Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be a Kähler manifold. This answer on MO quotes the exact sequence $$0 \to H^1(X, \mathbb{Z}) \to H^{0,1}(X) \to \operatorname{Pic}^0(X) \to 0$$ where $H^{0,1}(X) = H^1(X, \mathscr{O}_X)$ (I think).

Question: Is there an analogous exact sequence when $X$ is a smooth projective variety over an arbitrary field? If so, where can I find a derivation of this result?

share|cite|improve this question
This result almost certainly uses Hodge theory and or Lefschetz 1-1, so arbitrary smooth projective over arbitrary field is probably too much to ask for. Maybe one can get away with a variety for which the Hodge-de Rham spectral sequence degenerates ... but I even doubt that is strong enough? – Matt Sep 5 '12 at 22:59

2 Answers 2

up vote 6 down vote accepted

I'll show where the sequence comes from over $\mathbb{C}$ and then point out why some of the obvious ways people try to generalize to other settings don't yield anything immediately.

The sequence expressed above comes from the exponential sequence $0\to 2\pi i\mathbb{Z}\to \mathcal{O}_X\stackrel{exp}{\to} \mathcal{O}_X^*\to 0$. We take the long exact sequence associated to this and get

$$H^1(X, \mathbb{Z})\hookrightarrow H^1(X, \mathcal{O}_X)\to H^1(X, \mathcal{O}_X^*)\stackrel{c_1}{\to} H^2(X, \mathbb{Z})\to $$

Now $H^1(X,\mathcal{O}_X^*)=Pic(X)$ and $c_1$ is the first Chern class which in cases where it makes sense corresponds to the degree of the line bundle. Thus taking $Pic^0(X)$ to be line bundles of Chern class $0$, we get the sequence you wanted $0\to H^1(X, \mathbb{Z})\to H^{0,1}(X)\to Pic^0(X)\to 0$.

Suppose $X/k$ is smooth and projective. We have lots of issues in generalizing right off. What do we want $Pic^0(X)$ to mean? How should we replace cohomology with $\mathbb{Z}$ coefficients? What do we do about the exponential sequence which no longer exists?

The standard fix for the exponential sequence is to use the Kummer sequence in etale cohomology. Suppose $char(k)=p$ and choose a prime $\ell\neq p$. We have an exact sequence $$0\to \mathbb{Z}/\ell^n \to \mathbb{G}_m\stackrel{\ell^n}{\to} \mathbb{G}_m\to 0$$

The long exact sequence in etale cohomology gets us the "right" Betti numbers and "integral" coefficients, but fails to produce something that looks like the sequence you want. We get by taking the inverse limit over all $n$:

$$\to H^1_{et}(X_{\overline{k}}, \mathbb{Z}_\ell(1))\to Pic(X)\to Pic(X)\stackrel{c_1}{\to} H^2_{et}(X_{\overline{k}}, \mathbb{Z}_\ell(1))\to $$

Where $c_1$ is now the first etale Chern class. We could do a similar thing and make $Pic^0(X)$ the line bundles with Chern class $0$ to get surjective on the end. As you see we won't get any sort of Hodge decomposition though.

Another attempted fix would be to assume the Hodge-de Rham spectral sequence degenerates and hence $H_{dR}(X/k)\simeq H^{1,0}\bigoplus H^{0,1}$ to get some sort of Hodge part to the sequence, but I don't see how to get it into a sequence without an exponential map because the standard projection is clearly going to have kernel $H^{1,0}$ and not be injective.

This doesn't answer your question, but I thought writing it out might help someone else figure it out.

share|cite|improve this answer
So, what is the map in your limit sequence then? – David Speyer Sep 6 '12 at 14:50 I've massively confused myself. Something fishy is going on there. Maybe they are supposed to be $\otimes \mathbb{Z}_\ell$ after the limit so that you can interpret it as a system in which the $n$-th component gets raised to the $\ell^n$ power? I wish I knew a reference for this. – Matt Sep 6 '12 at 15:50
Don't know if it pinged you @DavidSpeyer – Matt Sep 6 '12 at 15:58

I don't think you can ask for anything like this, because the map $H^1(X, \mathcal{O}) \to Pic^0(X)$ is not algebraic. Think about the case of an elliptic curve. The left hand side is a one dimensional complex vector space and the right hand side is an elliptic curve; the map is the Weierstrass $\wp$ map, which is given by a transcendental function.

Thinking about the points of these groups also makes it pretty clear that they shouldn't match up. If $X$ is an elliptic curve over $\mathbb{Q}$, then $H^1(X, \mathcal{O})$ is a one dimensional $\mathbb{Q}$ vector space, and $Pic^0(X)(\mathbb{Q})$ is a finitely generated abelian group (by Mordell-Weil). How are you going to map the former to the latter? If $k$ has characteristic $p$ and $X$ is again an elliptic curve, then every element of $H^1(X, \mathcal{O})$ is $p$-torsion, but $Pic^0(X)[p]$ is either $\mathbb{Z}/p$ or $(0)$.

The closest thing I can think of is that the tangent space to $Pic^0(X)$ at the identity is naturally isomorphic to $H^1(X, \mathcal{O})$. That's true over any field, for $X$ smooth and projective.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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