An explicit representation of a free nilpotent group by unitriangular matrices

Question

P. Hall proved that every finitely generated torsion-free nilpotent group can be faithfully represented by upper unitriangular matrices over $\mathbb Z$. The most famous example is the integral Heisenberg group, which is free nilpotent, of rank $2$ and class $2$, and is isomorphic to $UT(3,\mathbb{Z})$.

I would like to know what is the easiest such representation for the free nilpotent group $F(3,2)$, of rank $3$ and class $2$. Its group presentation is $$F(3,2)=\langle x,y,z \,\|\, [x,[x,y]],[y,[x,y]],[z,[x,y]],[x,[y,z]],[y,[y,z]],[z,[y,z]],[x,[x,z]],[y,[x,z]],[z,[x,z]] \rangle.$$

An algorithm to compute this representation was implemented in GAP, based on the work of de Graaf and Nickel [de Graaf, Willem A.; Nickel, Werner, Constructing faithful representations of finitely-generated torsion-free nilpotent groups, J. Symb. Comput. 33, No. 1, 31-41 (2002). ZBL1021.20006.] Unfortunately I am not proficient with GAP to run it myself.

In the above paper the authors state that the algorithm produces a representation for $F(3,2)$ in $UT(6,\mathbb{Z})$.

Is there a faithful representation of $F(3,2)$, over integer matrices, of degree less than $6$?

Answer 1

I think that I have found an embedding of $F(3,2)$ into ${\rm UT}(5,{\mathbb Z})$. I used a combination of GAP and Magma to do this, but here is the (putative) result in GAP.

The images of the three generators of $F(3,2)$ are

a := [
    [ 1, 0, 1, 0, -1 ],
    [ 0, 1, 0, 0, 0 ],
    [ 0, 0, 1, 0, -1 ],
    [ 0, 0, 0, 1, 0 ],
    [ 0, 0, 0, 0, 1 ]
];;
b := [
    [ 1, 0, 0, 0, 0 ],
    [ 0, 1, 0, 1, 0 ],
    [ 0, 0, 1, 0, -1 ],
    [ 0, 0, 0, 1, 0 ],
    [ 0, 0, 0, 0, 1 ]
];;
c := [
    [ 1, 0, 0, 0, 0 ],
    [ 0, 1, 1, 1, -1 ],
    [ 0, 0, 1, 1, -1 ],
    [ 0, 0, 0, 1, 0 ],
    [ 0, 0, 0, 0, 1 ]
];;

If you calculate the three commutators $[a,b]$, $[a,c]$ and $[b,c]$, you will find that they clearly generate a subgroup isomorphic to ${\mathbb Z}^3$, and they are all centralized by $a$,$b$, and $c$, which I think is enough to show that the group they generate is isomorphic to $F(3,2)$.

gap> x:=Comm(a,b);; y := Comm(a,c);; z := Comm(b,c);;
gap> Display(x); Display(y); Display(z);
[ [   1,   0,   0,   0,  -1 ],
  [   0,   1,   0,   0,   0 ],
  [   0,   0,   1,   0,   0 ],
  [   0,   0,   0,   1,   0 ],
  [   0,   0,   0,   0,   1 ] ]
[ [   1,   0,   0,   1,  -1 ],
  [   0,   1,   0,   0,   1 ],
  [   0,   0,   1,   0,   0 ],
  [   0,   0,   0,   1,   0 ],
  [   0,   0,   0,   0,   1 ] ]
[ [  1,  0,  0,  0,  0 ],
  [  0,  1,  0,  0,  1 ],
  [  0,  0,  1,  0,  0 ],
  [  0,  0,  0,  1,  0 ],
  [  0,  0,  0,  0,  1 ] ]

Answer 2

Here is how GAP can be used to construct a 6-dimensional faithful representation of $F(3,2)$.

gap> LoadPackage("Polycyclic");;
gap> LoadPackage("nq");;
gap> f:=FreeGroup(3);;
gap> x:=f.1;;y:=f.2;;z:=f.3;;
gap> rels:=[Comm(x,Comm(x,y)),Comm(y,Comm(x,y)),Comm(z,Comm(x,y)),Comm(x,Comm(y,z)),Comm(y,Comm(y,z)),Comm(z,Comm(y,z)),Comm(x,Comm(x,z)),Comm(y,Comm(x,z)),Comm(z,Comm(x,z))];
[ f1^-1*f2^-1*f1^-1*f2*f1*f2^-1*f1*f2, f2^-2*f1^-1*f2*f1*f2*f1^-1*f2^-1*f1*f2, f3^-1*f2^-1*f1^-1*f2*f1*f3*f1^-1*f2^-1*f1*f2, f1^-1*f3^-1*f2^-1*f3*f2*f1*f2^-1*f3^-1*f2*f3, f2^-1*f3^-1*f2^-1*f3*f2*f3^-1*f2*f3, 
f3^-2*f2^-1*f3*f2*f3*f2^-1*f3^-1*f2*f3, f1^-1*f3^-1*f1^-1*f3*f1*f3^-1*f1*f3, f2^-1*f3^-1*f1^-1*f3*f1*f2*f1^-1*f3^-1*f1*f3, f3^-2*f1^-1*f3*f1*f3*f1^-1*f3^-1*f1*f3 ]
gap> G:=f/rels;
<fp group on the generators [ f1, f2, f3 ]>
gap> p:=NilpotentQuotient(G);
Pcp-group with orders [ 0, 0, 0, 0, 0, 0 ]
gap> IsPcpGroup(p);
true
gap> rep:=UnitriangularMatrixRepresentation(p);
[ g1, g2, g3, g4, g5, g6 ] -> <6 6x6-matrices>

Here NilpotentQuotient is used just to convert a finitely presented group G into a nilpotent group, maybe there is a more straightforward way. p got 6 generators, I guess, they are the original generators and pairwise commutators (it's probably in the docs somewhere). Now one can look at the generators, e.g. the 1st and the 6th:

 gap> m:=Image(rep);
 <matrix group of size infinity with 6 generators>
 gap> mg:=GeneratorsOfGroup(m);;
 gap> Display(mg[1]);
 [ [   1,   0,   0,   0,   0,   0 ],
   [   0,   1,   0,   0,   0,   0 ],
   [   0,   0,   1,   0,   0,   0 ],
   [   0,   0,  -1,   1,   0,   0 ],
   [   0,   1,   0,   0,   1,   0 ],
   [   0,   0,   1,   0,   0,   1 ] ]
 gap> Display(mg[6]);
 [ [  1,  0,  0,  0,  0,  0 ],
   [  0,  1,  0,  0,  0,  0 ],
   [  0,  0,  1,  0,  0,  0 ],
   [  0,  0,  0,  1,  0,  0 ],
   [  0,  0,  0,  0,  1,  0 ],
   [  1,  0,  0,  0,  0,  1 ] ]

For better control, one can use explicit homomorphisms:

  gap> F:=FreeGroup(3);
  <free group on the generators [ f1, f2, f3 ]>
  gap> f:=NqEpimorphismNilpotentQuotient(F,2); # this gives F(3,2) as the image
  [ f1, f2, f3 ] -> [ g1, g2, g3 ]
  gap> Image(f);
  Pcp-group with orders [ 0, 0, 0, 0, 0, 0 ]
  gap> Image(f,F.3*F.1);
  g1*g3*g5
  gap> m:=UnitriangularMatrixRepresentation(Image(f));
  [ g1, g2, g3, g4, g5, g6 ] -> <6 6x6-matrices>
  gap> Image(m,Image(f,F.1));
  [ [ 1, 0, 0, 0, 0, 0 ], [ 0, 1, 0, 0, 0, 0 ], [ 0, 0, 1, 0, 0, 0 ], [ 0, 0, -1, 1, 0, 0 ], [ 0, 1, 0, 0, 1, 0 ], [ 0, 0, 1, 0, 0, 1 ] ]
  gap> Display(Image(m,Image(f,F.3*F.1)));
  [ [   1,   0,   0,   0,   0,   0 ],
    [   0,   1,   0,   0,   0,   0 ],
    [   0,   0,   1,   0,   0,   0 ],
    [   0,   0,  -1,   1,   0,   0 ],
    [   0,   1,   0,   0,   1,   0 ],
    [   0,   1,   1,   0,   1,   1 ] ]

Answer 3

The group $F(r,c)$ embeds in the $\mathbf{Q}$-group associated to the free $c$-step nilpotent Lie $\mathbf{Q}$-algebra on $r$ generators. The latter possesses a natural grading in $\{1,\dots,r\}$.

More generally if a finite-dimensional Lie algebra $g$ over a field $K$ is graded in $\mathbf{Z}$ with $g_0=\{0\}$ [this can be shown to imply $g$ nilpotent], then it has a natural faithful representation in dimension $\dim(g)+1$.

Indeed, first define $D$ as the (invertible) derivation of $g$ acting as $\times k$ on $g_k$. Let $h$ be the semidirect product $K\ltimes g$, where $1\in K$ acts as $D$. Then $h$ has trivial center, so its adjoint representation is faithful, hence is faithful in restriction to $g$. With a little effort and exponentiation one should be able to provide explicit matrices for the corresponding representation of $F(r,c)$.

This is the easiest faithful representation I can figure out, which doesn't mean it's the easiest. And doesn't mean it's the smallest one (indeed for $F(3,2)$ this yields dimension 7, while other answers provide a 5-dimensional one — for $F(r,2)$ it yields dimension $1+r+r(r-1)/2$).