# A question regarding isomorphism in cohomology for moduli space of stable bundles over a compact Riemann surface

Let $$N(n,k)$$ denote the moduli space of stable vector bundles of rank $$n$$ and degree $$k$$ over a compact Riemann surface $$X$$, and let $$N_0(n,k)$$ denote the moduli space where we fix rank $$n$$ and some fixed determinant bundle of degree $$k$$. We know that the determinant map $$det: N(n,k)\rightarrow Pic^k(X)$$ is a proper submersion with fibers isomorphic to $$N_0(n,k)$$.

In the paper 'The Yang-Mills equations over Riemann surfaces' by M. F. Atiyah and R. Bott, (Phil. Trans. R. Soc. Lond. A 308, 523-615 (19)) the authors prove the following:

Proposition 9.7. (page 578) For rational cohomology we have $$H^*(N(n,k)) \simeq H^*(N_0(n,k))\otimes H^*(Pic^0(X)).$$

Immediately after this, the authors mention the following :

"This proposition, which is equivalent to the statement that $$\Gamma_n := H^1(X,\mathbb{Z}_n)$$ acts trivially on the rational cohomology of $$N_0(n,k)$$, ..."

( $$\Gamma_n$$ actually corresponds to the $$n$$-torsion line bundles on $$X$$, and a line bundle $$L\in \Gamma_n$$ acts on $$N_0(n,k)$$ by sending $$E\mapsto E\otimes L$$.)

I want to understand the above equivalence , at least the direction why the statement implies the proposition. I believe that it is somehow related to the monodromy action of $$\pi_1(Pic^k(X))\simeq H^1(X,\mathbb{Z})$$ on the cohomology of fibers via the $$det$$ map I mentioned above. If this action is trivial, then one can deduce that the map on cohomologies induced by the inclusion of a fiber is surjective, which would imply the proposition thanks to Leray-Hirsch theorem. But the authors state that instead the action of $$\Gamma_n=H^1(X,\mathbb{Z}_n)$$ is trivial. Of course, applying UCT we have $$H^1(X,\mathbb{Z}_n)=H^1(X,\mathbb{Z})\otimes \mathbb{Z}_n \simeq \pi_1(Pic^k(X))\otimes \mathbb{Z}_n,$$ does this imply that the monodromy action of $$\pi_1(Pic^k(X))$$ factors through the action of $$\Gamma_n$$? Is this the reason?

Things are actually simpler. View $$\Gamma _n=H^1(X,\mathbb{Z}/n)$$ as the group of line bundles $$L\in \operatorname{Pic}^{0}(X)$$ with $$L^{{\tiny \otimes }n}=\mathscr{O}_X$$. The map $$N_0(n,k) \times \operatorname{Pic}^{0}(X) \rightarrow N(n,k)\$$ given by $$\ (E,L)\mapsto E\otimes L\$$ identifies $$N(n,k)$$ to the quotient of $$N_0(n,k) \times \operatorname{Pic}^{0}(X)$$ by $$\Gamma _n$$. Thus $$H^*(N(n,k))$$ is the subgroup of $$H^*(N_0(n,k))\otimes H^*(\operatorname{Pic}^{0}(X) )$$ invariant by $$\Gamma _n$$. But since $$\Gamma _n$$ acts by translation on $$\operatorname{Pic}^{0}(X)$$, it acts trivially on cohomology; so $$H^*(N(n,k))$$ is isomorphic to the whole tensor product if and only if $$\Gamma _n$$ acts trivially on $$H^*(N_0(n,k))$$.