A_b[12]) = e^{i \pi (A_b[01] + A_b[12] - [A_b[01] + A_b[12]])}$. For a flat $\textrm{U}(1)$ gauge field, we have $A_b[01] + A_b[12] -A_b[13] \in \mathbb Z$. Therefore we can write $A_b[01] + A_b[12] = n + A_b[13]$, and $[ A_b[01] + A_b[12] ] = A_b[13]$. Therefore

$$\begin{aligned} \omega _2(A_b[01], A_b[12]) = e^{i \pi (A_b[01] + A_b[12] - A_b[13])} = (-1)^{dA}, \end{aligned}$$

(C17)

which is precisely the mod 2 reduction of the first Chern class.

1.3 $\mathcal {G}_f$ structure derivation for anti-unitary symmetries

The generalization to the case where $G_b$ contains anti-unitary symmetries is given by

$$\begin{aligned} \delta \xi _{\mathcal {G}} = w_2 + w_1^2 + A_b^* \omega _2. \end{aligned}$$

(C18)

Let us go through the derivation of this in the case where $G_b$ is a split extension of $\mathbb Z_2^{\textbf{T}}$ by the unitary subgroup $G_b^u$: $G_b = G_b^u \times \mathbb Z_2^{\textbf{T}}$ or $G_b = G_b^u \rtimes \mathbb Z_2^{\textbf{T}}$. In this case, the two cases above imply that

$$\begin{aligned} \mathcal {G}_b = {\left\{ \begin{array}{ll} O(d) < imes G_b^u = \mathbb Z_2^{\textbf{T}} < imes [G_b^u \times \textrm{SO}(d)] = G_b < imes \textrm{SO}(d)\\ O(d) \times G_b^u \end{array}\right. } \end{aligned}$$

(C19)

Next, we need to lift the $\mathcal {G}_b$ bundle to its double cover $\mathcal {G}_f$. Here, there is an important subtlety. We are interested in the Euclidean space-time symmetry group for the fermions. This requires us to do a Wick rotation, which changes ${\textbf{T}}^2 = 1$ to ${\textbf{T}}^2 = (-1)^F$ in $\mathcal {G}_f$ and vice versa [29, 58]. Therefore, we consider a different extension $[\omega _2^E] \in \mathcal {H}^2(G_b, \mathbb Z_2)$, given by

$$\begin{aligned} \omega _2^E = \omega _2 + \epsilon _2. \end{aligned}$$

(C20)

Here $\epsilon _2 = s^{*}\overline{\epsilon }_2$, where $[\overline{\epsilon }_2]$ is the non-trivial class in $\mathcal {H}^2(\mathbb Z_2^{\textbf{T}}, \mathbb Z_2)=\mathbb Z_2$ and $s:G_b \rightarrow \mathbb Z_2^{\textbf{T}}$ is non-trivial on anti-unitary symmetries. If we let $\phi : M^d \rightarrow B\mathbb Z_2$ specify the orientation bundle on $M^d$, then $\phi ^*\epsilon _2 = w_1^2$.

Then, $\mathcal {G}_f$ should be considered to be a $\mathbb Z_2$ extension of $\mathcal {G}_b$ specified by²¹

$$\begin{aligned} \overline{w}_2(\mathcal {G}_b) \in \mathcal {H}^2(\mathcal {G}_b, \mathbb Z_2) = \mathcal {H}^2(G_b,\mathbb Z_2) \oplus \mathcal {H}^2(\textrm{SO}(d), \mathbb Z_2). \end{aligned}$$

(C21)

We thus have

$$\begin{aligned} \overline{w}_2(\mathcal {G}_b) = \overline{w}_2(\textrm{SO}(d)) + \omega _2^E \end{aligned}$$

(C22)

Then, to specify the lift of the $\mathcal {G}_b$ bundle to a $\mathcal {G}_f$ bundle, we need a choice of $\xi \in \mathbb Z_2$, such that

$$\begin{aligned} \delta \xi&= \mathcal {A}_b^* \overline{w}_2(\mathcal {G}_b) = w_2 + A_b^* (\omega _2 + \epsilon _2)\nonumber \\&= w_2 + w_1^2 + A_b^* \omega _2 \end{aligned}$$

(C23)

As an example, consider $G_b = \mathbb Z_2^{\textbf{T}}$ with trivial $\omega _2$, so that $G_f = \mathbb Z_2^{\textbf{T}} \times \mathbb Z_2^f$. In this case Eq. (C23) gives $\delta \xi _{\mathcal {G}} = w_2 + w_1^2$, which is what we expect for $\textrm{Pin}^-$ structures. Next consider $G_b = \mathbb Z_2^{\textbf{T}}$ with $G_f = \mathbb Z_4^{ {\textbf{T}}, f}$. In this case $\omega _2({\textbf{T}}, {\textbf{T}}) = 1$. Then we identify $A_b = w_1$, so $\omega _2(A_b[01], A_b[12]) = w_{01}w_{12}$, so $A_b^*\omega _2 = w_1^2$. It then follows that Eq. (C23) reduces to $\delta \xi _{\mathcal {G}} = w_2$, which is what we expect for a $\textrm{Pin}^+$ structure.

As another example, consider $G_b = \textrm{U}(1)\rtimes \mathbb Z_2^{\textbf{T}}$ and $G_f = [\textrm{U}(1)^f \rtimes \mathbb Z_4^{{\textbf{T}},f}]/\mathbb Z_2$. Here we have $\omega _2((z,t), (z',t') ) = z + \,^{t'}z' - [z + \,^{t'}z'] + tt'$, where $z \in [0,1)$ parameterizes the $\textrm{U}(1)$ part, $t = 0,1$ parameterizes the $\mathbb Z_2^{\textbf{T}}$ part, and $[a] = a {\text { mod }} 1$. Note that $\,^t z = -z$ if $t = 1$ to account for the non-trivial action of $\mathbb Z_2^{\textbf{T}}$ on $\textrm{U}(1)$. The gauge field $A_b[ij] = (A_b^u[ij], w_1[ij])$ where $A_b^u$ is the unitary part and $w_1$ sets the anti-unitary part to coincide with the orientation bundle. In this case, Eq. (C23) reduces to $\delta \xi _{\mathcal {G}} = w_2 + (w_1^2 + \tilde{F} \mod 2) = w_2 + w_1^2 + \tilde{F} \mod 2$. Here $\tilde{F}[012] = A_b^u[01] + w_1[01]A_b^u[12] - A_b^u[02]$ is an element of the local cohomology $H^2(M, \mathbb Z)$ which is twisted by $w_1$.

The case where $G_b = \textrm{U}(1)\times \mathbb Z_2^{\textbf{T}}$ and $G_f = \textrm{U}(1)^f \rtimes \mathbb Z_2^{\textbf{T}}$ is similar except $\omega _2((z,t), (z',t')) = z + \,^{t'}z' - [z + \,^{t'}z']$, so that $\delta \xi _{\mathcal {G}} = w_2 + \tilde{F} \mod 2$, which defines a $\textrm{Pin}_-^{\tilde{c}}$ structure.

Let us consider the case $G_b = G_b^u \times \mathbb Z_2^{\textbf{T}}$ and $G_f = G_b^u \times \mathbb Z_2^{\textbf{T}} \times \mathbb Z_2^f$. In this case, $\omega _2$ is trivial, so we expect $\delta \xi _{\mathcal {G}} = w_2 + w_1^2$, which gives a $\textrm{Pin}^-$ structure. If we instead consider $G_f = G_b^u \times \mathbb Z_4^{{\textbf{T}}, f}$, then $A_b^* \omega _2$ reduces to $w_1^2$, and we get $\delta \xi _{\mathcal {G}} = w_2$, which is a $\textrm{Pin}^+$ structure. The above examples of $G_f$ and ${\mathcal {G}}_f$ are summarized in Table 1, along with additional examples when $G_b$ consists of internal unitary symmetries and time reversal.

Table 1 Some possible cases for $G_f$ and $\mathcal {G}_f$ for $G_b = G_b^u \times \mathbb Z_2^{\textbf{T}}$ and $G_b = G_b^u \rtimes \mathbb Z_2^{\textbf{T}}$

We can also consider the case of the symmetry groups associated with the “relativistic” 10-fold way, as shown in Table 2, in order to connect our results to periodic table of free fermion topological insulators and superconductors.

Table 2 Relativistic 10-fold way and the relevant $\mathcal {G}_f$ groups. Note that $\textrm{Pin}^c = [\textrm{Pin}^+ \times \textrm{U}(1)]/\mathbb Z_2 \simeq [\textrm{Pin}^- \times \textrm{U}(1)]/\mathbb Z_2$. $\textrm{U}(1)^f$ and $\textrm{SU}(2)^f$ denote the $\textrm{U}(1)$ and $\textrm{SU}(2)$ groups with the $-1$ element identified with fermion parity. A similar table was also given in Ref. [98]

The $f_\infty$ Map and Turning Cochains into Chains

Here we describe the $f_\infty$ map of [59] that shows up in several places in our constructions.

1.1 Definition of $f_\infty$

On a manifold $M^d$ equipped with a branched triangulation, there is a map

$$\begin{aligned} f_\infty : Z^k(M^d,\mathbb Z_2) \rightarrow Z_{d-k}(M^d,\mathbb Z_2) \end{aligned}$$

(D1)

that turns a $\mathbb Z_2$ cocycle on the triangulation of $M^d$ to a cycle on the triangulation of $M^d$. Note that this is distinct from the usual cochain-level Poincaré duality, which maps k-cocycles on the triangulation to $(d-k)$-cycles on the dual cellulation. One can thus think of $f_\infty$ as implementing a somewhat different cochain-level Poincaré duality. Since $\mathbb Z_2$ cocycles are dual to submanifolds living on the dual cellulation of M, one can equivalently think of this map as taking submanifolds living on the dual cellulation to submanifolds living on the original triangulation of a manifold. This map readily generalizes to more general coefficient groups and non-closed cochains, but we will not need it for our purposes.

Recall that chains $C_{d-k}(M,\mathbb Z_2)$ uniquely represent the most general kind of linear function on the cochains $C^{d-k}(M,\mathbb Z_2)$. In particular, any linear functional $C^{d-k}(M,\mathbb Z_2) \rightarrow \mathbb Z_2$ can be represented as $\alpha \mapsto \int _c \alpha$ for some choice of chain $c \in C_{d-k}(M,\mathbb Z_2)$. The main idea of the $f_\infty$ map is that the cup product pairing $C^{d-k}(M,\mathbb Z_2) \times C^{k}(M,\mathbb Z_2): (\alpha , \beta ) \mapsto \int _M \alpha \cup \beta$ for some fixed $\beta$ is also a linear functional on $\alpha$. This observation leads us to the definition of $f_\infty \beta$. Linearity tells us that $\beta$ can be mapped to a unique chain, which we will call $f_\infty \beta$, for which $\int \alpha \cup \beta = \int _{f_\infty \beta } \alpha$ for all $\alpha \in C^{d-k}(M,\mathbb Z_2)$. The fact that (away from $\partial M$) $f_\infty \beta$ is a closed submanifold for $\beta$ closed is because $\int \delta \lambda \cup \beta = 0$ for any $\lambda \in C^{d-k-1}(M,\mathbb Z_2)$, and a chain c representing a nonclosed collection of $(d-k)$-simplices will always have a $\lambda$ for which $\int _c \delta \lambda = \int _{\partial c}\lambda \ne 0$.

To actually compute $f_\infty \beta$, one needs to find all of the $(d-k)$-simplices S for which the indicator cochain $\alpha _S$ satisfies $\int \alpha _S \cup \beta = 1$. We illustrate a $d=2$ example for this in Fig. 29. Note that the end result of the chain can roughly be visualized as ‘flowing’ the dual of $\beta$ opposite to the branching structure until it reaches the original triangulation. This perspective is expanded on in [59] in reference to a so-called ‘Morse Flow’ associated to the branching structure, and we reference it in our discussions about how the geometry is encoded in the diagrammatics.

Fig. 29

Illustration of $f_\infty \beta$ in $d=2$. The triangle on the bottom right illustrates a way to think of $(\alpha \cup \beta )(012)=\alpha (01)\beta (12)$ as an intersection of the dual of $\alpha$ with a shifted version of $\beta$. Note that the only intersection of $\alpha$ with $\beta$ occurs with $\alpha (01)$ and $\beta (12)$. To compute $f_\infty \beta$ for $\beta \in Z^1(M^2,\mathbb Z_2)$, we collect all the dual 1-simplices for which their shifted version intersects the dual of $\beta$. Then, the dual of this dual collection becomes $f_\infty \beta$. In the bottom-right panel, a simpler example illustrates how $f_\infty$ can be visualized as flowing the dual of a cochain opposite to the branching structure

1.2 Use of $f_\infty$ to define orientation-reversing walls

As discussed in Sect. 3, the $G_b$ gauge field determines an element $A_b^{*}s \in Z^1(M,\mathbb Z_2)$, where $s:G_b \rightarrow \mathbb Z_2$ is non-trivial on anti-unitary group elements. The interpretation of $A_b^{*}s$ is that it is nonzero on 1-simplices across which the local orientation is reversed. The object that plays a crucial role in our constructions of both $z_c$ and $Z_b$ is the orientation-reversing wall $w_1 \in Z^1(M^{\vee },\mathbb Z_2) \sim Z_{d-1}(M,\mathbb Z_2)$. The $f_{\infty }$ map is a natural way to obtain a cochain representative of $w_1$ from a cochain representative of $A_b$ via

$$\begin{aligned} w_1 := f_\infty A_b^*s. \end{aligned}$$

(D2)

As mentioned in Sect. 6.1.2, using this representative is important to ensure that the bosonic shadow is independent of various choices. It also plays a role in specifying the anomaly of $Z_b$, as described in Appendix G 2. In particular, to ensure that the shadow is independent of our trivalent resolution of the five objects that meet at a 3-simplex, we need the F- and R-moves relating different trivalent resolutions to cancel. This cancellation requires consistency with the way the diagrams are drawn in both 4-simplices’ 15j symbols, which is related to the orientations induced on the 3-simplex from the 4-simplices as in Fig. 30.

We claim that the above representative of $w_1$ causes the following fact to be true: a 3-simplex lies on the orientation-reversing wall of a closed manifold M (i.e., it has the same induced orientation from both 4-simplices which contain it), if and only if a domain wall for an anti-unitary group element surrounds that 3-simplex in exactly one 15j symbol. That is, the 3-simplices would then appear in 15j symbols in the way shown in Fig. 31. By inspection, this fact means that F- and R-moves do indeed cancel and the shadow is independent of trivalent resolution.

Fig. 30

Induced orientations on a 3-simplex and how they appear in a 15j symbol

Fig. 31

Induced orientations for a 3-simplex $\langle abcd \rangle$ (ordered $a \rightarrow b \rightarrow c \rightarrow d$) compared to the group elements ${\textbf{g}}_1, {\textbf{g}}_2$ surrounding them in 15j symbols of the 4-simplices $\langle a b c d e \rangle ,\langle a b c d f \rangle$ (unordered, where e,f can be in any position relative to $a \rightarrow b \rightarrow c \rightarrow d$). (Left) In the case ${\textbf{g}}_1 {\textbf{g}}_2^{-1}$ is unitary, $\langle abcd \rangle$ has opposite induced orientations on $\langle a b c d e \rangle ,\langle a b c d f \rangle$. (Right) In the case ${\textbf{g}}_1 {\textbf{g}}_2^{-1}$ is anti-unitary, $\langle abcd \rangle$ has the same induced orientation on $\langle a b c d e \rangle ,\langle a b c d f \rangle$, either both $+$ or both − as in Fig. 30. The shared induced orientation means that $w_1(\langle abcd \rangle ) \ne 0 \in \mathbb Z_2$, so that $\langle abcd \rangle$ is labeled as a ± $(d-1)$-simplex on the orientation-reversing wall

To see this, first examine the 15j symbols. Suppose that the 3-simplex $\langle abcd \rangle$, with vertices ordered $a \rightarrow b \rightarrow c \rightarrow d$ according to the branching structure, appears in the two 4-simplices $\langle abcde \rangle$ and $\langle abcdf \rangle$, where this notation does not specify how e or f are ordered in the branching structure. Then in the 15j symbol for $\langle abcde \rangle$, $\langle abcd \rangle$ appears surrounded by a ${\textbf{g}}_1$ domain wall, where

$$\begin{aligned} {\textbf{g}}_1 = {\left\{ \begin{array}{ll} {\textbf{g}}_{ed} &{}\quad {\text {branching structure orders }}a \rightarrow b \rightarrow c \rightarrow d \rightarrow e \\ {\textbf{1}} &{}\quad {\text {else}} \end{array}\right. }. \end{aligned}$$

(D3)

Likewise, in the 15j symbol for $\langle abcdf \rangle$, $\langle abcd \rangle$ appears surrounded by a ${\textbf{g}}_2$ domain wall where

$$\begin{aligned} {\textbf{g}}_2 = {\left\{ \begin{array}{ll} {\textbf{g}}_{fd} &{}\quad {\text {branching structure orders }}a \rightarrow b \rightarrow c \rightarrow d \rightarrow f \\ {\textbf{1}} &{}\quad {\text {else}} \end{array}\right. }. \end{aligned}$$

(D4)

Note that $s({\textbf{g}}_1 {\textbf{g}}_2^{-1})$ is the non-trivial element of $\mathbb Z_2$ if and only if exactly one of ${\textbf{g}}_1$ or ${\textbf{g}}_2$ is anti-unitary. Thus a domain wall for an anti-unitary group element surrounds $\langle abcd \rangle$ in exactly one 15j symbol if and only if $s({\textbf{g}}_1 {\textbf{g}}_2^{-1})$ is the non-trivial element of $\mathbb Z_2$.

Although they are motivated by the 15j symbols, the group elements ${\textbf{g}}_{1,2}$ are well-defined independent of the diagrams. With that definition, one may directly compute for a general branching structure that

$$\begin{aligned} f_\infty A_b^*s (\langle abcd \rangle ) = s({\textbf{g}}_1 {\textbf{g}}_2^{-1}). \end{aligned}$$

(D5)

By definition $w_1(\langle abcd \rangle )$ is nonzero if and only if $\langle abcd \rangle$ is on the orientation-reversing wall; with our choice of representative of $w_1$ given by Eq. (D2), $\langle abcd \rangle$ is on the orientation-reversing wall if and only if $s({\textbf{g}}_1 {\textbf{g}}_2^{-1})$ is the non-trivial element of $\mathbb Z_2$. As shown in the previous paragraph, these are also exactly the 3-simplices which are surrounded by a domain wall for an anti-unitary group element in exactly one 15j symbol; this proves our assertion.

1.3 Perturbation of orientation-reversing wall and $w_1^2$ in relation to $f_\infty$

Now that we have explained how the gauge field determines a representative orientation-reversing wall $w_1 \in Z_3(M,\mathbb Z_2)$ and how this representative of $w_1$ is encoded in the diagrammatics, we explain the same for a representative of $w_1^2 \in Z_2(M,\mathbb Z_2)$. Given Eq. (D2), there is a natural representative

$$\begin{aligned} w_1^2 = f_{\infty } \left( A_b^*s \cup A_b^*s \right) . \end{aligned}$$

(D6)

Recall that $w_1^2$ is the intersection of $w_1$ with a perturbed version of itself. Note that there are two distinct perturbations of $w_1$ compatible with $w_1^2$: any given one and its reversal. It is important to distinguish these and fix one, since the Grassmann integral depends not only on a cochain representative of $w_1^2$, but also the particular perturbation. We claim that $f_{\infty }$ actually encodes such a perturbation. Specifically, take a particular 3-simplex $\langle abcd \rangle$, which appears in two 4-simplices. Suppose that in the 15j symbols, it is surrounded by group elements ${\textbf{g}}_1,{\textbf{g}}_2$ as in Fig. 31. If $\langle abcd \rangle$ is part of the orientation-reversing wall, then exactly one of the ${\textbf{g}}_1,{\textbf{g}}_2$ will have an anti-unitary action. We claim that $f_{\infty }$ encodes a perturbation of $w_1$ into the 4-simplex whose surrounding bubble is anti-unitary.

This follows from the interpretation in [59] of $f_\infty$ corresponding to a ‘Morse flow’ of the dual of the cochain to the output chain (see Fig. 29). We can imagine flowing the dual of $A_b^* s \in Z^1(M,\mathbb Z_2)$ along some vector fields until it reaches the orientation-reversing wall $f_\infty A_b^* s$. The Morse flow works as follows. For each edge e for which $A_b^*s(e) \ne 0$, break its dual 3-cell into pieces, each of which is contained in a single 4-simplex. For a given piece contained in, say, 4-simplex $\langle 01234 \rangle$ (where the vertices are given in order of the branching structure), there are two possibilities. If $e=\langle 34 \rangle$ for this 4-simplex, then the piece flows within $\langle 01234 \rangle$ to the 3-simplex $\langle 0123 \rangle$. Otherwise, the piece degenerates under the flow. This connects to the diagrammatics because as shown in Sect. D 2, a 15j symbol for $\langle 01234 \rangle$ contains an anti-unitary domain wall loop if and only if ${\textbf{g}}_{34}$ is anti-unitary, i.e. if $A_b^*s({\langle 34 \rangle }) \ne 0$. Hence, the piece of the 3-cell which has a nondegenerate flow is, at any intermediate point in the flow, always contained inside the 4-simplex whose 15j symbol contains an anti-unitary domain wall loop. Therefore, any 3-simplex in $f_{\infty } A_b^* s$ originates from the flow of portions of 3-cells contained in 4-simplices whose 15j symbol contains an anti-unitary domain wall loop.²²

For completeness, we give an algorithmic way to see if a given 2-simplex $\langle abc \rangle$ is part of $w_1^2 \in Z_2(M,\mathbb Z_2) = f_\infty \left( A_b^*s \cup A_b^*s \right)$ using the diagrammatics. We first collect the loop of 3-simplices that contain it, which forms ${\text {Link}}(\langle abc \rangle )$. Recall these are the 3-simplices for which $\delta \alpha _{\langle abc \rangle }$ acting on them is nonzero, where $\alpha _{\langle abc \rangle }$ is the indicator cochain. Then draw out all the 15j symbols that contain $\langle abc \rangle$. For all the $\langle a'b'c'd' \rangle \ni \langle abc \rangle$ in $w_1$, one can draw a thin blue circle around $\langle a'b'c'd' \rangle$ in all the 15j symbols where the group element surrounding it is unitary. Likewise draw a thick orange circle around it in a 15j symbol if the group element is anti-unitary. Then, as in Fig. 32 we draw a directed loop along ${\text {Link}}(\langle abc \rangle )$ by going between the two 3-simplices containing $\langle abc \rangle$ in each 15j symbol in sequence, drawing an arrow within each 15j symbol. (Note that we have two choices for the direction of the loop, of which we arbitrarily pick one.) We sum up a quantity going around the loop that adds up to $w_1^2(\langle abc \rangle ) {\text { (mod 2)}}$ as follows. For a particular 15j symbol, add 0 or $+1/2$ if the tail of the arrow is adjacent to a thin-blue or thick-orange circle respectively, and add 0 or $-1/2$ if the head of the arrow is adjacent to a thin-blue or thick-orange circle. Since $\delta \alpha _{\langle abc \rangle }$ is cohomologically trivial, this loop of 3-simplices crosses $w_1$ an even number of times and will therefore pick up an even number of $\pm 1/2$ signs, giving us something that adds up to $w_1^2(\langle abc \rangle ) {\text { (mod 2)}}$. Conceptually, since each 3-simplex on $w_1$ has been perturbed towards the anti-unitary group element, the above algorithm associates a $+1/2$ to a crossing of Link$(\langle abc \rangle )$ in the direction of the perturbation and a $-1/2$ when the crossing is opposite the perturbation. The only way that the loop can cross $w_1$ twice in the same direction relative to the perturbation is if the perturbed $w_1$ crosses the unperturbed version on $\langle abc \rangle$, that is, if $w_1^2(\langle abc \rangle )=1$.

As an example, we can compute $w_1^2$ on the triangulation of ${\mathbb {R}}{\mathbb {P}}^4$ as in Appendix J with 15j symbols drawn out in Fig. 52. The orientation-reversing wall consists of the 3-simplices labeled , and the only 2-simplices $\langle abc \rangle$ which could possibly have $w_1^2(abc)$ non-trivial are labeled . Following the procedure in Fig. 32 gives:

(D7)

Fig. 32

The link of a 2-simplex $\langle abc \rangle$ can be seen from the 15j symbols, consisting of simplices $s_1 \rightarrow s_2 \rightarrow \cdots$ that contain $\langle abc \rangle$. In the top left and bottom left portions of the figure, we drawn part of each 15j symbol that contains $\langle abc \rangle$. The one total solid orange line around each $s_k, s_\ell$ mean that $w_1(s_k) = w_1(s_\ell ) = 1$ and that $w_1$ is perturbed into the 4-simplex consisting of that 15j symbol. The thick red line on $s_k$ and $s_{\ell }$ means $w_1(s_k) = w_1(s_{\ell }) = 1$. One can compute $w_1^2(\langle abc \rangle )$ by adding up $\pm 1/2$ for every crossing of the arrows with an anti-unitary (thick-orange) circle and 0 for every unitary (thin-blue) circle. (Top) Crossing $w_1$ in the perturbing direction twice giving $w_1^2(\langle abc \rangle ) = 1$. (Bottom) Crossing in the perturbing direction once and in the opposite direction once gives $w_1^2(\langle abc \rangle ) = 0$

Lemmas About Pachner Moves with Branching Structure and Background Gauge Fields

1.1 Introduction

Pachner’s theorem states that any two triangulations that define PL-equivalent manifolds are connected by a series of combintorial moves, called “Pachner moves”. In our state sum, it would be useful to have more general results than this because the triangulations we deal with are decorated with more general objects, like branching structures and gauge fields.

In this appendix, we first extend Pachner’s result to include branching structures, in particular showing that any two branched triangulations of a PL-manifold can be connected by Pachner moves which respect branching structures. Then, we add background flat gauge fields, showing that any two gauge fields on the triangulation that are gauge equivalent can be connected by Pachner moves. This is proved for general groups in the case of 1-form gauge fields and for Abelian higher-form gauge fields.

Note that in this appendix, we will often use the symbols $v_{\cdots },w_{\cdots }$ to refer to vertices. These symbols will not have anything to do with the vector fields v and Stiefel-Whitney classes w from earlier.

1.2 Pachner-connectedness of branching structures

In this section, we will show that any two branching structures are connected by Pachner moves. The strategy is to connect an arbitrary branched triangulation to a canonical “inwards” branching structure on the barycentric subdivision of the triangulation. This proves the theorem since any two branching structures can be connected using this common refinement.

First, we describe in some detail what we mean by branched Pachner moves and the barycentric subdivision. Then we prove that the barycentric subdivision can be obtained from branched Pachner moves. We start the proof in $d=2$ and $d=3$, which will then prepare us for the proof in arbitrary dimensions.

1.2.1 Preliminaries

i. Branched Pachner moves Recall that given a triangulation of a d-dimensional manifold and $k \in \{1 \cdots d+1\}$, a $(k,d+2-k)$ Pachner move is a transformation that can be thought of as a transformation of the triangulation consisting of the following steps:

• Attach a $(d+1)$-simplex to the manifold along k adjacent simplices.

• Remove the original k simplices and replace them with the other ‘leftover’ $(d+2-k)$ simplices

This can be thought of as an elementary bordism of the manifold with itself. See for example Fig. 33.

Fig. 33

Pachner moves in $d=2$ as attaching a 3-simplex to a 2-manifold and discarding the attaching region. The red part represents the original 2-manifold, the blue part are the ‘new’ 2-simplices of the 3-simplex, and the hatched regions are the shared triangles of the 3-simplex and the manifold, which get discarded. The final triangulation is the red part plus blue part. (Left) A branched (1,3) Pachner move. (Right) A branched (2,2) move

A branched Pachner move is entirely analogous, except the original triangulation and the attaching $(d+1)$-simplex are endowed with branching structures, and the branching structure of the manifold must agree with that of the attaching $(d+1)$-simplex. All branched Pachner moves in $d=2$ are given in Fig. 34, and some $d=3$ ones are given in Fig. 35.

Fig. 34

All possible branched Pachner moves for 2-manifold triangulations. The vertex labels and the 3-simplex corresponding to them are in the top-left cell

Fig. 35

Some examples of branched $d=3$ Pachner moves, which involve a 4-simplex $\langle 01234 \rangle$. All $d=3$ Pachner moves will be $\{(1,4),(2,3),(3,2),(4,1)\}$-moves, so will look like the ones drawn, except with different vertex orders and branching structures. The differently-colored triangles illustrate boundaries between distinct 3-simplices. (Left) A (2,3) move $\{\langle 0123 \rangle ,\langle 0124 \rangle \} \rightarrow \{\langle 1234 \rangle ,\langle 0234 \rangle ,\langle 0134 \rangle \}$. (Right) A (1,4) move $\{\langle 0124 \rangle \} \rightarrow \{\langle 1234 \rangle ,\langle 0234 \rangle ,\langle 0134 \rangle ,\langle 0123 \rangle \}$

ii. Barycentric subdivision Given a triangulation of a manifold, there is a well-known canonical refinement of the triangulation called the barycentric subdivision. A property of this subdivision is that for each $n \in \{0 \cdots d\}$, every n-simplex of the triangulation gets split into $(n+1)!$ different n-simplices. We give a sketch of its definition as follows.

First, for $n\in \{0 \cdots d\}$ define $S_n$ as the n-simplices of the original triangulation. We can define the vertex set $\tilde{S}_0$ of the barycentric subdivision as having one vertex for every simplex of the original:

$$\begin{aligned} \tilde{S}_0 := \bigsqcup _{m=0}^d \{ \langle m_{a} \rangle | a \in S_m \} \end{aligned}$$

(E1)

Each element of $\tilde{S}_0$ represents the barycenter of a simplex in $S_m$. To an m-simplex a, we label an element of $\tilde{S}_0$ as $\langle m_{a} \rangle$, so the barycenters of the original m-simplices all have a label specifying the dimension of their simplex.

The n-simplices $\tilde{S}_n$ of a barycentric subdivision are labelled by ascending sequences of simplices of length-$(n+1)$, i.e. sequences of simplices $\{a_0 \subsetneq \cdots \subsetneq a_n\}$ with dimensions $\{k_0< \cdots < k_n\}$, where $a \subsetneq b$ means that simplex a is a subsimplex of b not equal to b. We can write this as:

$$\begin{aligned} \tilde{S}_n = \bigsqcup _{0 \le k_0< \cdots < k_n \le d} \{\langle (k_0)_{a_0} \cdots (k_n)_{a_n} \rangle | a_0 \subsetneq \cdots \subsetneq a_n \} \end{aligned}$$

(E2)

where each $(k_i)_{a_i}$ is a vertex in $\tilde{S}_0$ coming from a $k_i$-simplex $a_i \in S_{k_i}$.

Every top-dimensional d-simplex of the barycentric subdivision is given a canonical vertex ordering coming from the inclusion ordering, $(k_0)_{a_0} \rightarrow \cdots \rightarrow (k_d)_{a_d}$. it is clear that these vertex orderings are consistent and define a branching structure on the subdivision. See Fig. 36 for illustration of the barycentric subdivision and the induced branching structures in $d=2$ and $d=3$.

Fig. 36

A barycentric subdivision of a simplex in $d=2$/$d=3$ and its canonical branching structure. (Left) $d=2$, vertices are $\{v,w,x\}$, edges are $\{a,b,c\}$ and the face is $\{A\}$. (Right) $d=3$, vertices are $\{v,w,x,y\}$, edges are $\{a,b,c,d,e,f\}$, 2-faces are $\{A,B,C,D\}$, and the 3-simplex is $\{T\}$. For visual clarity, we do not draw the edges going towards the barycenter vertex $3_T$ (shown as a yellow star). The reader should imagine an edge going from each of the shown vertices directed towards $3_T$

1.2.2 Proof of branched barycentric subdivision from Pachner moves

The idea of our proof borrows heavily from [99], which illustrates, without branching structures, how to obtain a barycentric subdivision from Pachner moves in $d=2$ and $d=3$. We essentially fill in arrows to their $d=2$ and $d=3$ pictures to prove the desired statement in $d=2$/$d=3$. The structure of their construction (and decorating branching structures) readily generalize to arbitrary dimensions. Since the dimension we care about is $d=4$, we will not be able to prove the statement pictorially and will need to translate the construction into symbols.

We distinguish the original simplices of dimensions between $0 \rightarrow 4$ as follows. 0-simplices are lower-case roman $\{v,w,x,...\}$ in the upper-alphabet. 1-simplices are lower-case roman $\{e,f,...\}$ near the lower alphabet. 2-simplices are upper-case Roman $\{A,B,C,...\}$ near the lower-alphabet (although in the $d=3$ proof we find it convenient to deviate from this convention). 3-simplices are upper-case roman $\{S,T,...\}$ near the upper-alphabet. 4-simplices are lower-case Greek $\{\alpha ,\beta \}$ near the lower-alphabet.

i. Proof in $d=2$ First we will reproduce their pictorial $d=2$ argument making sure to carefully account for the branching structure. Then we will translate the $d=2$ proof into symbols in a way that will make it apparent how to generalize to arbitrary dimensions.

The proof in $d=2$ is summarized in Fig. 37. First, rename every vertex ‘v’ as $0_v$. Then for every 2-simplex ‘A’, we perform a (1, 3)-move which creates a new vertex $2_A$ such that all new edges point towards $2_A$. Then, the link of any of the original edges ‘e’ will consist of exactly two triangles. For each of these e, another (1, 3)-move on one of these triangles creates an additional vertex $1_e$, and another (2, 2) move completes the subdivision.

Fig. 37

Obtaining the branched barycentric subdivision of a $d=2$ triangulation using branched Pachner moves. We illustrate it with a specific initial choice of branching structure, but the same argument applies for any such choice. The last two moves are restricted to the yellow regions surrounding the original edges, and can they can be done independently for each such region

Symbolically, we can express this in the following multistep process. The branching structure is accounted for through the ordering of the vertices of each simplex.

• Step 0: For all the vertices of the original triangulation, change notation $v \rightarrow 0_v$.

• Step 1: For every 2-simplex $A=\langle 0_v 0_w 0_x \rangle$, perform a (1, 3)-move creating a new vertex $2_A$:

  $$\begin{aligned} \langle 0_v 0_w 0_x \rangle \rightarrow \{\langle 0_v 0_w 2_A \rangle ,\langle 0_v 0_x 2_A \rangle ,\langle 0_w 0_x 2_A \rangle \} \end{aligned}$$

• Step 2: This step has two sub-parts.

  • (a) After Step 1, for each of the original 1-simplices $e=\langle 0_w 0_y \rangle$, there are exactly two 2-simplices $\langle 0_w 0_y 2_A \rangle ,\langle 0_w 0_y 2_B \rangle$ adjacent to e. Do a (1, 3)-move on one of these, creating a new vertex $1_e$:

    $$\begin{aligned} \langle 0_w 0_y 2_A \rangle \rightarrow \{\langle 0_w 1_e 2_A \rangle ,\langle 0_y 1_e 2_A \rangle ,\langle 0_w 0_y 1_e \rangle \} \end{aligned}$$

  • (b) Do the (2, 2)-move:

    $$\begin{aligned} \{\langle 0_w 0_y 1_e \rangle ,\langle 0_w 0_y 2_A \rangle \} \rightarrow \{\langle 0_w 1_e 2_A \rangle ,\langle 0_w 1_e 2_A \rangle \} \end{aligned}$$

At the end of this process every 2-simplex will be of the form $(0_v 1_e 2_f)$ for each possible $v \subsetneq e \subsetneq f$ of the original triangulation, so we have produced a barycentric subdivision with the correct branching structure.

ii. Proof in $d=3$ The argument in $d=3$ will proceed similarly, but will be a bit more work. We will omit drawing some of the pictures in $d=3$, which is already done nicely in [99] (and which we recommend the reader look at while reading our symbolic argument). Instead we translate their pictorial argument into a symbolic argument parallel to the $d=2$ case, allowing us to track the branching structure using the vertex orderings. We note the top row of the bottom panel of Fig. 40 depicts a dualized version of the process although in the context of a 2-form gauge field.

• Step 0: For all the vertices of the original triangulation, change notation $v \rightarrow 0_v$.

• Step 1: For every 3-simplex $T=\langle 0_v 0_w 0_x 0_y \rangle$, perform a (1, 4)-move creating a new vertex $3_T$:

  $$\begin{aligned} \langle 0_v 0_w 0_x 0_y \rangle \rightarrow \{\langle 0_v 0_w 0_x 3_T \rangle ,\langle 0_v 0_w 0_y 3_T \rangle ,\langle 0_v 0_x 0_y 3_T \rangle ,\langle 0_w 0_x 0_y 3_T \rangle \} \end{aligned}$$

• Step 2: This step again has two sub-parts.

  • (a) After Step 1, for each of the original 2-simplices $A=\langle 0_v 0_w 0_x \rangle$, there are exactly two 3-simplices $\langle 0_v 0_w 0_x 3_S \rangle ,\langle 0_v 0_w 0_x 3_T \rangle$ adjacent to e. Do a (1, 4)-move on one of these, creating a new vertex $2_A$:

    $$\begin{aligned} \langle 0_v 0_w 0_x 3_T \rangle \rightarrow \{\langle 0_v 0_w 2_A 3_T \rangle ,\langle 0_v 0_x 2_A 3_T \rangle ,\langle 0_w 0_x 2_A 3_T \rangle ,\langle 0_v 0_w 0_x 2_A \rangle \} \end{aligned}$$

  • (b) Do the (2, 3)-move:

    $$\begin{aligned} \{\langle 0_v 0_w 0_x 2_A \rangle ,\langle 0_v 0_w 0_x 3_S \rangle \} \rightarrow \{\langle 0_v 0_w 2_A 3_S \rangle ,\langle 0_v 0_x 2_A 3_S \rangle ,\langle 0_w 0_x 2_A 3_S \rangle \} \end{aligned}$$

  At this point every 3-simplex in the triangulation will be of the form $\langle 0_v 0_w 2_A 3_T \rangle$ for the original edges $e=\langle 0_v 0_w \rangle$. And, we will have one of these 3-simplices for every sequence $e \subsetneq A \subsetneq T$.

• Step 3: The goal of this step is, for each $e=\langle 0_v 0_w \rangle$, to transform the set of the $\{\langle 0_v 0_w 2_A 3_T \rangle |e \subsetneq A \subsetneq T\}$ into the set $\{\langle 0_x 1_e 2_A 3_T \rangle |x \subsetneq e \subsetneq A \subsetneq T\}$, which would complete the branched barycentric subdivision. We first note that the link of the edge $e=\langle 0_v 0_w \rangle$ consists of 1-simplices $\{\langle 2_T 3_A \rangle |e \subsetneq A \subsetneq T\}$. These 1-simplices taken together will be homeomorphic to a circle, and will in fact be the barycentric subdivision of the link of e in the original triangulation.

  • (a) There are an even number, call it 2m, of edges and 2m vertices in the link of e. Of these vertices, m of them will be of the form $3_T$ for a 3-simplex T, and m of them will be of the form $2_{(S T)}$ where (ST) is the 2-simplex adjacent to 3-simplices S, T. We label the 3-simplices as $T_1,\cdots ,T_m$, and the edges in the link of e will be:

    $$\begin{aligned} \langle 2_{(T_1 T_2)} 3_{T_1} \rangle ,\langle 2_{(T_1 T_2)} 3_{T_2} \rangle ,\langle 2_{(T_2 T_3)} 3_{T_2} \rangle ,\langle 2_{(T_2 T_3)} 3_{T_3} \rangle ,\cdots ,\langle 2_{(T_m T_1)} 3_{T_m} \rangle ,\langle 2_{(T_m T_1)} 3_{T_1} \rangle \end{aligned}$$

  • (b) For the first 3-simplex $\langle 0_v 0_w 2_{(T_1 T_2)} 3_{T_1} \rangle$ perform a (1, 4)-move, creating a new vertex $1_e$:

    $$\begin{aligned} \langle 0_v 0_w 2_{(T_1 T_2)} 3_{T_1} \rangle&\rightarrow \{\langle 0_v 1_e 2_{(T_1 T_2)} 3_{T_1} \rangle ,\langle 0_w 1_e 2_{(T_1 T_2)} 3_{T_1} \rangle ,\langle 0_v 0_w 1_{e} 3_{T_1} \rangle ,\\&\qquad \langle 0_v 0_w 1_{e} 2_{(T_1 T_2)} \rangle \} \end{aligned}$$

    Then, do a (2, 3)-move replacing $\langle 0_v 0_w 1_{e} 2_{(T_1 T_2)} \rangle ,\langle 0_v 0_w 2_{(T_1 T_2)} 3_{T_2} \rangle$:

    $$\begin{aligned} \{\langle 0_v 0_w 1_{e} 2_{(T_1 T_2)} \rangle ,\langle 0_v 0_w 2_{(T_1 T_2)} 3_{T_2} \rangle \}&\rightarrow \{\langle 0_v 0_w 1_e 3_{T_2} \rangle ,\langle 0_w 1_e 2_{(T_1 T_2)} 3_{T_2} \rangle ,\\&\qquad \langle 0_v 1_e 2_{(T_1 T_2)} 3_{T_2} \rangle \} \end{aligned}$$

  • (c) Do two more (2, 3)-moves. First, replace:

    $$\begin{aligned} \{\langle 0_v 0_w 1_{e} 3_{T_2} \rangle ,\langle 0_v 0_w 2_{(T_2 T_3)} 3_{T_2} \rangle \}&\rightarrow \{\langle 0_v 0_w 1_e 2_{(T_2 T_3)} \rangle ,\langle 0_v 1_e 2_{(T_2 T_3)} 3_{T_2} \rangle ,\\&\qquad \langle 0_2 1_e 2_{(T_2 T_3)} 3_{T_2} \rangle \} \end{aligned}$$

    Then, replace:

    $$\begin{aligned} \{\langle 0_v 0_w 1_{e} 2_{(T_2 T_3)} \rangle ,\langle 0_v 0_w 2_{(T_2 T_3)} 3_{T_3} \rangle \}&\rightarrow \{\langle 0_v 0_w 1_e 3_{T_3} \rangle ,\langle 0_v 1_e 2_{(T_2 T_3)} 3_{T_3} \rangle ,\\&\qquad \langle 0_w 1_e 2_{(T_2 T_3)} 3_{T_3} \rangle \} \end{aligned}$$

  • (d) Inductively repeat this process until the replacement

    $$\begin{aligned} \{\langle 0_v 0_w 1_{e} 2_{(T_m T_1)} \rangle ,\langle 0_v 0_w 2_{(T_m T_1)} 3_{T_m} \rangle \}&\rightarrow \{\langle 0_v 0_w 1_e 2_{(T_m T_1)} \rangle ,\langle 0_v 1_e 2_{(T_m T_1)} 3_{T_m} \rangle ,\\&\qquad \langle 0_v 1_e 2_{(T_m T_1)} 3_{T_m} \rangle \} \end{aligned}$$

    Now, there are three 3-simplices $\{\langle 0_v 0_w 1_e 2_{(T_m T_1)} \rangle ,\langle 0_v 0_w 1_e 3_{T_1} \rangle , \langle 0_v 0_w 2_{(T_m T_1)} 3_{T_1}\rangle \}$ on which we perform (3, 2)-move:

    $$\begin{aligned}&\{\langle 0_v 0_w 1_e 2_{(T_m T_1)} \rangle ,\langle 0_v 0_w 1_e 3_{T_1} \rangle ,\langle 0_v 0_w 2_{(T_m T_1)} 3_{T_1} \rangle \} \\&\quad \rightarrow \{\langle 0_v 1_e 2_{(T_m T_1)} 3_{T_1} \rangle ,\langle 0_w 1_e 2_{(T_m T_1)} 3_{T_1} \rangle \} \end{aligned}$$

  At this point we have done all the desired replacements, and have completed the barycentric subdivision. And, the branching structure matches the canonical one.

iii. Proof in $d=4$ and higher The proof in $d=4$ is again quite similar. The Steps 0-3 of the $d=3$ proof have a direct analog in $d=4$. Except instead, Steps 1-3 will begin with (1, 5)-moves:

$$\begin{aligned} \langle 0_v 0_w 0_x 0_y 0_z \rangle&\rightarrow \{\langle 0_v 0_w 0_x 0_y 4_\alpha \rangle ,\langle 0_v 0_w 0_x 0_z 4_\alpha \rangle ,\langle 0_v 0_w 0_y 0_z 4_\alpha \rangle ,\\&\qquad \langle 0_v 0_x 0_y 0_z 4_\alpha \rangle ,\langle 0_w 0_x 0_y 0_z 4_\alpha \rangle \} \\&\dots {\text {etc.}} \end{aligned}$$

where $4_\alpha$ is a new vertex created for an original 4-simplex $\alpha$. And in general of the $(a,5-a)$-moves done for $a=1,2,3$ will be replaced with $(a,6-a)$ moves.

At the end of Step 3, we will have that all of the 4-simplices in the triangulation will be of the form:

$$\begin{aligned} \{\langle 0_v 0_w 2_A 3_T 4_\alpha \rangle | e=\langle vw \rangle {\text { was originally a 1-simplex and }} e \subsetneq A \subsetneq T \subsetneq \alpha \} \end{aligned}$$

The goal of Step 4 would be similar, in trying to use Pachner moves to replace the above set (with fixed 1-simplex e) with the set $\{\langle 0_x 1_e 2_A 3_T 4_\alpha \rangle |x \subsetneq e \subsetneq A \subsetneq T \subsetneq \alpha \}$. After Step 3, we will have a similar statement that the link of e is the set $\{\langle 2_T 3_A 4_\alpha \rangle |e \subsetneq A \subsetneq T\}$. But instead, $\textrm{Link}(e)$ will be homoemorphic to a 2-sphere, and will be the barycentric subdivision of the link of e in the original triangulation.

The strategy to do these replacements is basically the same as in the $d=3$ case (or Step 3 in general) where we instead tried to do the replacements with on a circle instead of on a sphere. The only complication is that it was more straightforward to do on a circle, since there was a natural cyclic order to do the replacements. But the same argument still applies.

We illustrate $\textrm{Link}(e)$ and more generally the argument for Step 4 in Fig. 38. The idea is as follows. For some $\langle 2_A 3_T 4_\alpha \rangle \in \textrm{Link}(e)$, perform a (1, 5)-move on $\langle 0_v 0_w 2_A 3_T 4_\alpha \rangle$:

$$\begin{aligned} \langle 0_v 0_w 2_A 3_T 4_\alpha \rangle&\rightarrow \{\langle 0_v 1_e 2_A 3_T 4_\alpha \rangle ,\langle 0_w 1_e 2_A 3_T 4_\alpha \rangle ,\langle 0_v 0_w 1_e 3_T 4_\alpha \rangle ,\\&\qquad \langle 0_v 0_w 1_e 2_A 4_\alpha \rangle ,\langle 0_v 0_w 1_e 2_A 3_T \rangle \} \end{aligned}$$

Fig. 38

Step 4 of barycentric subdivision via Pachner moves in $d=4$. (Top-left) $\textrm{Link}(e)$ for a particular 1-simplex $e=\langle v, w \rangle$ of the original triangulation after the first three steps of subdivision. $\textrm{Link}(e)$ is homeomorphic to a 2-sphere and has been barycentrically subdivided in the other steps. (Other diagrams) Each large arrow corresponds to a subdivision process involving a particular 2-simplex in $\textrm{Link}(e)$. After a 2-simplex’s corresponding 4-simplex is subdivided, we shade it yellow and hatch it, so $\mathcal {R}_{{\text {used}}}$ (see main text) corresponds to the shaded region. Red edges are in correspondence with “leftover 4-simplices", which are described in the main text and listed at the beginning of each step. For each such subdivision process, a leftover simplex is changed for each edge. If the edge was black before subdivision, a leftover simplex is produced, while if the edge was red before subdivision, the corresponding leftover simplex is removed. The leftover simplices are in correspondence with the boundary of the yellow hatched region, i.e., with $\partial \mathcal {R}_{{\text {used}}}$. At the end, the entirety of $\textrm{Link}(e) \cong S^{2}$ will be part of $\mathcal {R}_{{\text {used}}}$ which completes the barycentric subdivision

Note that the first two of these simplices after the subdivision $\{\langle 0_v 1_e 2_A 3_T 4_\alpha \rangle , \langle 0_w 1_e 2_A 3_T 4_\alpha \rangle \}$ are of the final form we want, so we can keep them and forget about them. The other three $\{\langle 0_v 0_w 1_e 3_T 4_\alpha \rangle ,\langle 0_v 0_w 1_e 2_A 4_\alpha \rangle ,\langle 0_v 0_w 1_e 2_A 3_T \rangle \}$ are not in the final form we seek, so we will need to keep subdividing them away. We will call these other three simplices ‘leftover simplices’ that we will need to deal with.

Note that the leftover simplices are of the form $\{\langle 0_v 0_w 1_e \eta \, \xi \rangle | \langle \eta \, \xi \rangle \in \partial \langle 2_A 3_T 4_\alpha \rangle \}$ and correspond directly to the 1-simplices on the boundary of the 2-simplex $\langle 2_A 3_T 4_\alpha \rangle$.

The procedure will be to iterate this subdivision process by subdividing 4-simplices corresponding to neighboring 2-simplices in $\textrm{Link}(e)$. In general, we will keep track of the set $\mathcal {R}_{{\text {used}}}$ of 2-simplices $\langle 2_A 3_T 4_\alpha \rangle$ in $\textrm{Link}(e)$ for which $\langle 0_v 0_w 2_A 3_T 4_\alpha \rangle$ has been subdivided. This will turn out to be the same as the set of 2-simplices in $\textrm{Link}(e)$ for which both $\langle 0_v 1_e 2_A 3_T 4_\alpha \rangle ,\langle 0_w 1_e 2_A 3_T 4_\alpha \rangle$ have been created. And, it will turn out that the ‘leftover simplices’ will always be of the form $\langle 0_v 0_w 1_e \eta \, \xi \rangle$ for $\langle \eta \, \xi \rangle \in \partial {\mathcal {R}_{{\text {used}}}}$, which are the 1-simplices $\langle \eta \, \xi \rangle$ on the boundary of the set $\mathcal {R}_{{\text {used}}}$ of ‘used’ 2-simplices.

In general, a (2, 4)-move will correspond to adding a simplex to the set $\mathcal {R}_{{\text {used}}}$ so that the size of $\partial \mathcal {R}_{{\text {used}}}$ increases by one. A (3, 3)-move will cause $\#(\partial \mathcal {R}_{{\text {used}}})$ to decrease by one. And a (4, 2)-move will cause $\#(\partial \mathcal {R}_{{\text {used}}})$ to decrease by three. And, every ‘leftover’ 4-simplex that is removed from $\mathcal {R}_{{\text {used}}}$ at a step will be an input to the Pachner move. See, for example Fig. 38. In that Figure, the Pachner moves corresponding to each of the arrows is:

• 1. $\langle 0_v 0_w 2_A 3_T 4_\alpha \rangle {\rightarrow }\{\langle 0_v 1_e 2_A 3_T 4_\alpha \rangle ,\langle 0_w 1_e 2_A 3_T 4_\alpha \rangle ,\langle 0_v 0_w 1_e 3_T 4_\alpha \rangle ,\langle 0_v 0_w 1_e 2_A 4_\alpha \rangle , \langle 0_v 0_w 1_e 2_A 3_T \rangle \}$

• 2. $\{\langle 0_v 0_w 1_e 2_A 3_T \rangle ,\langle 0_v 0_w 2_A 3_T 4_\beta \rangle \}{\rightarrow }\{\langle 0_v 1_e 2_A 3_T 4_\beta \rangle ,\langle 0_w 1_e 2_A 3_T 4_\beta \rangle ,\langle 0_v 0_w 1_e 3_T 4_\beta \rangle , \langle 0_v 0_w 1_e 2_A 4_\beta \rangle \}$

• 3. $\{\langle 0_v 0_w 1_e 3_T 4_\alpha \rangle ,\langle 0_v 0_w 2_B 3_T 4_\alpha \rangle \}{\rightarrow }\{\langle 0_v 1_e 2_B 3_T 4_\alpha \rangle ,\langle 0_w 1_e 2_B 3_T 4_\alpha \rangle ,\langle 0_v 0_w 1_e 2_B 3_T \rangle , \langle 0_v 0_w 1_e 2_A 4_\alpha \rangle \}$

• 4. $\{\langle 0_v 0_w 1_e 2_B 3_T \rangle ,\langle 0_v 0_w 1_e 3_T 4_\alpha \rangle ,\langle 0_v 0_w 2_B 3_T 4_\beta \rangle \}{\rightarrow }\{\langle 0_v 1_e 2_B 3_T 4_\beta \rangle ,\langle 0_w 1_e 2_B 3_T 4_\beta \rangle , \langle 0_v 0_w 1_e 2_B 4_\beta \rangle \}$

At the end of this whole process, we indeed complete the barycentric subdivision with the correct branching structure!

We note that essentially the same process holds in higher dimensions. These Steps 0-4 will go through exactly the same in higher dimensions. Except in dimension d, we will need a series of Steps 0-d. A step k will work in the exact same way, except we are dealing with the link of a $(d-k+1)$-simplex which is a barycentrically subdivided $(k-2)$-sphere. This process of subdividing the d-simplices corresponding to the $(k-2)$ simplices in the link, one simplex at a time, goes through in the exact same manner. (In fact, we could have used this picture in $d=2$ and $d=3$ as well.)

1.3 Pachner connectedness of equivalent flat gauge connections

Now we consider triangulations decorated with flat gauge connections. Consider a triple $(M^d,T,\alpha )$, where T is a triangulation of d-manifold $M^d$ and $\alpha$ is some flat gauge connection. We show that any gauge equivalent gauge fields on $(M^d,T)$ can be connected by Pachner moves, in the case of 1-form gauge fields for general groups and higher-form gauge fields for Abelian groups.

First, we explain the preliminaries of defining gauge fields on triangulations. Then we prove the connected of equivalent gauge fields.

1.3.1 Preliminaries

i. Gauge fields on a triangulation Let us review how to formulate gauge fields and higher-form gauge fields on a triangulation. Here, we will be assuming all gauge fields are flat. Throughout this section, we will work with a pair (M, T) of a manifold M and triangulation T.

A flat 1-form gauge field with gauge group G on M is an assignment of group elements ${\textbf{g}}_{ij} \in G$ on all of the directed 1-simplices $\langle ij \rangle$ satisfying certain consistency constraints. The first constraint is that ${\textbf{g}}_{ij} = {\textbf{g}}_{ji}^{-1}$, that the group elements are inverse in opposite directions. Second, for every 2-simplex $\langle i j k \rangle$, there is a ‘flatness’ condition ${\textbf{g}}_{ij}{\textbf{g}}_{jk}={\textbf{g}}_{ik}$. This definition does not depend on whether G is Abelian or non-Abelian. Note that if G is Abelian, we can express the gauge field ${\textbf{g}}_{ij}$ as a cochain ${\textbf{g}} \in C^1(M,G)$. The flatness condition reduces to $\delta {\textbf{g}} = 1$ (using the multiplicative notation that $\delta {\textbf{g}} (\langle ijk \rangle ) = {\textbf{g}}_{ij}{\textbf{g}}_{jk}{\textbf{g}}_{ik}^{-1}$). For Abelian G this means that ${\textbf{g}}$ would be a closed cocycle, that ${\textbf{g}} \in Z^1(M,G)$. Two 1-form gauge fields ${\textbf{g,g}}'$ are gauge equivalent if there exists a function ${\textbf{h}}_i$ on vertices of the triangulation such that ${\textbf{g}}'_{ij} = {\textbf{h}}_i {\textbf{g}}_{ij} {\textbf{h}}_j^{-1}$. For Abelian gauge fields, this equivalence condition is the same as transforming the cocycle $a \rightarrow a + \delta \lambda$ where $\lambda \in C^0(M,A)$ is some function on vertices. So, equivalence classes of Abelian gauge fields are given by elements of $H^1(M,A)$. For non-Abelian gauge fields, the more general statement is that gauge fields are in correspondence with $\textrm{Hom}(\pi _1(M), G) / G$.²³

Higher-form gauge fields for Abelian groups A are defined similarly. A k-form gauge field with gauge group A can be identified with an element $a \in Z^{k}(M,A)$. So, there would be group elements on each k-simplex with analog of the ‘flatness’ condition being $\delta a = 0$ (using the additive notation for $\delta a$). Two higher-form gauge fields $a,a'$ are equivalent if $a' = a + \delta \lambda$ where $\lambda \in C^{k-1}(M,A)$ is some $(k-1)$-cochain. This means that equivalence classes of higher-form A-gauge fields are in bijection with $H^{k}(M,A)$.

ii. Pachner moves with background gauge fields First, we describe Pachner moves in the presence of background gauge fields. The idea is that since a Pachner move changes the simplices in the triangulations, we can assign gauge fields to the new simplices that are consistent with the flatness conditions and the fields on the other unchanged simplices. Another way to think about a gauge field on series of Pachner moves on M is that it is a gauge field on the bordism geometry that gives a triangulation of $M \times I$.

We give some examples of Pachner moves in the presence of a background gauge field in Fig. 39 for 1-form fields and Fig. 40 for higher-form ones. Note that for a 1-form gauge field, the only situation in which there are several possibilities for the final gauge field after the Pachner move is the $(1,d+1)$ move in which a new vertex is created. This is because if all the vertices involved are in the triangulation both before and after the move, then the flatness condition ${\textbf{g}}_{ij}{\textbf{g}}_{jk}={\textbf{g}}_{ik}$ fixes all the links’ group elements. But if a new vertex v is created inside a simplex $\langle 0 \cdots d \rangle$, then there are different choices of gauge field are parameterized by any choice $g_{0 v} \in G$. However, note as in Fig. 40 that for higher-form gauge fields, there are usually more possibilities.

Fig. 39

General Pachner move in $d=2$ with background gauge field. The 3-simplex with the gauge field group elements is on top, and the gauge fields in the presence of a Pachner move is on the bottom

Fig. 40

Example Pachner moves in $d=3$ with background 2-form $\mathbb Z_2$ gauge field, a. A 2-simplex $\langle ijk \rangle$ with a(ijk) non-trivial is shaded in red. The Poincaré dual pictures are below, where a is dual to a closed set of lines

Note that the branching structure plays no role in these gauge transformations. As such, for the remainder of this section, any simplex $\langle 0 \cdots n \rangle$ we write down will be unordered.

1.3.2 Gauge transformations via Pachner moves for 1-form gauge fields

First, we explain the procedure to do a gauge transformation for 1-form gauge fields. The method and picture for the proof will be quite similar to the last step of the construction of the barycentric subdivision in Sect. E 2 b iii.

First, let’s phrase more precisely what we want to achieve. Given a 1-form gauge field ${\textbf{g}}_{ij}$, we want to implement the gauge transformations ${\textbf{g}}_{ij} \rightarrow {\textbf{h}}_i {\textbf{g}}_{ij} {\textbf{h}}_j^{-1}$ for functions ${\textbf{h}}_i$ on vertices $\{i\}$. All such transformations can be implemented by elementary transformations such that ${\textbf{h}}_i = {\textbf{h}} \ne {\textbf{1}}$ on exactly a single vertex i. This means that for any given vertex i and some ${\textbf{h}} \in G$, we want to change ${\textbf{g}}_{ik} \rightarrow {\textbf{h}} {\textbf{g}}_{ik}$ for each neighboring vertex k.

Fig. 41

Pachner moves that will implement a gauge transformation of a 1-form gauge field on a vertex i in $d=3$ as in Sect. E 3 b. The $w_{\cdots }$ are in $\textrm{Link}(i)$ (note i is not shown), which is homeomorphic to a 2-sphere. Each 3-simplex $\langle i w_{\alpha } w_{\beta } w_{\gamma } \rangle$ is in correspondence with a 2-simplex $\langle w_\alpha w_\beta w_\gamma \rangle \in \textrm{Link}(i)$. We perform a set of Pachner moves for each 3-simplex, one at a time, and shade the corresponding 2-simplex after completing those moves. The first move creates a vertex j, and the last move removes the original i; suitable decorations of gauge fields (described in the main text) effectively implement the gauge transformation. Each set of Pachner moves creates and removes “leftover” 3-simplices in correspondence with the (depicted red) edges of the 2-simplex such that the set of leftover 3-simplices correspond with the boundary of the yellow shaded region

To implement this process, we consider the link of the vertex i, $\textrm{Link}(i)$, which consists of the $(d-1)$-simplices $\langle v_0 \cdots v_{d-1} \rangle$ for which $i \not \in \{v_0 \cdots v_{d-1}\}$ and that $\langle i \, v_0 \cdots v_{d-1} \rangle$ is a simplex in the triangulation. In general, we will have that $\textrm{Link}(i)$ is homeomorphic to a $(d-1)$-sphere.

For some arbitrary choice of $\langle v_0 \cdots v_{d-1} \rangle \in \textrm{Link}(i)$, we will implement the $(1,d+1)$-move creating a new vertex j

$$\begin{aligned} \langle i \, v_0 \cdots v_{d-1} \rangle&\rightarrow \{\langle j \, v_0 \cdots v_{d-1} \rangle , \langle i \, j \, v_1 \cdots v_{d-1} \rangle , \dots , \langle i \, j \, v_0 \cdots \hat{v}_k \cdots v_{d-1} \rangle , \\&\quad \dots \langle i \, j \, v_0 \cdots v_{d-2} \rangle \} \end{aligned}$$

where $\langle i \, j \, v_0 \cdots \hat{v}_k \cdots v_{d-1} \rangle$ is a $(d-1)$-simplex consisting of all of vertices $\{i, j, v_0, \cdots , v_{d-1}\}$ except for $v_k$. Along with this move, we will choose the gauge field ${\textbf{g}}'$ at the end of this move so that ${\textbf{g}}_{j i} = {\textbf{h}}$ which will enforce that ${\textbf{g}}_{j v_k} = {\textbf{g}}_{j i} {\textbf{g}}_{i v_k} = {\textbf{h}} {\textbf{g}}_{i v_k}$.

From here, the strategy will be similar to Sect. E 2 b iii. However, the end goal will be to replace the vertex i with j in all connections involving vertex i and at the end of the day remove i with a $(d+1,1)$-move. If we are able to accomplish this without doing any more $(1,d+1)$-moves (i.e. without creating any new vertices), then the fact that the gauge fields are uniquely determined at the end of each move will mean that the final gauge fields ${\textbf{g}}_{j v_k}^{\text {final}}$ will all be ${\textbf{g}}_{j v_k}^{\text {final}} = {\textbf{h}} {\textbf{g}}_{i v_k}$. In other words, we will have accomplished the desired gauge transformation.

Note that at the end of the first move above, there is one simplex $\langle j \, v_0 \cdots v_{d-1} \rangle$ that is of the final form we want, and the rest $\langle i \, j \, v_0 \cdots \hat{v}_k \cdots v_{d-1} \rangle$ are ‘leftover’ d-simplices that we want to eliminate via Pachner moves.

To complete the process, we will iteratively perform Pachner moves, one for each $(d-1)$-simplex in $\textrm{Link}(i)$. At each step, define $\mathcal {R}_{{\text {used}}}$ to be the set of $(d-1)$-simplices $\{\langle w_0 \cdots w_{d-1} \rangle \}$ for which we have performed a Pachner move involving $\langle i \, w_0 \cdots w_{d-1} \rangle$. Next, choose a $(d-1)$-simplex $\langle w_0 \cdots w_{d-1} \rangle$ such that the next $\mathcal {R}_{{\text {used}}}' = \mathcal {R}_{{\text {used}}} \sqcup \{\langle w_0 \cdots w_{d-1} \rangle \}$ is connected. Perform a $(k+1,d+1-k)$ Pachner move, where k is the number of $(d-2)$-simplices that the boundary $\partial \langle w_0 \cdots w_{d-1} \rangle$ shares with $\partial \mathcal {R}_{{\text {used}}}$. The $(k+1)$ “input" d-simplices consist of $\langle i w_0 \cdots w_{d-1} \rangle$ and the “leftover simplices" corresponding to the shared $(d-2)$-simplices. The outputs of the Pachner move are the $(d+1-k)$ “leftover simplices" corresponding to the unshared $(d-2)$-simplices of $\partial \{\langle w_0 \cdots w_{d-1} \rangle \}$. This Pachner move produces some new leftover simplices so that the new set of all leftover simplices corresponds to $\partial \mathcal {R}_{{\text {used}}}'$. Repeat this process until $\mathcal {R}_{{\text {used}}} = \textrm{Link}(i)$; the last step will be a $(d+1,1)$-move that entirely eliminates the vertex i.

We illustrate the steps of Pachner moves in $d=3$ in Fig. 41, where the link of i is some triangulation of the sphere $S^2$. Note the similarity of the process to Fig. 38. The first two moves written out are:

• 1. $\langle i \, w_1 w_2 w_3 \rangle \rightarrow \{\langle j \, w_1 w_2 w_3 \rangle ,\langle i \, j \, w_2 w_3 \rangle ,\langle i \, w_1 \, j \, w_3 \rangle ,\langle i \, w_1 w_2 \, j \rangle \}$

• 2. $\{\langle i \, w_1 w_3 w_4 \rangle ,\langle i \, j \, w_1 w_3 \rangle \} \rightarrow \{\langle j \, w_1 w_3 w_4 \rangle ,\langle i \, j \, w_1 w_4 \rangle ,\langle i \, w_3 \, j \, w_4 \rangle \}$

1.3.3 Higher-form gauge transformations via Pachner moves

Now, we will explain how a similar process applies to implement gauge transformations of higher-form gauge fields. Given an n-form gauge field $\alpha \in Z^n(M,A)$, we want to implement the transformation $\alpha \rightarrow \alpha + \delta \lambda$ for $\lambda \in C^{n-1}(M,A)$. If we can implement the transformation for the basic gauge transformations for which $\lambda$ is nonzero on a single $(n-1)$-simplex, then any general gauge transformation can be implemented.

The set-up will be similar in that we will implement the transformation for $\lambda = a$ on $(n-1)$-simplex $s_{n-1} = \langle i_0 \cdots i_{n-1} \rangle$ by considering $\textrm{Link}(s_{n-1})$, which will be some triangulation of the sphere $S^{d-n}$. First we will describe the series of Pachner moves and then describe how to decorate them with gauge fields. We recommend the reader to look at the Fig. 42 for an illustration of the moves and the 2-form $\mathbb Z_2$ gauge fields in $d=3$ while reading the rest of the argument.

Fig. 42

Implementing $\alpha \rightarrow \alpha + \delta \lambda$ for 2-form $\mathbb Z_2$ gauge fields in $d=3$. (Top) The original link of an $(n-1)$-simplex and its dualized version. No gauge field are specified in this part. (Bottom) Decoration with gauge fields and (as viewed from the dual persepective) Pachner moves implementing $\alpha \rightarrow \alpha + \delta \lambda$ for $\lambda (s_{n-1}) = 1$ and $\lambda (t)=0$ for $t \ne s_{n-1}$. Note that throughout the top line of moves, all dual edges on newly created edges on ‘bottom hexagon’ are always zero. And, at the end of the top line that the dual edges on the ‘top hexagon’ are equal to the ‘orignal hexagon’ except gauge transformed. This is illustrating the $\alpha (t) = 0$ for n-simplices $t = \langle v \, j \, i_0 \cdots \hat{i}_k \cdots i_{n-1} \rangle , k \ne 0$ and how at the end of the first half of moves $\alpha (v \, j \, i_1 \cdots i_{n-1}) = a + \alpha ^{\text {original}}(v i_0 \cdots i_{n-1})$ as described in the main text. The bottom line of moves brings us back to the original triangulation and shows that $\alpha \rightarrow \alpha + \delta \lambda$ has indeed been implemented

The procedure will be similar to the 1-form case and the barycentric subdivision. First pick some $\langle w_0 \cdots w_{d-n} \rangle \in \textrm{Link}(s_n)$ and do a $(1,d+1)$-move creating a new vertex j.

$$\begin{aligned} \begin{aligned}&\langle i_0 \cdots i_n \, w_0 \cdots w_{d-n} \rangle \\&\quad \rightarrow \{\langle j \, i_1 \cdots i_{n-1} \, w_0 \cdots w_{d-n} \rangle ,\langle i_0 \, j \, i_2 \cdots i_{n-1} \, w_0 \cdots w_{d-n} \rangle , \dots ,\\&\qquad \langle i_0 \, i_1 \cdots i_{n-2} \, j \, w_0 \cdots w_{d-n} \rangle , \\&\qquad \langle i_0 \cdots i_{n-1} \, j \, w_1 \cdots w_{d-n} \rangle , \langle i_0 \cdots i_{n-1} \, w_0 \, j \, w_2 \cdots w_{d-n} \rangle , \dots ,\\&\qquad \langle i_0 \cdots i_{n-1} \, w_0 w_1 \cdots w_{d-n-1} j \rangle \} \end{aligned} \end{aligned}$$

(E3)

And proceeding in the same fashion as for the 1-form case, we can keep doing moves one at a time and collect all of the ‘leftover’ simplices that involve both j and the entire simplex $s_n$; leftover simplices contain all of $\{j, i_0, \dots , i_{n-1}\}$ and are to be gotten rid of one at a time. After each step, there will be a ‘used’ region $\mathcal {R}_{{\text {used}}} \subset \textrm{Link}(s_n)$ that corresponds to those $(d-n)$-simplices $\langle v_0 \cdots v_{d-n} \rangle$ for which the d-simplices

$$\begin{aligned} \{\langle j \, i_1 \cdots i_{n-1} \, v_0 \cdots v_{d-n} \rangle ,\langle i_0 \, j \, i_2 \cdots i_{n-1} \, v_0 \cdots v_{d-n} \rangle , \dots ,\langle i_0 \, i_1 \cdots i_{n-2} \, j \, v_0 \cdots v_{d-n} \rangle \} \end{aligned}$$

have all been created. And, each step will leave ‘leftover’ d-simplices that are in bijection with $\partial \mathcal {R}_{{\text {used}}}$. So for some $\langle x_0 \cdots x_{d-n-1} \rangle \in \partial \mathcal {R}_{{\text {used}}}$, the d-simplex

$$\begin{aligned} \langle j i_0 \cdots i_{n-1} \, x_0 \cdots x_{d-n-1} \rangle \end{aligned}$$

is a ‘leftover’ simplex which still exists and which we will eliminate. After every step, we do a Pachner move on a simplex neighboring the ‘used’ region $\mathcal {R}_{{\text {used}}}$ and eventually, $\mathcal {R}_{{\text {used}}}$ will consist of the entire $\textrm{Link}(s_{n-1})$. We can collect all of the terms at the end of this process and they will all be of the form:

$$\begin{aligned} \langle j \, i_0 \, \cdots \hat{i}_k \cdots i_{n-1} \, v_0 \cdots v_{d-n} \rangle \end{aligned}$$

where $\hat{i}_k$ is the vertex that is not included in the simplex and $\langle v_0 \cdots v_{d-n} \rangle$ is any $(d-n)$-simplex in the link. Note that the number of d-simplices in the link at this point is n times the number of d-simplices originally, so our triangulation is not the same as it was originally. To remedy this, we will reverse the order of the Pachner moves to bring us back to the original triangulation. The hope is that at the end of this process, if we are careful about which group elements we place on the links, then the change $\alpha \rightarrow \alpha + \delta \lambda$ can be implemented, for $\lambda (s_{n-1}) = a \in A$ and $\lambda = 0$ on all other $(n-1)$-simplices.

To do this, we first specify the group elements on n-simplices after the first move (E3) in the sequence. Consider a move for which the n-simplex $\langle j \, i_0 \cdots i_{n-1} \rangle$ is given value $\alpha (\langle j \, i_0 \cdots i_{n-1} \rangle )=a$ and for which $\alpha (t) = 0$ for every n-simplex t of the form $t = \langle v \, j \, i_0 \cdots \hat{i}_k \cdots i_{n-1} \rangle , k \ne 0$. Note that this condition together with $\alpha (\langle j \, i_0 \cdots i_{n-1} \rangle )=a$ and the values of the unchanged n-simplices completely specify each $\alpha (\langle v j \, i_1 \cdots i_{n-1} \rangle )$ as

$$\begin{aligned} \alpha (\langle v j \, i_1 \cdots i_{n-1} \rangle ) = \alpha ^{{\text {original}}}(\langle v i_0 \, i_1 \cdots i_{n-1} \rangle ) + a \end{aligned}$$

after this first move because of the condition $\delta \alpha = 0$. (The values on other n-simplices can be fixed arbitrarily and do not matter.)

Then, we proceed by doing Pachner moves as described above. In general, for each vertex $v \in \textrm{Link}(s_{n-1})$ and each n-simplex $t = \langle v \, j \, i_0 \cdots \hat{i}_k \cdots i_{n-1} \rangle , k \ne 0$, we will require that $\alpha (t) = 0$. Note that as long as $\partial \mathcal {R}_{{\text {used}}}$ is nonempty $\langle j \, i_0 \cdots i_{n-1} \rangle$ will still exist in the simplicial complex with $\alpha (\langle j \, i_0 \cdots i_{n-1} \rangle ) = a$, and it only disappears at the last step when $\partial \mathcal {R}_{{\text {used}}}$ vanishes and there are no ‘leftover’ simplices. This together with the values of $\alpha$ on the other n-simplices will again specify $\alpha (\langle v j \, i_1 \cdots i_{n-1} \rangle ) = \alpha ^{{\text {original}}}(\langle v i_0 \cdots i_{n-1} \rangle ) + a$ since $\delta \alpha = 0$.

Now consider the end of this first half of the steps when the d-simplices are all of the form $\langle j \, i_0 \, \cdots \hat{i}_k \cdots i_{n-1} \, v_0 \cdots v_{d-n} \rangle$ for $k \in \{0 \cdots n-1\}$ and $\langle v_0 \cdots v_{d-n} \rangle \in \textrm{Link}(s_{n-1})$. We will have that for each vertex $v \in \textrm{Link}(s_{n-1})$, $\alpha (v j \, i_1 \cdots i_{n-1})$ will be $a + \alpha ^{\text {original}}(v i_0 \cdots i_{n-1})$, where $\alpha ^{\text {original}}(v i_0 \cdots i_{n-1})$ was the value of the cochain at the beginning of this process.

Now the second half of the steps consists in doing the Pachner moves backwards to the original triangulation. Since $\langle j i_0 \cdots i_{n-1} \rangle$ was removed at the last Pachner move, we will be free to reassign $\alpha (\langle j i_0 \cdots i_{n-1} \rangle ) = 0$ this time around when it reappears. Continuing this backwards process with this value of $\alpha (\langle j i_0 \cdots i_{n-1} \rangle ) = 0$ together with $\delta \alpha = 0$ will completely specify the cochain everywhere. And at the end, we will have that for each vertex $v \in \textrm{Link}(s_{n-1})$, $\alpha (v i_0 \cdots i_{n-1}) = a + \alpha ^{\text {original}}(v i_0 \cdots i_{n-1})$. This is the change $\alpha \rightarrow \alpha + \delta \lambda$ we seek and we are done.

Effect of Vertex-Basis Transformation $\Gamma ^{\psi \psi }_1$

Here, we examine the effect of vertex-basis transformations $\Gamma ^{a b}_c$ (see Eq. (45)) on the bosonic shadow $Z_b(M,A_b,f)$ for closed M. As explained in Sects. 6.1.2, 6.1.3, the restriction of only considering $\Gamma ^{\psi \psi }_1 = +1$ leaves the path integral invariant using the same proof as in [30]. In this section we will review the proof and examine how the path integral changes under more general $\Gamma ^{\psi \psi }_1 \ne 1$.

First, we will see that the presence of an anti-unitary symmetry action constrains $\Gamma ^{\psi ,\psi }_1$. Gauge transformations must preserve the constraint Eq. (89), and therefore must leave $U_{\textbf{g}}(\psi ,\psi ;1)$ invariant. Using the transformation rules Eq. (86) for U under a vertex basis transformation, we find $U_{\textbf{g}}$ is invariant provided

$$\begin{aligned} \Gamma ^{\psi \psi }_1 [(\Gamma ^{\psi \psi }_1)^{-1}]^{s({\textbf{g}})}=1. \end{aligned}$$

(F1)

If ${\textbf{g}}$ has a unitary action (i.e. if $s({\textbf{g}}) = +1$), then the above equation is automatically satisfied for any $\Gamma ^{\psi \psi }_1 \in \textrm{U}(1)$. However, if ${\textbf{g}}$ is anti-unitary (i.e. $s(\textbf{g}) = *$), then the above constraint gives $(\Gamma ^{\psi \psi }_1)^2 = +1$, which is only true if

$$\begin{aligned} \Gamma ^{\psi \psi }_1 = \pm 1. \end{aligned}$$

(F2)

The only nontrivial transformation allowed in the presence of anti-unitary symmetries is thus $\Gamma ^{\psi \psi }_1 = -1$.

Now let the amplitude after a vertex basis gauge transformation be $Z_b'(M,A_b,f)$. As above, if there are no anti-unitary symmetry actions in the category, we may consider $\Gamma ^{\psi \psi }_1$ to be any phase. In this case, we will show that

$$\begin{aligned} Z_b'(M,A_b,f) = Z_b(M,A_b,f), \end{aligned}$$

(F3)

i.e. that the amplitude is fully invariant under any $\Gamma ^{\psi \psi }_1$ transformation. However, the case of an anti-unitary action will give the change

$$\begin{aligned} Z_b'(M,A_b,f) = Z_b(M,A_b,f) \cdot (-1)^{\int w_1 \cup f}, {\text { if }} \Gamma ^{\psi \psi }_1 = -1 \end{aligned}$$

(F4)

for the only nontrivial choice of $\Gamma ^{\psi \psi }_1$.

Now, we explain the argument for the above statements. In the process, we will review how the path integral is totally invariant under all transformations $\Gamma ^{a b}_c$ that are not $\Gamma ^{\psi \psi }_1$. The relevant parts of the diagrammatics to consider are given in Fig. 43, which show the directions that the background fermion lines exit from a 3-simplex and how the diagrams look with respect to relative induced orientations on 3-simplices.

There are two potential classes of fusion vertices that we need to consider. First are those on a 3-simplex $\Delta _3$ that involve the anyons $b_{\Delta _3},b_{\Delta _3} \times \psi ^{f(\Delta _3)}$. There are three such fusion vertices for each 3-simplex in each diagram. For example, this class of fusion vertices for $\langle 0123 \rangle$ would be $\{(012 , 023 \leftrightarrow 0123), (0123 , \psi ^{f(0123)} \leftrightarrow 0123 \times \psi ^{f(0123)}), (0123 \times \psi ^{f(0123)} \leftrightarrow 013 , 123\}$. Since M is closed, $\Delta _3$ occurs in exactly two 4-simplices. If $\Delta _3$ is not part of $w_1$, then each fusion vertex in this class appears twice, but with opposite orientation. Under a vertex basis transformation, one appearance contributes $\Gamma ^{a b}_c$ and the other contributes $(\Gamma ^{a b}_c)^{-1}$, which cancel out in the product over 4-simplices. The same argument also applies if $\Delta _3$ is part of $w_1$, since even though the fusion vertices appear with the same induced orientation, one factor will be complex conjugated by being in a region of the diagram with an anti-unitary twist.

Fig. 43

Relevant parts of the diagrammatics needed to analyze the effects of vertex-basis transformations $\Gamma ^{a b}_c$. (Left) The orientations of the background fermion lines emanating from the 3-simplices, where each $\hat{i}$ represents the 3-simplex missing $i \in \{0,1,2,3,4\}$. On a $+$-oriented [−-oriented] 4-simplex, the lines emanating from $\hat{i}$ for i even go up [down] and those from $\hat{i}$ for i odd go down [up]. (Right) The relative orientations of the 3-simplices on its two adjacent 4-simplices. The left-side shows how it looks away from $w_1$ and the right-side shows how they look when the 3-simplex is part of $w_1$. See Fig. 31 and the surrounding text for more discussion regarding induced orientation on the 3-simplices

The second class of fusion vertices involve the junctions where the five potential background fermion lines on a 4-simplex fuse into each other. Here, the only relevant basis transformations are $\Gamma ^{\psi \psi }_1$ since we only consider gauges such that all transformations $\Gamma ^{\psi 1}_\psi ,\Gamma ^{1 \psi }_\psi$ are 1.²⁴ First, we will use the trivalent resolution in the diagrams to split the background fermion lines into a set of distinct loops which enter and exit a 4-simplex through two different 3-simplices. This splits the loop up in the same way as it would the trivalent resolution of the dual 1-skeleton, as in Fig. 5. Without loss of generality, we will consider the case that f is dual to a single loop (note the final result given by the multiplicative factor $(-1)^{\int w_1 \cup f}$ is linear in f). Since the factors can be considered independently on different loops, the proof for a single loop implies the general case.

In Fig. 44, the different cases of when the amplitude gets multiplied by $\Gamma ^{\psi \psi }_1$ or $(\Gamma ^{\psi \psi }_1)^{-1}$ are enumerated. It is convenient to think in terms of the vector $v_d$ along the dual 1-skeleton, as explained in the text below Fig. 44.

Fig. 44

Factors of $\Gamma ^{\psi \psi }_1$ that multiply the amplitude $Z_b$ and how they relate to the parity of the 4-simplex and the 3-simplices in question. All other cases of $\hat{i},\hat{j}$ give a trivial factor of $\Gamma ^{\psi 1}_\psi = 1$ or $\Gamma ^{1 \psi }_\psi = 1$. The top case of the amplitude being multiplied by $\Gamma ^{\psi \psi }_1$ corresponds to the case where the vector $v_d$ along the dual 1-skeleton (see Sect. A 2) points as between $\hat{i}$ and $\hat{j}$. The bottom case of a factor $(\Gamma ^{\psi \psi }_1)^{-1}$ corresponds to $v_d$ pointing as along this leg. As such, the only times when such factors appear correspond to these ‘direction switches’ of $v_d$ along the loop. The top/bottom cases can be referred to as a “$+ {\text { to }} +$ switch” / “$- {\text { to }} -$ switch” because they interpolate between two 3-simplices both with induced $+$-orientation / both −-orientation (see Fig. 30)

From here, we can compute the contribution of the amplitude change from this loop. The logic is summarized pictorially in Fig. 45, and is explained below.

Fig. 45

Pictorial argument for the changes in the amplitude under $\Gamma ^{\psi \psi }_1$ in the orientable and non-orientable cases. The pink line is a single fermion loop considered with ends at the black rectangles being identified, and the red arrows are the $v_d$ vector field along the loop. refer to the number of direction switches of $v_d$ that occur inside a 4-simplex; in the non-orientable case this means they do not count those coming from crossing $w_1$

Let us first consider the case with no anti-unitary symmetries. Denote as the number of $+ {\text { to }} +$ switches and as the number of $- {\text { to }} -$ switches of $v_d$ in the loop (see Fig. 44). The amplitude change from the loop is

Note that since there are no anti-unitary symmetries, the manifold must be orientable. All of the direction switches that occur inside a loop are inside a 4-simplex and accounted for above. This means that . So, the change in amplitude is , which is invariant as we wanted.

In the anti-unitary case, the above argument needs modification due to the presence of $w_1$. In the orientable case, the $v_d$ vector always points in the same direction going across a 3-simplex. In the non-orientable case, the $v_d$ vector changes directions going across $w_1$, like in Fig. 23. However, only direction switches inside a 4-simplex contribute to the amplitude change: direction switches across $w_1$ are not accounted for in the diagrammatics and don’t contribute any factors. As such, we will define to be the number of such direction switches occurring only inside a 4-simplex. Since the total number of direction switches (i.e. including those coming from $w_1$) is even, we have the equality . And if we consider the only possible nontrivial $\Gamma ^{\psi \psi }_1 = -1$ in the antiunitary case, we find

(F5)

which is what we set out to show.

${{\textrm{Sq}}}^2 + A_b^{*} \omega _2$ Anomaly of $Z_b(M,f)$

Here, we demonstrate how the ${{\textrm{Sq}}}^2 + A_b^{*} \omega _2$ anomaly of the bosonic shadow can be derived. We derive this by considering a Pachner move $M \rightarrow M'$ in which M and $M'$ differ by gluing on a 5-simplex to M on some collection of 4-simplices and removing the original 4-simplices, analogously to Fig. 34 in two dimensions higher.

The Pachner move $M \rightarrow M'$ can thought of as a 5-manifold W with boundary $\partial W = M \bigsqcup \overline{M'}$ and consists of a single 5-simplex $\langle 012345 \rangle$. In addition, after decorating the 3-form field f dual fermion lines and the $G_b$ gauge field $A_b$ onto these Pachner moves, the fields f and $A_b$ can be thought to be extended onto this 5-manifold W.

The invariance property we want to show is:

(G1)

We can alternatively express the sign difference as:

$$\begin{aligned} (-1)^{\int _{\langle 012345 \rangle } {{\textrm{Sq}}}^2(f) + f \cup A_b^{*} \omega _2} = (-1)^{\int _{\langle 012345 \rangle } ({{\textrm{Sq}}}^2 + f_\infty A_b^{*} \omega _2 )(f)} \end{aligned}$$

First we will describe how the ${{\textrm{Sq}}}^2(f)$ part of the anomaly arises in the Pachner calculation. This part is ‘universal’ in the sense that it would occur even in the absence of symmetry domain walls.

Then we will describe how the U and $\eta$ symbols coming from the symmetry domain walls in the Pachner move gives a contribution of $f \cup A_b^{*} \omega _2$ to the $Z_b$ anomaly.

Technically, we need to show this for all possible Pachner moves with all possible decorated branching structures and fermion and gauge field decorations. We will only describe these calculations in detail for the case of the 3-3 Pachner move for which we change the simplices from $\langle 12345 \rangle ,\langle 01345 \rangle ,\langle 01235 \rangle$ in M to $\langle 01234 \rangle ,\langle 01245 \rangle ,\langle 02345 \rangle$ in $M'$. However the arguments will indeed translate to the other cases. We will explain how at the end of each sub-part of the calculation.

1.1 The ${{\textrm{Sq}}}^2$ part

Here we describe the ${{\textrm{Sq}}}^2$ part of the anomaly. The present calculation is valid when $A_b=0$, implying there is no orientation-reversing wall in the vicinity. We will see in the next subsection that the calculation for when there are orientation-reversing walls and nonzero $A_b$ can in fact mapped onto this case.

Since there is no orientation-reversing wall, the move $\langle 12345 \rangle ,\langle 01345 \rangle ,\langle 01235 \rangle$ to $\langle 01234 \rangle ,\langle 01245 \rangle ,\langle 02345 \rangle$ can be checked to take three 4-simplices of the same orientation (say all $+$’s) to three other $+$ 4-simplices. It will eliminate the 2-simplex $\langle 024 \rangle$ and the 3-simplices $\langle 0124 \rangle ,\langle 0234 \rangle ,\langle 0245 \rangle$ on M and create the new 2-simplex $\langle 135 \rangle$ and the 3-simplices $\langle 0135 \rangle ,\langle 1235 \rangle ,\langle 1345 \rangle$ on $M'$. This means that the equation we have to verify is roughly

$$\begin{aligned}{} & {} {\text {``}} \sum _{024,0124,0234,0245} Z_b^+(02345)Z_b^+(01245)Z_b^+(01234) \\{} & {} \quad = \sum _{135,0135,1235,1345} Z_b^+(12345)Z_b^+(01345)Z_b^+(01235) {\text {''}} \end{aligned}$$

where the quotation marks mean we have not accounted for quantum dimension factors. The normalizing quantum-dimension factors for the amplitudes in $Z_b(M)$ in front of each product of diagram evaluations on 4-simplices are:

$$\begin{aligned} \mathcal {D}^{2(N_0 - N_1) - \chi (M)} \frac{\prod _{{\text {all 2-simplices }} \Delta _2} d_{\Delta _2}}{\prod _{{\text {all 3-simplices }} \Delta _3} d_{\Delta _3}} \prod _{{\text {4-simplices }} \Delta _4} \sqrt{ \frac{\prod _{{\text {3-simplices }} \Delta _3 \subset \Delta _4} d_{\Delta _3}}{\prod _{{\text {2-simplices }} \Delta _2 \subset \Delta _4} d_{\Delta _2}} } \end{aligned}$$

where the product over 4-simplices involves $\mathcal {N}_{\Delta _4}$ (see Eq. (95)). Define the above as $\mathcal {N}_{\text {q-dims}}$. Note that every 3-simplex is present in exactly two 4-simplices. This means that the second product over 4-simplices gives a factor of $d_{\Delta _3}$ for every 3-simplex. So the two products over 3-simplices cancel out. The right-hand product over 2-simplices gives a factor of $(\frac{1}{\sqrt{d_{\Delta _2}}})^{|{\text {Link}}(\Delta _2)|}$, since $|{\text {Link}}(\Delta _2)|$ is the number of 3-simplices that $\Delta _2$ is a part of, which is also the number of 4-simplices $\Delta _2$ is a part of (recalling that the link of a $(d-2)$-simplex is always a circle). This gives

$$\begin{aligned} \mathcal {N}_{\text {q-dims}} = \mathcal {D}^{2(N_0 - N_1) - \chi (M)} \prod _{{\text {all 2-simplices }} \Delta _2} \sqrt{d_{\Delta _2}}^{2-|{\text {Link}}(\Delta _2)|} \end{aligned}$$

(G2)

For the 3-3 move $\langle 01234 \rangle ,\langle 01245 \rangle ,\langle 02345 \rangle$ to $\langle 12345 \rangle ,\langle 01345 \rangle ,\langle 01235 \rangle$, ${\text {Link}} (\langle 024 \rangle )$ consists of only the 4-simplices $\langle 01234 \rangle ,\langle 01245 \rangle ,\langle 02345 \rangle$ which are in the original M and that ${\text {Link}}(\langle 135 \rangle )$ consists of only the 4-simplices $\langle 01235 \rangle ,\langle 01345 \rangle , \langle 12345 \rangle$ on the transformed $M'$. This is because those 4-simplices all border their respective 2-simplices and form a loop, so they must be the entire link.

In addition, one can check (by counting the number of 3-simplices containing them before and after) that the sizes of the links of the 2-simplices $\{\langle 045 \rangle ,\langle 245 \rangle ,\langle 025 \rangle ,\langle 014 \rangle , \langle 124 \rangle , \langle 012 \rangle , \langle 034 \rangle , \langle 023 \rangle , \langle 234 \rangle \}$ will all decrease by one whereas those of $\{\langle 123 \rangle , \langle 125 \rangle , \langle 235 \rangle , \langle 013 \rangle , \langle 035 \rangle , \langle 015 \rangle ,\langle 134 \rangle , \langle 145 \rangle ,\langle 345 \rangle \}$ increase by one.

Finally, note that the number of 2-,3-,4-simplices in the triangulation stays the same before and after the Pachner move. So $N_0-N_1$ stays the same before and after since $N_0 - N_1 + N_2 - N_3 + N_4 = \chi (M)$ is a topological invariant throughout the process.

All together means that the precise equation we need to check is

$$\begin{aligned} \begin{aligned} LHS'&:= \frac{1}{\sqrt{d_{045}d_{245}d_{025}d_{014}d_{124}d_{012}d_{034}d_{023}d_{234}}}\\&\quad \sum _{024,0124,0234,0245}\frac{1}{\sqrt{d_{024}}}\frac{Z_b^+(01234)Z_b^ +(01245)Z_b^+(02345)}{\mathcal {N}_{01234}\mathcal {N}_{01245}\mathcal {N}_{02345}}\\&=\frac{(-1)^{f(0345)f(0123)+f(0145)f(1234)+f(0125)f(2345)}}{\sqrt{d_{123} d_{125}d_{235}d_{013}d_{035}d_{015}d_{134}d_{145}d_{345}}}\\&\quad \sum _{135,0135,1235,1345} \frac{1}{\sqrt{d_{135}}}\frac{Z_b^+(01235)Z_b^+(01345) Z_b^+(12345)}{\mathcal {N}_{01235}\mathcal {N}_{01345}\mathcal {N}_{12345}}\\&=: RHS' \cdot (-1)^{{{\textrm{Sq}}}^2(f)(012345)} \end{aligned} \end{aligned}$$

(G3)

where each $\frac{Z_{{\text {4-simplex}}}}{\mathcal {N}_{{\text {4-simplex}}}}$ is an unnormalized 15j symbol and all the quantum dimension factors are there to compensate for the deletion of 024, the addition of 135, and the changes in sizes of other 2-simplex links as stated above. The factor $(-1)^{f(0345)f(0123)+f(0145)f(1234)+f(0125)f(2345)} = (-1)^{(f \cup _1 f)(012345)} = (-1)^{{{\textrm{Sq}}}^2(f)(012345)}$ is the anomaly factor that we wanted to find in this subsection.

To verify this, we will first need the “merging lemmas" in Fig. 46 that allow us to convert products of diagrams into single diagrams, slightly modifying the ones in [30, 100] to account for the fermion lines. Using the merging lemmas, Eq. (G3) follows from the diagrammatic calculation in Figs. 47, 48 and 49.

Fig. 46

(Top) Merging Lemma I and (bottom) Merging Lemma II with diagrammatic proof. The second equality of Merging Lemma II uses Merging Lemma I. Here, we take $\tilde{\psi }$ to be either the identity anyon or the fermion $\psi$. Each step uses resolutions of the identity as in Eq. (35). In addition, at one point the fermion lines can be ‘peeled off’ using the same resolution of identity as well as the freedom to ‘bend’ the fermion lines discussed in Sect. 5.2

Fig. 47

Diagrammatic simplification of $LHS'$ of Eq. (G3). The first step uses Merging Lemma I for $x = 0245$ and the second uses Merging Lemma II for $x=0124,y=0234,z=024$

Fig. 48

Diagrammatic simplification of $RHS'$ of Eq. (G3). The first step uses Merging Lemma I for $x = 1235$ and the second uses Merging Lemma II for $x=1345,y=0135,z=135$

Fig. 49

Comparison of the fermion lines of Figs. 47 and 48 shows that $LHS'/RHS' = (-1)^{{{\textrm{Sq}}}^2(f)(012345)}$, which is the ${{\textrm{Sq}}}^2$ anomaly. Note that one has to arrange the fermion lines in matching orders to do this comparison, which entails rearranging the fusion channels in this diagram to go in the order shown here

This calculation generalizes to more general Pachner moves, and we leave it to the reader to verify this explicitly. The idea is the same for each possible move. First, one has to check that the quantum dimension factors all match up as before. Then one can use the merging lemmas to verify the ${{\textrm{Sq}}}^2$ as before. In all cases, the final comparison of fermion lines will look like Fig. 49. This is because to produce that figure, we reflected the fermion lines fusing from the RHS, which means that their orientation in that fused figure is the same as how it would be for a − simplex. For general moves, the fusion of lines from $\langle 01234 \rangle ,\langle 01245 \rangle ,\langle 02345 \rangle$ will appear as they do in a $+$ 4-simplex and those from $\langle 12345 \rangle ,\langle 01345 \rangle ,\langle 01235 \rangle$ will appear as they do in a − 4-simplex.

1.2 The $A_b^{*} \omega _2$ part

We will find that the $f \cup A_b^{*} \omega _2$ part of the anomaly arises by considering how the U and $\eta$ symbols coming from the diagrams all differ. We again consider the same 3-3 Pachner move. We denote ${\textbf{g}}_{ij} = A_b(ij)$ in this subsection.

We now have to deal with how potential orientation-reversing walls coming from the gauge fields interplay with the orientations of the 4-simplices in the move. The relative orientations of a 3-simplex can be related to the gauge fields surrounding each of the two 4-simplices a 3-simplex via the $f_\infty$ map as in Sect. D 2.

Examining the orientations of the simplices (for which one can look at Figs. 47 and 48 along with Fig. 10) shows that for all the orientations of simplices to be consistent in the 3-3 Pachner move $\langle 01234 \rangle ,\langle 01245 \rangle ,\langle 02345 \rangle$ to $\langle 12345 \rangle ,\langle 01345 \rangle ,\langle 01235 \rangle$, we will need the orientations of $\langle 01245 \rangle ,\langle 02345 \rangle ,\langle 12345 \rangle ,\langle 01345 \rangle ,\langle 01235 \rangle$ to be the same orientation (say $+$) and that the orientation of $\langle 01234 \rangle$ should be $s({\textbf{g}}_{45}) = s(45)$.

Contract all of the domain walls in the 15j symbols, collect all the U and $\eta$ symbols from the 15j symbols on the LHS, $\langle 01234 \rangle ,\langle 01245 \rangle ,\langle 02345 \rangle$, and define their product as

$$\begin{aligned} {\text {symm}}_{\text {LHS}}&= \bigg (\frac{1}{\eta _{012}(23,34)^{s(24)}} \Big (\frac{U_{34}(013,123 ; 0123 \times f_{0123})}{U_{34}(023, {^{32}}012 ; 0123 ) U_{34}(f_{0123},0123 ; 0123 \times f_{0123})} \Big )^{s(34)} \bigg )^{s(45)}\nonumber \\&\quad \times \frac{1}{\eta _{012}(24,45)^{s(25)}} \Big (\frac{U_{45}(014,124 ; 0124 \times f_{0124})}{U_{45}(024, {^{42}}012 ; 0124 ) U_{45}(f_{0124},0124 ; 0124 \times f_{0124})} \Big )^{s(45)} \nonumber \\&\quad \times \frac{1}{\eta _{023}(34,45)^{s(35)}} \Big (\frac{U_{45}(024,234 ; 0234 \times f_{0234})}{U_{45}(034, {^{43}}023 ; 0234 ) U_{45}(f_{0234},0234 ; 0234 \times f_{0234})} \Big )^{s(45)} \end{aligned}$$

(G4)

where the exponent s(45) on the first row corresponds to the the orientation of $\langle 01234 \rangle$. And similarly, we define the corresponding product on the RHS, $\langle 01235 \rangle ,\langle 01345 \rangle , \langle 12345 \rangle$, as

$$\begin{aligned} \begin{aligned} {\text {symm}}_{\text {RHS}}&= \frac{1}{\eta _{012}(23,35)^{s(25)}} \Big (\frac{U_{35}(013,123 ; 0123 \times f_{0123})}{U_{35}(023, {^{32}}012 ; 0123 ) U_{35}(f_{0123},0123 ; 0123 \times f_{0123})} \Big )^{s(35)} \\&\quad \times \frac{1}{\eta _{013}(34,45)^{s(35)}} \Big (\frac{U_{45}(014,134 ; 0134 \times f_{0134})}{U_{45}(034, {^{43}}013 ; 0134 ) U_{45}(f_{0134},0134 ; 0134 \times f_{0134})} \Big )^{s(45)}\\&\quad \times \frac{1}{\eta _{123}(34,45)^{s(35)}} \Big (\frac{U_{45}(124,234 ; 1234 \times f_{1234})}{U_{45}(134, {^{43}}123 ; 1234 ) U_{45}(f_{1234},1234 ; 1234 \times f_{1234})} \Big )^{s(45)} \end{aligned} \end{aligned}$$

(G5)

The hope is that after contracting the domain walls, we will be able to apply the argument of the previous subsection to the resulting diagrams. However, following [30], there are two problems. First is that the diagram on $\langle 01234 \rangle$ is twisted by s(45) relative to the calculation in the previous subsection. Second, looking at the resulting diagrams shows that the anyon lines on $\langle 01234 \rangle$ differ from the lines on the rest of the diagrams up to group multiplication by ${\textbf{g}}_{45}$. To fix these issues, we “sweep” a ${\textbf{g}}_{45}$ domain wall across the entire $\langle 01234 \rangle$ diagram, as in Fig. 50, which has two effects. First, the domain wall sweep gives us an extra factor of

$$\begin{aligned} \begin{aligned} x&= \frac{U_{54}({^{54}}014,{^{54}}124;{^{54}}0124 \times f_{0124})}{U_{54}({^{54}}024,{^{52}}012;{^{54}}0124) U_{54}(f_{0124},{^{54}}0124 \times f_{0124};{^{54}}0124)} \\&\quad \times \frac{U_{54}({^{54}}024,{^{54}}234;{^{54}}0234 \times f_{0234}) U_{54}(f_{0234},{^{54}}0234 \times f_{0234};{^{54}}0234)}{U_{54}({^{54}}034,{^{53}}023;{^{54}}0234)} \\&\quad \times \frac{U_{54}({^{53}}023,{^{52}}012;{^{53}}0123) U_{54}(f_{0123},{^{53}}0123; {^{53}}0123 \times f_{0123})}{U_{54}({^{53}}013,{^{53}}123;{^{53}}0123 \times f_{0123})} \\&\quad \times \frac{U_{54}({^{54}}034,{^{53}}013;{^{54}}0134) U_{54}(f_{0134},{^{54}}0134; {^{54}}0134 \times f_{0134})}{U_{54}({^{53}}014,{^{54}}134;{^{54}}0134 \times f_{0134})} \\&\quad \times \frac{U_{54}({^{54}}134,{^{53}}123;{^{54}}1234) U_{54}(f_{1234},{^{54}}1234; {^{54}}1234 \times f_{1234})}{U_{54}({^{54}}124,{^{54}}234;{^{54}}1234 \times f_{1234})} \end{aligned} \end{aligned}$$

(G6)

that will multiply ${\text {symm}}_{\text {LHS}}$. This formula assumes that the locality constraint Eq. 89 is obeyed, or else additional factors of U may appear. Then, note that after the sweep the diagram lies in a region of group element ${\textbf{g}}_{45}$. This means by the graphical calculus rules in Fig. 8b that the diagram evaluation in this region should be raised to the power of s(45) as compared to when the diagram lies in a region with the identity group element. So in effect, this twist of s(45) from the ${\textbf{g}}_{45}$ region cancels out the twist s(45) from the orientation compatibility, leaving a diagram in the identity region without any total twist by s(45).

Fig. 50

Sweeping a domain wall across the $\langle 01234 \rangle$ to make the anyon labels match the rest of those in the Pachner calculation gives us an extra factor x of U and $\eta$ symbols

Now, we need to evaluate this big product of U and $\eta$ in $\frac{{\text {symm}}_{\text {LHS}} \cdot x}{{\text {symm}}_{\text {LHS}}}$. The result is that this large product simplifies to

$$\begin{aligned} \frac{{\text {symm}}_{\text {LHS}} \cdot x}{{\text {symm}}_{\text {RHS}}} = \eta _{f_{0123}}({\textbf{g}}_{34},{\textbf{g}}_{45}) \end{aligned}$$

(G7)

which is exactly the evaluation of $f \cup A_b^{*} \omega _2 = (f_\infty A_b^*\omega _2)(f)$ on the 5-simplex $\langle 012345 \rangle$.

The first step is to organize all of the U symbols coming from the same fusion vertices. Then, insert several copies of unity in the form $U_{{\textbf{1}}}(a,b;c) = 1$ to put the combination of U symbols from each fusion vertex into the form $\left( \frac{U_{\textbf{g}}(a,b;c)U_{\textbf{h}}({^{\bar{\textbf{g}}}}a,{^{\bar{\textbf{g}}}}b;{^{\bar{\textbf{g}}}}c)}{U_{\textbf{gh}}(a,b;c)} \right) ^{\pm s({\textbf{k}})}$ where sometimes ${\textbf{gh}} = {\textbf{1}}$. The relation $s(12)s(23)=s(13)$ is important. Then, repeatedly use the relation Eq. (82) to obtain

$$\begin{aligned}&\frac{{\text {symm}}_{\text {LHS}} \cdot x}{{\text {symm}}_{\text {RHS}}} \\&\quad = \left( \frac{\eta _{013}(35,54) \eta _{123}(35,54)}{\eta _{0123 \times f_{0123}}(35,54)} \frac{\eta _{0123}(35,54)}{\eta _{023}(35,54) \eta _{{^{32}}012}(35,54)} \frac{\eta _{0123 \times f_{0123}}(35,54)}{\eta _{0123}(35,54) \eta _{f_{0123}}(35,54)} \right) ^{s(35)} \\&\qquad \times \left( \frac{\eta _{0124 \times f_{0124}}(45,54)}{\eta _{014}(45,54) \eta _{124}(45,54)} \frac{\eta _{024}(45,54) \eta _{{^{42}}012}(45,54)}{\eta _{0124}(45,54)} \frac{\eta _{f_{0124}}(45,54) \eta _{0124}(45,54)}{\eta _{0124 \times f_{0124}}(45,54)} \right) ^{s(45)} \\&\qquad \times \left( \frac{\eta _{014}(45,54) \eta _{134}(45,54)}{\eta _{0134 \times f_{0134}}(45,54)} \frac{\eta _{0134}(45,54)}{\eta _{{^{43}}013}(45,54) \eta _{034}(45,54)} \frac{\eta _{0134 \times f_{0134}}(45,54)}{\eta _{f_{0134}}(45,54) \eta _{0134}(45,54)} \right) ^{s(45)} \\&\qquad \times \left( \frac{\eta _{0234 \times f_{0234}}(45,54)}{\eta _{024}(45,54) \eta _{234}(45,54)} \frac{\eta _{034}(45,54) \eta _{{^{43}}023}(45,54)}{\eta _{0234}(45,54)} \frac{\eta _{0234}(45,54)}{\eta _{f_{0234}}(45,54) \eta _{0234 \times f_{0234}}(45,54)} \right) ^{s(45)} \\&\qquad \times \left( \frac{\eta _{124}(45,54) \eta _{234}(45,54)}{\eta _{1234 \times f_{1234}}(45,54)} \frac{\eta _{1234}(45,54)}{\eta _{134}(45,54) \eta _{{^{43}}123}(45,54)} \frac{\eta _{1234 \times f_{1234}}(45,54)}{\eta _{1234}(45,54)\eta _{f_{1234}}(45,54)} \right) ^{s(45)} \\&\qquad \times \left( \frac{\eta _{012}(23,35)^{s(25)}\eta _{013} (34,45)^{s(35)}\eta _{123}(34,45)^{s(35)}}{\eta _{012}(23,34)^{s(24)} \eta _{012}(24,45)^{s(25)}\eta _{023}(34,45)^{s(35)}} \right) , \end{aligned}$$

where the first five rows were obtained using Eq. (82) and the last row is the remaining $\eta$ symbols from the original terms.

The next step is is to cancel many of the above $\eta$ symbols. In addition, we insert unity a few times in the form of

$$\begin{aligned} \frac{1 \cdot 1}{1 \cdot 1} \cdot (1)^{s(25)} = \frac{\eta _{023}(34,{\textbf{1}})^{s(35)}\eta _{012}(24,{\textbf{1}})^{s(25)}}{\eta _{013}(34,{\textbf{1}})^{s(35)} \eta _{123}(34,{\textbf{1}})^{s(35)}} \left( \frac{\eta _{012}(25,54)}{\eta _{012}(25,54)}\right) ^{s(25)} \end{aligned}$$

to obtain

$$\begin{aligned} \begin{aligned} \frac{{\text {symm}}_{\text {LHS}} \cdot x}{{\text {symm}}_{\text {RHS}}}&= \left( \frac{\eta _{013}(35,54)\eta _{013}(34,45)}{\eta _{{^{43}}013} (45,54)^{s(34)}\eta _{013}(34,{\textbf{1}})} \right) ^{s(35)} \left( \frac{\eta _{123}(35,54)\eta _{123}(34,45)}{\eta _{{^{43}}123} (45,54)^{s(34)}\eta _{123}(34,{\textbf{1}})} \right) ^{s(35)} \\&\quad \times \left( \frac{\eta _{{^{43}}023}(45,54)^{s(34)}\eta _{023}(34,{\textbf{1}})}{\eta _{023}(35,54)\eta _{023}(34,45)} \right) ^{s(35)} \\&\quad \times \left( \frac{\eta _{{^{42}}012}(45,54)^{s(24)}\eta _{012}(24,{\textbf{1}})}{\eta _{012}(24,45)\eta _{012}(25,54)} \frac{\eta _{012}(23,35)\eta _{012}(25,54)}{\eta _{{^{32}}012} (35,54)^{s(23)}\eta _{023}(23,34)} \right) ^{s(25)} \\&\quad \times \frac{1}{\eta _{f_{0123}}(35,54)^{s(35)}} \left( \frac{\eta _{f_{0124}}(45,54)\eta _{f_{1234}}(45,54)}{\eta _{f_{0234}}(45,54)\eta _{f_{0134}}(45,54)}\right) ^{s(45)} \end{aligned} \end{aligned}$$

The first three rows of the above vanish due to the identity Eq. (83). In the final row, all the terms $\eta _{f_{\text {3-simplex}}}$ are all $\pm 1$ so we may as well ignore the twists $s({\textbf{g}})$. We can also use the fact that $\delta f = 0$ implying $f_{0124} \times f_{1234} \times f_{0234} \times f_{0134} = f_{0123}$. These mean we can write the above as

$$\begin{aligned} \begin{aligned} \frac{{\text {symm}}_{\text {LHS}} \cdot x}{{\text {symm}}_{\text {RHS}}} = \frac{\eta _{f_{0123}}(45,54)}{\eta _{f_{0123}}(35,54)} = \eta _{f_{0123}}(34,45)\eta _{f_{0123}}(34,{\textbf{1}}) = \eta _{f_{0123}}(34,45) \end{aligned} \end{aligned}$$

where the second equality again uses Eq. (83). This is exactly what we wanted.

For the other Pachner moves, this calculation is identical, except which terms go into the LHS and which go into the RHS get switched around. However, they get switched around in a way that keeps the overall expression $\frac{{\text {symm}}_{\text {LHS}} \cdot x}{{\text {symm}}_{\text {RHS}}}$ the same because the relative orientations ± of the 4-simplices will add powers of $\pm 1$ which will make the LHS and RHS ratio be the same.

Multiplicativity of Z Under Connected-Sums

It is a quick calculation to show that Z is multiplicative under taking connected-sums, as follows. Let $M, M'$ be manifolds equipped with background fields $A_{b},A_{b}'$, $f,f'$ and with twisted spin structures $\xi _{\mathcal {G}},\xi _{\mathcal {G}}'$. Furthermore, suppose WLOG that the triangulations of $M,M'$ have been subdivided sufficiently finely so that we can choose attaching spheres on $M,M'$ without any background gauge fields or spin structures living on them. This entails choosing a subdivision so that there is at least one 4-simplex on M and one on $M'$ for which all $A_b = f = \xi = 0$, which can always be done for flat gauge fields $A_b$, f. Then we can form a manifold $M \# M'$ with background gauge fields $A_{b} \# A_{b}'$, $f \# f'$ and spin structure $\xi _{\mathcal {G}} \# \xi _{\mathcal {G}}'$. To show that Z is multiplicative, we first use the fact that each bosonic shadow piece $Z_b$ is multiplicative, i.e. that:

$$\begin{aligned} Z_b(M \# M', A_{b} \# A_{b}', f \# f') = Z_b(M, A_{b}, f) Z_b(M', A_{b}', f') \end{aligned}$$

(H1)

This follows from the similar well-known fact that the Crane–Yetter sum (even for non-modular $\mathcal {C}$) is multiplicative under connected-sums²⁵ together with locality so that one can focus on the region where the background structures vanish to use the Crane–Yetter result.

Since the connected-sum can be chosen on 4-simplices with no representatives of $f,f'$ or $\xi _{\mathcal {G}}, \xi _{\mathcal {G}}'$, it also follows that

$$\begin{aligned} z_c(M \# M', f \# f', \xi _{\mathcal {G}} \# \xi _{\mathcal {G}}') = z_c(M, f, \xi _{\mathcal {G}}) z_c(M', f', \xi _{\mathcal {G}}') \end{aligned}$$

(H2)

since the loop decompositions of $f,f'$ are disjoint and unaffected by the connected-sum. We can also use for $*=0 \cdots 4$ the canonical identifications $H_{*}(M,\mathbb Z_2) \oplus H_{*}(M',\mathbb Z_2) \equiv H_{*}(M \# M',\mathbb Z_2)$ with respect to the natural inclusion maps . Note these are equivalent to $H^{4-*}(M,\mathbb Z_2) \oplus H^{4-*}(M',\mathbb Z_2) \equiv H^{4-*} (M \# M',\mathbb Z_2)$. From here, multiplicativity under connected-sum is proved as:

$$\begin{aligned} \begin{aligned}&Z(M \# M', A_{b} \# A_{b}', \xi _{\mathcal {G}} \# \xi _{\mathcal {G}}') \\&\quad = \frac{1}{\sqrt{|H^2(M \# M', \mathbb Z_2)|}} \\&\qquad \sum _{f = f \# f' \in H^3(M \# M',\mathbb Z_2)} Z_b(M \# M', A_{b} \# A_{b}', f \# f') z_c(M \# M', f \# f', \xi _{\mathcal {G}} \# \xi _{\mathcal {G}}') \\&\quad = \left( \frac{1}{\sqrt{|H^2(M, \mathbb Z_2)|}} \sum _{f \in H^3(M,\mathbb Z_2)} Z_b(M, A_{b}, f) z_c(M, f, \xi _{\mathcal {G}}) \right) \\&\qquad \cdot \left( \frac{1}{\sqrt{|H^2(M', \mathbb Z_2)|}} \sum _{f' \in H^3(M',\mathbb Z_2)} Z_b(M', A_{b}', f') z_c(M', f', \xi _{\mathcal {G}}') \right) \\&\quad = Z(M, A_{b},\xi _{\mathcal {G}})Z(M', A_{b}',\xi _{\mathcal {G}}') \end{aligned} \end{aligned}$$

(H3)

Explicit Data for Semion-Fermion Theory and $\textrm{SO}(3)_3$ with Time-Reversal Symmetry $\textbf{T}^{2} \,\hbox {=}\, (-1)^F$

In this appendix, we state the relevant BTC and symmetry fractionalization data for the semion-fermion and $\textrm{SO}(3)_3$ theories, which are the input categories that produce the $\nu =2,\nu =3$ phases in the $\mathbb Z_{16}$ classification for $G_f=\mathbb Z_4^{{\textbf{T}},f}$.

1.1 Semion-fermion

The anyon content of the semion-fermion theory can be labeled $1,s,\tilde{s},\psi$. Although this label set matches that of $\textrm{SO}(3)_3$, semion-fermion is an Abelian theory. It is a direct product of $(1,\psi )$ and the semion theory $\textrm{U}(1)_2$. The fusion rules are generated by the fusion rules:

$$\begin{aligned} s \times s = \tilde{s} \times \tilde{s} = \psi \times \psi&= 1\end{aligned}$$

(I1)

$$\begin{aligned} s \times \psi&= \tilde{s} \end{aligned}$$

(I2)

The non-trivial F-symbols are $F^{abc}=-1$ when (a, b, c) is any combination of only s and $\tilde{s}$. The R-symbols, with the anyons ordered $(1,s,\psi , \tilde{s})$, are

$$\begin{aligned} R^{ab} = \begin{pmatrix} 1 &{} 1 &{} 1 &{} 1\\ 1 &{} i &{} 1 &{} i\\ 1 &{} 1 &{} -1 &{} -1\\ 1 &{} i &{} -1 &{} -i \end{pmatrix} \end{aligned}$$

(I3)

Time-reversal must exchange s and $\tilde{s}$.

Gauge-fixing $U_{\textbf{T}}(s,s;1)=U_{\textbf{T}}(\psi ,\psi ;1)=U_{\textbf{T}}(s,\psi ;\tilde{s})=1$, we find that, with the same anyon ordering as before,

$$\begin{aligned} U_{\textbf{T}}(a,b;a\times b) = \begin{pmatrix} 1 &{} 1 &{} 1 &{} 1\\ 1 &{} 1 &{} 1 &{} 1\\ 1 &{} -1 &{} 1 &{} -1\\ 1 &{}-1 &{} 1 &{} -1 \end{pmatrix} \end{aligned}$$

(I4)

Solving the consistency equations for $\eta$ in this gauge, we find $\eta _{\psi }({\textbf{T}},{\textbf{T}}) = \pm 1$ and $\eta _{s,\tilde{s}}({\textbf{T}},{\textbf{T}}) = \pm i$, subject to the constraint $\eta _s \eta _{\tilde{s}}/\eta _{\psi }=-1$. (We drop the explicit ${\textbf{T}}$ from now on - there is only one non-trivial $\eta$ per anyon.)

Inputting the $\eta _{\psi }=-1$, $\eta _s = \pm i$, $\eta _{\tilde{s}}=\mp i$ case into our fermionic state sum, we can evaluate Eq. (129) to find that

$$\begin{aligned} Z({\mathbb {R}}{\mathbb {P}}^4) = \exp \left( \mp 2 \times \frac{2\pi i}{16}\right) \end{aligned}$$

(I5)

1.2 $\textrm{SO}(3)_3$

We list for completeness the explicit F and R symbols of $\textrm{SO}(3)_3$. The fractionalization data of $U,\eta$ symbols is given in Sect. 8.3.

In the usual labeling of anyons of $\textrm{SU}(2)_6$ where anyons are labeled by integers from 0 to 6, we identify $s \sim 2$, $\tilde{s} \sim 4$, and $\psi \sim 6$. With these identifications, the F- and R-symbols can be obtained by standard formulas in, e.g. [101].

Let $q=e^{\pi i/4}$. The R-symbols are simple to write:

$$\begin{aligned} R^{a, b}_c = (-1)^{(a+b-c)/2}q^{\frac{1}{8}\left( c(c+2)-a(a+2)-b(b+2)\right) } \end{aligned}$$

(I6)

Note that by halving the labels (i.e., instead label $s \sim 1$, $\tilde{s} \sim 2$, $\psi \sim 3$) we could eliminate some factors of 2, but we retain these factors of 2 for consistency with $\textrm{SU}(2)_6$ labelings.

The F-symbols require a set of auxiliary functions

$$\begin{aligned} \lfloor n \rfloor&= \sum _{m=1}^n q^{(n+1)/2-m}\end{aligned}$$

(I7)

$$\begin{aligned} \lfloor n \rfloor !&= \lfloor n \rfloor \lfloor n-1 \rfloor \cdots \lfloor 1 \rfloor \end{aligned}$$

(I8)

$$\begin{aligned} \Delta (a,b,c)&= \sqrt{\frac{\lfloor (a+b-c)/2\rfloor !\lfloor (a-b+c)/2\rfloor !\lfloor (-a+b+c)/2\rfloor !}{\lfloor (a+b+c+2)/2\rfloor !}} \end{aligned}$$

(I9)

for $n\ge 1$, and with $\Delta$ defined when a, b, c satisfy the triangle inequality. We also define $\lfloor 0\rfloor !=1$.

With this definition, we can define the F-symbols by the following rather complicated formula:

$$\begin{aligned} F^{abc}_{def}&= (-1)^{(a+b+c+d)/2}\Delta (a,b,e)\Delta (c,d,e)\Delta (b,c,f)\Delta (a,d,f) \sqrt{\lfloor e+1 \rfloor \lfloor f+1 \rfloor } \nonumber \\&\quad \times \sum _{n}' \frac{(-1)^{n/2}\lfloor (n+2)/2\rfloor !}{\lfloor (a+b+c+d-n)/2\rfloor !\lfloor (a+c+e+f-n)/2\rfloor !\lfloor b+d+e+f-n\rfloor !} \nonumber \\&\quad \times \frac{1}{\begin{matrix}\lfloor (n-a-b-e)/2\rfloor ! \lfloor (n-c-d-e)/2\rfloor ! \lfloor (n-b-c-f)/2\rfloor ! \\ \lfloor (n-a-d-f)/2\rfloor !\end{matrix}} \end{aligned}$$

(I10)

These formulae match the gauge choices of [14].

$\mathbb Z_{16}$ Anomaly Indicator for $\mathbf{{T}}^{2} \,\hbox {=}\, (-1)^F$ Symmetry

Here we perform an explicit calculation using our state sum for the case of time-reversal symmetry squaring to $(-1)^F$.

This corresponds to the total symmetry group $G_f=\mathbb Z_4^{{\textbf{T}},f}$ which corresponds to the group extension $\mathbb Z_2^f \rightarrow \mathbb Z_4^{{\textbf{T}},f} \rightarrow \mathbb Z_2^{\textbf{T}}$ described by the cocycle $\omega _2$ with $\omega _2({\textbf{T}},{\textbf{T}})=-1$ and all others $+1$. This means that $A_b^{*} \omega _2 = w_1^2$ in cohomology and given by $A_b^*s \cup A_b^*s \in Z^2(M,\mathbb Z_2)$ on the cochain level with $A_b^*s \in Z^1(M,\mathbb Z_2)$ the anti-unitary flux representing $[w_1]$, with $w_1 = f_\infty A_b^* s \in H^1(M^\vee ,\mathbb Z_2^{\textbf{T}})$.

The (dual of the) $\mathcal {G}_f$-structure is a cycle $\xi \in Z_{d-1}(M,\mathbb Z_2)$ that trivializes $w_2 + w_1^2 + f_\infty A_b^{*} \omega _2 = w_2 \in Z_{d-2}(M,\mathbb Z_2)$. Note the $f_\infty$ map will push $A_b^{*} \omega _2 = A_b^*s \cup A_b^*s \in Z^2(M,\mathbb Z_2)$ to the same cycle representative of $w_1^2 \in Z_{d-2}(M,\mathbb Z_2)$ since the representative of $[w_1^2]$ coming from $A_b^{*} \omega _2$ is the same as $A_b^*s \cup A_b^*s$.

As discussed in the main text, the outline of the calculation is as follows. First, we give an explicit cellulation that we call $T_{\star }$ of $M={\mathbb {R}}{\mathbb {P}}^4$ and determine a set of cochain representatives for $w_2, A_b$, and $w_1$. Next, we perform the diagrammatic calculations to explicitly evaluate the state sum associated to the triangulation and give an analytic expression for it for any general super-modular category and symmetry fractionalization class of time-reversal symmetry. This has the interpretation of being an ‘anomaly-indicator’ for the symmetry fractionalization class. In particular, we will evaluate the shadow $Z_b(({\mathbb {R}}{\mathbb {P}}^4 ,T_{\star }),A_b,f)$ for two choices $f = 0 \in Z^3(({\mathbb {R}}{\mathbb {P}}^4 ,T_{\star }),A_b,\mathbb Z_2)$ and f with $[f] \ne 0 \in H^3({\mathbb {R}}{\mathbb {P}}^4 ,\mathbb Z_2)$. The bosonic shadow results given in Eq. (126) are reproduced below,

$$\begin{aligned} \begin{aligned} Z_b(({\mathbb {R}}{\mathbb {P}}^4 ,T_{\star }),A_b,f=0)&= \frac{1}{\mathcal {D}} \sum _{x | x = {^{\textbf{T}}}x} d_x \theta _x \eta ^{\textbf{T}}_x \\ Z_b(({\mathbb {R}}{\mathbb {P}}^4 ,T_{\star }),A_b,[f]\ne 0)&= \pm \frac{1}{\mathcal {D}} \sum _{x | x = {^{\textbf{T}}}x \times \psi } d_x \theta _x \eta ^{\textbf{T}}_x \end{aligned} \end{aligned}$$

(J1)

where the ± depends on the specific representative of $[f] \ne 0 \in H^3({\mathbb {R}}{\mathbb {P}}^4 ,\mathbb Z_2)$. The FSPT partition function will be the result

$$\begin{aligned} Z(({\mathbb {R}}{\mathbb {P}}^4 ,T_{\star }),A_b,\xi _{{\text {pin}}^+})=\frac{1}{\sqrt{2}\mathcal {D}} \left( \sum _{x | x = {\,^{\textbf{T}}}x} d_x \theta _x \eta ^{\textbf{T}}_x + i \sum _{x | x = {\,^{\textbf{T}}}x \times f} d_x \theta _x \eta ^{\textbf{T}}_x \right) \end{aligned}$$

(J2)

Here, the fact that the last term has a $+i$ indicates our choice of pin$^+$ structure and is independent of the specific representatives of f. The other pin$^+$ structure would have a $-i$ there. This is exactly the $\mathbb Z_{16}$ indicator formula giving $e^{2 \pi i \nu / 16}$.

1.1 Compact cellulation of ${\mathbb {R}}{\mathbb {P}}^4$

Now, we list the cellulation $T_{\star }$ of ${\mathbb {R}}{\mathbb {P}}^4$. This cellulation comes from the gluing procedure of generating manifolds from [30]. In particular we use a cellulation generated from $\mathbb Z_2^{\textbf{T}}$ in that paper. One technical note is that the gluing procedure in [30] actually produces two plausible ways of gluing things together. One of them is not a manifold because it corresponds to a degenerate gluing in the sense that links of 2-simplices are not circles. We choose to work with the one spelled out explicitly below, which is indeed a manifold.

There are four 4-simplices $\langle a_1 b_1 c_1 d_1 e_1 \rangle , \langle a_2 b_2 c_2 d_2 e_2 \rangle , \langle a_3 b_3 c_3 d_3 e_3 \rangle , \langle a_4 b_4 c_4 d_4 e_4 \rangle$ that are all $+$-oriented and branched as $a \rightarrow b \rightarrow c \rightarrow d \rightarrow e$. Then there are ten 3-simplices given by some identifications of the simplices. The identifications of 3-simplices must be compatible with the branching structure. Those identifications then uniquely fix the identifications of all the lower-dimensional simplices via the ordering induced by the branching structure. In total, there are twelve 2-simplices, eight 1-simplices, and three 0-simplices. This means that there is a sum over $10+12=22$ different anyon labels. Below we list all the simplices together with an arbitrary number (in the column $\#$) from $1,\dots ,22$ as shorthand for the anyon label we chose. This list includes the identifications of 3-simplices and 2-simplices, in that each $\{{\text {simplex-1}}, \cdots , {\text {simplex-k}}\}$ in braces represents that all simplices in that list are identified. One can also look at Fig. 52 which lists all the relevant 15j symbols and also marks the anyon label.

(J3)

For reference, the 1-simplices and 0-simplices are:

(J4)

Inputting this cellulation into the program Regina [102] shows that this cellulation is a manifold whose double-cover is simply-connected with $H^2(-,\mathbb Z) = 0$, which means that its double-cover is homeomorphic to $S^4$ by the topological Poincaré conjecture in $d=4$. As such, this indicates that the original manifold is homeomorphic to ${\mathbb {R}}{\mathbb {P}}^4$.

1.2 Representatives of $w_2$ and ${\text {pin}}^+$ structure, $w_1$ and gauge fields, and $[f] \ne 0$ and $z_c(f)$

We want to find explicit cochain representatives of the ${\text {pin}}^+$ structure and of a non-trivial background fermion line $[f] \ne 0$ at hand, for which we will need representatives of the orientation-reversing wall $w_1$ and of $w_2$.

1.2.1 $w_2$ and ${\text {pin}}^+$ structure

Recall from the beginning of this section that the cochain representative of $w_2 \in Z^{2}(({\mathbb {R}}{\mathbb {P}}^4,T_{\star })^{\vee },\mathbb Z_2)$ is trivialized by $\xi _{{\text {pin}}^+}$. On the cochain-level, we can compute the canonical representative of $w_2$ using Eq. (A18).

We apply this formula to the above cellulation of ${\mathbb {R}}{\mathbb {P}}^4$. One can compute $w_2$ on the dual of each 2-simplex, slightly abusing the labels $1,\cdots ,22$ from the previous subsection to label the dual 2-cells of the corresponding 2-simplices. The result is

$$\begin{aligned} \begin{aligned} w_2(2)&= w_2(5) = w_2(7) = w_2(9) = w_2(11) = w_2(15) = w_2(16) = w_2(19) = +1 \\ w_2(3)&= w_2(12) = w_2(14) = w_2(22) = -1 \\ \end{aligned} \end{aligned}$$

(J5)

From here, we want to find a cochain $\xi _{{\text {pin}}^+} \in Z^1(({\mathbb {R}}{\mathbb {P}}^4,T_{\star })^\vee ,\mathbb Z_2) \cong Z_{d-1}(({\mathbb {R}}{\mathbb {P}}^4,T_{\star }),\mathbb Z_2)$ such that $\delta \xi _{{\text {pin}}^+} = w_2$ (dual to $\partial \xi = w_2$). Again slightly abusing the labels of 3-simplices to refer to their dual 1-cells, this can be accomplished by choosing $\xi$ to act non-trivially as:

$$\begin{aligned} \xi _{{\text {pin}}^+}(4) = \xi _{{\text {pin}}^+}(6) = \xi _{{\text {pin}}^+}(8) = \xi _{{\text {pin}}^+}(10) = \xi _{{\text {pin}}^+}(17) = \xi _{{\text {pin}}^+}(20) = -1 \end{aligned}$$

(J6)

with $\xi _{{\text {pin}}^+}=+1$ on all other dual 1-cells. Note that the only non-trivial homology class is dual to $w_1$ and consists of the 3-simplices 17, 20. So the other ${\text {pin}}^+$ structure will have $\xi '_{{\text {pin}}^+}(17)=\xi '_{{\text {pin}}^+}(20)=+1$ as opposed to $-1$.

1.2.2 $w_1$ and ${\textbf{T}}$ Gauge Fields

The procedure in [30] also tells us what the gauge fields ${\textbf{g}}_{ij}$ should be on the 1-simplices. All 1-simplices that have a ${\textbf{T}}$-valued gauge field are

$$\begin{aligned} A_b(\langle b_1 c_1 \rangle ) = A_b(\langle c_1 e_1 \rangle ) = A_b(\langle b_2 c_2 \rangle ) = {\textbf{T}}. \end{aligned}$$

(J7)

All 1-simplices $\langle ij \rangle$ not identified with the above via Eq. (J4) will have ${\textbf{g}}_{ij} = {\textbf{1}}$.

One can also check that the $f_\infty$ map makes the orientation-reversing wall $w_1$ consist of the two 3-simplices labeled as 17, 20.

1.2.3 $[f] \ne 0$ and $z_c(f)$

We will work with a cochain $f \in C^3(({\mathbb {R}}{\mathbb {P}}^4,T_{\star }),\mathbb Z_2)$ defined with respect to the above 3-simplex labels as

$$\begin{aligned} \begin{aligned} f(17) = f(21) = f(10)&= 1 \\ f({\text {other 3-simplices}})&= 0 \\ \end{aligned} \end{aligned}$$

(J8)

One can show that it is closed, which will be more apparent once we write out the 15j symbols in Fig. 52. Also, we can see that it corresponds to the non-trivial class in $H^3({\mathbb {R}}{\mathbb {P}}^4,\mathbb Z_2)$. To see it is non-trivial, note that it is nonzero on exactly one 3-simplex, 20, from the dual of $w_1$, so $\int _{w_1} f = 1$.

Now we want to compute $z_c(f) = (-1)^{\int \xi _{{\text {pin}}^+}(f)} \sigma (f)$. The ${\text {pin}}^+$ structure part can be quickly computed using Eq. (J6). In particular, we will have $(-1)^{\int \xi _{{\text {pin}}^+}(f)} = +1$ because $\xi _{{\text {pin}}^+}$ and f share two nonzero 3-simplices, labeled as 10 and 17. The $\sigma (f)$ part can be calculated in two ways. First, we can use the winding definition. In this gauge, f encodes a single fermion loop which takes the path shown in Fig. 51. Following the discussion in Sect. D 3 of how the perturbation of the $w_1$ wall is encoded in the 15j symbols and using the winding definition from Eqs. (B6,B7) will give

$$\begin{aligned} \sigma (f) = -\begin{pmatrix}1&0&1&0\end{pmatrix} \underbrace{\begin{pmatrix} -iX &{} 0 \\ 0 &{} iX \end{pmatrix}}_{\begin{array}{c} \hat{2} \rightarrow \hat{0} {\text { on}} \\ + {\text { 4-simplex }} \\ \langle a_1 b_1 c_1 d_1 e_1 \rangle \end{array}} \underbrace{\begin{pmatrix} 0 &{} iY \\ -iY &{} 0 \end{pmatrix}}_{\begin{array}{c} {\text {crossing }} w_1 {\text { in the }} \\ {\text {perturbing direction on }} \\ - {\text { 3-simplex }} 17 \end{array}} \underbrace{\begin{pmatrix} iX &{} 0 \\ 0 &{} -iX \end{pmatrix}}_{\begin{array}{c} \hat{0} \rightarrow \hat{4} {\text { on}} \\ + {\text { 4-simplex }} \\ \langle a_3 b_3 c_3 d_3 e_3 \rangle \end{array}} \underbrace{\begin{pmatrix} -iX &{} 0 \\ 0 &{} iX \end{pmatrix}}_{\begin{array}{c} \hat{3} \rightarrow \hat{1} {\text { on}} \\ + {\text { 4-simplex }} \\ \langle a_4 b_4 c_4 d_4 e_4 \rangle \end{array}} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0\end{pmatrix} = -i \end{aligned}$$

Alternatively, we can use the Grassmann integral to give the same answer

$$\begin{aligned} \sigma (f) = \int d\theta _{10} d\overline{\theta }_{10} d\theta _{17} d\overline{\theta }_{17} d\theta _{21} d\overline{\theta }_{21} \underbrace{(\overline{\theta }_{17} \theta _{10})}_{\begin{array}{c} {\text {from}} \\ \langle a_1 b_1 c_1 d_1 e_1 \rangle \end{array}} \underbrace{(\theta _{17} \theta _{21})}_{\begin{array}{c} {\text {from}} \\ \langle a_3 b_3 c_3 d_3 e_3 \rangle \end{array}} \underbrace{(\overline{\theta }_{10} \overline{\theta }_{21})}_{\begin{array}{c} {\text {from}} \\ \langle a_4 b_4 c_4 d_4 e_4 \rangle \end{array}} \cdot \underbrace{(-i)}_{\begin{array}{c} {\text {crosssing}} \\ - {\text { oriented}} \\ 17 \end{array}} = -i \end{aligned}$$

As such, we have $z_c(f) = (-1)^{\int \xi _{{\text {pin}}^+}(f)} \sigma (f) = (+1) \cdot (-i) = -i$.

Fig. 51

The dual path for the representative f for $[f] \ne 0 \in H^3({\mathbb {R}}{\mathbb {P}}^4,\mathbb Z_2)$. (Left) How the fermion path looks in the 15j symbols. The arrows on the fermion lines give the loops a direction. The thin blue circle around 17 on the leftmost diagram indicates that that 3-simplex is part of $w_1$, but it is not perturbed into the 4-simplex $\langle a_1 b_1 c_1 d_1 e_1 \rangle$. (Right) Another depiction of the path of f along with some useful data used to compute $\sigma (f)$. In particular, the orientations of the 4-simplices it passes through, the paths $\hat{i} \rightarrow \hat{j}$ it takes within each 4-simplex, the corresponding assignments of Grassmann variables, and the vector $v_d$ along the dual 1-skeleton are all shown. For the Grassmann variables, the solid circles represent $\theta$ while the open circles represent $\overline{\theta }$. The arrows between the variables indicate the order in which they appear in the computation of $\sigma (f)$, as ${\text {first}} \rightarrow {\text {second}}$ or ${\text {second}} \leftarrow {\text {first}}$

Fig. 52

Result of step 1 of ${\mathbb {R}}{\mathbb {P}}^4$ calculation

Fig. 53

Result of steps 2–3 of ${\mathbb {R}}{\mathbb {P}}^4$ calculation

Fig. 54

Result of steps 4–5 of ${\mathbb {R}}{\mathbb {P}}^4$ calculation

Fig. 55

Result of steps 6–7 of ${\mathbb {R}}{\mathbb {P}}^4$ calculation

Fig. 56

Result of steps 8–9 of ${\mathbb {R}}{\mathbb {P}}^4$ calculation

Fig. 57

Result of steps 10–11 of ${\mathbb {R}}{\mathbb {P}}^4$ calculation

Fig. 58

Result of steps 12–13 of ${\mathbb {R}}{\mathbb {P}}^4$ calculation

Fig. 59

Result of steps 14–15 of ${\mathbb {R}}{\mathbb {P}}^4$ calculation. NOTE: starting Step 15 we restrict our attention to the specific cochain representative with $f(10)=f(17)=f(21)=1$. The pink line here now refers to the anyon $\psi$

Fig. 60

Result of steps 16–17 of ${\mathbb {R}}{\mathbb {P}}^4$ calculation

Fig. 61

Result of steps 18–19 of ${\mathbb {R}}{\mathbb {P}}^4$ calculation

Fig. 62

Result of steps 20–22 of ${\mathbb {R}}{\mathbb {P}}^4$ calculation

1.3 Diagrammatic calculation

Now, we do the diagrammatic calculation on $({\mathbb {R}}{\mathbb {P}}^4,T_{\star })$. We are summing over 22 different labelings of anyon lines.

Note that the links of the 2-simplices $\{2,5,7,9,11,15,16,19\}$ have size 4 whereas the rest of them have size 2. Also, $(N_0 - N_1) = 3-8 = -5$, $\chi (M) = N_0 - N_1 + N_2 - N_3 + N_4 = 3-8+12-10+4=1$. Then the quantum dimensions out in front of the diagrams, as in Eq. (G2), will be

$$\begin{aligned} \mathcal {N}_{\text {q-dim}} = \frac{\mathcal {D}^{2(3-8) - 1}}{d_2 d_5 d_7 d_9 d_{11} d_{15} d_{16} d_{20}} \end{aligned}$$

(J9)

The rest of the calculation is evaluating the 15j symbols and proceeds as follows. The steps described below correspond to the diagrams of Figs. 52, 53, 54, 55, 56, 57, 58, 59, 60, 61 and 62.

• Step 1: Draw out all the necessary 15j symbols and write quantum dimensions.

• Step 2: Remove the domain walls on the first two 4-simplices since they contribute nothing. Then use Merging Lemma I to sum over anyon labels 1, 6. This adds a factor $\times \sqrt{d_2 d_3 d_{15} d_{16}} \sqrt{d_2 d_7 d_{15} d_{22}}$.

• Step 3: Resolve the identity with sum over anyon 3. Adds a factor $\sqrt{\frac{d_8 d_{11}}{d_3}}$. Note that we have to use the bending rules in Eqs. (39,40). But since the bending factors are unitary, they cancel out. This will happen several more times in the remaining steps although we will not explicitly say it.

• Step 4: Resolve the identity with sum over anyon 8. Adds a factor $\sqrt{\frac{d_{9} d_{7}}{d_{8}}}$.

• Step 5: Use Merging Lemma I to sum over anyon 13. Adds a factor $\sqrt{d_{9} d_{12} d_{15} d_{19}}$.

• Step 6: Resolve the identity with sum over anyon 22. Adds a factor $\sqrt{\frac{d_{9} d_{21}}{d_{22}}}$.

• Step 7: Resolve the identity with sum over anyon 15. Adds a factor $\sqrt{\frac{d_{14} d_{10}}{d_{15}}}$.

• Step 8: Resolve the identity with sum over anyon 10 and remove a leftover $d_9$ loop. Together, these add factors $\sqrt{\frac{d_{5} d_{9}}{d_{10}}} \cdot d_9$.

• Step 9: Resolve the identity with sum over anyon 14. Adds a factor $\sqrt{\frac{d_{2} d_{18}}{d_{14}}}$.

• Step 10: Resolve the identity with sum over anyon 18. Adds a factor $\sqrt{\frac{d_{11} d_{19}}{d_{18}}}$. Then sum over 9 gives a factor $\sum _{9} (d_9)^2 = \mathcal {D}^2$.

• Step 11: Resolve the identity with sum over anyon 2. Adds a factor $\sqrt{\frac{d_{4} d_{5}}{d_{2}}}$.

• Step 12: Resolve the identity with sum over anyon 4 and gain $d_{12}$ from the leftover 12 loop. Adds a factor $d_{12} \times \sqrt{\frac{d_{11} d_{12}}{d_{4}}}$. Then sum over 12 gives a factor $\sum _{12} (d_{12})^2 = \mathcal {D}^2$.

• Step 13: Resolve the identity with sum over anyon 11 and gain $d_{21}$ from the leftover 21 loop. Adds a factor $d_{21} \times \sqrt{\frac{d_{16} d_{21}}{d_{11}}}$. Then sum over 21 gives a factor $\sum _{21} (d_{21})^2 = \mathcal {D}^2$.

• Step 14: Expand the domain walls. Factors of $\eta _{5}({\textbf{T,T}}),\eta _{19}({\textbf{T,T}}),U_{\textbf{T}}({\overline{16}},16)$ each appear twice with opposite exponents and cancel out. Two factors of $\eta _\psi ({\textbf{T,T}})$ become $(\eta _\psi ({\textbf{T,T}}))^{f(17)+f(20)}$. Also resolve the $\,^{\textbf{T}}19$ and $\,^{\textbf{T}}5$ lines into a sum over anyons x.

The above steps work for any choice of background f lines. The rest of the calculation will be for the particular choice of background $f \in Z^3(({\mathbb {R}}{\mathbb {P}}^4,T_{\star }),\mathbb Z_2)$ from Eq. (J8) that corresponds to the nonzero class $[f] \in H^3({\mathbb {R}}{\mathbb {P}}^4,\mathbb Z_2)$. We will see that this gives the expected formula

$$\begin{aligned} \begin{aligned} Z_b(({\mathbb {R}}{\mathbb {P}}^4,T_{\star }),A_b,f)&= -\frac{\eta _\psi ({{{\textbf {T,T}}}})}{\mathcal {D}} \sum _{x | x = \,^{\textbf{T}}x \times \psi } d_x \theta _x \eta ^{{{\textbf {T}}}}_{\,^{{{\textbf {T}}}}x} = -\frac{1}{\mathcal {D}} \sum _{x | x = \,^{{{\textbf {T}}}}x \times \psi } d_x \theta _x \eta ^{{{\textbf {T}}}}_x \end{aligned} \end{aligned}$$

for this path integral. The calculation with $f=0$ is quite similar and slightly simpler; the only difference in the result is a minus sign and that the sum is restricted to $\{x | x = \,{^{\textbf{T}}}x\}$.

A key tool in the rest of the diagrammatics is the fact that there are two ‘regions’ of the diagram containing the same anyon lines but one ‘half’ is inside a time-reversed region labeled by the group element ${\textbf{T}}$. The diagrammatic calculus Fig. 8b then states that all F, R moves done inside the T-region differ by complex conjugation with respect to the same moves in the identity-group-element region. In particular, unitarity means that these conjugated F and R symbols are the inverses of the symbols in the identity-region. This means we can do F and R moves in pairs, one in the identity-region and one in the ${\textbf{T}}$-region. Unitary implies that these conjugated F,R moves cancel out and allow us to diagrammatically simplify things without explicitly needing to write out the respective F and R symbols.

Step 15::

Gain a factor of $(-1)$ from the crossing $f(17)=1$ with $f(21)=1$. Cancelling F,R moves bring the remaining fermion line from the 17 3-simplices onto the 19 2-simplices.

Step 16::

Cancelling F moves turn the sum over 20 lines into a sum over $\widetilde{20}$ lines.

Step 17::

Cancelling F moves turn the sum over 7 lines into a sum over $\widetilde{7}$ lines. But the x line coming from the bottom necessitate $\widetilde{7} = \,^{\textbf{T}}x$ while the top gives $\widetilde{7} = x \times \psi$. This in turn implies that the sum over x restricts to $\,^{\textbf{T}}x = x \times \psi$.

Step 18::

Cancelling F oves turn the sum over 16 lines into a sum over $\widetilde{16}$ lines. But charge conservation gives that the only contributing term is $\widetilde{16} = {^T}x = x \times \psi$.

Step 19::

Cancelling F,R moves turn the sum over 17 lines into a sum over $\widetilde{17}$ lines. But $\widetilde{17}$ gets attached to $\widetilde{20}$ as a tadpole so $\widetilde{17}$ must be the identity. Then we get two $\widetilde{20}$ loops whose contraction give a factor of $d_{\widetilde{20}}^2$. The sum over $\widetilde{20}$ then gives $\sum _{\widetilde{20}} (d_{\widetilde{20}})^2 = \mathcal {D}^2$.

Step 20::

Remove some of the anyon lines’ ‘twists’ using R moves to give a factor of $R^{\,{^{\textbf{T}}}5 \,{^{\textbf{T}}}19}_{x}R^{\,{^{\textbf{T}}}19 \,{^{\textbf{T}}5}}_{x} \theta _{\,{^{\textbf{T}}}19} \theta _{\,{^{\textbf{T}}}5}$. Then we can use the ribbon identity Eq. (50) to equate this to a factor of $\theta _x$. Alternatively, one can manipulate the diagram into a twist on the x anyon.

Step 21::

Contracting the remaining ${\textbf{T}}$-domain-wall bubble gives a factor $(U_{\textbf{T}}(5,19;x \times \psi )U_{\textbf{T}}(5,19;x \times \psi )^{-1}) \cdot \eta ^{\textbf{T}}_{\,^{\textbf{T}}x} \cdot U_{\textbf{T}}(\,^{\textbf{T}}x,\psi ;\,^{\textbf{T}}x \times \psi ) = \eta _{\,^{\textbf{T}}x}({\textbf{T,T}}) \cdot U_{\textbf{T}}(\,^{\textbf{T}}x,\psi ;\,^{\textbf{T}}x \times \psi )$. This is exactly the gauge-invariant quantity $\eta ^{\textbf{T}}_{{\,^{\textbf{T}}} x}$ defined in Eq. (127). The U-$\eta$ consistency condition Eq. (82) shows this equals the $\eta _\psi ({\textbf{T,T}})\eta ^{\textbf{T}}_x$ that goes into the sum.

Step 22::

Evaluate the rest of the diagram. First remove the fermion line using canceling F-moves. Then evaluate all of the inner products to produce three factors of $\sqrt{\frac{d_{19} d_{5}}{d_x}}$ and $N_{19,5}^{x}$. Finally, evaluate the remaining x loop for another factor of $d_x$. See below for the algebra giving the final simplification.

The sums on 5 and 19 can be performed with a little algebra:

$$\begin{aligned} \sum _{5,19} d_{19}d_5 N_{19,5}^x = \sum _{5,19} d_{19}d_5 N_{5,\overline{x}}^{\overline{19}} = \sum _{19} d_x d_{19}^2 = \mathcal {D}^2 d_x \end{aligned}$$

(J10)

And from here we get the result

$$\begin{aligned} Z_b(({\mathbb {R}}{\mathbb {P}}^4,T_{\star }),A_b,f_{\text {non-trivial}}) = -\frac{1}{\mathcal {D}} \sum _{x | \,^{{{\textbf {T}}}}x = x \times \psi } d_x \theta _x \eta ^{{{\textbf {T}}}}_x \end{aligned}$$

And since $z_c(f_{\text {non-trivial}}) = -i$ for the representative we chose, we obtain the total partition function exactly corresponding to Eq. (126).