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Imagine you have a curve $X$ (integral scheme of dimension 1, proper over $k$ (algebraically closed) whose local rings are regular) of genus $g$. What can be said about $H^1(X, ( \Omega^{1}_{X / k})^n)$?

Greetings

Marc

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What do you want to know and what do you already know about these vectors spaces ? –  user18119 Nov 10 '12 at 18:09
    
I want to use a base-change theorem by grothendieck which needs the first cohomology to vanish, but I'm not quite sure why it should vanish... –  marc Nov 10 '12 at 18:28
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1 Answer

up vote 2 down vote accepted

By Serre duality, your $H^1$ is isomorphic to the dual (as vector space) of $H^0(X, (\Omega_X)^{\otimes (1-n)})$. Let $e_n$ be its dimension. Then:

  1. If $n=1$, $e_n=1$.
  2. If $n\ge 2$ and $g\ge 2$, then $e_n=0$ because $(\Omega_X)^{\otimes (1-n)}$ has degree $(1-n)(2g-2)<0$.
  3. If $n\ge 2$ and $g=1$, then $e_n=1$ because $\Omega_X=O_X$.
  4. If $n\ge 2$ and $g=0$, then $e_n=2n-2+1=2n-1$ again by Riemann-Roch for the projective line.
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You mean Serre duality? –  Sanchez Nov 10 '12 at 19:43
    
wonderful explanation. Thanks a lot –  marc Nov 10 '12 at 22:06
    
And yes, with "dual" he means Serre duality –  marc Nov 10 '12 at 22:07
    
You are right @Sanchez. Thanks ! –  user18119 Nov 10 '12 at 22:29
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@Sanchez: no, I'm not QLC. Just QiL :) –  user18119 Nov 10 '12 at 23:18
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