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Let $X$ be a smooth proper geometrically integral variety over a field of characteristic zero (no assumptions are made on the closedness of the field).

How should I interpret the vanishing of $H^i(X, \mathcal{O}_X)$, for, say, $i = 1,2,3$?

Martin Brandenburg
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kyrilmath
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  • Cohomology can often be interpreted as obstructions to something, though I assume an algebraic geometer could tell you more about this specific case. – Najib Idrissi Nov 21 '14 at 21:15
  • Do you know about gerbes and stacks? Higher sheaf cohomology correspond to equivalence classes of higher grebes. Just make some computations in Cech cohomology to see what this means. – user40276 Nov 21 '14 at 21:20
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    It doesn't matter if you work over $k$ or $\bar{k}$--$H^1(X,\mathcal{O}X)\otimes\bar{k}=H^1(X{\bar{k}},\mathcal{O}_{\bar{k}})$. The intuition for, say, $H^1$ is that the space is really, really bad about finding functions which agree on overlaps. Depending on what the dimension of $X$ is, say it's $2$, then $H^2(X,\mathcal{O}_X)$ is the same as $H^0(X,\omega_X)$, and so this says that space has an obstruction to creating global differentials. This is usually interpreted as some generalization of a genus obstruction. In fact, $h^0(\omega_X)$ is the geometric genus of $X$. Something like that? – Alex Youcis Nov 21 '14 at 21:28
  • @AlexYoucis That's really helpful, thanks!! Do you know any references where I can learn more? – kyrilmath Nov 21 '14 at 21:51
  • @user40276 Unfortunately I don't know much about gerbes and stacks – kyrilmath Nov 21 '14 at 21:52
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    Vanishing of $H^1$ is equivalent to the vanishing of the Albanese variety. Equivalently, $H^1$ is zero iff any morphism to an abelian variety is constant. – Ariyan Javanpeykar Nov 21 '14 at 22:44
  • Does $H^i$ mean sheaf cohomology with respect to the Zariski topology or with respect to the étale topology? The latter will give more interesting answers, I think. – Martin Brandenburg Nov 21 '14 at 22:53
  • Do you know Hodge theory? $H^p(X,\Omega^q_X)$ gives you a "piece" of $H^{p+q}(X(\mathbf C),\mathbf C)$, where I've chosen an embedding of $k$ in the complex numbers and the latter denotes usual singular cohomology of the analytification of $X$. – Dan Petersen Nov 21 '14 at 23:01
  • I would say: $H^1(X,\mathcal O_X)=0$ means line bundles do not deform. – Brenin Nov 21 '14 at 23:57
  • @MartinBrandenburg Etale topology – kyrilmath Nov 22 '14 at 00:13
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    @Martin Brandenburg: For quasi-coherent sheaves (like here $\mathcal O_X$ ), étale cohomology coincides with Zariski cohomology. – Georges Elencwajg Nov 22 '14 at 08:52
  • You could also think of $H^1(X,\mathcal{O}X)$ as $\text{Ext}^1(\mathcal{O}_X,\mathcal{O}_X)$, and so its measuring extensions of $\mathcal{O}_X$ by $\mathcal{O}_X$. @kyrilmath The standard references I think. As Georges said, since $\text{Qcoh}(X\text{et})=\text{Qcoh}(X_\text{zar})$ we can compute cohomology on the small Zariski site. What I stated was then 1)[gluing functions] the mere definition (with some thought) you get from Cech cohomology and 2)[differentials] Serre duality, with $\omega_X=\Omega^1_X$. There are a lot of responses now, are any of them particularly interesting to you? – Alex Youcis Nov 22 '14 at 09:58
  • @kyrilmath Just to clarify, when I said the standard references I meant something like Hartshorne. I'm sure you know Cech cohomology, and if you want to freshen up on Serre duality for $X/k$ smooth, you can look at 3.7 of Hartshorne. I don't know if this are the type of references you were expecting/wanted. Let me know! – Alex Youcis Nov 22 '14 at 10:02

1 Answers1

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I'll assume in order not to have complicated statements that $k$ is algebraically closed: your question is interesting enough even under that hypothesis!

Rationality
A smooth rational variety has all $H^i(X, O_X)=0$ for $i\gt0$.
So the vanishing of one of those groups is the vanishing of some obstruction to rationality.
For example:
If $X$ is a curve, rationality of $X$ is equivalent to $H^1(X, O_X)=0$.
If $X$ is a surface Castelnuovo's criterion for rationality is the conjunction of $H^1(X, O_X)=0$ and $H^0(X, \omega^2_X)=0$.

Topology
Over $\mathbb C$ Hodge theory implies that $H^1(X, O_X)=\frac 1 2b_1(X)$, half the first Betti number.
So the vanishing of $H^1(X, O_X)$ is the vanishing of an obstruction to $X$ being simply connected.
For example if $X$ is a curve we have:
$X$ simply connected $\iff$ $H^1(X,\mathcal O_X)=0$ $\iff$ $X\cong \mathbb P^1$

Picard variety
a) The tangent space at the origin of the Picard variety of $X$ is $H^1(X,\mathcal O_X)$.
So if $H^1(X,\mathcal O_X)=0$ the Picard group of line bundles on $X$ reduces to the Néron-Severi group and is thus a finitely generated abelian group.
For a nice example, take a smooth cubic surface in $\mathbb P^3_k$. It is rational and its Picard group is isomorphic to $\mathbb Z^7$.

b) The condition $H^1(X, O_X)=0$ implies the nice formula valid for any variety $Y$: $$\operatorname {Pic}(X\times Y)=\operatorname {Pic}(X)\times \operatorname {Pic}(Y) $$ Such a formula is completely false in general.

Community
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Georges Elencwajg
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  • I think there are quite a few more one could say, but these certainly are a great start! Of course, if $k\ne\bar{k}$, then the fact that $H^1(X_{\bar{k}},\mathcal{O}{\bar{k}})=0$ just says that $X{\bar{k}}$ satisfies the properties you state. So, $V(x^2+y^2+z^2)\subseteq\mathbb{P}^2_\mathbb{R}$ has $h^1(\mathcal{O}_X)=0$, but that's because it's geometrically rational.

    Also, just to make sure I understand. For your last statement, you're just saying that $\dim \text{Pic}^0=h^1(\mathcal{O}_X)=0$, so $\mathrm{NS}(X)=\text{Pic}(X)/\text{Pic}^0(X)=\text{Pic}(X)$, correct? Thanks!

     – Alex Youcis Nov 22 '14 at 12:20
  • Dear @Alex, yes there is more to say of course: I hope you will add some of your insights and I might come back to this answer. As for the Picard variety, it is smooth in characteristic zero ( Cartier's thesis) so that indeed the dimension of the Picard variety is the same as the dimension of its tangent space at the origin, namely $h^1(X,\mathcal O_X)$. [Igusa gave an example of a non-reduced Picard scheme in char. $p$] I didn't want to enter these technicalities in my post, which I tried to make readable also by beginning algebraic geometers. – Georges Elencwajg Nov 22 '14 at 13:03