## A simple question about the degree of some vector bundle over rational curve.

Let $E$ be a holomorphic vector bundle (infact complex vector bundle is enough) over $\mathbb{P}^1$. Let $c: \mathbb{P^1} \rightarrow \mathbb{P^1}$ be the anti-holomorphic involution, $c(z)=\frac{-1}{\bar{z}}$, and after all suppose we have a commutative diagram (I couldn't draw it here, imagine it yourself)

$\pi: E \rightarrow \bar{E}$

where $\pi$ is an anti-holomorphic (or anti-complex linear) involution covering $c$. In this situation, is it true that $c_1(E)$ should be even!

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 If you think it is not true, you may try to find such a $\pi$ for $E = \mathcal{O}(1)$. – Mohammad F.Tehrani Jan 26 2012 at 2:38

Complex curves with anti-holomorphic involutions correspond to real algebraic curves. Your involution has no fixed points, so your curve corresponds to a real curve $C$ of genus 0 with no real points (there is only one). Equivariant vector bundles come from vector bundles on $C$, so the first Chern class of your equivariant vector bundle comes from a 0-cycle on $C$, hence it is enough to see check that $C$ only has 0-cycles of even degree. This is easy: suppose that it had a cycle of odd degree. Since it is has a point of degree 2, it would have a cycle of degree 1; by Riemann-Roch, this would correspond to a real point of $C$, which does not exist.

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 Dear Angelo, did you mean to say that $C$ has a point of degree 2, rather than 0? – MP Jan 26 2012 at 8:16 To MP: yes, sorry, I fixed it. – Angelo Jan 26 2012 at 8:30