# Showing the associated graded group of a separated, filtered group $G$ is an $R$-Lie Algebra

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.

• Sorry, can you make a bit clearer what you have already proved and what your actual question is? – Torsten Schoeneberg Apr 2 at 17:17
• @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. – user366818 Apr 2 at 17:26
• 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 $RY$. – Torsten Schoeneberg Apr 2 at 18:27
• @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. – user366818 Apr 2 at 20:34
• 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 $G$. – Torsten Schoeneberg Apr 2 at 22:45

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.