This is already posted here <https://math.stackexchange.com/questions/3695584/a-computation-in-a-commutative-frobenius-algebra> but I didn't get any answers.


Given a commutative Frobenius algebra (in the category of vector spaces over $\mathbb C$) $A$ with multiplication $m\colon A\otimes A\to A$, comultiplication $c\colon A\to A\otimes A$, identity $1\in A$ and trace $\mathrm{tr}\colon A\to\mathbb C$, does the following expression simplify
$$Z\left[g\right]=\mathrm{tr}\left(f^{\circ g}\left(1\right)\right)$$
where $m\circ c\equiv\_\text{def} f\colon A\to A$ and $f^{\circ g}=f\circ f\dots\circ f$ ($g$ times) where $g\in\mathbb N$. Also, why does $Z[1]=\dim A$?


I'm thinking about this a trying to evaluate a 2d TQFT on a closed 2-manifold of genus $g$. Lurie claims that $Z\left[1\right]=\dim A$. Any help in calculating any other specific values would be very helpful. ($Z[0]=\mathrm{tr}\,1$ obviously.)



A computation in a commutative Frobenius algebra
I don't know what you mean when you ask if the expression simplifies. The generating relations for a commutative Frobenius algebra are equivalent to the fact that the only thing that matters is the diffeomorphism type of the corresponding surface (possibly with boundary) that you draw. (I like to think of this as the idea that the algebraic relations of a Frobenius algebra are precisely those encoding the handlebody theory for surfaces coming from Morse theory.) If you have a genus $1$ surface (so you're computing $Z[1]$), there are many possible expressions, but you need at least one occurrence of each of the four generating elementary operations: the unit, the counit (which you call the trace), multiplication, and comultiplication, and the expression you've written for $Z[1]$ is the simplest expression you could get, in some sense. (There aren't many Morse functions on a genus $1$ surface with only four critical points! This expression corresponds to the only one, in a way you can make precise.) For $Z[g]$, there are other expressions with the same number of each of the elementary terms, which again is minimal; the one you've given is one of the more natural choices, but I could also imagine, for example, writing $$Z[2] = \epsilon \circ m \circ (m \otimes \mathrm{id}\_A) \circ (c \otimes \mathrm{id}\_A) \circ c \circ \eta.$$


I'm using $\eta \colon \mathbb{C} \rightarrow A$ for the unit (which is more natural) and $\epsilon \colon A \rightarrow \mathbb{C}$ for the counit, for clarity, since I will use the word trace in a more standard manner. It is arguably more complicated, in that you have to include these extra identity morphisms (corresponding to trivial cylinders).


As for verifying $Z[1] = \dim A$, intuitively, you're taking a trivial cylinder, which corresponds to the identity map $\mathrm{id}\_A \colon A \rightarrow A$, and gluing the two ends together. This tells you, as a general rule, that you need to take the trace of this map, so that


$$Z[1] = \mathrm{tr}(\mathrm{id}\_A) = \dim A.$$


Here are a few more details about how to show this purely algebraically. From drawing surfaces, you'll see that the trivial cylinder can be decomposed in such a way that one finds


$$\mathrm{id}\_A = (\mathrm{id}\_A \otimes \epsilon) \circ (\mathrm{id}\_A \otimes m) \circ (c \otimes \mathrm{id}\_A) \circ (\eta(1) \otimes \mathrm{id}\_A)$$


If you had a basis $\{e^i\}$ (NB: any Frobenius algebra is automatically finite-dimensional!), and you wrote everything out in this basis accordingly (e.g. $\eta(1) = \eta\_i e^i$ and $c(e^i) = c^i\_{jk}e^j \otimes e^k$), this tells you that $$\eta\_i c^i\_{jk}\mu^{k\ell}\_{m}\epsilon^m = \delta\_j^{\ell}$$ (where I'm assuming Einstein summation). Now we take the input and output and glue them together. What that does, in a basis, is to force the indices $j$ and $\ell$ to be the same. To be concrete, in a basis, we have


$$Z[1] = (\epsilon \circ m \circ c \circ \eta)(1) = \eta\_ic^{i}\_{jk}\mu^{jk}\_{m}\epsilon^{m} = \delta\_{j}^{j} = \dim A,$$


where we have used the commutativity, in the form $\mu^{jk}\_m = \mu^{kj}\_m$.


