# Torsors and Central Extensions

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 $\alpha^* \mathcal L\simeq \mathcal L \boxtimes \mathcal L$) )

• 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. – Jason Starr Jun 22 '15 at 17:10
• 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 . . . – 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.