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 Ktheory, where this can be read between the lines)

4$\begingroup$ 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. $\endgroup$– André HenriquesSep 5, 2011 at 16:56

1$\begingroup$ @Andre': that's precisely the strategy I had in mind. And MO is essential for step1! :) completely offtopic: have you recived my email message from a few days ago? I'm not sure I was able to correcltly decipher your email address from your home page ;) $\endgroup$– domenico fiorenzaSep 5, 2011 at 18:56

$\begingroup$ 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? $\endgroup$– Marcel BischoffDec 16, 2013 at 9:26
2 Answers
I don't know a reference but it's not clear to me you need one for this statement, since it's close to definitional. More precisely you just need the fact that compactification on a circle corresponds to taking Hochschild homology, or in this case just complexified Ktheory, of a category, for which there are lots of references (my kneejerk reaction is to quote Lurie's TFT manuscript though I'm sure for this you can find many older references). Then you're simply asserting that the Kgroups tensor C inherit a commutative multiplication, a unit and a trace from the braided tensor category you started from (that's the definition of the Verlinde algebra), and that from the field theory these are given on the category as the pair of pants and (in or outgoing) disc, hence by the same pictures times S^1 on the Verlinde algebra, hence by the same pictures again in the dimensionally reduced theory.

3$\begingroup$ Here you're assuming that you already know that there's a orpo 321 theory with values in the 2category of categories attached to any MTC which assigns the circle to the MTC and where the structural morphisms are assigned to the obvious bordisms (tensor to pants, etc.). That's the folklore result. $\endgroup$ Sep 5, 2011 at 17:53

1$\begingroup$ I would think everything I need here is already in say the book of BakalovKirillov  eg I don't need 123 or even all of 12, just that a braided tensor category is the same as an algebra over the E_2 operad, plus some finiteness of dimensions to allow me to close up discs, and I automatically have enough finiteness to be able to take product with a circle.. $\endgroup$ Sep 5, 2011 at 18:05

1$\begingroup$ Hi David, thanks. Yes, I know the statement is almost tautological, but being able to cite a reference for this precise statement would be convenient to me. Otherwise I can follow Andre''s avdice and try to write a proof of this myself. $\endgroup$ Sep 5, 2011 at 18:33

3$\begingroup$ Oh, I see. If I understand correctly, you're saying that this really has very little to do with 3dimensions at all. Dimensional reduction takes you from a 321 theory to a 210 theory, but the real calculation (going from the category that the 210 theory assigns to a point to the Frobenius algebra that it assigns to a circle) takes place entirely inside the 210 theory. $\endgroup$ Sep 5, 2011 at 19:05
A somehow detailed answer could be as follows (thanks to Alessandro Valentino, who is a coauthor of this answer (but me alone is to blame for mistakes and inaccuracies in it)). Kevin Walker's notes have been an essential source of inspiration.
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 3dimensional 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.