Timeline for Any holomorphic vector bundle over a compact Riemann surface can be defined by only one transition function?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 25, 2019 at 20:33 | comment | added | Alexander Chervov | @zudumazics yes that is correct | |
| Apr 18, 2019 at 22:51 | vote | accept | zudumazics | ||
| Apr 18, 2019 at 22:30 | comment | added | zudumazics | 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? | |
| Apr 17, 2019 at 18:06 | comment | added | Alexander Chervov | @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. | |
| Apr 17, 2019 at 18:01 | comment | added | Alexander Chervov | @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 ... | |
| Apr 16, 2019 at 15:37 | comment | added | zudumazics | 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 $g: D\cap (X-\{p\}) \rightarrow G$? | |
| Apr 16, 2019 at 15:24 | vote | accept | zudumazics | ||
| Apr 16, 2019 at 15:25 | |||||
| Apr 14, 2019 at 19:45 | history | edited | Alexander Chervov | CC BY-SA 4.0 |
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| Apr 14, 2019 at 19:26 | comment | added | Qfwfq | 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. | |
| Apr 14, 2019 at 19:18 | history | answered | Alexander Chervov | CC BY-SA 4.0 |