Are extensions of linear groups linear? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T12:50:56Z http://mathoverflow.net/feeds/question/22814 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/22814/are-extensions-of-linear-groups-linear Are extensions of linear groups linear? Guntram 2010-04-28T06:52:14Z 2010-04-28T12:06:19Z <p>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?</p> <p>Background: <a href="http://en.wikipedia.org/wiki/Arithmetic_group" rel="nofollow">Arithmetic groups</a> 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).</p> http://mathoverflow.net/questions/22814/are-extensions-of-linear-groups-linear/22816#22816 Answer by Angelo for Are extensions of linear groups linear? Angelo 2010-04-28T07:00:20Z 2010-04-28T07:10:53Z <p>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.</p> http://mathoverflow.net/questions/22814/are-extensions-of-linear-groups-linear/22817#22817 Answer by Robin Chapman for Are extensions of linear groups linear? Robin Chapman 2010-04-28T07:01:41Z 2010-04-28T07:01:41Z <p>The universal cover $G$ of $SL_2(\mathbb{R})$ has no <strong>continuous</strong> 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 $SL_2(\mathbb{R})$.</p> <p>See wikipedia <a href="http://en.wikipedia.org/wiki/SL%E2%82%82%28R%29" rel="nofollow">http://en.wikipedia.org/wiki/SL%E2%82%82%28R%29</a> for more details.</p> http://mathoverflow.net/questions/22814/are-extensions-of-linear-groups-linear/22823#22823 Answer by Tom Church for Are extensions of linear groups linear? Tom Church 2010-04-28T08:16:57Z 2010-04-28T08:16:57Z <p>The universal central extension <code>$\widetilde{\text{Sp}_{2n}}\mathbb{Z}$</code> is the preimage of $\text{Sp}_{2n}\mathbb{Z}$ in the universal cover of <code>$\text{Sp}_{2n}\mathbb{R}$</code>, and fits into the sequence</p> <p><code>$1\to \mathbb{Z}\to \widetilde{\text{Sp}_{2n}}\mathbb{Z}\to \text{Sp}_{2n}\mathbb{Z}\to 1$</code>.</p> <p>Deligne proved that <code>$\widetilde{\text{Sp}_{2n}}\mathbb{Z}$</code> is not residually finite; the intersection of all finite-index subgroups of is <code>$2\mathbb{Z}&lt;\widetilde{\text{Sp}_{2n}}\mathbb{Z}$</code>. In particular, this implies that $\widetilde{\text{Sp}_{2n}}\mathbb{Z}$ is not linear. But certainly $\mathbb{Z}$ and <code>$\text{Sp}_{2n}\mathbb{Z}$</code> are. If you want an arithmetic group, you can take the corresponding $\mathbb{Z}/k\mathbb{Z}$-extension of <code>$\text{Sp}_{2n}\mathbb{Z}$</code>, which will not be linear as long as $k\neq 2$.</p> <p>I learned the proof of this theorem from Dave Witte Morris, who has written up his fairly-accessible notes as "<a href="http://people.uleth.ca/~dave.morris/talks.shtml" rel="nofollow">A lattice with no torsion-free subgroup of finite index</a> (after P. Deligne)" (<a href="http://people.uleth.ca/~dave.morris/talks/deligne-torsion.pdf" rel="nofollow">PDF</a> link).</p> http://mathoverflow.net/questions/22814/are-extensions-of-linear-groups-linear/22826#22826 Answer by Keivan Karai for Are extensions of linear groups linear? Keivan Karai 2010-04-28T08:49:41Z 2010-04-28T08:49:41Z <p>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.</p> <p>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.</p> http://mathoverflow.net/questions/22814/are-extensions-of-linear-groups-linear/22835#22835 Answer by Simon Thomas for Are extensions of linear groups linear? Simon Thomas 2010-04-28T11:21:30Z 2010-04-28T12:06:19Z <p>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 $\mathbb{C}$.</p> <p>A. Erschler, Not residually finite groups of intermediate growth, commensurability and non-geometricity, J. Alg. 272 (2004), 154--172.</p>