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:



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> **Proposition 9.7. (page 578)** For rational cohomology we have $H^\*(N(n,k)) \simeq H^\*(N\_0(n,k))\otimes H^\*(Pic^0(X)).$
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Immediately after this, the authors mention the following :



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> "This proposition, which is equivalent to the statement that $\Gamma\_n := H^1(X,\mathbb{Z}\_n)$ acts trivially on the rational cohomology of $N\_0(n,k)$, ..."
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( $\Gamma\_n$ 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.



A question regarding isomorphism in cohomology for moduli space of stable bundles over a compact Riemann surface
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 $H^\*(N\_0(n,k))$.


