MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Assume M affine algebraic manifold over C and we know that H^1(M,Z)=H^2(M,Z) = 0.

Question Does it imply Pic^algebraic(M) = 0 ?

Pic^algebraic means group of algebraic line bundles = H^1(M, O^*) in Zariskky topology.

This is follow up to:

Are there other ways to show Pic(G)is trivial when G is a simple-connected semisimple algebraic groups over C?

Question 2 If it is true what should be analogue of this statement for arbitrary algebraic closed field ? Can we substitute topological cohomology by etale ?

Let me point out the following subtlety:

If I ask about analytic line bundles, then the answer is YES, by exponential sequence argument and vanishing of H^i(M, O) for affine manifolds. If manifold would be compact then by GAGA they coincide, but for non-compact algebraic and analytic are essentially different. If you take elliptic curve and drop out 1 point - you get affine curve - so by exponential sequence Pic^analytic=0, while Pic^algebraic = curve itself+point (as far as I remember). This was quite strange and surprising for me when I learn it.

In the comments Daniel Loughran suggested that Kummer sequence might work, I think it is reasonable, but I am not so experience with etale-cohomology, so I am asking

share|cite|improve this question
up vote 9 down vote accepted

Let us assume that we work over an algebraically closed base field $k$, and fix a prime $\ell$ different from the characteristic of $k$. First of all, the Kummer-sequence $$0\rightarrow \mu_{\ell^n}\rightarrow \mathbb{G}_m\xrightarrow{\ell^n}\mathbb{G}_m\rightarrow 0$$ for $\ell$ prime to the characteristic of $k$, shows that $Pic(X)$ contains no $\ell$-power torsion. In particular, if $char(k)=0$, then $Pic(X)$ is torsion free.

Next, recall that for étale cohomology we have $H^1(X,\mathbb{Z}_\ell(1))=\hom^{cont}(\pi^{et}_1(X),\mathbb{Z}_{\ell})$. Hence, if $H^1(X,\mathbb{Z}_\ell(1))=0$, then the maximal abelian pro-$\ell$-quotient of $\pi_1^{et}$ is $0$.

Now assume that we have a smooth proper scheme $X'$ containing $X$ as a dense open subset. In your characteristic $0$ situation this can always be achieved. Then the abelian maximal pro-$\ell$-quotient of $\pi_1^{et}(X)$ surjects onto the maximal abelian pro-$\ell$ quotient of $\pi_1(X')$, so this group is also trivial. Again, using the Kummer sequence, this implies that $Pic(X')$ contains no $\ell$-power torsion. Since $X'$ is proper, there is a Picard scheme $Pic_{X'/k}$, such that $Pic(X')=Pic_{X'/k}(k)$. In particular, if $Pic^0_{X'/k}$ denotes the connected component of the origin, then $Pic^0_{X'/k}(k)$ has no $\ell$-power torsion. But $Pic^0_{X'/k}$ (or rather its reduced closed subscheme, if $k$ has positive characteristic) is an abelian variety. It follows that $Pic^0_{X'/k}$ is a $0$-dimensional abelian variety (as otherwise there would be nontrivial $\ell$-torsion). This implies that $Pic(X')=NS(X')=Pic_{X'/k}(k)/Pic^0_{X'/k}(k)$, which is a finitely generated group without prime-to-$p$-torsion, i.e. if $char (k)=0$, then $Pic(X')$ is free of finite rank.

Since $X'$ was smooth, we have a surjection $Pic(X')\twoheadrightarrow Pic(X)$, so $Pic(X)$ is free of finite rank (because we already knew it is torsion free).

Next, the Kummer sequence gives an exact sequence $$0\rightarrow Pic(X)\xrightarrow{(-)^{\ell^n}}Pic(X)\xrightarrow{c_{1,\ell^n}}H^2(X,\mathbb{Z}/\ell^n\mathbb{Z}(1))$$ for every $n$, so $Pic(X)/\ell^n\subset H^2(X,\mathbb{Z}/\ell^n\mathbb{Z}(1))$, and passing to the limit gives $Pic(X)\otimes \mathbb{Z}_\ell\subset H^2(X,\mathbb{Z}_\ell(1))$, because we know that $Pic(X)$ is finitely generated.

Thus, if we assume that $H^2(X,\mathbb{Z}_\ell(1))=0$, then $Pic(X)=0$.

share|cite|improve this answer
Thank you very much! If we are over C are there some routes that avoid etsle cohomology? – Alexander Chervov May 18 '12 at 19:37
Hi, since $X$ is smooth over $\mathbb{C}$, it is true that there is a connical ismomorphism between etale cohomology with coefficients in $\mathbb{Z}/\ell^n\mathbb{Z}$ and classical cohomology with coefficients in $\mathbb{Z}/\ell^n\mathbb{Z}$. – Lars May 20 '12 at 8:08

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.