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It is known that any holomorphic bundle of any rank over a noncompact Riemann surface is trivial. A proof can be found in Forster's "Lectures on Riemann surfaces", section 30.

Let $E$ be a holomorphic vector bundle over a compact Riemann surface $X$ with gauge group $G$. A consequence of the above theorem is the restriction $E|_{X-\{p\}}$ for any point $p\in X$ is a trivial bundle. Thus $E$ can be recovered by specifying the transition function $g: D\cap (X-\{p\}) \rightarrow G$ where $D$ is a small disk containing $p$.

Is this correct? If not, could you give a counter-example? I am mainly interested in learning about the moduli space of holomorphic bundles over $X$ in a concrete way, e.g. using transition functions.

If the argument is correct, then there is another issue. Consider a one-parameter family of transition functions $g_{\alpha\beta}(t)$. Imposing the cocycles condition on $g_{\alpha\beta}' := g_{\alpha\beta}+\epsilon\dot{g}_{\alpha\beta}$ where $\epsilon$ is infinitesimal, one finds $\dot{g}_{\alpha\beta}$ defines a class in $H^1(X,\mathfrak{g})$. Thus the tangent space at $[E]$ to the moduli space of bundles $Bun_G(X)$ is $H^1(X,\mathfrak{g})$; equivalently, $$T_{[E]}^\ast Bun_G(X) \cong H^0(X,\mathfrak{g}\otimes K_X)$$ by Serre's duality. This is the standard argument in constructing e.g. the Hitchin's system.

But now as we minimally only have one transition function, we have no cocycle conditions to impose. How do I still see that $T_{[E]}^\ast Bun_G(X) \cong H^0(X,\mathfrak{g}\otimes K_X)$?

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  • $\begingroup$ Yes that is correct. However it is not a way to work with bundles 'explicitly', as beginner may hope. $\endgroup$ – Alexander Chervov Apr 14 at 12:40
  • $\begingroup$ @AlexanderChervov could you please elaborate? for me one transition function over an annulus is as explicit a description one could hope for for the moduli space of bundles. at least, locally. $\endgroup$ – zudumazics Apr 14 at 12:44
  • $\begingroup$ @AlexanderChervov I have also added another issue in my question. It would be great if you could have a look. $\endgroup$ – zudumazics Apr 14 at 13:16
  • $\begingroup$ The cocycle condition for $H^1$ on a 2 set cover is trivial. But the coboundaries are not-- these tell you when your perturbed bundle is isomorphic to the original. $\endgroup$ – Phil Tosteson Apr 14 at 14:05
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It is known that any holomorphic bundle of any rank over a noncompact Riemann surface is trivial.

The idea of proof is the following: non-compact curve is actually affine manyfold. On affine manyfolds cohomologies of coherent sheaves vanishes (Serre). Extensions of one bundle by another is controlled by Ext^1(V,W) which is coherent and hence vanishes. So non-triviality of vector bundles reduces to linear bundles. Linear bundles are trivial by exponential sequence (Pay attention here is difference between holomorphic and algebraic situation - algebraic linear bundles can be non-trivial on affine curves).

Let $E$ be a holomorphic vector bundle over a compact Riemann surface $X$ with gauge group $G$. A consequence of the above theorem is the restriction $E|_{X-\{p\}}$ for any point $p\in X$ is a trivial bundle. Thus $E$ can be recovered by specifying the transition function $g: > D\cap (X-\{p\}) \rightarrow G$ where $D$ is a small disk containing $p$.

Is this correct? If not, could you give a counter-example? I am mainly interested in learning about the moduli space of holomorphic bundles over $X$ in a concrete way, e.g. using transition functions.

Yes, that is correct. Such point of view on bundles became very popular since end 1980-ies. It is widely used in reseach related to confomal field theory (Verlinde formula), integrbale systems (Hitchin system) and Langlands correspondence over C.

However it is not something which makes things "explicit".

The goal of that technique is to translate geometric problems about vector bundles to Lie group/algebra questions about loop groups. The benefit is that you have "uniformization" - you can think of transition function function which are actually loop group G((t)) as a kind of "universal" moduli space of vector bundles - universal in very strong way since curves of all genuses "sit inside". In a sense that is in a spirit of global to local principle - geometric questions (global) reduced to questions on G((t)) - which are local.

The Verlinde formula story, Knizhnik-Zamolodchikov equation, Hitchin system etc benefit much from it, see e.g. https://mathoverflow.net/a/316733/10446

But now as we minimally only have one transition function, we have no cocycle conditions to impose. How do I still see that $T_{[E]}^\ast > Bun_G(X) \cong H^0(X,\mathfrak{g}\otimes K_X)$?

I am not sure I understand question correctly. But let me try to answer.

Let us think in terms of group theory. So as we discussed above above G((t)) is a kind of universal moduli space. Tanget space at "e" is Lie algebra g((t)). But we are interested in cotanget space ! The point is that natural pairing is given by $ \int f d g $ , so g((t)) are functions, but $g^*((t))$ are 1-forms ! That is how "K_x" appears on the level of Lie algebra.

(Going further for central extension of g((t)) dual space will be identified with space of connections).

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  • $\begingroup$ I assume when you say "noncompact Riemann surface" you are leaving "algebraic" as understood, right? Because, if I'm not mistaken, there are analytic manifolds of dimension $1$ (such as infinite genus curves) that aren't (affine) complex algebraic manifolds. $\endgroup$ – Qfwfq Apr 14 at 19:26
  • $\begingroup$ Thanks for your answer. I am now mainly working in the complex-analytic approach, so it would take me some time to digest your answer. Let me clarify what I mean in my second question anw. I want to show a tangent vector on $Bun_G(X)$ is the same as a class in $H^1(X,\mathfrak{g})$. Can I do this by analyticly deforming the transition function $g: D\cap (X-\{p\}) \rightarrow G$? $\endgroup$ – zudumazics Apr 16 at 15:37
  • $\begingroup$ @Qfwfq Well, I was keeping in mind compact surface minus finite number of points or disks. Which are Stein manyfolds. It might be cerain care should be imposed. But does "infinite genus curves" exist in a sence of complex manyfolds ? For me it was kind if informal analogy between Schrodinger operators as a infinite genus curve ... $\endgroup$ – Alexander Chervov Apr 17 at 18:01
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    $\begingroup$ @zudumazics Well, it is simple to construct tangent vector from H^1(X,g). Element H^1(X,g) is a Lie algebra-valued function on the inresection of two charts, so we can roughly speaking exponentiate it to get a Lie group-value function on the same charts intersection and multply that function on the transition function defining the vector bundle - so you get deformation. More formally we should not speak about exponent but just consider (1+\epsiloin*g) , \epsilon^2 = 0 - and it gives defomation of the Lie group-value transition fiunction. $\endgroup$ – Alexander Chervov Apr 17 at 18:06
  • $\begingroup$ Aha! So very explicitly the deformation for the Lie group-value $g(x)$ at $x$ by a Lie algebra-value $X(x)$ is just $X(x)$ push-forwarded from $e\in G$ to $g(x)$, correct? Then that gives $H^1(X,\mathfrak{g}) \subset T_{[E]} Bun_G(X)$. The other direction comes from the fact that on a 2-set cover, any 1-cocycle automatically lies in the kernel of the boundary map? $\endgroup$ – zudumazics Apr 18 at 22:30

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