Spin-Cobordisms, Surgeries and Fermionic Modular Bootstrap

Abstract

We consider general fermionic quantum field theories with a global finite group symmetry G, focusing on the case of 2 dimensions and torus spacetime. The modular transformation properties of the family of partition functions with different backgrounds are determined by the ’t Hooft anomaly of G and fermion parity. For a general possibly non-abelian G we provide a method to determine the modular transformations directly from the bulk 3d invertible topological quantum field theory (iTQFT) corresponding to the anomaly by inflow. We also describe a method of evaluating the character map from the real representation ring of G to the group which classifies anomalies. Physically the value of the map is given by the anomaly of free fermions in a given representation. We assume classification of the anomalies/iTQFTs by spin-cobordisms. As a byproduct, for all abelian symmetry groups G, we provide explicit combinatorial expressions for corresponding spin-bordism invariants in terms of surgery representation of arbitrary closed spin 3-manifolds. We work out the case of $G={\mathbb {Z}}_2$ in detail, and, as an application, we consider the constraints that ’t Hooft anomaly puts on the spectrum of the infrared conformal field theory.

Appendices

Details on the Derivation of S and T Matrix Elements

Here we report the computation for the S and T matrices.

Let us start from the class (0, 0) and explain what are the ${\textrm{NS}}0$ directions we chose to contract in order to define the basis elements of $Z^{(0,0)}_i$. Under S the only element of this vector mapped to itself is $Z^{(0,0)}_3=Z^{{\textrm{NS}}0}_{{\textrm{NS}}0}$, while the others are mapped to themselves under $S^2$. This means that we can choose deliberately the direction of contraction for 4 out of these 8 tori in such a way that they are mapped to themselves under $S^2$ with no additional phase. The basis of the 4 remaining ones will follow by using as reference bordism the S transformation itself. As a consequence, we are able to fix all the non-zero entries of $S_{(0,0)}$ to 1 with the exception of $\{{\textrm{NS}}0,{\textrm{NS}}0\}$. However, this particular case has no ${\mathbb {Z}}_2$ holonomies and no pin$^-$ surface. Therefore $(S_{(0,0)})^3_3=1$ as well, which completes $S_{(0,0)}$. The basis we chose that satisfies this property is

(A.1)

(A.2)

(A.3)

where the red dashed lines represent the direction of each torus contracted to a point in the bounding solid 3-torus.

Next, we turn to the computation of $T_{(0,0)}$. One can see that for the set (A.1)–(A.3) each element of basis $e_{a}$ ends up to the proper basis element $e_{T\cdot a}$ with the exception of two cases, namely for $(T_{(0,0)})^9_6$ and $(T_{(0,0)})^4_8$.

Let us start with the first. By applying Eq. (3.27) it follows that the bounding 3-manifold $Y=MT((T_{(0,0)})^9_6)$ is given by joining the two solid tori:

(A.4)

At this point one needs to find a surface ${\textrm{PD}}(a_g)$ associated to the 1-cocycle $a_g\in H^1(Y,{\mathbb {Z}}_2)$, which is determined by appropriately extending over the $3^{\textrm{rd}}$ direction the curve Poincaré dual of the restriction $a_g|_{\{{\textrm{NS}}1,{\textrm{R}}1\}}$. This curve ${\textrm{PD}}(a_g|_{\{{\textrm{NS}}1,{\textrm{R}}1\}})$ is found by a T transformation of ${\textrm{PD}}(a_g|_{\{{\textrm{NS}}0,{\textrm{R}}1\}})$, i.e.

(A.5)

Therefore the 3-manifold is the one on the left in Fig. 21. One can reach analogous conclusions for $(T_{(0,0)})^4_8$ and find the surface on the right of Fig. 21.

Fig. 21

Left: a mapping torus for which evaluation of $\nu$ gives the element $(T_{(0,0)})^9_6$. Right: a mapping torus for which evaluation of $\nu$ gives the element $(T_{(0,0)})^4_8$

The closed 3-manifold in discussion is (in both cases) $\mathbb{R}\mathbb{P}^3$. Since $\mathbb{R}\mathbb{P}^2$ is not orientable, there is no possibility to draw the full $\textrm{pin}^-$ surface, but only some open subsets of it. To understand which kind of surface we are talking about, we need to inspect how it twists once we approach the radial directions of the tori that compose the full 3-manifold.

Fig. 22

Representation of the two solid tori describing the manifold (A.4). On the torus on the right is represented the generator b of $H^1(\mathbb{R}\mathbb{P}^2,{\mathbb {Z}}_2)$, given by the contraction of the path going from $x'$ to $x''$ to the core

We start from the first of (A.5). We can draw the solid torus as on the left of Fig. 22, where the symplectic basis depicted of the 2-torus sections changed, so that one of the generators is the ${\textrm{NS}}0$ direction which is being contracted to a point. In the figure the radial direction is represented by the two near-vertical lines, which are identified. It is clear that in this case there is no twist and thus this particular subset of the surface can be represented as a disk. By repeating the same reasoning with the second solid torus one arrives at the different situation represented on the right of Fig. 22. Here there is a clear twist of the surface so this part of its 2-skeleton is represented by a Möbius band with core circle $\mathbb{R}\mathbb{P}^1$. This means that the surface we are working with is $\mathbb{R}\mathbb{P}^2$, for which ${\textrm{ABK}}$ is determined by the value q(b) of its generator b. To compute it, we look more carefully at the contraction of b to the core of this solid torus. We can consider as even framing along b a framing which does a full $2 \pi$ rotation while going around the loop. As one can see from Fig. 23, the normal bundle on the surface does a single negative half twist with respect to such framing, so that

$$\begin{aligned} q(b)=-1 \quad \Rightarrow \quad {\textrm{ABK}}= 7\quad \textrm{mod}\,8. \end{aligned}$$

(A.6)

By repeating the same reasoning we see that $\mathbb{R}\mathbb{P}^2$ for $(T_{(0,0)})^4_8$ has instead ${\textrm{ABK}}=1\mod 8$. Indeed the procedure is almost the same, with the main difference being that the core of the torus now has ${\textrm{NS}}$ periodicity, which is equivalent to changing the quadratic enhancement of the generator as $q(b)\rightarrow q(b) +2\mod 4$.

Fig. 23

The Möbius strip embedded into the solid torus depicted on the right side of Fig. 22

Next, we continue with the computation of $S_{(0/1,1)}$. Since these classes behave almost identically with the exception of changing the periodicity ${\textrm{NS}}1 \leftrightarrow {\textrm{R}}1$, we will use from now on a special notation to identify the two simultaneously. Depending on the class we are working with, the surfaces $\{X_1,\,X_2,\,X_3\}$ will stand for $\{\{{\textrm{R}}0,{\textrm{NS}}1\},\{{\textrm{NS}}1,{\textrm{R}}0\},\{{\textrm{NS}}1,{\textrm{NS}}1\}\}$ or $\{\{{\textrm{R}}0,{\textrm{R}}1\},\{{\textrm{R}}1,{\textrm{R}}0\},\{{\textrm{R}}1,{\textrm{R}}1\}\}$ for (0, 1) and (1, 1) respectively. The pin$^-$-surfaces for which ${\textrm{ABK}}$ invariant determines the value of the matrix entries $(S_{(0/1,1)})^i_j$ and $(T_{(0/1,1)})^i_j$ will instead be denoted as $\Sigma ^{S/T}_{ij}$.

The first entry one has to compute is $(S_{(0/1,1)})^1_2$. The mapping torus $MT((S_{(0/1,1)})^1_2)$ associated to it is given by applying S twice to $X_1$, which is our reference 2-manifold, and then identifying the boundaries $X_1 \times \{0\} \sim X_1 \times \{1\}$ (remembering of the additional $(-1)^{{\mathcal {F}}}$ action); see Fig. 24. By the same argument of before one can see that the $\textrm{pin}^-$ surface we are interested in is just a curve parallel to the time direction of $X_1$ spanned along the vertical direction. Therefore the surface $\Sigma ^S_{12}$ in question is a Klein bottle $K=\mathbb{R}\mathbb{P}^2_b\#\mathbb{R}\mathbb{P}^2_{c=a+b}$. Here b and c denote the generators of $H_1(K,{\mathbb {Z}}_2)$ related to each component of its connected sum.

After identifying the surface, we have to fix the transition function from the top slice $X_1\times \{1\}$ to the bottom one $X_1\times \{0\}$. In this case the two bases are related by

$$\begin{aligned} \begin{pmatrix} \partial '_\chi&\partial '_\theta \end{pmatrix} = \begin{pmatrix} \partial _\chi&\partial _\theta \end{pmatrix} \begin{pmatrix} 0 &{} -1 \\ 1&{}0 \\ \end{pmatrix}^T,\quad \begin{pmatrix} \theta '\\ \chi ' \end{pmatrix} = \begin{pmatrix} -1 &{} 0 \\ 0&{}-1 \\ \end{pmatrix} \begin{pmatrix} \theta \\ \chi \end{pmatrix}. \end{aligned}$$

(A.7)

Here we denoted $(\theta ,\chi )$ and $(\theta ',\chi ')$ the canonical coordinates on the tori $X_1\times \{0\}$ and $X_1\times \{1\}$ respectively. They are determined by identifying $(\partial _\theta ,\partial _\chi )=(\omega _1,\omega _2)$ and $(\partial '_\theta ,\partial '_\chi )=(\omega '_1,\omega '_2)$, where $\omega _1$ and $\omega _2$ are the lattice generators that define the modular parameter $\tau =\omega _2/\omega _1$.

Therefore if we start from a vector $v \in T_p MT((S_{(1,0/1)})^1_2)$ for any point ${p \in X_1\times \{0\}}\cong X_1 \times \{1\}$ and write it in the canonical basis of $X_1\times \{1\}$, then its entries in the canonical basis used for $X_1\times \{0\}$ are given by applying the transition function ${\textsf{T}}_{S^2} = {\textsf{T}}_S {\textsf{T}}_S =R_3(\pi /2) R_3(\pi /2)$,

$$\begin{aligned} \begin{pmatrix} v_\theta \\ v_\chi \\ v_z \end{pmatrix}= R_3(\pi /2)R_3(\pi /2)\begin{pmatrix} v'_\theta \\ v'_\chi \\ v'_z \end{pmatrix}. \end{aligned}$$

(A.8)

Here z denotes the third direction (i.e. the vertical $\partial _z$, with the positive sign going from bottom to top in Fig. 24) and $R_i(\alpha )$ denotes the rotation around i-th axis by angle $\alpha$.

Fig. 24

Mapping torus $MT((S_{(0/1,1)})^1_2)$: the bottom slice is $X_1\times \{0\}$, while the top one is the image of $S^2$, i.e. $X_1\times \{1\}$. The figure also displays cycles a and b along which one moves to determine the ${\textrm{ABK}}$ invariant of the Klein bottle

The next step is choosing a lift for the transition function to ${\textrm{Spin}}(3)\cong SU(2)$. Obviously, we have

$$\begin{aligned} {\textsf{T}}_S= R_3(\pi /2) \xrightarrow {\textrm{lift}} \widetilde{ {\textsf{T}}}^\pm _S =\pm e^{-i\frac{\pi }{4}\sigma _3}. \end{aligned}$$

(A.9)

Since it will be useful also for the next computations,²⁴ we choose from now on to use the lift $\widetilde{{\textsf{T}}}_S{:}{=}\widetilde{{\textsf{T}}}_S^+$.

Lastly, for (0, 1) the transition functions in ${\textrm{Spin}}(2)$ for the identifications of the points $(\theta +1,\chi )\sim (\theta ,\chi )$ and $(\theta ,\chi +1)\sim (\theta ,\chi )$ are respectively $-\textrm{id}$ and $+\textrm{id}$, which are properly lifted to ${\mp }\textrm{id}\in {\textrm{Spin}}(3)\cong SU(2)$ for the mapping torus. For (1, 1) we do not have to keep track of this since it is always $+\textrm{id}$.

At this point we can compute the value of the enhancement q for the 1-cycles a and b.

• For the cycle a a framing is given by doing a $2\pi n$ rotation while going from the point $x\in X_1\times \{0\}$ with coordinates $(\theta ,\chi )=(1/2,0)$ to itself along the direction depicted. This means that a vector $v\in T_x MT((S_{(0/1,1)})^1_2)$ goes back to itself via

  $$\begin{aligned} v'=R_2(2\pi n)v=v. \end{aligned}$$

  (A.10)

  The lift in SU(2) is $(-\textrm{id})^n$, which implies that $n=1$ is needed for the framing to be even. In this way the normal bundle is doing two negative half twists around the framing chosen and $q(a)=2\mod 4$.

• For the cycle b we start again from the point x. Going along b the generic vector does the transformation

  $$\begin{aligned} v'=\underbrace{R_3(\pi /2)^2}_{{\textsf{T}}_{S^2}}R_3((2n+1)\pi )v. \end{aligned}$$

  (A.11)

  The ending point is $x'$, with coordinates $(\theta ',\chi ')=(1/2,0)$, equivalent to $(\theta ,\chi )=(-1/2,0)$. Thus to be back at x one has an additional $-\textrm{id}$ transformation to apply in the case of $\{{\textrm{R}}0,{\textrm{NS}}1\}$. We will keep track of it by writing it inside square brackets. In SU(2) this means that the transformation is

  $$\begin{aligned}{}[-\textrm{id}]\underbrace{(-\textrm{id})}_{(-1)^{\mathcal {F}}}\underbrace{e^{-i \frac{\pi }{2}\sigma _3}}_{\widetilde{{\textsf{T}}}_{S^2}}e^{-i(2n+1)\frac{\pi }{2}\sigma _3}= (-\textrm{id})^{n[+1]}. \end{aligned}$$

  (A.12)

  Consider now the two different classes. For (1, 1) the even framing is found for $n=1$. In this case the framing is then defined by doing three positive $\pi$ rotations, which means the normal bundle is doing three negative half rotations and $q(b)=-3\mod 4=1\mod 4$. For (0, 1) instead we have $n=0$ and accordingly $q(b)=3\mod 4$.

Now we can use (3.8) and find that $q(a)=q(b)$. Therefore we arrive to the values

$$\begin{aligned} (S_{(1,1)})^1_2= i^\nu , \qquad (S_{(0,1)})^1_2=(-i)^\nu . \end{aligned}$$

(A.13)

The next matrix element we are going to compute is $(S_{(0/1,1)})^3_3$. In this case the manifold describes a single S transformation of $X_3$. Proceeding as usual, the $\textrm{pin}^-$ surface is a smooth curve Poincaré dual to the ${\mathbb {Z}}_2$ holonomies spanned along the vertical direction. In this case one has to pay attention when choosing such a curve. Indeed, this amounts to choosing a resolution of its singular representative, which in general will not be invariant under S. From consistency with (3.22), (3.24) we have

(A.14)

With this fixed, it follows that ${\textrm{PD}}(a_g)$ is the saddle surface in Fig. 25, represented by the string of 1-chains

$$\begin{aligned} acb^{-1}db^{-1}cad\sim eeffhh, \end{aligned}$$

(A.15)

where $e=cb^{-1}db^{-1}$, $f=bd^{-1}$ and $h=da$. Thus $\Sigma ^S_{33}=\mathbb{R}\mathbb{P}^2_e\#\mathbb{R}\mathbb{P}^2_f\#\mathbb{R}\mathbb{P}^2_h$, where again the indices represent the generators of the first ${\mathbb {Z}}_2$ homology group of each $\mathbb{R}\mathbb{P}^2$ component. By knowing the value of q for three independent elements in $H^1(\Sigma _{33}^S,{\mathbb {Z}}_2)$ we can determine the value of its ${\textrm{ABK}}$ invariant.

Fig. 25

Left: the mapping torus $MT((S_{(0/1,1)})^3_3)$ with the embedded surface $\Sigma _{33}^S$. It is understood we are using the same convention of the other figures. Right: the polygon representation of $\Sigma _{33}^S$ and its generators

In this case the change of coordinates in ${\mathbb {T}}^2\times \{0\}\sim {\mathbb {T}}^2\times \{1\}$ is

$$\begin{aligned} \begin{pmatrix} \partial '_\chi&\partial '_\theta \end{pmatrix} = \begin{pmatrix} \partial _\chi&\partial _\theta \end{pmatrix} \begin{pmatrix} 0 &{} -1 \\ 1&{}0 \\ \end{pmatrix}^T, \quad \begin{pmatrix} \theta '\\ \chi ' \end{pmatrix} = \begin{pmatrix} 0 &{} 1 \\ -1&{}0 \\ \end{pmatrix} \begin{pmatrix} \theta \\ \chi \end{pmatrix}, \end{aligned}$$

(A.16)

while the transition functions in SU(2) for the identifications $(\theta +1,\chi ) \sim (\theta ,\chi )$ and $(\theta ,\chi +1)\sim (\theta ,\chi )$ are both $-\textrm{id}$ for (0, 1) and $+\textrm{id}$ for (1, 1).

We now compute the value of q for the cycles $h,\, f$ and $e+f$.

• We can move along the cycle h in 4 steps. First we start at x with coordinates $(\theta ,\chi )=(1/2,0)$ and arrive to the point y with $(\theta ,\chi )=(0,1/2)$ by moving along $a^{-1}$ and doing a $\pi /2$ rotation around the $3^{\textrm{rd}}$ axis $\partial _z$. Then at y the cycle can be smoothed so that by following it we do a rotation of $\pi /2$ around $\partial _\chi$. We then go to $x'$ with coordinates $(\theta ',\chi ')=(0,1/2)$ along $d^{-1}$ by doing a $(2n+1)\pi$ rotation around $\partial _z$. Finally, we do a $\pi /2$ rotation around $\partial '_\chi$ before applying the transition function. Since $x'=(\theta =-1/2,\chi =0)$, in order to go completely back to the starting basis of the spin bundle on x there is an additional $[-\textrm{id}]\in SU(2)$ for the manifold $\{{\textrm{NS}}1,{\textrm{NS}}1\}$. The total transformation for a vector v is

  $$\begin{aligned} v' = \underbrace{R_3(\pi /2)}_{{\textsf{T}}_S}R_2(\pi /2)R_3((2n+1)\pi )R_2(\pi /2)R_3(\pi /2)v = v, \end{aligned}$$

  (A.17)

  so that the requirement of an even framing is equivalent to the condition

  $$\begin{aligned}{}[-\textrm{id}]\underbrace{(-\textrm{id})}_{(-1)^{\mathcal {F}}}\underbrace{ e^{-i \frac{\pi }{4}\sigma _3}}_{\widetilde{ {\textsf{T}}}_S}e^{-i\frac{\pi }{4}\sigma _2}e^{-i\frac{2n+1}{2}\pi \sigma _3}e^{-i\frac{\pi }{4}\sigma _2}e^{-i\frac{\pi }{4}\sigma _3} =(-\textrm{id})^{n[+1]} \overset{!}{=}\ -\textrm{id}.\nonumber \\ \end{aligned}$$

  (A.18)

• Dividing in similar steps the framing of cycle f, a generic $v\in T_x MT((S_{0/1,1})^2_2)$ transforms as

  $$\begin{aligned} v' = \underbrace{R_3(\pi /2)}_{{\textsf{T}}_S}R_2(-\pi /2)R_3((2n+1)\pi )R_2(-\pi /2)R_3(\pi /2)v = v. \end{aligned}$$

  (A.19)

  With the aid of Fig. 25 it is clear that by starting at x considered with coordinates²⁵$(\theta =1/2, \chi =1)$, then we arrive at $x'$ with coordinates $(\theta ',\chi ')=(1,1/2)\sim (\theta ,\chi )=(-1/2,1)$. Thus the lift of the full rotation in ${\textrm{Spin}}(3)$ gets again an additional $[-\textrm{id}]$ in the case $\{{\textrm{NS}}1,{\textrm{NS}}1\}$. Therefore the even framing condition for f is

  $$\begin{aligned}{}[-\textrm{id}]\underbrace{(-\textrm{id})}_{(-1)^{\mathcal {F}}}\underbrace{ e^{-i \frac{\pi }{4}\sigma _3}}_{\widetilde{ {\textsf{T}}}_S}e^{i\frac{\pi }{4}\sigma _2}e^{-i\frac{2n+1}{2}\pi \sigma _3}e^{i\frac{\pi }{4}\sigma _2}.e^{-i\frac{\pi }{4}\sigma _3} =(-\textrm{id})^{n[+1]} \overset{!}{=}\ -\textrm{id}. \end{aligned}$$

  (A.20)

• Finally we look at the cycle $e+f\sim cb^{-1}$. Starting at x with coordinates $(\theta ,\chi )=(1/2,0)$ a vector v transforms as

  $$\begin{aligned} v'=\underbrace{R_3(\pi /2)}_{{\textsf{T}}_S}R_2(-\pi /2)R_3(-\pi /2)R_1(-\pi /2)R_3(2 \pi n) = v. \end{aligned}$$

  (A.21)

  The arrival point of this cycle now is $x'$ with coordinates $(\theta ',\chi ')=(1,1/2)\sim (\theta ,\chi )=(-1/2,1)$, so there is no additional sign to keep track of. Therefore in ${\textrm{Spin}}(3)$ we have

  $$\begin{aligned} \underbrace{(-\textrm{id})}_{(-1)^{\mathcal {F}}}\underbrace{e^{-i \frac{\pi }{4}\sigma _3}}_{\widetilde{ {\textsf{T}}}_S}e^{i\frac{\pi }{4}\sigma _2}e^{i\frac{\pi }{4} \sigma _3}e^{-i\frac{\pi }{4}\sigma _1}e^{i\pi n \sigma _3} = (-\textrm{id})^{n+1} \overset{!}{=}\ (-\textrm{id}). \end{aligned}$$

  (A.22)

From these results we see that for the class (1, 1) one must set $n=1$ for the loops h and f. Thus, in both cases the normal bundle does three negative half twists with respect to the framing chosen, or, in other words, $q(f)=q(h)=-3\mod 4=1\mod 4$. Instead, $n=0$ is necessary for $e+f$, so that on this cycle the normal bundle does not do any twist and $q(e)=q(f)+2\mod 4$.

By the same logic, for the class (0, 1) one must impose $n=0$ for all the loops $h,\,f$ and $e+f$. This means that in this case the enhancement has the values $q(h)=q(f)=3\mod 4$ and $q(e)=q(f)+2\mod 4$.

Therefore the final result is

$$\begin{aligned} (S_{(0,1)})^2_2 = e^{-i\frac{\pi }{4}\nu }, \qquad (S_{(1,1)})^2_2 = e^{i \frac{\pi }{4}\nu }. \end{aligned}$$

(A.23)

The last two mapping tori we need to look at are the ones related to the entries $(T_{(0/1,1)})^2_2$ and $(T_{(0/1,1)})^3_1$. By applying (3.27) it follows that the manifold describing the first of these two entries is given by the ${\mathcal {T}}$-bordism of $X_2$ with the identification of the boundaries $X_2\times \{0\}\sim X_2\times \{1\}$ like in the previous cases. We can see such a manifold in Fig. 26, where we depicted also $\Sigma ^T_{22}$, found with the same reasoning of before. In this case the surface is just a torus represented by the string of 1-cycles

$$\begin{aligned} acb^{-1}a^{-1}c^{-1}b\sim e^{-1}de d^{-1}, \end{aligned}$$

(A.24)

where $d=a^{-1}c^{-1}$ and $e=bc^{-1}$ are a symplectic basis of $H^1(\Sigma ^T_{(0/1,1)},{\mathbb {Z}}_2)$.

Fig. 26

Left: the mapping torus $MT((T_{(0/1,1)})^2_2)$ with the embedded surface $\Sigma _{22}^T$. Right: the plain representation of $\Sigma _{22}^T$ and its generators

The next step is defining the transition function which changes the canonical basis of the tangent space given by the coordinates used in the top slice $X_2 \times \{1\}$ to the ones of the bottom slice $X_2 \times \{0\}$. Here the transition function will be lifted to elements of $\widetilde{{\textrm{SL}}}(3,{\mathbb {R}})$, the double cover of ${\textrm{SL}}(3,{\mathbb {R}})$. Under a T transformation

$$\begin{aligned} \begin{pmatrix} \partial '_\chi&\partial '_\theta \end{pmatrix} = \begin{pmatrix} \partial _\chi&\partial _\theta \end{pmatrix} \begin{pmatrix} 1 &{} 1 \\ 0&{}1 \\ \end{pmatrix}^T,\quad \begin{pmatrix} \theta '\\ \chi ' \end{pmatrix} = \begin{pmatrix} 1 &{} -1 \\ 0&{}1 \\ \end{pmatrix} \begin{pmatrix} \theta \\ \chi \end{pmatrix}. \end{aligned}$$

(A.25)

If we start from a vector $v \in T_p MT((T_{(0/1,1)})^2_2)$ with $p \in X_2 \times \{0\} \simeq X_2 \times \{1\}$ and write it in the canonical basis of $X_2\times \{1\}$, then by applying ${\textsf{T}}_T = H(1)$, where

$$\begin{aligned} H(t):=\ \begin{pmatrix} 1&{} \quad t \quad 0\\ 0 \quad 1 \quad 0\\ 0 \quad 0 \quad 1 \end{pmatrix} \end{aligned}$$

(A.26)

we have:

$$\begin{aligned} \begin{pmatrix} v_\theta \\ v_\chi \\ v_z \end{pmatrix}= \begin{pmatrix} 1 \quad 1 \quad 0\\ 0 \quad 1 \quad 0\\ 0 \quad 0 \quad 1 \end{pmatrix}\begin{pmatrix} v'_\theta \\ v'_\chi \\ v'_z \end{pmatrix}. \end{aligned}$$

(A.27)

For the lift ${\textsf{T}}_T\rightarrow \widetilde{{\textsf{T}}}_T\in \widetilde{{\textrm{SL}}}(3,{\mathbb {R}})$ one notes that ${\textrm{SL}}(3,{\mathbb {R}})$ can be continuously retracted to SO(3) via the Gram-Schmidt procedure. If $\gamma :{\textrm{SL}}(3,{\mathbb {R}})\rightarrow SO(3)$ is such retraction, then its lifting $\widetilde{\gamma }$ has to make the following diagram commute:

(A.28)

Here $\pi$ denotes the projections. It follows that ${\widetilde{\gamma }}$ maps the center $\{\pm \textrm{id} \}$ to itself.²⁶ From the fact that $\gamma (H(t))=\textrm{id}$ $\forall t\in {\mathbb {R}}$, then the lifting of the transition function can be $\widetilde{{\textsf{T}}}_{T}={\widetilde{H}}_{\pm }(1)$, defined by the property $\widetilde{\gamma }({\widetilde{H}}_{\pm }(1))=\pm \textrm{id} \in SU(2)$. For consistency with the choice made for S, we choose²⁷ the lift $\widetilde{{\textsf{T}}}_{T}={\widetilde{H}}_+(1)$.

We can now turn to computing the value of q for the cycles d and e.

• For the loop e we start from x, which has the coordinates $(\theta ,\chi )=(0,1/2)$. We then go to a point p in the middle of the segment c connecting x to $y=(\theta '=0,\chi '=1/2)$ while doing a $2 \pi n$ rotation along the direction $\partial _z$. From here we reach y by doing a continuous transformation $H(-t)$ with $t\in [0,1]$. At this point the cycle can be smoothed so that by following it we do a $\pi$ rotation along the $\partial '_\chi$. Then $x'=(\theta '=1/2,\chi '=1/2)$ is reached without doing any rotation to the reference frame. Applying the transition function we are back to x with the original basis of the tangent space. So taking everything into account we see that a vector v goes to itself by

  $$\begin{aligned} v' = \underbrace{H(1)}_{{\textsf{T}}_T}R_2(-\pi )R_2(\pi )H(-1)R_3(2 \pi n)v = v. \end{aligned}$$

  (A.29)

  The lift of this transformation is²⁸

  $$\begin{aligned} \underbrace{(-\textrm{id})}_{(-1)^{\mathcal {F}}}\underbrace{{\widetilde{H}}_+(1)}_{\widetilde{{\textsf{T}}}_T} e^{i \frac{\pi }{2}\sigma _2}e^{-i \frac{\pi }{2}\sigma _2}{\widetilde{H}}(-1)e^{- i n \pi } \xrightarrow {\widetilde{\gamma }}\underbrace{\widetilde{\gamma }({\widetilde{H}}_-(1))}_{\widetilde{\gamma }((-1)^{\mathcal {F}}\widetilde{{\textsf{T}}}_T)}\widetilde{\gamma }({\widetilde{H}}(-1)(-\textrm{id})^n)= (-\textrm{id})^{n+1}. \nonumber \\ \end{aligned}$$

  (A.30)

  For the deformation contraction one has to group together all the components that define the transition function between the two bases of the tangent space of $X_2$, so in this case $(-1)^{\mathcal {F}}$ and $\widetilde{{\textsf{T}}}_T$, and the transformations done by moving along the loop. The equality on the right side follows by remembering that ${\widetilde{H}}(-1)$ is the ending point of a continuous lift ${\widetilde{H}}(-t)$ with $t\in [0,1]$, ${\widetilde{H}}(0)=\textrm{id}$ and using also the fact that for any element $A\in \widetilde{{\textrm{SL}}}(3,{\mathbb {R}})$ it holds ${\widetilde{\gamma }}(AB)={\widetilde{\gamma }}(A){\widetilde{\gamma }}(B)$ if $B\in Z(\widetilde{{\textrm{SL}}}(3,{\mathbb {R}}))$. At this point we can simply set $n=0$ to have an even framing, which means the normal bundle does not twist with respect to it and $q(e)=0\mod 4$.

• With the same approach along the cycle c any vector v rotates by

  $$\begin{aligned} \underbrace{H(1)}_{{\textsf{T}}_T}R_2(\pi )R_2(-\pi )H(-1)R_3(2\pi n) = \textrm{id}. \end{aligned}$$

  (A.31)

  The deformation contraction of its lift gives again the same element of (A.30), so we can conclude that $q(d)=q(e)$.

This means that for both classes (0/1, 1) we have

$$\begin{aligned} (T_{(0,1)})^2_2=(T_{(1,1)})^2_2=1. \end{aligned}$$

(A.32)

Finally, we are left with the computation of $(T_{(0/1,1)})^3_1$. In this case the 3-manifold is given by identifying the two boundaries $X_1$ of the $T^2$-bordism ${\mathcal {T}}_{\{{\textrm{R}}0,{\textrm{NS}}/{\textrm{R}}1\}}^{\{{\textrm{R}}0,{\textrm{NS}}/{\textrm{R}}1\}}:\{{\textrm{R}}0,{\textrm{NS}}/{\textrm{R}}1\}$ $\rightarrow \{{\textrm{R}}0,{\textrm{NS}}/{\textrm{R}}1\}$. The surface $\Sigma _{13}^T$ is found by looking for a smooth interpolation between its projection on $X_1\times \{0\}$ and $X_1\times \{1\}$. A schematic representation is given by

(A.33)

with $0<t<t'<1$, while in Fig. 27 we can see its embedding in the 3-manifold. As usual, we can use some chains to write the string presentation of the surface, which in this case is

$$\begin{aligned} abcdc^{-1}ebea^{-1}d^{-1} \sim rr uu ll, \end{aligned}$$

(A.34)

with $r=abea^{-1}d^{-1},\,u=dae^{-1}c$ and $l=c^{-1}ea^{-1}$.

Fig. 27

Left: the mapping torus $MT((T_{(0/1,1)}^T)^3_1)$ with the embedded surface $\Sigma _{31}^T$. Right: the polygon representation of $\Sigma _{31}^T$ and its generators

Therefore $\Sigma ^T_{13}=\mathbb{R}\mathbb{P}^2_{r}\#\mathbb{R}\mathbb{P}^2_{u}\#\mathbb{R}\mathbb{P}^2_{l}$.

To find the value of its ${\textrm{ABK}}$ invariant we compute the enhancement q for $l,\,l+u$ and $l+u+r$. For our purpose it comes at hand to consider the portions of these cycles on $X_1\times \{1\}$ as if they were instead on the slice $X_1\times \{t'\}$. This does not pose any problem, because the value of q is robust under any continuous deformation of the surface.

• We begin with $l\sim a e^{-1}c$, which we divide as usual in various pieces. From the starting point p on $X_1\times \{1\}$ with coordinates $(\theta ',\chi ')=(1/2,0)$ we go to $q=(\theta '=0,\chi '=1/4)$ while doing a $\pi /2$ rotation around $\partial _z$. At q we smooth the cycle and do a $-\pi /2$ rotation around $\partial '_\chi$. From q to $r=(\theta '=0,\chi '=3/4)$ along $e^{-1}$ we do first a $\pi$ rotation around the $1^{\textrm{st}}$ axis describing a u-turn, followed by a $(2n+1)\pi$ rotation around $\partial _z$. Like before, at r we have to do a second $-\pi /2$ rotation around $\partial '_\chi$ and, while going back to p, one last $-\pi /2$ rotation around $\partial '_z$. This means that $v\in T_p MT((T_{0/1,1})^3_1)$ transforms as

  $$\begin{aligned} v'= R_3(-\pi /2)R_2(-\pi /2)R_3((2n+1)\pi )R_1(\pi )R_2(-\pi /2)R_3(\pi /2)v=v.\nonumber \\ \end{aligned}$$

  (A.35)

  The lift is given by

  $$\begin{aligned}{}[-\textrm{id}]e^{+i\frac{\pi }{4}\sigma _3}e^{i\frac{\pi }{4}\sigma _2}e^{-i\frac{(2n+1)\pi }{2}\sigma _3}e^{-i\frac{\pi }{2}\sigma _1}e^{i\frac{\pi }{4}\sigma _2}e^{-i\frac{\pi }{4}\sigma _3}=(-\textrm{id})^{n[+1]}, \end{aligned}$$

  (A.36)

  where the first $[-\textrm{id}]$ is present only for the bordism class (0, 1). This means that $n=1$ and $q(l)=-3\mod 4=1\mod 4$ for (1, 1) and $n=0,\,q(l)=3\mod 4$ for (0, 1).

• For the cycle $l+u+r\sim e^{-1}b^{-1}$ we start from $q=(\theta '=0,\chi '=1/4)$ on $X_1\times \{1\}$. Here running along $e^{-1}$ we do the same rotations as before. The difference in the computation is that once we arrive at the point r we have instead to do a $\pi /2$ rotation around $\partial '_\chi$, then go back to q and conclude repeating doing a second $\pi /2$ rotation. The final result is that from $v \in T_q MT((T_{0/1,1})^3_1)$ we arrive to

  $$\begin{aligned} v'=R_2(\pi /2)R_2(\pi /2)R_3((2n+1)\pi )R_1(\pi )v=v. \end{aligned}$$

  (A.37)

  This is essentially the exact expression of (A.35) up to changing the orientation of the rotations around the $2^{\textrm{nd}}$ axis, so it is no surprise that the lift in SU(2) gives $(-\textrm{id})^{n+1[+1]}$. Therefore $q(l+u+r)=3\mod 4$ for (1, 1) and $q(l+u+r)=1\mod 4$ for (0, 1).

• Finally it is the turn of $l+u\sim d$. In this case going from the $X_1\times \{0\}$ to $X_1\times \{1\}$ we do a $2\pi n$ rotation around $\partial _z$, followed by a continuous transformation $H(-2t),\, t\in [0,1]$. Thus a vector transforms as

  $$\begin{aligned} v'= \underbrace{H(1)H(1)}_{{\textsf{T}}_{T^2}}H(-2)R_3(2\pi n)v = v. \end{aligned}$$

  (A.38)

  The lift of the transformation in $\widetilde{{\textrm{SL}}}(3,{\mathbb {R}})$ is then

  $$\begin{aligned} \underbrace{(-\textrm{id})}_{(-1)^{\mathcal {F}}}{\widetilde{H}}_+(1){\widetilde{H}}_+(1){\widetilde{H}}(-2)e^{-i\pi n \sigma _3}=(-\textrm{id})^{n+1}. \end{aligned}$$

  (A.39)

  We conclude that $n=0$ and $q(l+u)=0\mod 4$.

With these explicit computations one can arrive to the last results that determine the T matrix, i.e.

$$\begin{aligned} (T_{(0,1)})^3_1 =e^{i\frac{\pi }{4}\nu },\qquad (T_{(1,1)})^3_1 =e^{-i\frac{\pi }{4}\nu }. \end{aligned}$$

(A.40)

Details on Modular Bootstrap

Here we review in detail the linear functional method used to find the bounds presented in Sect. 5.3, based on the original papers [23, 75]. We also present the free fermion CFTs that almost saturate the bounds found for the kinks of Figs. 15, 18.

In order to apply the linear functional method to (5.12) one needs to expand the partition functions in terms of Virasoro characters:

$$\begin{aligned} \sum _{h,{\bar{h}},j} n_{h,{\bar{h}},j}\left( \delta _i^j \chi _h(-1/\tau ){\bar{\chi }}_{{\bar{h}}}(-1/{\bar{\tau }})-S^j_i\chi _h(\tau ){\bar{\chi }}_{{\bar{h}}}({\bar{\tau }})\right) =0,\qquad \forall \,i. \end{aligned}$$

(B.1)

Recall that in the proper basis the degeneracies $n_{h,{\bar{h}},j}$ are positive. Then, by defining $\textbf{M}^j$ the vectors with entries

$$\begin{aligned} M^j_i = \delta _i^j \chi _h(-1/\tau ){\bar{\chi }}_{{\bar{h}}}(-1/{\bar{\tau }})-S^j_i\chi _h(\tau ){\bar{\chi }}_{{\bar{h}}}({\bar{\tau }}), \end{aligned}$$

(B.2)

we simply look for a functional $\alpha$ which returns a real function of $\Delta$ and s that satisfies

$$\begin{aligned} {\left\{ \begin{array}{ll} \alpha [\textbf{M}^1](0,0)>0,&{}\\ \alpha [\textbf{M}^j](|s|,s)\ge 0,&{}\qquad \forall j,\,s\ne 0,\\ \alpha [\textbf{M}^j](\Delta ,s)\ge 0,&{}\qquad \forall j,\,s,\, \Delta >\Delta ^*_j.\\ \end{array}\right. } \end{aligned}$$

(B.3)

Here s is supposed to be any admitted value of the spins on ${\mathcal {H}}_j$, with the exception of $s=0$ for the second condition if it is a priori allowed in the corresponding ${\mathcal {H}}_j$. Instead, $\Delta ^*_j$ will vary depending on the bound we want to find.

The first condition asks for the functional to be strictly positive when evaluated on the vacuum, so that applying it to the modular crossing equation always guarantees to rule out the corresponding spectra.

In our analysis we will consider a derivative basis for it around the point $\tau =i$. By introducing the variable z such that $\tau =i \exp z$ we can then expand it as follows:

$$\begin{aligned} \alpha [\textbf{M}^j](\Delta ,s)= \sum _{n,m,i} \gamma _{n,m,i} \partial _z^n \partial _{{\bar{z}}}^m M^j_i|_{z={\bar{z}}=0}. \end{aligned}$$

(B.4)

We now turn into explaining what are the values $\Delta ^*_j$ for which we try to find a functional $\alpha$ that satisfies (B.3):

• 1. In order to find the lightest non-degenerate primary on the spectrum of ${\mathcal {H}}_{j_0}$, the definition is

  $$\begin{aligned} \Delta ^*_j= {\left\{ \begin{array}{ll} \textrm{max}\left( \Delta ^{j_0}_{\textrm{gap}},|s|\right) ,&{}\textrm{if}\, j = j_0,\\ |s|,&{}\textrm{otherwise}. \end{array}\right. } \end{aligned}$$

  (B.5)

• 2. In order to find the same bound, but for scalar primaries, we have

  $$\begin{aligned} \Delta ^*_j= {\left\{ \begin{array}{ll} \Delta ^{j_0}_{\textrm{scal}},&{}\textrm{if}\,j=j_0,\,s=0,\\ |s|,&{}\textrm{otherwise}. \end{array}\right. } \end{aligned}$$

  (B.6)

We mention here that numerically it is convenient to allow for the possibility of having the limit $\Delta ^j_{\mathrm {gap/scal}}\rightarrow |s|$ for non-degenerate Virasoro characters. In this case the contribution to the partition function (assuming for example $s>0$) is

$$\begin{aligned} \lim _{\Delta \rightarrow s} \chi _{\frac{\Delta +s}{2}}(\tau ) {\bar{\chi }}_{\frac{\Delta -s}{2}}({\bar{\tau }})= \chi _s(\tau )({\bar{\chi }}_0({\bar{\tau }})+{\bar{\chi }}_1({\bar{\tau }})), \end{aligned}$$

(B.7)

i.e. it is equivalent to the contribution of a primary with $\Delta =s+1$ and a conserved current. We call this particular limit case a non-degenerate Virasoro character of generic type.

With these premises, the algorithm that determines the bound on the primaries is simple: one starts with some fixed value of $\Delta ^{j_0}_{{\textrm{gap}}/{\textrm{scal}}}$ and then searches for a functional that has the properties discussed. Whether one exists or not then determines if the bound considered can be lowered or raised. The procedure is then repeated until one reaches the wanted precision.

To numerically implement this search, we must truncate the basis of the linear functional up to some derivative order $\Lambda$, i.e. $n+m\le \Lambda$. For us the choice will be set at $\Lambda =10$.

Moreover, we note that the usual procedure to utilize parity invariant partition functions is of no use here, with the exception of when $\nu =0,4\mod 8$. Indeed, these are the only values for which the spin selection rules for the sectors ${\mathcal {H}}_i$ are symmetric under the reflection $s\mapsto -s$. Thus, if for other values of $\nu$ we were to restrict to parity invariant partition functions, the bounds we would find would not be as strict as possible, since we would mix ${\widetilde{Z}}_i$ and $\widetilde{{\bar{Z}}}_i$, that have different spin selection rules. This means that computationally we are able to reduce the basis of functionals to a symmetric one only for the aforementioned cases $\nu =0,4\mod 8$.

Finally, we recall that by performing the computations with the SDPB solver, while we are actually able to consider for each i a continuum spectrum of operators with $\Delta > \Delta ^*_i$, we can instead enforce the conditions (B.3) only up to some value of spin $s\le s_{\textrm{max}}$. Albeit a priori this might be a significant problem, one usually finds that at fixed $\Lambda$ for a reasonable value of $s_{\textrm{max}}$ numerical stability is reached and the bounds converge within some error $\Delta _\delta$. Of course $s_{\textrm{max}}$ will vary depending on the precision we want to meet, but generally is found to be of the same order of magnitude of $\Lambda$. In our case we decided to set $\Delta _\delta =0.01$, for which numerical stability is reached by considering values of the spins up to $s_{\textrm{max}}=40$.

We report also the other relevant parameters with which the numerical analysis has been performed, i.e.

precision	700
primalErrorThreshold	$10^{-30}$
dualErrorThreshold	$10^{-30}$
maxComplementarity	$10^{100}$
feasibleCenteringParameter	0.1
infeasibleCenteringParameter	0.3
stepLengthReduction	0.7

1.1 Free fermions kinks

Here we recall some basic facts about stacks of free fermion CFTs, in order to show that these theories saturate the bounds at the kinks found from the numerical analysis of Sect. 5.3.

A generic stack of Majorana fermions

$$\begin{aligned} {\mathcal {L}} = \sum _{a=1}^n \psi _a {\bar{\partial }} \psi _a + {\bar{\psi }}_a \partial {\bar{\psi }}_a \end{aligned}$$

(B.8)

describes a spin-CFT with the value of central charge $c=n/2$. For us, it is sufficient to consider fermions on a torus with periodicity conditions

$$\begin{aligned} \psi (\theta +1,\chi )=e^{2\pi i \alpha }\psi (\theta ,\chi ),\qquad \psi (\theta ,\chi +1)=e^{2\pi i \beta }\psi (\theta ,\chi ), \end{aligned}$$

(B.9)

where $\alpha ,\,\beta =0,1/2$ are the sum$\mod 1$ of the periodicity conditions defined by the spin structure and the ${\mathbb {Z}}_2$ global symmetry. We can treat these together as an element in $H^1({\mathbb {T}}^2,{\mathbb {Z}}_2)$. Then the holomorphic partition function contribution of each Majorana fermion is of the form²⁹

(B.10)

where $\vartheta _i(\tau )$ are the Jacobi Theta functions. We remember they satisfy the following identities:

$$\begin{aligned} \begin{aligned}&\vartheta _1(\tau )=0,\\&\frac{\vartheta _2(-1/\tau )}{\eta (-1/\tau )}= \frac{\vartheta _4(\tau )}{\eta (\tau )},{} & {} \quad \frac{\vartheta _2(\tau +1)}{\eta (\tau +1)}= e^{i\pi /6}\frac{\vartheta _2(\tau )}{\eta (\tau )},\\&\frac{\vartheta _3(-1/\tau )}{\eta (-1/\tau )}= \frac{\vartheta _3(\tau )}{\eta (\tau )},{} & {} \quad \frac{\vartheta _3(\tau +1)}{\eta (\tau +1)}= e^{-i\pi /12}\frac{\vartheta _4(\tau )}{\eta (\tau )},\\&\frac{\vartheta _4(-1/\tau )}{\eta (-1/\tau )}= \frac{\vartheta _2(\tau )}{\eta (\tau )},{} & {} \quad \frac{\vartheta _4(\tau +1)}{\eta (\tau +1)}= e^{-i\pi /12}\frac{\vartheta _3(\tau )}{\eta (\tau )}. \end{aligned} \end{aligned}$$

(B.11)

If one considers a single Majorana fermion charged under both $(-1)^F$ and $(-1)^Q=(-1)^{F_L}$, its total contribution to the partition function in ${\mathcal {H}}_{{\textrm{NS}}1}$ will be

(B.12)

signaling that under stacking it increases the anomaly $\nu \mapsto \nu +1 \mod 8$. Instead, by considering it to be charged under $(-1)^Q=(-1)^{F_R}$ rather than $(-1)^{F_L}$, one gets a decrease in the anomaly $\nu \mapsto \nu -1 \mod 8$. Therefore, for a free fermion theory with $c=n/2$ we can have arbitrary anomaly $\nu ={\widetilde{\nu }}_L-{\widetilde{\nu }}_R \mod 8$, where ${\widetilde{\nu }}_{L,R}\le n$ are the number of left/right-moving fermions charged under $(-1)^{Q}$.

We start to look at the case $c=1$. This means the possible anomalies are $\nu =0,\pm 1,\pm 2\mod 8$. The partition functions (5.5) for $\nu =0,1,2$ are the following:

$$\begin{aligned} \nu =0\,:&\qquad \left\{ \begin{aligned}&Z_{{\textrm{NS}}0}^{+F+Q} =\frac{1}{2}\frac{|\vartheta _3(\tau )|^2+|\vartheta _4(\tau )|^2}{|\eta (\tau )|^2},\\&Z_{{\textrm{NS}}0}^{+F-Q}=0, \end{aligned}\right. \end{aligned}$$

(B.13)

$$\begin{aligned} \nu =1\,:&\qquad \left\{ \begin{aligned}&Z_{{\textrm{NS}}0}^{+F+Q} =\frac{1}{4}\frac{(\sqrt{\vartheta _3(\tau )}+\sqrt{\vartheta _4(\tau )})(\overline{\vartheta _3(\tau )}\sqrt{\vartheta _3(\tau )}+\overline{\vartheta _4(\tau )}\sqrt{\vartheta _4(\tau )})}{|\eta (\tau )|^2},\\&Z_{{\textrm{NS}}0}^{+F-Q} =\frac{1}{4}\frac{(\sqrt{\vartheta _3(\tau )}-\sqrt{\vartheta _4(\tau )})(\overline{\vartheta _3(\tau )}\sqrt{\vartheta _3(\tau )}-\overline{\vartheta _4(\tau )}\sqrt{\vartheta _4(\tau )})}{|\eta (\tau )|^2}, \end{aligned}\right. \end{aligned}$$

(B.14)

$$\begin{aligned} \nu =2\,:&\qquad \left\{ \begin{aligned}&Z_{{\textrm{NS}}0}^{+F+Q} =\frac{1}{4}\left|\frac{\vartheta _3(\tau )+\vartheta _4(\tau )}{\eta (\tau )}\right|^2,\\&Z_{{\textrm{NS}}0}^{+F-Q} =\frac{1}{4}\left|\frac{\vartheta _3(\tau )-\vartheta _4(\tau )}{\eta (\tau )}\right|^2. \end{aligned}\right. \end{aligned}$$

(B.15)

The partition functions for $\nu =-1,-2$ cases are complex conjugated partition functions for $\nu =1,2$ respectively.

Expanding them in q we learn that the lightest primary states are:

$\nu$	${\mathcal {H}}_{{\textrm{NS}}0}^{+F+Q}$	${\mathcal {H}}_{{\textrm{NS}}0}^{+F-Q}$
0	(1/2, 1/2)	$\times$
$\pm 1$	(1/2, 1/2)	(1/2, 1/2)
$\pm 2$	(1, 0), (0, 1)	(1/2, 1/2)

As anticipated, these values almost saturate the bounds found from the numerical analysis presented in Figs. 15 and 18. Note that for $\nu =2$ we have a conserved current state. However, this is not unexpected, as its mix with the vacuum $(h,{\bar{h}})=(0,0)$ defines a non-degenerate Virasoro character of generic type.

Next we focus on the kinks we found at $c=n/2=4\pm \nu /2$ for ${\mathcal {H}}_{{\textrm{NS}}0}^{+F+Q}$. These are simply a set of $n=8\pm \nu$ Majorana fermions charged under $(-1)^Q=(-1)^{F_L}$ for $n=8+\nu$ and $(-1)^Q=(-1)^{F_R}$ for $n=8-\nu$. For example, in light of what we said, the partition function for $c=4+\nu /2$ is just

$$\begin{aligned} Z_{{\textrm{NS}}0}^{+F+Q}=\frac{1}{4}\left|\frac{\vartheta _3^c(\tau )+\vartheta _4^c(\tau )}{\eta (\tau )}\right|^2 \end{aligned}$$

(B.16)

and its expansion confirms that the lightest scalar is indeed a marginal operator with $\Delta =2$.

Finally, we note that the set of $n=4$ real fermions charged under $(-1)^Q=(-1)^{F_L}$ describes also the kink at $c=2$ for $v=4 \mod 8$. In fact in this case the partition function is

$$\begin{aligned} Z_{{\textrm{NS}}0}^{+F-Q}=\frac{1}{4}\left|\frac{\vartheta _3^2(\tau )-\vartheta _4^2(\tau )}{\eta (\tau )}\right|^2, \end{aligned}$$

(B.17)

so the lightest non-degenerate primary has again $(h,{\bar{h}})=(1/2,1/2)$.