Extending the Trace of a Pivotal Monoidal Functor

Abstract

We consider a pivotal monoidal functor whose domain is a modular tensor category (MTC). We show that the trace of such a functor naturally extends to a representation of the corresponding tube category. As irreducible representations of the tube category are indexed by pairs of simple objects in the underlying MTC, the simple multiplicities of this representation form a candidate modular invariant matrix. In general, this matrix will not be modular invariant, however it will always commute with the T-matrix. Furthermore, under certain additional conditions on the original functor, it is shown that the corresponding representation of the tube category is a haploid, symmetric, commutative Frobenius algebra. Such algebras are known to be connected to modular invariants, in particular a result of Kong and Runkel implies that the matrix of simple multiplicities commutes with the S-matrix if and only if the dimension of the algebra is equal to the dimension of the underlying MTC. Finally, this procedure is applied to certain pivotal monoidal functors arising from module categories over the Temperley–Lieb category and the associated MTC.

Appendix A. Lemmas on Frobenius Algebras

The purpose of this section is to provide the necessary definitions on Frobenius algebras followed by certain technical results required in Sect. 6.

Definition A.1

Let ${\mathcal {B}}$ be a monoidal category. A Frobenius algebra A in ${\mathcal {B}}$ is an algebra and a coalgebra in ${\mathcal {B}}$ such that

$$\begin{aligned} ({{\,\mathrm{id}\,}}_A \otimes \nabla ) \circ (\Delta \otimes {{\,\mathrm{id}\,}}_A) = \Delta \circ \nabla = (\nabla \otimes {{\,\mathrm{id}\,}}_A) \circ ({{\,\mathrm{id}\,}}_A \otimes \Delta ) \end{aligned}$$

(24)

where $\nabla$ is the product and $\Delta$ is the coproduct. Using the graphical notation

we can rewrite Condition (24) as

(25)

We also use

to denote the unit and

to denote the counit. A is called haploid if is satisfies ${{\,\mathrm{Hom}\,}}_{{\mathcal {B}}}({\mathbf{1 }}, A) = {{\mathbb {C}}}$. If ${\mathcal {B}}$ is braided then A is called commutative if the underlying algebra structure is commutative i.e.

(26)

If ${\mathcal {B}}$ is pivotal then A is called symmetric if is satisfies

(27)

Remark A.2

Let ${\mathcal {B}}$ be a fusion category with complete set of simples ${{\,\mathrm{Irr}\,}}({\mathcal {B}})$ and let A be an object in ${\mathcal {B}}$. Any morphism $\nabla$ from $A \otimes A$ to A gives rise to the following morphisms,

where X, Y, Z are in ${\mathcal {B}}$ and $A_X :={{\,\mathrm{Hom}\,}}_{{\mathcal {B}}}(X,A)$. The full map $\nabla$ is determined by $\nabla _R^{S,T}$ for $R,S,T \in {{\,\mathrm{Irr}\,}}({\mathcal {B}})$. Indeed we can recover it via

$$\begin{aligned} \bigoplus \limits _{RST} \sum \limits _{g,h,\alpha } \nabla _R^{S,T} (\alpha )(g \otimes h) \circ \alpha ^* \circ (g^* \otimes h^*) = \nabla \end{aligned}$$

where g ranges over a basis of $A_S$, h ranges over a basis of $A_T$ and $\alpha$ ranges over a basis of ${{\,\mathrm{Hom}\,}}_{{\mathcal {B}}}(R,ST)$. Similarly any morphism from A to $A \otimes A$ can also be decomposed in the following way

and then recovered via

$$\begin{aligned} \sum \limits _{\begin{array}{c} RST \\ \beta , f \end{array}} \Delta ^R_{ST}(\beta )(f) \circ \beta ^* \circ f^* = \Delta . \end{aligned}$$

Lemma A.3

Let A be an object in ${\mathcal {B}}$ and let $\nabla$ be in ${{\,\mathrm{Hom}\,}}_{{\mathcal {B}}}(A \otimes A,A)$. Then $\nabla$ is associative if

$$\begin{aligned} \nabla ^{RS,T}_{RST} ({{\,\mathrm{id}\,}}) \big ( \nabla ^{R,S}_{RS} ({{\,\mathrm{id}\,}}) (f \otimes g) \otimes h \big ) = \nabla ^{R,ST}_{RST} ({{\,\mathrm{id}\,}}) \big ( f \otimes \nabla ^{S,T}_{ST}({{\,\mathrm{id}\,}})(g \otimes h) \big ) \end{aligned}$$

(28)

for all $R,S,T \in {{\,\mathrm{Irr}\,}}({\mathcal {B}})$, $\alpha \in {{\,\mathrm{Hom}\,}}_{{\mathcal {B}}}(R,ST)$, $f \in A_R$, $g \in A_S$ and $h \in A_T$. An element $u \in A_{{\mathbf{1 }}}$ is a unit for $\nabla$ if

$$\begin{aligned} \nabla ^{{\mathbf{1 }},S}_S ({{\,\mathrm{id}\,}}) (u \otimes g) = g \quad \mathrm {and} \quad \nabla ^{S,{\mathbf{1 }}}_S ({{\,\mathrm{id}\,}}) (g \otimes u) = g \end{aligned}$$

(29)

Furthermore, if ${\mathcal {B}}$ is braided then $\nabla$ is commutative if

(30)

Proof

The first claim follows from the fact that, by decomposing the top of each strand (as in, for example, [HK19, Lemma 3.3]), we have

and

Similarly the second claim follows from

and the third claim from

$\square$

Lemma A.4

Let ${\mathcal {B}}$ be a spherical fusion category and let A be an object in ${\mathcal {B}}$ together with a collection of perfect pairings

$$\begin{aligned} \langle -, -\rangle _S :A_S \otimes A_{S^\vee } \rightarrow {{\mathbb {C}}} \end{aligned}$$

for all $S \in {{\,\mathrm{Irr}\,}}({\mathcal {B}})$. Let c be a map from ${{\,\mathrm{Irr}\,}}({\mathcal {B}})$ to ${{\mathbb {C}}} \setminus \{ 0 \}$. We consider the morphisms

and

where $\{b\}$ is a basis of $A_S$ and $\{b'\}$ is the corresponding dual basis of $A_{S^\vee }$ with respect to $\langle -, -\rangle _S$. Then is a dual object to A. Furthermore, with respect to this duality, we have

$$\begin{aligned} \langle f , (g^*)^\vee \rangle = c(S) g^*(f) \end{aligned}$$

(31)

for all $f,g \in A_S$.

Proof

We have

and, in the same way, we also have

To prove the second claim we simply compute

$\square$

Lemma A.5

Let ${\mathcal {B}}$ be a spherical fusion category and let A be an algebra object in ${\mathcal {B}}$ (with product $\nabla$) together with structure maps that make A self-dual. Then A satisfies (16) if and only if

(32)

for all $R,S,T \in {{\,\mathrm{Irr}\,}}({\mathcal {B}})$, $\beta \in {{\,\mathrm{Hom}\,}}_{{\mathcal {B}}}(ST,R)$, $h,f \in A_R$, $g,h \in A_T$ and $f,g \in A_S$.

Proof

Decomposing the coproduct given by the left-hand side of (16) gives

In an analogous way, we also have

which proves the proposition. $\quad \square$