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Let $M$ be the moduli space of semistable vector bundles of fixed determinant $L$ and rank $r$ over a smooth curve $X$. Assume that $gcd(r,deg(L))=1$. Let $\mathcal U$ be the universal bundle over $M\times X$ and let $x\in X$. Denote by $\mathcal U_x=\mathcal U|_{M\times\{x\}}$. I read on a paper without any reference or proof that the first chern class $c_1(\mathcal U_x)\in H^2(M,\mathbb Z)=\mathbb Z$ is coprime to $r$.

How to show that? any reference?

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    $\begingroup$ $c_1(\mathcal{U}_x)=c_1(L)=\deg L$. $\endgroup$
    – Chen Jiang
    Commented Dec 9, 2017 at 1:42
  • $\begingroup$ How do you see that? $\endgroup$
    – Z.A.Z.Z
    Commented Dec 9, 2017 at 12:38
  • $\begingroup$ This is just by definition. $\mathcal{U}_x$ is just one of the semistable vector bundle in the moduli. $\endgroup$
    – Chen Jiang
    Commented Dec 9, 2017 at 17:57
  • $\begingroup$ I don't see that! $\mathcal U_x$ is a bundle over the moduli space. If $e\in M$ then I agree that $\mathcal U_e$ is a bundle as you said?! $\endgroup$
    – Z.A.Z.Z
    Commented Dec 10, 2017 at 14:00
  • $\begingroup$ Oh, you are right. I thought you are asking about the fiber over $M$. Sorry. $\endgroup$
    – Chen Jiang
    Commented Dec 11, 2017 at 17:24

2 Answers 2

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Let $\pi:X\times M \to M$ be the projection.

Theorem 9.11 in Atiyah-Bott's "The Yang-Mills Equations over Riemann Surfaces" states that $c_1(\mathcal{U}_x)$ and $c_1(R\pi_*\mathcal{U})$ generate the second cohomology group (knowing that $M$ is simply connected). This is true an arbitrary choice of $\mathcal{U}$. By Grothendieck-Riemann-Roch, we know that $$c_1(R\pi_*\mathcal{U}) = (1-g)c_1(\mathcal{U}_x) + \pi_*\operatorname{ch}_2(\mathcal{U}) .$$ It follows that an alternative set of generators for the integral second cohomology is given by $c_1(\mathcal{U}_x)$ and $\pi_*\operatorname{ch}_2(\mathcal{U})$.

Note that tensoring $\mathcal{U}$ by $\pi^*H$ for a line bundle $H$ on $M$ changes $c_1(\mathcal{U})$ by $rc_1(H)$, and that $$\pi_*\operatorname{ch}_2(\mathcal{U}\otimes \pi^*H) =\pi_*\operatorname{ch}_2(\mathcal{U})+d c_1(H)$$

Now choose integers $a,b$ satisfying $ar - b d = 1$, and pick $H$ satisfying $c_1(H) = \pi_*\operatorname{ch}_2(\mathcal{U})$. Let $\mathcal{U'}:=\mathcal{U}\otimes \pi^*H$. Then $\pi_*\operatorname{ch}_2(\mathcal{U}')$ is divisible by $r$, but generates the second cohomology together with $c_1(\mathcal{U}'_x)$. Therefore, the latter must be coprime to $r$, in particular this is true for $c_1(\mathcal{U})$.

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It was shown in [Remark 2.9] of the paper The moduli spaces of vector bundles over an algebraic curve, by S Ramannan https://eudml.org/doc/162339

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