2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Sorry, can you make a bit clearer what you have already proved and what your actual question is? $\endgroup$ – Torsten Schoeneberg Apr 2 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 Apr 2 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 $RY$. $\endgroup$ – Torsten Schoeneberg Apr 2 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 Apr 2 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 $G$. $\endgroup$ – Torsten Schoeneberg Apr 2 at 22:45
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.