# Are higher dimensional Heisenberg groups free nilpotent?

I know that the Heisenberg group { x,y | [x,[x,y]]=[y,[x,y]]=1 } is free nilpotent; what about the higher dimensional ones? Do the higer dimensional Heisenberg groups have nice presentations? By higher dimensional Heisenberg groups I mean nxn upper triangular matrices with integral entries. Thanks.

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What do you mean by 'Heisenberg groups'? Upper triangular matrices with integral entries and ones along the diagonal? –  Mariano Suárez-Alvarez Mar 1 '10 at 22:56
It would be nice to give faithful representations of the free nilpotent groups in GL(n,Z) for minimal n. GAP's polycyclic package can very quickly produce linear representations, but I'm not sure if they are minimal, and the basis chosen is somewhat difficult for humans. As a specific version, the free group of rank 2 and class 3 is represented in dimension 6 (which I believe is the minimal dimension, by easy counting). –  Jack Schmidt Mar 2 '10 at 3:26
There is no free nilpotent group; there are instead e.g. free k-step nilpotent groups. So in asking your question, you should figure out what the degree of nilpotency would be (i.e. the value of k). Then it should be clear that the Heisenberg groups are not free k-step nilpotent. In particular, if you look at the ranks of the associated graded (quotients of successive terms in the lower central series), the ranks for the free k-step nilpotent group grow according to Hall's formula. In the Heisenberg case they get smaller by the end; I think the last term has rank 1 (the upper-right corner)! –  Tom Church Mar 2 '10 at 17:07

Let $U_n$ be the group of upper triangular integer matrices of size $n$ by $n$ with ones on the diagonal. Then $U_n$ is generated by the elements $x_i$, $i=1,...,n-1$, with Serre relations $$[x_i,[x_i,x_{i+1}]]=[x_{i+1},[x_i,x_{i+1}]]=1,$$ and $[x_i,x_j]=1$ if $|i-j|\ge 2$ (EDIT: one needs additional relations, see below). Indeed, the group $G_n$ generated in this way maps surjectively to $U_n$ by $x_i\mapsto 1+E_{i,i+1}$, and one can check that this map is injective (EDIT: with additional relations) by writing every element of $G_n$ as an ordered product of powers of $x_{ij}:=[x_i[x_{i+1}...x_j]]$, $i\le j$. The corresponding groups over $\Bbb Z/m\Bbb Z$ are then obtained by adding the relations $x_{ij}^m=1$.

EDIT: As Jack kindly pointed out, I erroneously ignored the treacherous 2-torsion. According to Theorem 1 of the paper

D. Biss, S. Dasgupta, "A presentation for the unipotent groups over rings with $1$", J. of Algebra, v. 237, pp.691-707, 2001,

the above statements are correct over $\Bbb Z/m\Bbb Z$ when $m$ is an odd integer. In general (i.e. for any integer $m$, including $m=2$ and $m=0$), it suffices to add the relation $$[[x_i,x_{i+1}],[x_{i+1},x_{i+2}]]=1$$ for $i=1,...,n-3$. Moreover, over $\Bbb Z/m\Bbb Z$ with $m\ne 0$ it is enough to impose the relations $x_i^m=1$ (the relations $x_{ij}^m=1$ will then follow).

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For n=3, this map is not injective. You mean the group generated by x,y,z subject to the relations [x,z], [x,[x,y]], [y,[x,y]], [z,[y,z]], [y,[y,z]], right? The group G3/γ5(G3) (of nilpotency class 4) has 2-torsion. Maybe you also need to enforce a nilpotency class condition? –  Jack Schmidt Mar 2 '10 at 3:40
Thanks, Jack! I fixed it - see the edited answer. –  Pavel Etingof Mar 2 '10 at 12:55