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

Question

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?

Thanks in advance.

Answer 1

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 $H^*(N_0(n,k))$.