Suppose I have a CohFT, which is basically a vector space $V$ and a collection of maps $V^{\otimes n} \rightarrow H^\*(\overline{M\_{g,n}})$ satisfying certain properties under pull-backs by the gluing maps. Under which assumptions is it completely determined by its high genus values? High means with respect to the cohomology degree. I.e. for each $k$ I know the map $V^{\otimes n} \rightarrow H^k(\overline{M\_{g,n}})$ for $g$ large enough for all $n$. And I want to reconstruct it for all $g$. I saw some argument in a paper by Teleman which begins by saying that we can assume the genus is large, and then use the Madsen-Weiss theorem. So I wonder what's behind this trick and how to justify it properly.



Is a CohFT completely determined by its high genus values?
Teleman's trick works as follows. Suppose I want to deduce $\Omega\_{g,n}$ from $\Omega\_{g+1,n}$.


Note that $\overline{\mathcal{M}}\_{g+1,n}$ contains a copy of $\overline{\mathcal{M}}\_{g,n}$. It is constructed as follows. Take any fixed elliptic curve $E$ with two marked points. Now consider the locus of curves obtained by gluing $E$ to a stable genus $g$ curve at its $n$th marked point.


The factorization property of CohFTs tells you that $\Omega\_{g,n}$ is equal to the restriction of $\Omega\_{g+1,n}$ to this locus, contracted with the value of $\Omega\_{1,2}$ on $E$, which is simply the cohomological degree 0 part of $\Omega\_{1,2}$. So the subtlety is this: is the bilinear form $\Omega\_{1,2}(E) \in V^\* \otimes V^\*$ invertible (nondegenerate)? If yes, you can deduce $\Omega\_{g,n}$ from $\Omega\_{g+1,n}$. If not, in general you cannot.


By some linear algebra one checks that the invertibility of $\Omega\_{1,2}(E)$ is equivalent to $\Omega$ being semisimple, in other words, the algebra structure on $V$ given by $\Omega\_{0,3}$ should be semisimple.


