# (3,2,1)-TQFTs and Verlinde algebras

Given a modular category $\mathcal{C}$ there are two natural ways to get a Frobenius algebra out of $\mathcal{C}$. One is to take the Verlinde algebra (or `fusion algebra') of $\mathcal{C}$. The other consist in considering the $(3,2,1)$-dimensional TQFT associated with $\mathcal{C}$, and to get out of it a $(2,1)$-dimensional TQFT by multiplication by $S^1$ (and a $(2,1)$-dimensional TQFT is the same thing as the datum of a Frobenius algebra). It is well known in fully extended TQFT folklore that these two constructions coincide. Is anyone aware of a reference I could cite as a source for this statement? (I know Dan Freed's The Verlinde algebra is twisted equivariant K-theory, where this can be read between the lines)

• If: 1) you can't find a reference, and 2) you know how to prove this "folklore fact", then you should include a proof of it in your writing, maybe as an appendix. – André Henriques Sep 5 '11 at 16:56
• @Andre': that's precisely the strategy I had in mind. And MO is essential for step1! :) completely off-topic: have you recived my e-mail message from a few days ago? I'm not sure I was able to correcltly decipher your email address from your home page ;) – domenico fiorenza Sep 5 '11 at 18:56
• While I have the rough idea in mind and probably could work it out by myself, can someone give me a reference to the construction of the Frobenius algebra starting with the Verlinde algebra? – Marcel Bischoff Dec 16 '13 at 9:26

Let $\mathcal{C}$ be a modular tensor category, and let $\{X_i\}$ a set of representatives for the isomorphism classes of its simple objects. We write $X^i$ for the dual of $X_i$. Then the element associated by the (3,2,1)-TQFT associated with $\mathcal{C}$ to the cylinder with two outgoing $S^1$'s is the element $$\mathrm{coev}_{\mathcal{C}}=\bigoplus_{i\in I} X^i\boxtimes X_i$$ of $\mathcal{C}\boxtimes \mathcal{C}$, while the element $$\mathrm{ev}_{\mathcal{C}} \colon \mathcal{C}\boxtimes \mathcal{C}\to Vect$$ associated to the cylinder with two ingoing $S^1$'s is $$\mathrm{ev}_{\mathcal{C}}(A\boxtimes B) = \mathrm{Hom}(A^*,B)$$ We then have $$Z(T^2)\equiv\mathrm{ev}_{\mathcal{C}}\circ \mathrm{coev}_{\mathcal{C}}= \bigoplus_{i\in I} \mathrm{End}(X_i)=\bigoplus_{i\in I} \mathbb{K} = \mathbb{K}^I.$$ In other words $Z(T^2)$ is isomorphic to the vector space with basis given by isomorphism classes of simple objects of $\mathcal{C}$. In particular we have $$\dim Z(T^2) = \#\{\text{isomorphism classes of simple objects of \mathcal{C}}\}$$ Let us now describe a canonical basis for $Z(T^2)$. To do so we look at the boundary conditions for the TQFT associated with $\mathcal{C}$. Any object $X$ in $\mathcal{C}$ defines a boundary condition, that is, a $Vect$-linear functor $Vect\to \mathcal{C}$ which we will denote by the same symbol, i.e., we write $X\colon Vect\to \mathcal{C}$. The TQFT assigns the functor $X$ to a cylinder with the incoming copy of $S^1$ decorated by the colour $X$. Multplying this cylinder by $S^1$ we get the 3-dimensional cylinder over the basis $T^2$, with the incoming basis decorated by the colour $X$. This cylinder is a morphism $\mathbb{K}\to Z(T^2)$, i.e., an element of $Z(T^2)$. We will denote this element by $v_X$. Since $\mathcal{C}$ is semisimple, every short exact sequence $0\to X\to Y\to W\to 0$ in $\mathcal{C}$ splits, i.e., $Y\cong X\oplus W$. This implies that $v_Y=v_X+v_W$ and so $v$ defines a linear map $$v\colon K(\mathcal{C})\to Z(T^2)$$ We are going to show that this amp is a linear isomorphism of vector spaces. The cylinder over the basis $T^2$, with both copies of $T^2$ incoming defines the inner product on $Z(T^2)$; let us denote this inner product by $\langle \,|\,\rangle$. Then $\langle v_X | v_Y\rangle$ is the element in $k$ that the TQFT associates to the cylinder over the basis $T^2$, with both copies of $T^2$ incoming, the first one decorated by $X$ and the second one decorated by $Y$. This cylinder is obtained by multiplying by $S^1$ the cylinder over the basis $S^1$, with both copies of $S^1$ incoming, the first one decorated by $X$ and the second one decorated by $Y$. By the TQFT rules, this cylinder over the base $S^1$ is evaluated to $\mathrm{ev}_{\mathcal{C}}(X\boxtimes Y)=\mathrm{Hom}(X^*,Y)$. Therefore $$\langle v_X | v_Y\rangle=\dim \mathrm{Hom}(X^*,Y).$$ In particular, if we write $v^i$ for $v_{X^i}$ and $v_i$ for $v_{X_i}$ we see that $$\langle v^i | v_j\rangle=\dim \mathrm{Hom}(X_i,X_j)=\delta^i_j,$$ and so the vectors $\{v_i\}_{i\in I}$ are linearly independent vectors of $Z(T^2)$. Since their number equals the dimension of $Z(T^2)$, we see that the collection $\{v_i\}_{i\in I}$ is a linear basis of $Z(T^2)$. Since $\{v_i\}_{i\in I}$ is a basis, mapping $v_i$ to the equivalence class of $X_i$ in $K(\mathcal{C})\otimes k$ defines a linear map $Z(T^2)\to K(\mathcal{C})\otimes k$ which is manifestly the inverse to $v$.
The tensor product on $\mathcal{C}$ induces a multiplication on $K(\mathcal{C})\otimes k$ by setting $$[X_i]\cdot [X_j]= [X_i\otimes X_j]$$ If we write $$X_i\otimes X_j\cong \bigoplus_{k\in I} \mathbb{K}^{n_{ij}^k}\otimes X_k$$ then we see that the multiplication induced by the tensor product on $K(\mathcal{C})\otimes \mathbb{K}$ is given by $$[X_i]\cdot [X_j]= \sum_{k\in I} n_{ij}^k [X_k].$$ Notice that from $X_i\otimes X_j\cong \bigoplus_{k\in I} \mathbb{K}^{n_{ij}^k}\otimes X_k$ we see that $$n_{ij}^k=\dim\mathrm{Hom}(X_k,X_i\otimes X_j).$$ On the other hand we have a multiplication induced on $Z(T^2)$ by the TQFT; namely, by the pair of pants times $S^1$. Since the pair of pants corresponds to the tensor product of $\mathcal{C}$ it is quite natural to expect that the multiplication on $Z(T^2)$ will coincide with the multiplication induced by the tensor product on $K(\mathcal{C})\otimes \mathbb{K}$, i.e., that $v$ is actually an isomorphism of commutative algebras. To see that it is indeed so, we only have to compute the product $v_i\cdot v_j$ in $Z(T^2)$ and check that this is $$v_i\cdot v_j= \sum_{k\in I} n_{ij}^k v_k.$$ This is equivalent to showing that $$\langle v^k | v_i\cdot v_j\rangle = n_{ij}^k$$ The element on the left hand side is what the TQFT assigns to a pair of pants times $S^1$ with all the three copies of $T^2$ incoming, decorated by the colours $X_i$, $X_j$ and $X_k$ respectively. This is the same as a pair of pants with all the three copies of $S^1$ incoming, decorated by the colours $X_i$, $X_j$ and $X^k$ respectively, multiplicated by $S^1$. Since the pair of pants with all the three copies of $S^1$ incoming, decorated by the colours $X_i$, $X_j$ and $X^k$ respectively is mapped by the TQFT to $\mathrm{Hom}(X_k,X_i\otimes X_j)$ we see that when we multiply it by $S^1$ we get $\dim \mathrm{Hom}(X_k,X_i\otimes X_j)$. Hence we have $$\langle v^k | v_i\cdot v_j\rangle=\dim \mathrm{Hom}(X_k,X_i\otimes X_j)=n_{ij}^k,$$ which is precisely the sought for identity.