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A group $G$ is said to be linear if there exists a field $k$, an integer $n$ and an injective homomorphism $\varphi: G \to \text{GL}_n(k).$ Given a short exact sequence $1 \to K \to G \to Q \to 1$ of groups where $K$ and $Q$ are linear (over the same field), is it true that $G$ is linear too?

Background: Arithmetic groups are by definition commensurable with a certain linear group, so they are finite extensions of a linear group, and finite groups clearly are linear (over any field).

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up vote 21 down vote accepted

The universal central extension $\widetilde{\text{Sp}_{2n}}\mathbb{Z}$ is the preimage of $\text{Sp}_{2n}\mathbb{Z}$ in the universal cover of $\text{Sp}_{2n}\mathbb{R}$, and fits into the sequence

$$1\to \mathbb{Z}\to \widetilde{\text{Sp}_{2n}}\mathbb{Z}\to \text{Sp}_{2n}\mathbb{Z}\to 1.$$

Deligne proved that $\widetilde{\text{Sp}_{2n}}\mathbb{Z}$ is not residually finite; the intersection of all finite-index subgroups of is $2\mathbb{Z}<\widetilde{\text{Sp}_{2n}}\mathbb{Z}$. In particular, this implies that $\widetilde{\text{Sp}_{2n}}\mathbb{Z}$ is not linear. But certainly $\mathbb{Z}$ and $\text{Sp}_{2n}\mathbb{Z}$ are. If you want an arithmetic group, you can take the corresponding $\mathbb{Z}/k\mathbb{Z}$-extension of $\text{Sp}_{2n}\mathbb{Z}$, which will not be linear as long as $k\neq 2$.

I learned the proof of this theorem from Dave Witte Morris, who has written up his fairly-accessible notes as "A lattice with no torsion-free subgroup of finite index (after P. Deligne)" (PDF link).

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"In particular, this implies..." -- because a finitely generated linear group is residually finite, right? – Pete L. Clark Apr 28 '10 at 8:32
Well, the above pdf by Witte Morris is not exactly a proof: it explains how it follows from the congruence subgroup property and from the computation of the universal central extension of $\text{Sp}_n(\mathbf{Q}_p)$. These are highly nontrivial results, whose proof is the big part of the theorem (although indeed these were known things when Deligne wrote his note around 1977). – YCor Jun 26 '13 at 7:45

Erschler has shown that there exists a central extension $G$ of $\mathbb{Z} \mathbin{wr} \mathbb{Z}$ by a finite group $F$ which is not residually finite. Thus the short exact sequence $1 \to F \to G \to \mathbb{Z} \mathbin{wr} \mathbb{Z} \to 1$ provides an example of a non-linear group which is an extension of two linear groups over $\mathbb{C}$.

A. Erschler, Not residually finite groups of intermediate growth, commensurability and non-geometricity, J. Alg. 272 (2004), 154--172.

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The universal cover $G$ of $SL_2(\mathbb{R})$ has no continuous injective homomorphism into any $GL_n(\mathbb{R})$. Whether it has a faithful representation into any $GL_n(k)$ is a different question, but seems unlikely to me. Note that $G$ is an extension of $\mathbb{Z}$ (linear by your definition) by $SL_2(\mathbb{R})$.

See wikipedia for more details.

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I think you implicitly assume that $char k$ is fixed, as otherwise there are trivial counnter-examples: take the direct sum of a countable number of cyclic groups of order $p$ and $q$. Each group is linear (but over different $k$), but the sum is not.

Now, let's suppose that $k$ is assumed to have zero characteristic. Then again the answer is no. There are solvable torsion free groups that are not linear (a solvable linear group is by virtually nilpotent by abelian by Lie-Kolchin's theorem). There are also examples of such extensions in which the group $G$ {\bf is} linear, but that is far from obvious. For instance the automorphism group of the free group on two generators is an extension of the free group (the inner automorphisms) by $SL(2,Z)$ (the abelianization). Both of these groups are linear, but the fact that $Aut(F_2)$ is linear is a difficult theorem.

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I don't know in general, but this is certainly true when $Q$ is finite. If $K$ has a faithful linear representation, it is very easy to see that the induced representation of $G$ is also faitful.

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