Let $G$ be the group of $3 \times 3$ upper unitriangular matrices with entries in $R$, a commutative ring:
$$G = \left\{ \left(\matrix{1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1}\right) \mid x,y,z \in R\right\}$$
I am asked to show that if we give $G$ the filtration:
$\omega(g) = \sup\{n \in \mathbb N \mid g \in \gamma_n(G)\}$, where the $\gamma_n(G)$ are the lower central series of $G$, then:
$(G, \omega)$ is a separated filtration and:
$grG$, the associated graded group of $(G,\omega)$ is an $R$-Lie Algebra: $RX \bigoplus RZ \bigoplus RY$, where:
$gr_1G = RX \bigoplus RY, \; gr_2G = RZ$,
and $[X,Y] = Z, [X,Z] = [Y,Z] = 0$
First of all, I considered the matrices: $X = \left(\matrix{1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}\right), Y = \left(\matrix{1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1}\right), Z = \left(\matrix{1 & 0 & 1 \\ 0 & 1 & 0 \\ 0 & 0 & 1}\right)$,
and taking the Lie Bracket to be the commutator, then we get the relations we want, which would mean that, if $X,Y,Z$ multiplicatively generate G, then:
$\gamma_1(G) = G$ by definition
$\gamma_2(G) = \langle Z \rangle$
$\gamma_3(G) = \{I\}$
Which means that:
- $(G, \omega)$ is separated as $G$ is nilpotent.
- $G_1 = \{g \in G \mid \omega(g) \geq 1\} = \langle X,Y,Z\rangle = G$
- $G_{1^+} = \{g \in G \mid \omega(g) > 1\} = G_2 = \langle Z \rangle $
- $G_3 = \{I\}$
Therefore:
- $gr_1G = G_1 / G_{1^+} = \langle X, Y\rangle$
- $gr_2G = G_2 / G_{2^+} = \langle Z \rangle $
So by the commutative properties of the associated graded group, we see that we must in fact have $gr_1G, gr_2G$ being commutative, so:
$gr_1G = RX \bigoplus RY, \; gr_2G = RZ$ as required.
That this is an $R$-Lie Algebra is, I believe, a simple check, so we are effectively done presuming we can show that $\langle X, Y, Z \rangle = G$.
In my working I noted that by multiplication on the left, using $X, Y, Z$, we can generate the subgroup of $G$ with entries taken from a subring of $R$, isomorphic to $\mathbb Z$. However, I don't know how I can extend this to include the rest of the entries from $R$ as $R$ is a general, commutative ring.
This then makes me think perhaps we need more than just $X, Y, Z $, but then there would be problems with showing that the filtration is separated.
I believe it might be possible that I am approaching this question from the wrong path, but I am unsure of how to fix it. Any help would be appreciated, thank you.
I believe I have solved this.
First, for a subset $S \subset G$, let $\langle S \rangle$ denote the subgroup of $G$ generated multiplicatively by the elements of $S$.
Now define the matrices $X_r = \pmatrix{1 & r & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}, Y_s = \pmatrix{1 & 0 & 0 \\ 0 & 1 & s \\ 0 & 0 & 1}, Z_t = \pmatrix{1 & 0 & t \\ 0 & 1 & 0 \\ 0 & 0 & 1}$ for $r,s,t \in R$.
Further, let $X_1 = X, Y_1 = Y, Z_1 = Z$, and define the additive group (also the $R$-module):
$RX = \{ rX \mid r \in R\}$ where $rX = \pmatrix{1 & r & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}$,
and define $RY, RZ$ similarly.
Then if $S = \{X_r, Y_s, Z_t \mid r,s,t \in R\}$, we have $G = \langle S \rangle$.
Additionally, if we define $(A,B) = A^{-1}B^{-1}AB$ for $A, B \in G$, then for $r,s \in R$:
$(X_r,Y_s) = \pmatrix{1 & -r & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}\pmatrix{1 & 0 & 0 \\ 0 & 1 & -s \\ 0 & 0 & 1}\pmatrix{1 & r & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}\pmatrix{1 & 0 & 0 \\ 0 & 1 & s \\ 0 & 0 & 1}$
$ = \pmatrix{1 & -r & rs \\ 0 & 1 & -s \\ 0 & 0 & 1}\pmatrix{1 & r & rs \\ 0 & 1 & s \\ 0 & 0 & 1} = \pmatrix{1 & 0 & rs \\ 0 & 1 & 0 \\ 0 & 0 & 1} = Z_{rs}$
And:
$(X_r, Z_s) = \pmatrix{1 & -r & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}\pmatrix{1 & 0 & -s \\ 0 & 1 & 0 \\ 0 & 0 & 1}\pmatrix{1 & r & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1}\pmatrix{1 & 0 & s \\ 0 & 1 & 0 \\ 0 & 0 & 1}$
$= \pmatrix{1 & -r & -s \\ 0 & 1 & 0 \\ 0 & 0 & 1}\pmatrix{1 & r & s \\ 0 & 1 & 0 \\ 0 & 0 & 1} = \pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1} = I$
$(Y_r, Z_s) = \pmatrix{1 & 0 & 0 \\ 0 & 1 & -r \\ 0 & 0 & 1}\pmatrix{1 & 0 & -s \\ 0 & 1 & 0 \\ 0 & 0 & 1}\pmatrix{1 & 0 & 0 \\ 0 & 1 & r \\ 0 & 0 & 1}\pmatrix{1 & 0 & s \\ 0 & 1 & 0 \\ 0 & 0 & 1}$
$= \pmatrix{1 & 0 & -s \\ 0 & 1 & -r \\ 0 & 0 & 1}\pmatrix{1 & 0 & s \\ 0 & 1 & r \\ 0 & 0 & 1} = \pmatrix{1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1} = I$
So we see, that:
$G_1 = \gamma_1(G) = G$, by definition,
$G_{1^+} = G_2 = \gamma_2(G) = \{Z_r \mid r \in R\}$,
$G_{2^+} = G_3 = \gamma_3(G) = \{I\}$, which tells us $G$ is nilpotent and thus the filtration is separated.
Hence:
$G_1/G_{1^+} = gr_1G = \langle \{X_r, Y_s, Z_t | r,s,t \in R\} \rangle / \langle \{Z_t \mid t \in R\}\rangle$
$\Rightarrow gr_1G \cong \left\{\pmatrix{1 & r & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1} \mid r \in R\right\} \oplus \left\{\pmatrix{1 & 0 & 0 \\ 0 & 1 & s \\ 0 & 0 & 1} \mid s \in R \right\} = RX \oplus RY $
$G_2 / G_{2^+} = gr_2G = \{Z_r \mid r \in R\} = RZ$
Hence:
$grG = gr_1G \oplus gr_2G \cong RX \oplus RY \oplus RZ $
Define now a bracket $[\cdot, \cdot]_{\lambda, \mu} : gr_\lambda G \times gr_\mu G \rightarrow gr_{\max(\lambda + \mu, 2)}G$ on homogenous elements by:
$[A + G_{\lambda^+}, B + G_{\mu^+}] = (A,B) + G_{\max (\lambda + \mu, 2) ^ + }$
and extend $R$-linearly to cover all of $gr_\lambda G \times gr_\mu G$. Combining all of these gives us a bracket:
$[\cdot, \cdot] : grG \times grG \rightarrow grG$, which we may easily check to see that it satisfies the properties of being a Lie bracket.
Additionally we see that with this bracket, the desired relations of $X,Y,Z$ are also satisfied, by the work done previously.
Thus I believe we have shown everything we wanted to show.