Conjugation in GL(n) (p-adic setting) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T06:37:15Z http://mathoverflow.net/feeds/question/90370 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/90370/conjugation-in-gln-p-adic-setting Conjugation in GL(n) (p-adic setting) Marc Palm 2012-03-06T15:33:58Z 2012-07-30T11:37:31Z <p>In $GL(n, \mathbb{Q}_p)$, what are the orbits under conjugation of $GL(n, \mathbb{Z}_p)$?</p> http://mathoverflow.net/questions/90370/conjugation-in-gln-p-adic-setting/90380#90380 Answer by vytas for Conjugation in GL(n) (p-adic setting) vytas 2012-03-06T16:29:42Z 2012-03-06T16:29:42Z <p>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?</p> http://mathoverflow.net/questions/90370/conjugation-in-gln-p-adic-setting/90397#90397 Answer by Tommaso Centeleghe for Conjugation in GL(n) (p-adic setting) Tommaso Centeleghe 2012-03-06T21:21:18Z 2012-07-30T11:37:31Z <p>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.</p> <p>Let $F\in M(2,\mathbf{Z}_p)$ be any matrix, let $f(x)$ be its characteristic polynomial, and let</p> <p>$n(F)=\sup_i(i\in\mathbf{Z}_{\ge 0},$ $F$ mod $p^i$ is multiplication by a scalar $)$.</p> <p>Then the $GL(2,\mathbf{Z}_p)$ conjugacy class of $F$ is uniquely determined by $f(x)$ and $n(F)$.</p> <p>If $n(F)$ is infinite, then $F$ is a scalar matrix and therefore central.</p> <p>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 <code>$\begin{pmatrix} \lambda&amp;0\\ 0&amp;\lambda \end{pmatrix}+p^{n(F)}\begin{pmatrix} 0&amp;-a_0\\ 1&amp;-a_1 \end{pmatrix}$</code>.</p> <p>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 $\mathbf{Z}_p$.</p> <p>We have that $p^{n(F)}$ is the index of the ring $\mathbf{Z}_p[F]$ inside the ring $R_F:=\mathbf{Q}_p[F]\cap M(2, \mathbf{Z}_p)$.</p> <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 $\mathbf{Q}_p^2$.</p> <p>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.</p> <p>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 $h\leq n(F)$.</p> <p>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).</p> <p>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 $\mathbf{Z}_p^2$.</p> <p>With respect to this basis the action of $F''$ is given by a matrix of the form <code>$\begin{pmatrix} 0&amp;-a_0\\ 1&amp;-a_1 \end{pmatrix}$</code>, 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.</p> <hr> <p>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.</p>