## presentation for GL(n,K)

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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 $GL_n(K)$?

here are some relations, which correspond to simple relations about row operations:

• $E_i(1)=1$
• $E_i(ab) = E_i(a) E_i(b)$
• $E_i(a) E_j(b) = E_j(b) E_i(a)$
• $(E_j(-1) E_{ij})^2=1$
• $E_j(a+b)^{-1} E_{ij} E_j(a+b) = E_j(a)^{-1} E_{ij} E_j(b) E_i(a)^{-1} E_{ij} E_j(a)$
• $(E_{ji} E_{ij} E_{ji} E_j(-1))^2=1$

are these all relations? how can we prove that?

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 $K=\mathbb{Q}$?

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One way of getting a presentation for a Lie group is exponentiating a presentation from the Lie algebra. For the Lie algebra, the presentation will not be hard to do (just take the Cartan subalgebra, $e_{i,i+1}$, and $e_{i+1,i}$), and you can work out the relations for them. It is not necessary to use the matrices corresponding to all $(i,j)$-s - that might overcomplicate things. – Vinoth Dec 30 2009 at 12:10

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It seems to me that this should be a presentation of the simply connected form of the Lie group. But GL_2(R) is not simply connected: the universal cover is basically the metaplectic group. I could imagine much worse things could happen with nonarchimedean fields, but I don't know. – David Speyer Dec 30 2009 at 21:40

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Heh. Is $\mathbb F_2$ not concrete? :) – Mariano Suárez-Alvarez Jan 4 2010 at 2:04
can you give a presentation for $GL_n(\mathbb{F}_2)$? – Martin Brandenburg Jan 4 2010 at 9:42

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See Steinberg, Lectures on Chevalley groups. For $n=2$ extra relations are needed. This is related to David's comment. – Victor Protsak Jun 6 2010 at 6:03

## 4 Answers

You might want to look at Cohn's paper "On the structure of the ${\rm GL}_{2}$ of a ring", MR0207856.

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### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.

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.

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 this 1971 article describes how to present the R-points of any semisimple simply connected Chevalley-Demazure group, for R any commutative ring.

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.

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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.

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 $\mathrm{GL}(2,\mathbb F_2)$.

Exactly the same reasoning shows that the same happens for all $n\geq 2$: you get free products of cyclic groups of order $2$.

NB: 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 GAP to check whether the group given by your generators and relations is or not $\mathrm{GL}$, at least for other small fields...

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This is a sideways answer.

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 not $E(n,A)$, but what we call the $n$-th unstable Steinberg group $\mathrm{St}(n, A)$ of $A$. In general, this is larger than (precisely, an extension of) $E(n,A)$.

(NB: The following paragraph has been edited to make it match reality. Thanks to Allen for pointing the mistake in the comment bellow)

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...)

A nice reference for all this is Jonathan Rosenberger's Algebraic $K$-theory and its applications, and there is John Milnor's Introduction to algebraic $K$-theory, which is also extremely nice.

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.

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The kernel of the stabilized map $St(\infty,A)\to E(\infty,A)$ is $K_2(A)$, not $K_3(A)$. Milnor showed that $K_2(\mathbb Z)$ is cyclic of order 2. – Allen Hatcher Dec 30 2009 at 16:37
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