In the setting of algebraic groups:

I understand that a central extension of a group $G$ by an abelian group $A$ is a exact sequence of groups :$0\rightarrow A\rightarrow \tilde{G}\rightarrow G\rightarrow 1$ such that $A\subset Z(\tilde{G})$ and I understand that a $G-$torsor in a scheme $X$ is an $G-$action on $X$ such that it's simply transitive.

But how do i see a torsor as a central extension, (or as a multiplicative local system if $X$ is a group (a locally constant sheaf $\mathcal L$ of rank 1 such that $\alpha^* \mathcal L\simeq \mathcal L \boxtimes \mathcal L$) )

  • 2
    $\begingroup$ You have the wrong definition of $G$-torsor. The definition you write is quite related to the definition of $G$-torsor, but it is different. $\endgroup$ – Jason Starr Jun 22 '15 at 17:10
  • 2
    $\begingroup$ I suspect that you are asking why every $A$-torsor over $G$ arises from a central extension. This is not true, for instance, if $G$ is a projective complex torus and $A$ is $\mathbb{G}_m$: only torsors in $\text{Pic}^0$ will arise from central extensions (consider the pullback to the universal cover). For semisimple algebraic groups, I believe it is true that every $\mathbb{G}_m$-torus arises from a central extension, i.e., the Picard group of a simply connected semisimple algebraic group is trivial. I am not certain what happens for general linear algebraic groups . . . $\endgroup$ – Jason Starr Jun 22 '15 at 17:43

For $A$ an abelian group, $A$-torsors on $X$ are classified by $H^1(X,A)$. (I'm just going to take this to be the definition of $H^1(X,A)$.)

There is a natural map:

$$ H^1(X,A) \times H^1(Y,A) \to H^1(X \times Y, A)$$

For certain types of $X,Y$ and certain $A$, we can show that this map is an isomorphism. Wehen that happens, it is easy to verify that $\alpha^* \mathcal L= \mathcal L \boxtimes \mathcal L$: Torsors on $X \times X$ can be identified by pullback to a horizontal and a vertical fiber, and the pullback of $\alpha^* \mathcal L$ to the identity horizontal fiber and the identity vertical fiber are both $\mathcal L$, so $\alpha^* \mathcal L = \mathcal L \boxtimes \mathcal L$.

So your statement is not really true in general, but holds whenever we have a particular cohomological isomorphism.

Some cases where this works:

For $A$ finite of order prime to the characteristic of the base field, this works by the Kunneth formula for tame etale cohomology.

For $A = \mathbb G_a$ and $X,Y$ projective and geometrically connected this works due to the Kunneth formula.

For $A = \mathbb G_m$, if we restrict to $Pic^0$, this works by multiplicativity of $Pic^0$, as Jason Starr points out.

Some cases where this doesn't work:

For $A = \mathbb Z/p$, $G = \mathbb G_a$ over $\mathbb F_p$ there are many torsors that are not central extensions - indeed the total space can have arbitrarily high genus and thus not be a group.

For $A = \mathbb G_a$, $G = E \times \mathbb G_a$ with $E$ an elliptic curve there are also too many torsors.

For $A = \mathbb G_m$, $G$ an abelian variety, the torsors with nontrivial Chern class are problematic, as Jason points out.

In the topological setting, for $A$ finite, the identity of that map follows from $\pi_1(X \times Y) = \pi_1(X) \times \pi_1(Y)$, and the argument I sketch is basically the Eckmann-Hilton argument that the universal cover of a Lie group is its universal central extension.


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.