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Let $G$ be the nilpotent Lie group consisting of matrices $$ \begin{pmatrix} 1 & a_{12} & \cdots & a_{1,n}\\ 0 & 1 & \ddots & \vdots\\ \vdots & \ddots & \ddots & a_{n-1,n}\\ 0 & \cdots & 0 & 1 \end{pmatrix} $$ where $a_{ij}\in\mathbb R$.

I would like to find a presentation of the group $\Gamma=G\cap\mathrm{GL}_n\mathbb Z$.

The entries on the superdiagonal play a crucial role, in that the $n-1$ matrices with a single $1$ on the superdiagonal ($1$s on the diagonal and $0$s elsewhere) generate all of $\Gamma$.

Trying to write down a presentation, I would start by choosing $n-1$ generators, $a_{12},\dotsc,a_{n-1,n}$ (named after the matrix entries). It is also clear that we need commutation relations, like $$ [\cdots[[[a_{12},a_{23}],[a_{23},a_{34}]],\ldots]\cdots]=\cdots=e. $$ (If $n=3$, this would just be $[[a_{12},a_{23}],a_{12}]=[[a_{12},a_{23}],a_{23}]=e$.) I am just not sure, though, if I explicitly need to require that, say, $$ [a_{12},a_{34}]=e, $$ which seems to be directly related to the embedding I am thinking of. I would like a presentation in which any generators can be mapped to any of the matrices with a $1$ in the $(i,i+1)$ position. Is this possible?

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The Chevalley commutator relations may be what you are looking for. If the presentation was proven a long time ago, it was likely proven by Chevalley or Steinberg. I believe they describe your group for any ring R that satisfies some nice conditions (that the integers do not satisfy when n=2 or 3). Presumably one just needs to add a few relations if the low dimensional cases are actually exceptional. –  Jack Schmidt Jul 18 '12 at 17:48
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2 Answers

There is a paper by D. Biss and S. Dasgupta (A Presentation for the Unipotent Group over Rings with Identity, J. Algebra, vol. 237, 691-707 (2001)) which gives a presentation of the group of upper triangular matrices over $R$ with diagonal entries equal to 1, where $R$ is any ring with identity.

(Moreover, in some cases they show that their presentations are minimal both in the sense of having no redundant relations, and in the stronger sense of having the fewest possible number of generators and relations.)

In the case $R=\mathbb{Z}$, their generators are the same as the ones in the question above, and the relations are $$[a_{i,i+1},a_{j,j+1}] =1\ \ (i<j-1\leq n-2)$$

$$[a_{i,i+1},[a_{i,i+1},a_{i+1,i+2}]]=[a_{i+1,i+2},[a_{i,i+1},a_{i+1,i+2}]]=1$$ $$[[a_{i,i+1},a_{i+1,i+2}],[a_{i+1,i+2},a_{i+2,i+3}]]=1.$$

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Are you sure about the last relation? If we choose $a_{ij}$ to be the identity with an extra $1$ in the $(i,j)$ position, we have that $[a_{12},a_{23}]=a_{13}$ and $[a_{23},a_{34}]=a_{24}$, but $[a_{13},a_{24}]=a_{14}\neq1$. –  Earthliŋ Jul 25 '12 at 2:22
    
But $[a_{13},a_{24}]$ is trivial (maybe you meant $[a_{13},a_{34}]=a_{14}?). The last relation boils down to $[a_{i,i+2},a_{i+1,i+3}]=1$, which is as we would expect since $\{i,i+2\}$ and $\{i+1,i+3\}$ are disjoint. Incidentally, I think Biss and Dasgupta show that the last relation is automatic for certain rings. –  Shane O Rourke Jul 25 '12 at 10:28
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I assume that when you are using $a_{ij}$ to denote an element of $\Gamma$, you mean the identity matrix with an extra 1 in the $(i,j)$ position.

It is easy to write down a presentation of $\Gamma$ on the $m := n(n-1)/2$ generators $a_{ij}$ with $i<j$. Just use the $m(m-1)/2$ commutator relations $[a_{ij},a_{jk}] = a_{ik}$, and $[a_{ij},a_{kl}]=1$ when $j \ne k$, $i \le k$, and $j<l$ when $i=k$. Of course many of these relations are redundant.

To get a presentation on the generators $a_{i,i+1}$, you can then use the equations $[a_{ij},a_{jk}] = a_{ik}$ to eliminate all of the other generators. Doing that with $n=3$ gives the presentation you have written down.

Although many relations in this presentation are redundant, those like $[a_{12},a_{34}]=1$ between the commuting generators $a_{i,i+1}$ are definitely not redundant, because if you left one of those out, then you could make the group bigger by putting that commutator equal to a new central generator.

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Thank you. I had hoped that one could embed $\Gamma$ as abstract group into $G$ by sending the generators $a_{i,i+1}$ to any element in $G$ as long as the entries on the superdiagonal were linearly independent. Do you see any way of understanding of how one is allowed to embed $\Gamma$ into $G$? –  Earthliŋ Jul 19 '12 at 1:55
    
You certainly cannot send the $a_{i,i+1}$ to an arbitrary sequence of elements of $G$ with linearly independent entries on the superdiagonal. For example, for $n=4$, you cannot have $(a_{12},a_{23},a_{34}) \to (a_{23},a_{12},a_{34})$ because $[a_{12},a_{34}]= 1$ but $[a_{23},a_{34}]\ne 1$. I am pretty sure that the only embeddings possible are those that either map each $a_{i,i+1}$ to a "multiple" of $a_{i,i+1}$ (i.e. the image can have any nonzero entry in the $(i,i+1)$-position) or those that do the same but reverse the order of the $a_{i,i+1}$. –  Derek Holt Jul 19 '12 at 9:13
    
Interesting, I hope you are wrong. For $n=3$, the images of $a_{12}$ and $a_{23}$ must not commute, which is the case precisely when they map to any two elements in $G$, as long as the values on the superdiagonal are linearly independent. For general $n$ the relations are not symmetric on the generators, which complicates things. –  Earthliŋ Jul 19 '12 at 14:58
    
Maybe 'neighbouring' generators should be allowed to vary inside $\mathrm{GL}_2\mathbb R$...? –  Earthliŋ Jul 19 '12 at 15:06
    
Yes, what I said was wrong for $n=3$, but I still believe it is true for $n>3$. Since I don't have the energy to try and write down a proof at present, let me challenge you to try and find a counterexample to my claim, say for $n=4$. –  Derek Holt Jul 19 '12 at 15:40
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