Let $M$ be a connected closed surface (possibly with non-zero genus) and let $P\subset M$ be a nonempty finite set of points. Set $\dot{M} = M \setminus P$. Let $\pi : E \rightarrow \dot{M}$ be a complex vector bundle of rank $2$. Is this vector bundle trivial?
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1$\begingroup$ Your question is a bit difficult to parse. However, for my best guess of your meaning, so long as $M$ is connected and $P$ is nonempty, this follows from the Grauert-Oka principle. $\endgroup$– Jason StarrJan 2, 2017 at 17:15
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4$\begingroup$ In the topological category, the frame bundle of $E$ is a Serre fibration over $\stackrel{\bullet}{M}$ with path connected fibers. Since $\stackrel{\bullet}{M}$ has the homotopy type of a bouquet of (finitely many) circles, there is a continuous section of the frame bundle, i.e., a trivialization of the vector bundle. In the holomorphic category, you need to use the Grauert-Oka principle (or one of the other related theorems, e.g., Gromov's h-principle, ...). $\endgroup$– Jason StarrJan 2, 2017 at 17:45
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1$\begingroup$ The vector bundle $E \rightarrow \dot{M}$ is a complex vector bundle, not a holomorphic one, i.e. $\pi : E \rightarrow \dot{M}$ is not supposed to be holomorphic. $\endgroup$– TimoJan 2, 2017 at 17:49
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2$\begingroup$ The bundle is classified by a map $\dot{M} \to BU(2)$, and the latter is simply connected. $\endgroup$– Steve CostenobleJan 2, 2017 at 21:08
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1$\begingroup$ Apologies for the cryptic comment. I don't have a chance to check this forum that often, but Michael Albanese gave a very nice, full answer. $\endgroup$– Steve CostenobleJan 4, 2017 at 0:27
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1 Answer
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For any topological group $G$, there is a classifying space $BG$ and a principal $G$-bundle $EG \to BG$ called the universal principal $G$-bundle which is determined up to isomorphism by the fact that $EG$ is weakly contractible. On a paracompact topological space $X$, any principal $G$-bundle $P \to X$ admits a map $f : X \to BG$, called a classifying map, so that $P \to X$ is isomorphic to $f^*EG \to X$. Moreover, two principal $G$-bundles $P_1, P_2 \to X$ are isomorphic if and only if their classifying maps $f_1, f_2 : X \to BG$ are homotopic. In particular, a principal $G$-bundle is trivial if and only if its classifying map is nullhomotopic.
Complex rank $n$ vector bundles can be identified with principal $U(n)$-bundles, so your problem reduces to showing that every map $f : \dot{M} \to BU(2)$ is nullhomotopic.
The long exact sequence in homotopy applied to the universal principal $G$-bundle, together with the weak contractibility of $EG$, shows that $\pi_{k+1}(BG) \cong \pi_k(G)$. In particular, $\pi_1(BU(2)) \cong \pi_0(U(2)) = 0$ as $U(2)$ is path-connected.
The surface $\dot{M}$ deformation retracts onto a bouquet of circles. Restricting $f$ to one of these circles, we get a map $S^1 \to BU(2)$ which is nullhomotopic as $BU(2)$ is simply connected. It follows that $f$ is nullhomotopic and therefore every rank two complex vector bundle on $\dot{M}$ is trivial.
As $U(n)$ is path-connected for every $n$, $BU(n)$ is always simply connected so the argument above would still work if we replace $BU(2)$ by $BU(n)$. Therefore we see that every complex vector bundle on $\dot{M}$ is trivial. In fact, as $SO(n)$ is connected for all $n$, the same argument shows that all real orientable vector bundles on $\dot{M}$ (which correspond to principal $SO(n)$-bundles on $\dot{M}$) are trivial.
answered Jan 3, 2017 at 11:10
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$\begingroup$ Thank you very much for the very detailed answer! This helped me a lot. $\endgroup$– TimoJan 3, 2017 at 15:24
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$\begingroup$ An alternate approach to proving that complex vector bundles on the circle are trivial ,is to notice that such vector bundles are classified by homotopy classes of maps from the two point set to GL(n) .Since GL(n) is connected this set of homotopy classes is just a point .This proves the result . $\endgroup$ Jan 4, 2017 at 21:02
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1$\begingroup$ for more details on the clutching construction and homotopy see M F Atiyah K-Theory page 24 $\endgroup$ Jan 4, 2017 at 21:12