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.
Let the $\{e_i|i=1,\cdots,n\}$ be a basis of $A$ and let $\{e^i|i=1,\cdots,n\}$ be the dual basis of $A$. That is: $${\rm tr}(e^ie_j)=\delta_{ij},$$ where $\delta_{ij}$ is the Kronecker delta.
Then for $a\in A$:$$c(a)=ae^i\otimes e_i,$$ where we follow the summation convention.
Now $f(b)=b(e^ie_i)$ and $f^{\circ g}(b)=b(e^ie_i)^g$. Thus $$Z[g]={\rm tr}((e^ie_i)^g).$$
In particular: $$Z[1]={\rm tr}(e^ie_i)=\delta_{ii}={\rm dim}(A).$$
For completeness here is a derivation of the identity $c(a)=ae^i\otimes e_i$:
Let $c(a)=a'\otimes a''$ (summation suppressed). Then by comparing the topology of the corresponding surfaces, we see that $${\rm tr}(xay)={\rm tr}(xa'){\rm tr}(a''y).$$
In particular $${\rm tr}(e^ie^ja)={\rm tr}(e^ia'){\rm tr}(e^ja'')=a'_ia''_j.$$
Thus $$c(a)={\rm tr}(e^ie^ja)(e_i\otimes e_j)=(e^ja)_ie_i\otimes e_j=ae^j\otimes e_j,$$ as required.