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In $GL(n, \mathbb{Q}_p)$, what are the orbits under conjugation of $GL(n, \mathbb{Z}_p)$?

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up vote 15 down vote accepted

The problem can be reduced to that of classifying $GL(n,\mathbf{Z}_p)$ conjugacy classes in $M(n,\mathbf{Z}_p)$. The situation for general $n$ is complicated, but for $n=2$ the problem is settled by the following.

Let $F\in M(2,\mathbf{Z}_p)$ be any matrix, let $f(x)$ be its characteristic polynomial, and let

$n(F)=\sup_i(i\in\mathbf{Z}_{\ge 0},$ $F$ mod $p^i$ is multiplication by a scalar $)$.

Then the $GL(2,\mathbf{Z}_p)$ conjugacy class of $F$ is uniquely determined by $f(x)$ and $n(F)$.

If $n(F)$ is infinite, then $F$ is a scalar matrix and therefore central.

If $n(F)$ is finite then there exists a unique integer $\lambda\in\mathbf{Z}$, with $0\le\lambda\le p^{n(F)}-1$ such that $F$ is conjugate to $\begin{pmatrix} \lambda&0\\ 0&\lambda \end{pmatrix}+p^{n(F)}\begin{pmatrix} 0&-a_0\\ 1&-a_1 \end{pmatrix}$.

Here $a_0$ and $a_1$ are the constant and linear term of the polynomial $f_0(x):=p^{-2n(F)}f(p^{n(F)}x+\lambda)$, which has coefficients in $\mathbf{Z}_p$.

We have that $p^{n(F)}$ is the index of the ring $\mathbf{Z}_p[F]$ inside the ring $R_F:=\mathbf{Q}_p[F]\cap M(2, \mathbf{Z}_p)$.

All rings are viewed as subrings of $M(2,\mathbf{Q}_p)$. We will sometime think of $F$ and of elements of $R_F$ as endomorphisms of the standard lattice $\mathbf{Z}_p^2$ inside $\mathbf{Q}_p^2$.

Proof: if $n(F)$ is infinite then there is not much to prove, therefore we assume $n(F)$ finite, and $F$ not central. The ring $R_F$ as defined above contains $\mathbf{Z}_p[F]$ with finite index, since they both are finite free $\mathbf{Z}_p$-modules of rank two. This is clear for $\mathbf{Z}_p[F]$, since $f(x)$ is the minimal polynomial of $F$, since $F$ is not central. For $R_F$ it follows from the fact that $\mathbf{Q}_p[F]\cap M(2, \mathbf{Z}_p)$ is open, compact, and non-empty in $\mathbf{Q}_p[F]$, which has rank two over $\mathbf{Q}_p$, since $F$ is not central.

The ring $R_F$ has a $\mathbf{Z}_p$-basis of the form $(1, F')$ where $F'=(a+bF)/p^{h}$, for some $a,b\in\mathbf{Z}_p$ not both divisible by $p$, and where $p^h$, with $h\ge 0$, is the index of $\mathbf{Z}_p[F]$ in $R_F$. The $p$-adic integer $b$ is a unit, for otherwise $p^{h-1}F'-b'F=a/p$, with $b'=b/p\in\mathbf{Z}_p$, would belong to $R_F$, which is not possible since $a/p$ is not a $p$-adic integer. This shows that the natural action of $F$ on $\mathbf{Z}_p^2/(p^h)$ is multiplication by $-ab^{-1}$ mod $p^h$. Therefore $h\leq n(F)$.

On the other hand, if $\lambda$ is any integer such that $F-\lambda$ is zero mod $p^{n(F)}$, then $F'':=(F-\lambda)/p^{n(F)}$ is an element of $R_F$, since $F-\lambda$ commutes with $F$ and it is divisible by $p^{n(F)}$ in $M(2,\mathbf{Z}_p)$. Therefore $n(F)\leq h$. Thus $n(F)=p^h$ and $(1, F'')$ is a $\mathbf{Z}_p$-basis of $R_F$ (since it spans a lattice of the correct index).

Now, by the maximality of $n(F)$ we see that $F''$ does not act via scalar multiplication on $\mathbf{Z}_p^2/p$. This implies that there is a $\mathbf{Z}_p$-basis $(e_1, e_2)$ of $\mathbf{Z}_p^2$ such that $F''(\mathbf{Z}_p\cdot e_1)\not\equiv \mathbf{Z}_p e_1$ mod $p$. It follows that $(e_1, F''(e_1))$ is also a $\mathbf{Z}_p$-basis of $\mathbf{Z}_p^2$.

With respect to this basis the action of $F''$ is given by a matrix of the form $\begin{pmatrix} 0&-a_0\\ 1&-a_1 \end{pmatrix}$, where $a_0$ and $a_1$ are the constant and the linear term of the characteristic polynomial of $F''$, which is $f_0(x):=p^{-2n(F)}f(p^{n(F)}x+\lambda)$ and has coefficients in $\mathbf{Z}_p$. By picking $\lambda$ in the range $0,\ldots,p^{n(F)}-1$, we see that the action of $F$ with respect to the basis $(e_1, F''(e_1))$ is that given by the statement.

Notice that this shows that $\mathbf{Z}_p^2$ is a free $R_F$-module of rank one, and classifying the action of $F$ on $\mathbf{Z}_p^2$ is roughly equivalent to finding $R_F$. I was interested exactly in this in the context of Tate modules of elliptic curves over finite fields ($F=$Frobenius). Probably there is a more conceptual/simpler proof. I would be interested to hear what you get in higher dimension. It won't be that easy, I expect. What makes this case simple is that orders of $\mathbf{Q}_p[F]$ containing $F$ are classified by the index with which $\mathbf{Z}_p[F]$ sits in them.

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Very nice answer! – Igor Rivin Mar 6 '12 at 21:32
The old trick of putting backticks ` ` around broken latex worked. – j.c. Mar 6 '12 at 21:53
For $n=3$, the similarity problem is solved in the paper Similarity classes for $3\times 3$ matrices over a principal local ring by Avni, Onn, Vaserstein and myself (see An argument to show that the general problem includes the matrix pair problem and is therefore wild is found in S. V. Nagornyi, Complex representations of the general linear group of degree three modulo a power of a prime. Zap. Nauˇcn. Sem. Leningrad. Otdel. Mat. Inst. Steklov. (LOMI), 75:143–150, 197–198, 1978. – Amritanshu Prasad Mar 8 '12 at 3:48
In your paper you mention that Appelgate-Onishi solve the conjugacy problem in $SL_n(\mathbf Z_p)$, at least algorithmically. I imagine the same should therefore be true for $GL_n(\mathbf Z_p)$ (although this doesn't seem compatible with what Vytas says below). If so, somehow conjugacy is much worse in $M_n(\mathbf Z_p)$ than in $GL_n(\mathbf Z_p)$... – fherzig Mar 8 '12 at 14:06
Thanks! I was wrong: if I understand correctly, even the matrix pair problem is solvable algorithmically in the sense that if you are given a pair $(A,B)$ of $n \times n$ matrices you can find a canonical form which uniquely characterises the conjugacy class. (See V. V. Sergeichuk, Canonical matrices for linear matrix problems, 2000 and also S. Friedland, Simultaneous Similarity of Matrices, 1983.) So it's possible that there's an algorithm that decides whether any two given elements of $M_n(\mathbf Z_p)$ are conjugate, but one can't hope for a description of all conjugacy classes. – fherzig Mar 17 '12 at 21:17

The answer to this question would also determine the conjugacy classes in $GL(n, \mathbb Z_p)$. I think this is known to be a "wild" question for $n>2$, i.e. there is no hope to find an answer. I am not sure about $n=2$. Have you tried to work this case out?

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Hi Vytas, do you have a reference for the wildness? – fherzig Mar 7 '12 at 2:03

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