2
$\begingroup$

Let $X/k$ be a smooth projective geometrically integral variety, perhaps over $k$ algebraically closed. What is the geometric interpretation of $H^1_{Zar}(X,\mathcal{O}_X)$? Does it have something to do with the tangent space? Does $H^1_{Zar}(X,\mathcal{O}_X) = 0$ imply $Pic^0(X) = 0$?

Edit: Another interpretation: $H^1_{Zar}(X,\mathcal{O}_X) = Ext^1(\mathcal{O}_X,\mathcal{O}_X)$.

$\endgroup$
  • 1
    $\begingroup$ Are you assuming that $X$ is smooth? Is it a curve? $\endgroup$ – Alex Youcis Sep 19 '14 at 9:35
  • $\begingroup$ I have edited the question. $\endgroup$ – user5262 Sep 19 '14 at 9:40
2
$\begingroup$

These notes by Brian Conrad show (assuming $X$ has a $k$-rational point) that $H^1(O_X)$ can be identified with the tangent space to $Pic$ at the identity.

So $H^1$ "has something to do with the tangent space", if we're talking about the tangent space to the Picard scheme. Moreover, if $H^1=0$, then $Pic$ must be reduced and zero-dimensional, hence $Pic^0=0$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.