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Let $X$ be a compact Riemann surface and $E \rightarrow X$ a holomorphic vector bundle of rank $n$. We can construct a projective bundle $\mathbb{P}(E) \rightarrow X$ by taking the projective spaces of the fibers of $E$. There is an exact sequence of sheaves $$1 \rightarrow \mathcal{O}^* \rightarrow GL(n,\mathcal{O}) \rightarrow PGL(n,\mathcal{O})\rightarrow 1$$ by considering $1 \rightarrow \mathbb{C}^* \rightarrow GL(n,\mathbb{C}) \rightarrow PGL(n,\mathbb{C}) \rightarrow 1$.

Since $H^2(X,\mathcal{O}^*) = 0$, we have $H^1(X,GL(n,\mathcal{O})) \rightarrow H^1(X,PGL(n,\mathcal{O}))$ is surjective, i.e. every projective bundle on Riemann surface arises from some vector bundle. By an analogous definition of flat vector bundle, we also have a notion of flat projective bundle. It is obvious that $\mathbb{P}(E)$ is flat if $E$ is flat. My question is

Let $\phi$ be a flat projective bundle on Riemann surface $X$, can we find a flat vector bundle $E$ on $X$ such that $\phi = \mathbb{P}(E)$?

I think the answer is "NO", but I can't give a proof by counterexample. For simplicity, we can consider $n=2$ i.e. $\phi$ is a flat $\mathbb{P}^1$ bundle. Let $\phi = \mathbb{P}(E)$ for some holomorphic vector bundle of rank 2. The question is equivalent to the degree of $E$ is even or odd under the isomorphism $H^2(X,\mathbb{Z}) \cong \mathbb{Z}$. Any comment is welcome, thank you.

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    $\begingroup$ For a Riemann surface, the group $H^2(X,\mathcal{O}_X^*)$ is zero, but the group $H^2(X,\mathbb{C}_X^*)$ is nonzero. This should give rise to many counterexamples. $\endgroup$ – Jason Starr Nov 16 '16 at 11:41
  • $\begingroup$ Let $(E, h)$ be a projectively flat vector bundle over $M$. Then, for any Hermitian metric $g$ on M, (E, h) satisfies the weak Hermitian-Einstein metric. $\endgroup$ – user21574 Apr 28 '17 at 7:51
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Equivalently, you are asking whether every homomorphism $\pi _1(X)\rightarrow \mathrm{PGL}(n,\mathbb{C})$ lifts to a homomorphism $\pi _1(X)\rightarrow \mathrm{GL}(n,\mathbb{C})$. It is easy to find counter-examples, already for $n=2$ and $g(X)=1$: then $\pi _1(X)=\mathbb{Z}^2$, so you are asking whether two commuting homographies can be lifted to commuting matrices in $\mathrm{GL}(2,\mathbb{C})$. It is an easy exercise to prove that this does not hold for the homographies $z\mapsto -z$ and $z\mapsto 1/z$.

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