# First Chern class of the universal bundle

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?

• $c_1(\mathcal{U}_x)=c_1(L)=\deg L$. Dec 9, 2017 at 1:42
• How do you see that? Dec 9, 2017 at 12:38
• This is just by definition. $\mathcal{U}_x$ is just one of the semistable vector bundle in the moduli. Dec 9, 2017 at 17:57
• 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?! Dec 10, 2017 at 14:00
• Oh, you are right. I thought you are asking about the fiber over $M$. Sorry. Dec 11, 2017 at 17:24

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 $$c_1(\mathcal{U})$$.