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An offbeat question involving Milnor's $K_2$ has come up recently. Start with an algebraically closed field $F$ (perhaps required to be of characteristic 0). Let $G$ be a connected, simply connected simple algebraic group over $F$, for instance $\mathrm{SL}_n(F)$, maybe of rank $\neq 1,2$ to be on the safe side. Then consider a linear group $H \subset \mathrm{GL}(V)$ with $V$ finite dimensional over $F$, together with an epimorphism $\pi:H \rightarrow G$ of abstract groups whose kernel lies in the center of $H$. (EDIT: Further assume that $H$ is equal to its derived group. I neglected to include this crucial condition in the question originally posed to me.)

Are these conditions enough to imply that $\pi$ has trivial kernel?

Note that when $G$ is a special linear group of rank at least 2, its abstract universal central extension is the Steinberg group (with generators and relations specified by Matsumoto) and the kernel of the resulting map is $K_2(F)$. Typically this is an infinite group, uncountable if $F$ is uncountable (Milnor). However, if we add the assumption that $H$ acts irreducibly on $V$, then by Schur's Lemma $\pi$ induces a sort of reverse projective representation of $G$ with image equal to the image of $H$ in $\mathrm{PGL}(V)$. Older work of Steinberg would allow us to lift this projective representation to an ordinary one, thus mapping $G$ onto $H$. (See sections 6,7 of Steinberg's Yale lectures here.)

However, it's apparently undesirable in the original question to make any assumption about complete reducibility of the action of $H$. So I'm unsure what can be said, given only that $H$ is a linear group.

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What is to prevent us from taking $H= Z \times G$ where $Z$ is an Abelian linear algebraic group over $F$? –  Venkataramana Jul 21 '13 at 17:48
    
@Aakumadula: Quite right. I left out one crucial part of the original question, that $H$ should be perfect (equal to its derived group). (This assumption is the abstract version of "connected" in the setting of central extensions.) –  Jim Humphreys Jul 21 '13 at 19:59
    
Would it be stupid to hope all this is controlled by $H^2(G,F)$ and $H^1(G,F)$? Which are trivial, even in positive characteristic ... That book of Serre's where he thinks about extensions of algebraic groups comes to mind. –  David Stewart Jul 21 '13 at 21:52
    
Is it too optimistic to expect that every linear representation of a perfect central extension $H$ of $G$ is trivial on the kernel of $H\to G$? –  YCor Jul 21 '13 at 23:41
    
Yes, it is too optimistic: $H$ has by definition a faithful linear representation. If you wish to prove that the kernel is trivial, it is equivalent to your statement. –  Venkataramana Jul 22 '13 at 0:00

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