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Equivalently, in the Lie Algebra $M(2,q)$, how can I show that there are precisely $q^2$ solutions M to $[M,X]=0,$ where $X$ is a non central element of $\mathop{GL}(2,q)$, where $q$ is an odd prime?

It is easy to show that there are at least $q^2$ solutions and less than $q^4$, as the set of solutions forms a Lie subalgebra of $M(2,q)$ with the identity matrix and $X$ itself in this set of solutions, so the dimension of this Lie subalgebra is at least $2$. Assuming the solution set is all of $M(2,q)$ leads to a contradiction that $X$ is non-central, as substituting the matrix of 1s for $M$ and then the matrix with 1s in the top row and 0s in the bottom row leads to $X$ being central in $M(2,q)$. So all that remains is to show that dimension of the subalgebra of solutions cannot be $3$.

Edward ffitch
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    What is the relationship between $p$ and $q?$ – Igor Rivin Dec 29 '13 at 18:22
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    $M(2,q)$ is the Lie Algebra of $2 \times 2$ matrices over $\mathbb F_q$? – Patrick Da Silva Dec 29 '13 at 18:25
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    Yes, I mean the Lie Algebra of 2x2 matrices over F_{q}. It is easy to show that there are at least q^{2} solutions and less than q^{4}, as the set of solutions forms a Lie subalgebra of M(2,q) with the identity matrix and X itself in this set of solutions. All that remains is to show that its dimension cannot be 3. – Edward ffitch Dec 29 '13 at 18:28
  • @Edwardffitch : MSE users generally appreciate when the people asking questions add what they have tried in their questions so that it is both easier for you and for us to help you. You should add that previous comment to your question! – Patrick Da Silva Dec 29 '13 at 18:37
  • My general idea is that if a question asks something for $M(2,q)$ (or a general linear algebra problem with $2 \times 2$ matrices), it is a bit boring to try to be smart because the very smart argument will often take longer than the brute force approach to find. I gave you the brute force approach answer. – Patrick Da Silva Dec 29 '13 at 19:15

2 Answers2

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(Look at page p68-69 in Fulton, Harris Representation Theory) for an explanation of what $\epsilon$ is. http://www.math.ucsb.edu/~bigelow/math227b/fultonharris.pdf

Proposition: $p$ is an odd prime. Let $X \in GL_{2}(\mathbb{Z}_{p})$, $X \notin Z(GL_{2}(\mathbb{Z}_{p}))$, $M \in M_{2}(\mathbb{Z}_{p})$. Then there are precisely $p^{2}$ solutions to $MX-XM=0$. Equivalently, in the Lie algebra $M_{2}(\mathbb{Z}_{p})$ with Lie bracket $[\cdot,\cdot]$ the Lie subalgebra $Y$ of solutions $M$ to $[M,X]=0$ has dimension $2$.

Proof: It is routine to check that these statements are equivalent and that $Y$ is a Lie subalgebra of the Lie algebra $M_{2}(\mathbb{Z}_{p})$. Observe also that for $X,M$ as above and $P \in GL_{2}(\mathbb{Z}_{p})$\, $[M,X]=0 \Longleftrightarrow [P^{-1}MP,P^{-1}XP]=0$. So the dimension of $Y$ is invariant under conjugation of $X$, because if $M_{1},M_{2},M_{3},M_{4}$ is a basis of $M_{2}(\mathbb{Z}_{p})$, then so is $P^{-1}M_{1}P,P^{-1}M_{2}P,P^{-1}M_{3}P,P^{-1}M_{4}P$. So it suffices to check that the statement holds for a representative $X$ from each of the non-central conjugacy classes of $GL_{2}(\mathbb{Z}_{p})$. We use the representatives given in the table on page p68-69 in Fulton, Harris Representation Theory, and have 3 cases:

Case 1: $X=b_{x}=\begin{pmatrix} x & 1 \\ 0 & x \end{pmatrix}$, $x \neq 0$. We have a basis of $M_{2}(\mathbb{Z}_{p})$ given by: \begin{align*}A= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} & B=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} & C=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} & D=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}\end{align*} Now: \begin{align*}[A,X]&= 0 \\ [B,X] & = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} x & 1 \\ 0 & x \end{pmatrix} - \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix} x & 1 \\ 0 & x \end{pmatrix} = \begin{pmatrix} 0 & 2 \\ 0 & 0 \end{pmatrix} \neq 0 \\ [C,X]& = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}\begin{pmatrix} x & 1 \\ 0 & x \end{pmatrix} - \begin{pmatrix} x & 1 \\ 0 & x \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix} = \begin{pmatrix} 0 & x \\ 0 & 0\end{pmatrix}-\begin{pmatrix} 0 & x \\ 0 & 0\end{pmatrix} = 0 \\ [D,X]&=\begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix} x & 1 \\ 0 & x \end{pmatrix}-\begin{pmatrix} x & 1 \\ 0 & x \end{pmatrix}\begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix} = \begin{pmatrix} -1 & 0 \\ 0 & 1\end{pmatrix} \neq 0\end{align*} So $A,C \in Y$ and $dim(Y)\geq 2$. Note that $[B,X],[D,X]$ are linearly independent in $M_{2}(\mathbb{Z}_{p})$, so for $M=\alpha A+\beta B+ \gamma C+ \delta D \in M_{2}(\mathbb{Z}_{p})$, $[M,X]=0\Longleftrightarrow \beta=\delta=0$, so $dim(Y)=2$.

Case 2: $X=c_{x,y}=\begin{pmatrix} x & 0 \\ 0 & y \end{pmatrix}$. $[A,X], [B,X]=0$ in this case. We have: \begin{align*}[C,X]&=\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}\begin{pmatrix} x & 0 \\ 0 & y \end{pmatrix}-\begin{pmatrix} x & 0 \\ 0 & y \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix} = \begin{pmatrix} 0 & y-x \\ 0 & 0\end{pmatrix} \neq 0 \\ [D,X]&= \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix}\begin{pmatrix} x & 0 \\ 0 & y\end{pmatrix}-\begin{pmatrix} x & 0 \\ 0 & y\end{pmatrix}\begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix} = \begin{pmatrix} 0 & 0 \\ x-y & 0\end{pmatrix} \neq 0\end{align*} So $A,B \in Y$ and $dim(Y)\geq 2$. Note that $[C,X],[D,X]$ are linearly independent in $M_{2}(\mathbb{Z}_{p})$, so for $M=\alpha A+\beta B+ \gamma C+ \delta D \in M_{2}(\mathbb{Z}_{p})$, $[M,X]=0\Longleftrightarrow \gamma=\delta=0$, so $dim(Y)=2$ in this case.

Case 3: $X=d_{x,y}=\begin{pmatrix} x & \epsilon y \\ y & x \end{pmatrix}$ For this case, we use a slightly different basis for $M_{2}(\mathbb{Z}_{p})$, given by:\begin{align*}A= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} & B=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} & C=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} & D=\begin{pmatrix} 0 & \epsilon \\ 1 & 0 \end{pmatrix}\end{align*} We have:\begin{align*}[A,X]&=0 \\ [B,X]&=\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix} x & \epsilon y \\ y & x \end{pmatrix}-\begin{pmatrix} x & \epsilon y \\ y & x \end{pmatrix}\begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix} = \begin{pmatrix}0 & 2\epsilon y \\ -2y & 0\end{pmatrix} \neq 0 \\ [C,X]&=\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}\begin{pmatrix} x & \epsilon y \\ y & x \end{pmatrix}-\begin{pmatrix} x & \epsilon y \\ y & x \end{pmatrix}\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}=\begin{pmatrix} y & 0 \\ 0 & -y\end{pmatrix} \neq 0 \\ [D,X]&=\begin{pmatrix} 0 & \epsilon \\ 1 & 0 \end{pmatrix}\begin{pmatrix} x & \epsilon y \\ y & x \end{pmatrix}-\begin{pmatrix} x & \epsilon y \\ y & x \end{pmatrix}\begin{pmatrix} 0 & \epsilon \\ 1 & 0 \end{pmatrix}=0\end{align*}So $A,D \in Y$ and $dim(Y)\geq 2$. Note that $[B,X],[C,X]$ are linearly independent in $M_{2}(\mathbb{Z}_{p})$, so for $M=\alpha A+\beta B+ \gamma C+ \delta D \in M_{2}(\mathbb{Z}_{p})$, $[M,X]=0\Longleftrightarrow \beta=\gamma=0$, so $dim(Y)=2$ in this case, and the statement holds in all 3 cases. $\square$

Edward ffitch
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  • Yes, when I tried to attack the conjugacy classes that was essentially my idea. Good work! I like these kinds of questions, where the answerer does half the work and OP does the rest :) this way you learn a lot! +1. – Patrick Da Silva Jan 04 '14 at 03:29
  • Note : You could've just mentioned that $\varepsilon$ was a generator for the group of units of $\mathbb Z_p$ instead of referring to Fulton-Harris... and for the record, in general $\mathbb Z_p$ is a very bad notation for $\mathbb Z / p\mathbb Z$ because it confuses with the ring of $p$-adic integers $\mathbb Z_p$ and the localization of $\mathbb Z$ at the prime ideal $(p)$, which can also be denoted by $\mathbb Z_p$ for short-hand. I am usually very explicit before using these notations. – Patrick Da Silva Jan 04 '14 at 03:32
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Write $M(2,q) = \langle I, J, D, E \rangle_{\mathbb F_q}$ where $$ I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}, \quad J = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}, \quad D = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \quad E = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}. $$ Since this problem is invariant under a change of basis (I mean the basis of $\mathbb F_q^2$, not of $M(2,q)$), we can assume without loss of generality that $X = J$ (note that $J$ is not central and that $[M,X] = 0$ if and only if $[PMP^{-1},PXP^{-1}] = 0$ for $P \in \mathop{GL}(2,q)$, so the "without loss of generality" makes sense). You can explicitly compute the Lie brackets $[J,I]=0$, $[J,J] = 0$, $[J,D] = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$ and $[J,E] = \begin{bmatrix} 0 & 2 \\ 2 & 0 \end{bmatrix}$ to notice that the subspace of matrices $Y$ satisfying $[J,Y]=0$ has dimension $2$.

Not exactly a smart argument, but it works.

Hope that helps,

Patrick Da Silva
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  • Hm. Apparently this answer is wrong ; too many conjugacy classes in $\mathop{GL}(2,q)$! Argh... I'll leave it there as something to think about... Maybe up to scaling we can kill some conjugacy classes and this argument still works, but now I'm not sure. – Patrick Da Silva Dec 29 '13 at 19:24
  • I can confirm that this argument works perfectly for the 3 types of conjugacy classes of GL(2,q) given in Fulton and Harris's book Representation Theory. All you have to do is to modify your initial basis for M(2,q) for each case. I was under the impression that the brute force approach would be a lot more work than it actually was. Thanks again! – Edward ffitch Dec 30 '13 at 01:10
  • Thanks for actually pulling it through, if it's not too much work I'd like to see the details! I looked at the conjugacy classes and I got a bit overwhelmed on wikipedia because some classes looked a bit vague in their description. http://groupprops.subwiki.org/wiki/Element_structure_of_general_linear_group_of_degree_two_over_a_finite_field – Patrick Da Silva Dec 30 '13 at 05:39
  • I have at last supplied my answer Patrick. Sorry for the wait! – Edward ffitch Jan 04 '14 at 02:12