Timeline for Geometric interpretation of sections $H^0(\Theta_X, X)$ of the Tangent sheaf over curve
Current License: CC BY-SA 4.0
9 events
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| Oct 8, 2019 at 18:06 | comment | added | user267839 | great, then this indeed works completly without Hurwitz's automorphisms theorem (up to now I thought that Hurwitz' was the only standard way to show that the automorphism group of curves with $g >1$ is finite). again, thank you a lot for explanations! | |
| Oct 8, 2019 at 17:49 | comment | added | Will Chen | @TimGrosskreutz I don't have your book in front of me, but from the text you quoted in your OP, the authors say "...is reflected in the fact that the group of automorphisms of such curves is finite". That doesn't sound like they're concluding finiteness from 0-dimensionality. They're just saying that the 0-dimensionality is reflected in the fact that it's finite. After all you are of course correct - discreteness does not imply finiteness in general, but it may be interesting to note that in algebraic geometry, noetherian + 0-dimensional implies the underlying top. space is finite discrete... | |
| Oct 8, 2019 at 17:24 | comment | added | user267839 | another detail confuses me a bit. going through the argumentation we obtain information about tangent space of automorphism groups. and the fact the the tangent sheaf of our curve $X$ of genus $g >1$ is zero says that the automorphism group scheme $G_X$ has dimension zero, i.e. it is discrete space. why the author concludes that automorphism group $G_X$ is finite for $g>1$? discrete spaces are in general obviosly not finite. | |
| Oct 8, 2019 at 16:03 | comment | added | Will Chen | @TimGrosskreutz Sorry I meant to write $exp(tz)(x)$. Basically you push the point $x$ along by the infinitesimal automorphism $exp(tz)$ for $t\in(-1,1)$, and that gives you a small curve in a neighborhood of $x$ which represents the desired tangent vector. | |
| Oct 8, 2019 at 16:02 | history | edited | Will Chen | CC BY-SA 4.0 |
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| Oct 8, 2019 at 14:50 | comment | added | user267839 | Thank you a lot for your detailed answer. one aspect I not understand: I learned that if $G$ is a Lie-group with Lie-algebra $\mathfrak{g}=Lie(G)$ (that equals the tangent space at the origin $e$ of $G$). then every element $ z \in Lie(G)$ induces a curve $exp_z: [-1,1] \to G, t \to exp_z(tz)$. evaluation at $t=0$ gives map $exp: Lie(G) \to G$. the point which I not understand why your curve $exp_z$ map to the manifold $X$ back and not to the Lie group of the automorphisms $G \subset Aut(X)$? do I missing something? | |
| Oct 8, 2019 at 14:25 | vote | accept | user267839 | ||
| Oct 8, 2019 at 1:59 | history | edited | Will Chen | CC BY-SA 4.0 |
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| Oct 8, 2019 at 1:50 | history | answered | Will Chen | CC BY-SA 4.0 |