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



Torsors and Central Extensions
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.




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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.




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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.


