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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.

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  • $\begingroup$ Sorry, can you make a bit clearer what you have already proved and what your actual question is? $\endgroup$ Commented Apr 2, 2019 at 17:17
  • $\begingroup$ @TorstenSchoeneberg how can I show that the associated graded group $grG$ is an $R$-Lie Algebra of the above form? I believe I have effectively proven it in the case that $R = \mathbb Z$, but when this is not the case I do not know what to do. $\endgroup$
    – user366818
    Commented Apr 2, 2019 at 17:26
  • $\begingroup$ But what fails in the case of general $R$? If the notation $\langle ? \rangle$ means "subgroup generated by $\lbrace ? \rbrace$", it is indeed not true in general that $\langle X,Y \rangle = G$, but you don't need that, as certainly still $G=G_1$ and $G_2 = RZ$ holds, as well as everything else -- assuming $RZ$ is short (and abused) notation for $\lbrace \pmatrix{1 & 0 &* \\0&1&0 \\0&0&1}: * \in R\rbrace$, and analogously for $RX$ and $RY$. $\endgroup$ Commented Apr 2, 2019 at 18:27
  • $\begingroup$ @TorstenSchoeneberg Here, $RZ$ was given in the question and I assumed that it meant $\{rZ \mid r \in R\}$. The problem is then that $Z$ multiplicatively only generates $\mathbb Z Z$ which is not all of $RZ$ under this interpretation. Additionally, I don't think $X,Y,Z$ multiplicatively generate all of $G$, thus the argument fails at the very start. $\endgroup$
    – user366818
    Commented Apr 2, 2019 at 20:34
  • $\begingroup$ But what is "$rZ$" supposed to mean (if not, as I suggest, by bad notation the element $\pmatrix{1 & 0 & r \\ 0 & 1 & 0 \\ 0 & 0 & 1}$)? There is no obvious $R$-action on $G$. $\endgroup$ Commented Apr 2, 2019 at 22:45

1 Answer 1

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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.

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