Conjugacy for p-adic matrices of finite order II

Question: 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 if their reductions mod $p$ are conjugate in $GL_n({\mathbb F}_p)$ then they are conjugate in $GL_n({\mathbb Q}_p)$?

This is a follow-up to the question whether $GL_n({\mathbb F}_p)$-conjugacy implies $GL_n({\mathbb Z}_p)$-conjugacy, for which the answer is "No".

For $p=2$ this is not true, although I would be very much interested to know whether conjugacy mod 4 implies conjugacy over ${\mathbb Q}_2$.

(Note that if the answer is "Yes", then by character theory any two representations $\rho, \rho': G\to GL_n({\mathbb Z}_p)$ that are equivalent mod $p$ are equivalent in $GL_n({\mathbb Q}_p)$.)

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This seems relevant: mathoverflow.net/questions/69578/… – Igor Rivin Aug 9 '11 at 18:53
I think my answer, for the other question (I misread) is now correct for the direction of $F_p$ conjugacy implying $Q_p$ conjugacy. Namely, given a general Jordan block $J$ of dimension $d$ in the decomposition for $A$, this $J$ is a companion matrix (ones above the diagonal), which remains the case upon mod $p$ reduction, and so the minimal polynomial of the reduced block $\bar J$ is also of degree $d$, so that this block does not split. This means (referring to the argument there) that $\bar J$ determines $J$, so two distinct $J$ and $J'$ cannot reduce the same mod $p$. – Junkie Aug 10 '11 at 4:40
@Junkie: This is indeed the direction that I want! But can you expand your argument a bit (and possibly post it as an answer)? I think you are using that every ${\mathbb Q}_p$-conjugate of your matrix in $GL_n({\mathbb Z}_p)$ has the same ${\bar J}$. I cannot see why this must be true. But if it is, I agree, this proves the statement. – Tim Dokchitser Aug 10 '11 at 7:32

In the case of a cyclic $p$-group $G$, $p$ odd, it still seems unlikely to me that the reduction (mod $p$) of a $\mathbb{Z}_pG$-lattice $L$ will determine the isomorphism type of $M = L \otimes_{\mathbb{Z}_{p}} \mathbb{Q}_p$ in general. I have not found a counterexample so far, though they may well exist in the literature. However, I will record a couple of comments in case they are of use to someone else, (possibly in a positive direction if my intuition is wrong).
When $G = \langle u \rangle$ is cyclic of order $p^n$, ($p$ still odd), there are clearly just $n+1$ isomorphism types of irreducible $\mathbb{Q}_pG$-modules. These are the trivial module, and the representations obtained by representing $u$ respectively as the companion matrix of the irreducible polynomial $\frac{x^{p^m}-1}{x^{p^{m-1}}-1}$ for $1 \leq m \leq n.$
Let's label these as $V_0,V_1, \ldots,V_n$, where $V_0$ denotes the trivial module. Since $\mathbb{Q}_p$ has characteristic zero, the isomorphism type of $M$ is determined by the character $\chi$ it affords. As was known to E. Artin and R. Brauer, and is easily checked, in this situation, knowledge of the character afforded by $m$ is equivalent to knowing the dimension of the fixed-point space of $x^{p^j}$ on $M$ for each $j$ with $0 \leq j \leq n.$ Clearly $\chi$ determines the dimension of these fixed point spaces. On the other hand, these dimensions determine $\chi$ inductively because of the fact that $p^n \langle \chi, 1 \rangle = p^{n-1}\langle {\rm Res}^{G}_{\langle u^{p} \rangle}(\chi),1 \rangle + (p^n - p^{n-1})\chi(u),$ since $\chi$ is rational valued.
Hence (for $p$ odd and a cyclic $p$-group $G$), the question is equivalent to " can we determine the rank of the fixed sublattice $L^{G}$ solely from knowledge of the reduction (mod $p$) of $L$?" For if we were working inductively, we could assume that we knew how the character restricted to every proper subgroup of $G$, since we certainly know the reduction $(mod$p$)$ of all these restrictions. Hence, as above, if we can determine the rank of $L^{G}$, the character is detemined completely, while if we know the character, we certainly know the rank of $L^{G}$.
However, I will give an "algorithmic" description of how to proceed which might be helpful in the context of the original question (at least for cyclic $p$-groups, $p$ odd). As was discussed in the previous version of this question, it is possible to solve this problem in the case of a cyclic group of order $p$, even at the integral level. The description of what to do is easy. If $G$ is a cyclic group of order $p$, there are three isomorphism types of indecomposable $\mathbb{Z}_{p}G$-lattices. If $L$ is a $\mathbb{Z}_pG$-lattice, then $L$ has $a$ trivial indecomposable summands, $b$ indecomposable summands of rank $p-1$, and $c$ (projective) indecompsable summands of rank $p$. Here, $a$, $b$ and $c$ are respectively the number of Jordan blocks of size $1$,$p-1$ and $p$ in the reduction (mod p).$Now suppose that$G = \langle u \rangle$is cyclic of order$p^n$($p$odd) and we have a$\mathbb{Z}_p G$-lattice$L$. Let$v$denote a generator of the unique subgroup of order$p$of$G$. We can determine the rank of the fixed sublattice$L^{\langle v \rangle}$, as just discussed. This is a pure submodule of the original$\mathbb{Z}_pG$-module. The minimum polynomial of$u$on the quotient lattice$L/L^{\langle v \rangle}$is$\Phi_{n}(x) = \frac{x^{p^n}-1}{x^{p^{n-1}}-1},$because of the choice of$v$. Let$N = \{ w \in L: w \Phi_n(x) = 0 \},$a pure submodule of$L.$Then the rank of$N$is rank($L$) - rank($L^{\langle v \rangle}),$which is determined by the reduction (mod$p$) of$L.$Now$N + L^{\langle v \rangle}$is a full (ie maximal rank) submodule of$L,$and the quotient of$L$by this submodule is a torsion module. But we can still form the quotient module$L/N$. Now$v$acts trivially on$L/N$by construction, so$L/N$is "really" a module for the smaller group$G/\langle v \rangle$, and the rank of the$G$-fixed points on this module is the same as the rank of the$G$-fixed points of$L$. It might appear, then, that we are finished by an inductive argument in our quest to find the rank of the$G$-fixed points on$L.$The issue, though, is in what sense we can claim to know the reduction (mod$p$) of$L/N$, and in what sense the submodule$N$itself (rather than just its rank) is determined by the reduction (mod$p$) of$L.$This is why I do not consider that this is a proper solution to the question. - No. Let$R = \mathbb{Z}_p[C_p]$and consider the$R$-modules$M = R$and$N = \mathfrak{m}$, the maximal ideal of the local ring$R$. Let$A,B$the matrices in$GL_p(\mathbb{Z}_p)$giving the action of a generator of the cyclic group$C_p$on$M$and$N$respectively. Then$M \otimes_{\mathbb{Z}_p} \mathbb{Q}_p \cong \mathbb{Q}_p[C_p] \cong N \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$as$\mathbb{Q}_p[C_p]$-modules, so$A$is conjugate to$B$in$GL_p(\mathbb{Q}_p)$. Write$R \cong \mathbb{Z}_p[T] / \langle (1 + T)^p - 1\rangle$and let$f$be the image of$\frac{1}{T}( (1 + T)^p - 1)$in$R$; then it is not difficult to see that$N \cong R/\langle T\rangle \oplus R / \langle f\rangle$as$R$-modules. This means that$N/pN$is not a cyclic$R/pR$module. However$M/pM = R/pR$is a cyclic$R/pR$module, so$M/pM$is not isomorphic to$N/pN$over$\mathbb{F}_p[C_p]$. This is because the$C_p$-coinvariant spaces of$M/pM$and$N/pN$are$1$and$2$-dimensional over$\mathbb{F}_p$, respectively. Hence the reductions of$A$and$B$cannot be conjugate in$GL_p(\mathbb{F}_p)$. - (only if there's a short answer, ) what happens if$m$is prime to$p$? – Dror Speiser Aug 9 '11 at 20:36 Well then the answer is "yes" because the algebra$\mathbb{Z}_p[C_m]$splits up into a direct sum of blocks corresponding to the$Gal(\overline{\mathbb{F}_p} / \mathbb{F}_p)$-orbits of linear characters of$C_m$. So the$\mathbb{F}_p$-representation theory and the$\mathbb{Q}_p$-representation theory of$C_m$is the same if$p \nmid m$. In fact this is true for any finite group$G$of order coprime to$p$. – Konstantin Ardakov Aug 9 '11 at 21:43 Can you write the matrices$A$and$B$down explicitly for$p=3$? – Junkie Aug 10 '11 at 4:07 Yes: for general$p$, you take a matrix for a$p$-cycle as one of the matrices, and the direct sum of the companion matrix for$1 + x + \ldots + x^{p-1}$with the$1 \times 1$matrix$(1)$. – Geoff Robinson Aug 10 '11 at 7:12 Sorry, Konstantin, the question was meant to ask whether conjugacy over${\mathbb F}_p$implies that over${\mathbb Q}_p$. I fixed it now. – Tim Dokchitser Aug 10 '11 at 7:25 Edit: Sorry, just realized this is in in the wrong direction. What was asked for was an example of matrices over$\mathbb Z_p$which are conjugate over$\mathbb F_p$but not over$\mathbb Q_p$; I swapped$\mathbb Q_p$and$\mathbb F_p$, but then of course the assertion is clear anyway (I'll look at it again later). After looking at Geoff's answer and §34C in "Curtis and Reiner: Methods of Representation Theory, Volume I" which deals with integral representations of$C_{p^2}$, I came up with a counterexample (I would recommend to use GAP or Maple to verify it; doing the computations by hand would be insane). So, here is the counterexample: Consider the following two matrices in$\mathbb Q_3^{10\times 10}$: $$A=\left(\begin{array}{rrrrrrrrrr} 1&0&0&0&1&0&0&0&0&0\newline 0&0&0&1&1&0&0&0&0&0\newline 0&1&0&0&1&0&0&0&0&0\newline 0&0&1&0&1&0&0&0&0&0\newline 0&0&0&0&0&0&0&0&0&-1\newline 0&0&0&0&1&0&0&0&0&0\newline 0&0&0&0&0&1&0&0&0&0\newline 0&0&0&0&0&0&1&0&0&-1\newline 0&0&0&0&0&0&0&1&0&0\newline 0&0&0&0&0&0&0&0&1&0\newline \end{array}\right)$$ and $$B = \left(\begin{array}{rrrrrrrrrr}% 1&0&0&0&0&0&0&0&0&0\newline 0&0&0&1&0&0&0&0&0&0\newline 0&1&0&0&0&0&0&0&0&0\newline 0&0&1&0&0&0&0&0&0&0\newline 0&0&0&0&0&0&0&0&0&-1\newline 0&0&0&0&1&0&0&0&0&0\newline 0&0&0&0&0&1&0&0&0&0\newline 0&0&0&0&0&0&1&0&0&-1\newline 0&0&0&0&0&0&0&1&0&0\newline 0&0&0&0&0&0&0&0&1&0\newline \end{array}\right)$$ Then$A$and$B$are conjugate over$\mathbb Q_3$but their reductions to$\mathbb F_3$are not conjugate. Moreover, both have finite order (their order is$9$). To see that their reductions mod 3 are not conjugate just compute the rank of$A-\textrm{id}$and$B-\textrm{id}$in$\mathbb F_3^{10\times 10}$. The rank is 8 in the first case and$7$in the second, so they cannot be conjugate. To see that they are conjugate over$\mathbb Q_3$just compute minimal and characteristic polynomial. The minimal polynomial is$x^9-1$in both cases, which implies that the matrices are semisimple. Therefore they are conjugate if and only if their characteristic polynomials coincide. But the characteristic polynomial is$(x-1)(x^9-1)$in both cases. - Let me try this. I think that the answer this time is positive. Step 1: We will first reduce to$p$power order. Let$M$,$M'$be matrices over$\mathbb{F}_p$of order$p^na$, where$p\nmid a$. Then,$M$can be uniquely written as a product of two commuting matrices of orders$p^n$and$a$, respectively,$M=NL$, and similarly$M'=N'L'$. Since this presentation as a product of commuting matrices is unique and since conjugation preserves orders, it follows that$M$is conjugate to$M'$if and only if$N$is conjugate to$N'$and$L$is conjugate to$L'$. Also, if$M$and$M'$are reductions of$\mathbb{Z}_p$matrices and so are the respective factors, then the lifts of$L$and$L'$are conjugate if and only if$L$and$L'$themselves are, since they have order coprime to$p$(see Gjergji Zaimi's answer to part I of this question). So the conjugacy of$M$and$M'$lifts if and only if that of$N$and$N'$lifts. Step 2: So now, let$G=C_{p^n}$. As Geoff notes, the non-trivial irreducible$\mathbb{Q}_p$representations$\rho_k$of$G$are of dimensions$p^k-p^{k-1}$,$1\leq k\leq n$, each being the direct sum of one-dimensional$\bar{\mathbb{Q}}_p$-representations that send a fixed generator of$G$to a primitive$p^k$-th root of unity. Fix a lattice in such a representation. I claim that its reduction is indecomposable. This will immediately imply that the only way the reductions of sums of such things can be conjugate is if the original representations were conjugate over$\mathbb{Q}_p$. Let$M$be the matrix associated to a generator of$G$under$\rho_k$. The characteristic polynomial of$M$over$\mathbb{Q}_p$is then$\frac{x^{p^k}-1}{x^{p^{k-1}}-1}$. Clearly, this polynomial must still kill the reduction$\bar{M}$of$M$modulo$p$. However, since$p$is totally ramified in the$p^k$-th cyclotomic extension, the same polynomial is still irreducible over$\mathbb{F}_p$, so it is also the minimal polynomial of$\bar{M}$. Thus,$\bar{M}$cannot be block diagonalisable, since otherwise its minimal polynomial would be the product of the min polys of the blocks, QED. It is worth noting that, although in general the isomorphism class of the reduction of a representation depends on the choice of lattice (note that talking up to semi-simplification is no good in this context), it follows from the above argument, and from Higman's theorem that there is exactly one indecomposable$\mathbb{F}_p[G]$-module of dimension$k$for each$1\leq k\leq p^n$, that in this particular case, the reduction is independent of the choice of lattice. - @ Alex: I am not completely convinced by this argument. It hinges on what you mean by "a sum of such things". A general lattice is not necessarily a direct sum of such lattices (consider for example a projective indecompsable lattice), so I am not sure what you are intending here. – Geoff Robinson Aug 11 '11 at 17:42 @Geoff I think you are right. I am only proving that direct sums of lattices sitting inside the irreducible$\mathbb{Q}_p[G]$-modules cannot have conjugate reductions, unless the modules are conjugate, but that is of course much too weak. – Alex B. Aug 11 '11 at 17:49 If I was looking for a counterexample, I'd cook up two explicit non-isomorphic$\mathbb{Z}_p[C_{p^3}]$-modules$M$and$N$(as in your answer to the first question) with$M/pM$isomorphic to$N/pN$and see if they become isomorphic over$\mathbb{Q}_p$. This should be doable in theory. – Konstantin Ardakov Aug 12 '11 at 8:20 Admittedly, my solution to the first question is not terribly explicit. The other problem in trying to find such a counterexample is that the reductions might not be isomorphic for the chosen$M$and$N$, but might be isomorphic for some other choices of lattice in the same$\mathbb{Q}_p[G]\$-modules. – Alex B. Aug 12 '11 at 9:03