Let $X$ be a (non-singular) complex surface and $(V,\nabla)$ be a vector bundle $V$ equipped with a flat connection $\nabla$ on $X$. Fix a point $x \in X$ and $v_0 \in V_x$ an element in the fiber over the point $x$ of the vector bundle $V$. Does there exist a global section $s$ of the vector bundle $V$ such that at the point $x$, it takes the value $v_0$? I think this is true if $X$ is simply connected.
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$\begingroup$ Do you think same question in setting of smooth manifolds have same answer? $\endgroup$– Praphulla KoushikCommented Oct 13, 2019 at 17:25
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$\begingroup$ Let $M$ be a smooth manifold and $E\rightarrow M$ be a vector bundle. Let $\nabla$ be a connection on the vector bundle $(E,\pi,M)$. Fix a point $m\in M$ and an element $v$ in the fibre of $m$... Does there exists a global section $s:M\rightarrow E$ such that $s(m)=v$?? $\endgroup$– Praphulla KoushikCommented Oct 13, 2019 at 17:29
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1$\begingroup$ @PraphullaKoushik non-singular is the same as smooth in this setup. The main point is the existence of a flat connection (not all vector bundles can be equipped with a flat connection), which enables us to define parallel transport. However, I am not sure if one can use this parallel transport to define a global section. $\endgroup$– RonCommented Oct 13, 2019 at 17:35
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$\begingroup$ I am reading your comment as "because of flat connection, one can define parallel transport".. Is this what you mean? For any connection on vector bundle we can talk about parallel transport.. did I misunderstand some things $\endgroup$– Praphulla KoushikCommented Oct 13, 2019 at 18:19
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1$\begingroup$ In general, flat connections on a manifold (in either the C^\infty or holomorphic setting) are in correspondence with representations of the fundamental group. The correspondence takes the bundle $V$ to its fiber $V_x$ at the basepoint, equipped with the action of $\pi_1$ via parallel transport. In these terms, you can see that one can lift $v\in V_x$ to a global flat section if and only if $v$ is invariant under the action of $\pi_1$. $\endgroup$– Sam GunninghamCommented Oct 13, 2019 at 18:28
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