presentation for GL(n,K) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T14:19:54Z http://mathoverflow.net/feeds/question/10123 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/10123/presentation-for-gln-k presentation for GL(n,K) Martin Brandenburg 2009-12-30T11:26:39Z 2010-06-06T05:42:01Z <p>let $K$ be a field, $n \geq 1$. denote $E_{i,j}$ the elementary matrix having $1$ on the diagonale and in the entry $(i,j)$, and $E_i(a)$ the elementary matrix $diag(1,...,a,...,1)$. you know that $GL_n(K)$ is generated by these matrices, but what relations do we need in order to get a presentation for $GL_n(K)$?</p> <p>here are some relations, which correspond to simple relations about row operations:</p> <ul> <li>$E_i(1)=1$</li> <li>$E_i(ab) = E_i(a) E_i(b)$</li> <li>$E_i(a) E_j(b) = E_j(b) E_i(a)$</li> <li>$(E_j(-1) E_{ij})^2=1$</li> <li>$E_j(a+b)^{-1} E_{ij} E_j(a+b) = E_j(a)^{-1} E_{ij} E_j(b) E_i(a)^{-1} E_{ij} E_j(a)$</li> <li>$(E_{ji} E_{ij} E_{ji} E_j(-1))^2=1$</li> </ul> <p>are these all relations? how can we prove that?</p> <p>EDIT: Mariano has given a counterexample when $K = \mathbb{F}_2$. well, how can we fix this? add more relations? incorporate the structure of $K$ as a ring? what about concrete examples such as $K=\mathbb{Q}$?</p> http://mathoverflow.net/questions/10123/presentation-for-gln-k/10142#10142 Answer by Mariano Suárez-Alvarez for presentation for GL(n,K) Mariano Suárez-Alvarez 2009-12-30T14:23:08Z 2009-12-30T19:18:14Z <p>This is a sideways answer.</p> <p>Let $E_{ij}(a)=I+a E_{ij}$, for $i\neq j$ and $a\in A$.These matrices generate the conmutator subgroup $$E(n, A)=[\mathrm{GL}(n, A),\mathrm{GL}(n, A)]\subseteq\mathrm{GL}(n, A).$$ One can easily check that the obvious relations satisfied by these elements are $$E_{ij}(a)E_{ij}(b)=E_{ij}(a+b),$$ $$[E_{ij}(a),E_{jk}(a)]=E_{ik}(a) \mbox{ if i\neq k,}$$ $$[E_{ij}(a),E_{kl}(b)]=1 \mbox{ if i\neq l and j\neq k.}$$ Yet the group presented by generators and this relations is <em>not</em> $E(n,A)$, but what we call the $<em>n$-th unstable Steinberg group</em> $\mathrm{St}(n, A)$ of $A$. In general, this is larger than (precisely, an extension of) $E(n,A)$.</p> <p>(<strong>NB:</strong> The following paragraph has been edited to make it match reality. Thanks to Allen for pointing the mistake in the comment bellow)</p> <p>This is seen, for example, because the map $\mathrm{St}(n, A)\to E(n,A)$ has a non-trivial kernel. Indeed, after passing to the direct limit as $n$ goes to infinity, the kernel of that map is precisely the second algebraic $K$-theory group of $A$, $K_2(A)$. Milnor shows in his book that $K_2(\mathbb{R})$ is uncountable, and describes $K_2(\mathbb Q)$ (he also shows that $K_2(\mathbb Z)$ is cyclic of order two, so this can be done for rings that are not fields too...)</p> <p>A nice reference for all this is Jonathan Rosenberger's <a href="http://www.ams.org/mathscinet-getitem?mr=MR1282290" rel="nofollow"><em>Algebraic $K$-theory and its applications</em></a>, and there is John Milnor's <a href="http://www.ams.org/mathscinet-getitem?mr=MR0349811" rel="nofollow"><em>Introduction to algebraic $K$-theory</em></a>, which is also extremely nice.</p> <p>A short intuitive description for $K_2(A)$ is: it measures how much more information is there in the elementary matrices of a ring which does not follow formally from the Steinberg relations.</p> http://mathoverflow.net/questions/10123/presentation-for-gln-k/10636#10636 Answer by Mariano Suárez-Alvarez for presentation for GL(n,K) Mariano Suárez-Alvarez 2010-01-03T23:44:08Z 2010-01-03T23:44:08Z <p>Suppose $R=\mathbb F_2$ is the field with two elements and $n=2$. Then we need not consider the matrices $E_i(a)$, for the only possible $a$ is $1$, so your group is generated by $\alpha=E_{12}$ and $\beta=E_{21}$. Your fourth relation implies that $$\alpha^2=\beta^2=1.$$ Your fifth equation is empty in this case (for it is only meaningful for when $a$ and $b$ are non-zero elements of the field which add up to a non-zero element of the field!) Finally, your sixth relation in this case tells us that $(\alpha\beta\alpha)^2=(\beta\alpha\beta)^2=1$, but these two equalities follow from the previous equation.</p> <p>We thus see that the group generated by the $E_i(a)$'s and the $E_{ij}$ subject to your relations is, in this case, $$\langle\alpha,\beta:\alpha^2=\beta^2=1\rangle.$$ This is an infinite group, so it is not $\mathrm{GL}(2,\mathbb F_2)$.</p> <p>Exactly the same reasoning shows that the same happens for all $n\geq 2$: you get free products of cyclic groups of order $2$.</p> <p><strong>NB:</strong> It wouldn't be the first time that $\mathbb F_2$ behaves differently from other fields... I doubt that is the case, and surely someone with enough determination will be able to use <a href="http://www.gap-system.org/" rel="nofollow">GAP</a> to check whether the group given by your generators and relations is or not $\mathrm{GL}$, at least for other small fields...</p> http://mathoverflow.net/questions/10123/presentation-for-gln-k/10727#10727 Answer by S. Carnahan for presentation for GL(n,K) S. Carnahan 2010-01-04T18:32:06Z 2010-01-04T18:32:06Z <p>I suggest you give Mariano a check mark. MathOverflow does not function properly when you change a question substantially after it has been correctly answered.</p> <p>Regarding your revised question, we can fix the presentation by throwing it away and using Steinberg's presentation, which works without problems. I couldn't find an online copy of his paper, but <a href="http://www.ams.org/bull/1971-77-02/S0002-9904-1971-12704-0/S0002-9904-1971-12704-0.pdf" rel="nofollow">this 1971 article</a> describes how to present the R-points of any semisimple simply connected Chevalley-Demazure group, for R any commutative ring.</p> <p>I have heard that the presentation in the case of the general linear group is due to Schur, but I don't know a reference.</p> http://mathoverflow.net/questions/10123/presentation-for-gln-k/27225#27225 Answer by Josh Roberts for presentation for GL(n,K) Josh Roberts 2010-06-06T05:41:09Z 2010-06-06T05:41:09Z <p>You might want to look at Cohn's paper "On the structure of the ${\rm GL}_{2}$ of a ring", MR0207856.</p>