Conjugacy for $p$-adic matrices of finite order - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T21:39:41Z http://mathoverflow.net/feeds/question/72163 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/72163/conjugacy-for-p-adic-matrices-of-finite-order Conjugacy for $p$-adic matrices of finite order Tim Dokchitser 2011-08-05T09:49:11Z 2011-08-08T13:24:50Z <p>Say $p$ is an odd prime, and take two matrices $A,B\in GL_n({\mathbb Z}_p)$ of finite order $m$. Is it true that they are conjugate in $GL_n({\mathbb Z}_p)$ if and only if their reductions mod $p$ are conjugate in $GL_n({\mathbb F}_p)$?</p> <p><strong>Edit:</strong> Thank you all for the answers and a very insightful discussion! As the answer is <strong>NO</strong>, it is important for me to know whether conjugacy in $GL_n({\mathbb F}_p)$ at least implies conjugacy in $GL_n({\mathbb Q}_p)$. I cannot see this from the answers, so I think I will ask this as a separate question. </p> http://mathoverflow.net/questions/72163/conjugacy-for-p-adic-matrices-of-finite-order/72172#72172 Answer by Gjergji Zaimi for Conjugacy for $p$-adic matrices of finite order Gjergji Zaimi 2011-08-05T11:43:25Z 2011-08-05T11:43:25Z <p>I don't have a counter-example at the moment but the result seems too strong to be true. Here are some partial results instead. </p> <p>Let $f\in \mathbb{Z}_p[x]$ be so that it's reduction $\pmod{p}$ doesn't have repeated roots. Two matrices $A,B\in GL_n(\mathbb{Z}_p)$, satisfying $f(A)=f(B)=0$, will be conjugate in $\mathbb{Z}_p$ provided that they are conjugate over $\mathbb{F}_p$. This is theorem 2 in <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.dmj/1077377456" rel="nofollow"> Certain matrix equations over rings of integers</a> by R.W. Davis.</p> <p>If the orders of two matrices $A,B\in GL_n(\mathbb Z/p^n\mathbb Z)$ are coprime to $p$, then they are similar in $GL_n(\mathbb Z/p^n\mathbb Z)$ if and only if their reductions are similar in $GL_n(\mathbb F_p)$. This is proved in the article "On the conjugacy of matrices over a ring of residues" by D.A. Suprunenko (<a href="http://www.ams.org/mathscinet-getitem?mr=0172886" rel="nofollow">MR0172886</a>). </p> http://mathoverflow.net/questions/72163/conjugacy-for-p-adic-matrices-of-finite-order/72177#72177 Answer by Geoff Robinson for Conjugacy for $p$-adic matrices of finite order Geoff Robinson 2011-08-05T13:22:56Z 2011-08-05T13:28:15Z <p>Here is an infinite collection of "cheating" counterexamples for $p=2$. Let $G$ be a cyclic group of order $2^n ,$ generated by $g$, say. Let $M$ be the augmentation ideal of the group ring <code>$\mathbb{Z}_{2}G,$</code> so that $M = \sum_{x \in G} \alpha_x x,$ where $\sum_{x \in G}\alpha_{x} = 0$. Regard $M$ as a (say, right) $\mathbb{Z}_{2}G$-module (of rank $2^{n}-1$, for example with $\mathbb{Z}_2$-basis <code>$\{ x - 1_G: 1 \neq x \in G \}$</code>). Note that the minimum polynomial of $g$ on $M$ is $\frac{t^{2^n}-1}{t-1}$, and note also that $g$ acts with determinant $-1$ on $M.$ Now let $V$ be a rank 1 $\mathbb{Z}_2G$-module on which $g$ acts as $-1$. Then $M \otimes V$ and $M$ have isomorphic reductions (mod 2), but are not isomorphic as $\mathbb{Z}_2G$-modules (since $g$ acts with determinant $1$ on the first, and determinant $-1$ on the second). </p> http://mathoverflow.net/questions/72163/conjugacy-for-p-adic-matrices-of-finite-order/72183#72183 Answer by F. Ladisch for Conjugacy for $p$-adic matrices of finite order F. Ladisch 2011-08-05T14:46:02Z 2011-08-08T10:36:25Z <p>While Geoff Robinson has solved the problem for $p=2$, it may be worth to point out that the answer is "No" for all primes, for the following reason: It is known (see Curtis-Reiner, Methods of Rep'n Theory, §33) that the group ring $\mathbb{Z}_p G$ has infinite representation type, if $G$ is a cyclic $p$-group of order $\geq p^3$ (or if $G$ is a non-cyclic $p$-group, but that is not relevant here). On the other hand, $\mathbb{F}_p G$ has finite representation type (for $G$ cyclic). So we find an indecomposable $\mathbb{Z}_p G$-lattice $M$ that decomposes when reduced mod $p$. <strike>On the other hand, we may lift the summands of $M/pM$ to $\mathbb{Z}_p G$-lattices and form their direct sum, $N$ (say). Then $M/pM\cong N/pN$, but $M\not\cong N$.</strike><br> EDIT: As has been poited out in the comments, the former is not correct. However, it follows from the proof of Dade's theorem given in Curtis-Reiner (33.8) that there are indecomposable, faithful $\mathbb{Z}_pG$-lattices of rank $k\left| G\right|$ for all $k$. As Alex Bartel points out in his answer, it follows from this fact that for some $k$ big enough, there must be non-isomorphic lattices of rank $k|G|$ reducing to isomorphic modules mod $p$. However, while I didn't check this, it seems to me that the indecomposable lattices of rank $k|G|$ constructed in the proof of Dade's theorem reduce to $(\mathbb{F}_pG)^k$ mod $p$, as does $(\mathbb{Z}_p G)^k$, of course. If correct, this gives concrete counterexamples, the smallest of dimension $2|G|$.<br> End EDIT.<br> More generally, the result is wrong if $p^3$ divides $m$. On the positive side, it is true when $p$ does not divide $m$. (Added later: This is elementary. Remember that $1+ p M_n(\mathbb{Z}_p) \subseteq GL_n(\mathbb{Z}_p)$, so units of $M_n(\mathbb{F}_p)$ lift to units of $M_n(\mathbb{Z}_p)$. After replacing $B$ with a conjugate, we may assume that $A\equiv B \mod p$. Then<br> <code>$$U:= \frac{1}{m} \sum_{k=0}^{m-1} A^{-k} B^k \equiv \frac{1}{m} \sum_{k=0}^{m-1} I \equiv I \mod p$$</code> which implies that $U$ is invertible. One computes that $AU=UB$, so it follows $A^U = B$.)</p> http://mathoverflow.net/questions/72163/conjugacy-for-p-adic-matrices-of-finite-order/72235#72235 Answer by Alex Bartel for Conjugacy for $p$-adic matrices of finite order Alex Bartel 2011-08-06T11:21:11Z 2011-08-06T11:47:11Z <p>I think I finally have a correct answer for arbitrary $p$.</p> <p>As F. Ladisch notes, $G=C_{p^3}$ has only finitely many indecomposable modular representations. For the following argument, I will not only need infinitely many integral representations, but I will need them to occur at reasonably regular intervals, as the rank goes up. Luckily, they do. In "<a href="http://www.jstor.org/stable/1970218?seq=1" rel="nofollow">Representations of Cyclic Groups in Rings of Integers, II</a>", Heller and Reiner exhibit indecomposable $\mathbb{Z}_p[G]$-modules of rank $kp^3$ for arbitrary $k\in \mathbb{N}$.</p> <p>Now, let us count the number of arbitrary <code>$\mathbb{F}_p[G]$</code>-modules, respectively of arbitrary <code>$\mathbb{Z}_p[G]$</code>-modules, of rank $np^3$. The former correspond to partitions of $np^3$ into summands that come from a fixed finite set of integers (possibly some repetitions), and their number is easily seen to be polynomial in $n$ (e.g. if the finite set is $A$, then a very crude upper bound on the number of partitions is $\prod_{i\in A} np^3/i$).</p> <p>On the other hand, even if we consider only direct sums of the above mentioned modules of rank $kp^3$, $k\in \mathbb{N}$, we get a number that is essentially the partition number of $n$, which is <a href="http://en.wikipedia.org/wiki/Partition_%28number_theory%29#Partition_function_formulas" rel="nofollow">exponential in $\sqrt{n}$</a>. So for sufficiently large $n$, we get much more <code>$\mathbb{Z}_p[G]$</code>-modules of rank $np^3$ than <code>$\mathbb{F}_p[G]$</code>-modules. This shows the even stronger claim that arbitrarily many non-conjugate matrices over $\mathbb{Z}_p$ can reduce to conjugate $\mathbb{F}_p$-matrices.</p> <p>It is also worth remarking that for $G=C_p$, it is true that two <code>$\mathbb{Z}_p[G]$</code>-modules are isomorphic if and only if their reductions are. That's because the only indecomposable <code>$\mathbb{Z}_p[G]$</code>-modules are the trivial module, the regular module and the $p-1$-dimensional augmentation ideal in the regular module (at the moment, I can't find a good reference, so I might edit it in later). For $G=C_{p^2}$, one could also go through the <a href="http://www.jstor.org/stable/2041351" rel="nofollow">finite classification</a> and see whether this still holds.</p> http://mathoverflow.net/questions/72163/conjugacy-for-p-adic-matrices-of-finite-order/72278#72278 Answer by Jeff Adler for Conjugacy for $p$-adic matrices of finite order Jeff Adler 2011-08-07T09:30:20Z 2011-08-07T09:30:20Z <p>Here is a proof in the case where $p \not |m$ that doesn't involve modular representations, though it does involve a result of Gelfand and Kazhdan whose proof I can't remember, so I have no idea whether it's easy or hard.</p> <p>A special case of their result is that two elements of $GL_n(\mathbb{Z}_p)$ are conjugate in that group if and only if they are conjugate in $GL_n(\mathbb{Q}_p)$. Moreover, two elements of $GL_n(\mathbb{Q}_p)$ are conjugate in that group if and only if they are conjugate in $GL_n(E)$ for every extension $E/\mathbb{Q}_p$. Thus, if we assume that the reductions $\bar A$ and $\bar B$ of $A$ and $B$ are conjugate in $GL_n(\mathbb{F}_p)$, it will be enough to show that $A$ and $B$ are conjugate in $GL_n(E)$ for some $E$. </p> <p>Let $E/\mathbb{Q}_p$ be an extension containing all $m$th roots of unity. These roots of unity, and thus the eigenvalues of $A$ and $B$, are completely determined by their images in the residue field of $E$, which are the eigenvalues of $\bar A$ and $\bar B$. That is, $A$ and $B$ have the same multi-set of eigenvalues if and only if $\bar A$ and $\bar B$ do.</p> <p>I believe that one could construct another proof from the result about $GL_n(\mathbb{Z}/p^n\mathbb{Z})$ mentioned in Gjergji's answer [at least if he really meant $GL_n(\mathbb{Z}_p/p^k\mathbb{Z})$], via some limiting process.</p> http://mathoverflow.net/questions/72163/conjugacy-for-p-adic-matrices-of-finite-order/72329#72329 Answer by Junkie for Conjugacy for $p$-adic matrices of finite order Junkie 2011-08-08T04:43:10Z 2011-08-08T13:24:50Z <p>Here is a trial proof for the question over $Q_p$.</p> <p>Write $J[f(x)^k]$ for the general Jordan form of a irreducible $f$, being $k$ identical blocks joined by 1's in general (minimal polynomial of block is $f^k$).</p> <p>Let $A$ be finite order over $Q_p$, so $A\sim\oplus J[f(x)]$ where the $f$ have $f|\Phi_m$ (cyclotomic polynomials) and finiteness implies the $f$ are irreducible (not powers).</p> <p>Note $\bar f$ determines $m$ up to $p$-powers, writing $m=up^v$ for $(u,p)=1$. Further note, if $\bar\Phi_u=\prod \bar g$ then $\bar\Phi_{up^v}=\prod\bar g^{\phi(p^v)}$, and what is more, the corresponding Jordan block to $\bar g^{\phi(p^v)}$ does not split, in other words this is the minimal polynomial. This follows since the reduction (mod $p$) of the companion matrix of $f$ is itself a companion matrix (ones above the diagonal) over a field $F_p$, and so has its minimal and characteristic polynomials equal to $\bar f=\bar g^{\phi(p^v)}$.</p> <p>So, every reduction to $\bar f$ from the $A\sim\oplus J[f]$ decomposition has $\bar f(x)=\bar g(x)^{\phi(p^v)}$ for some irreducible $\bar g|\bar\Phi_u$, that lifts to $g|\Phi_u$. What is more, $\bar A\sim\oplus J[\bar g(x)^{\phi(p^v)}]$.</p> <p>From this, $\bar A\sim\oplus J[\bar g(x)^{\phi(p^v)}]$ determines the general Jordan form of $A$ uniquely as something like $A\sim\oplus J[\Phi_{pu}^{g-part}(x^{p^{v-1}})]$. The general Jordan form classifies the conjugacy type over a field, as is $Q_p$.</p> <p>Note that, $\Phi_3\Phi_6$ and $\Phi_6^2$ give 4x4 matrices with order 6, failing for $p=2$.</p>