# $\text{mod} \, p^2$ trace identity

Let $p$ be a prime, and let $\text{GL}_n \big( \Bbb{Z} / p^2 \Bbb{Z} \big)$ be the group of $n \times n$ invertible matrices over the ring $\Bbb{Z} / p^2 \Bbb{Z}$. Does there exist a positive integer $N$ depending on $p$ and $n$ such that the following identity is valid for any matrix $g$ in $\text{GL}_n \big( \Bbb{Z} / p^2 \Bbb{Z} \big)$ ?

$$\text{tr} \big( g^N \big) - n \ = \ \text{det}^{p-1}(g) - 1$$

• this ought to be viewed as a $p$-analogue of the following trace identity for the $\text{GL_n}$-loop group, namely: $$\text{tr} \big( g^{-1} \, {dg \over {dt}} \big) \ = \ {d \over {dt}} \text{log} \, \text{det}(g)$$ – A. Leverkuhn Dec 9 '13 at 2:46
• also when $n=2$ the value of $N$ should be $q^2 -1$. – A. Leverkuhn Dec 9 '13 at 2:52
• ... sorry, that should be a $p^2 -1$ and not a $q^2 -1$ – A. Leverkuhn Dec 9 '13 at 3:15

Sorry, this is broken! Even when $n=2$, this doesn't work. Take $p=2$ and $A=\begin{pmatrix} 1 & 2 \\ 1 & 1 \end{pmatrix}$. For all $N$, $Tr(A^N) \equiv 2 \bmod 4$. (Specifically, it is OEIS A002203.) And $\det A=-1$. So $Tr(A^N) - n \equiv 2-2 \equiv 0 \bmod 4$ and $\det(A)^{p-1}-1 \equiv -1 -1 \equiv 2 \bmod 4$.

The proof below works as long as the eigenvalues of $A$ are unramified over $\mathbb{Z}_p$. At first I thought that tame ramification might also be enough (which would happen when $n<p$), but it doesn't. Take $p=5$ and $$A=\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 5 & 0 & 1 \end{pmatrix}.$$ Then $\det(A)^{p-1} \equiv 21 \bmod 25$ but $Tr(A^N)$ is $3$, $13$ or $18 \bmod 25$ for all $N$. In general, if $$A=\begin{pmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ p & 0 & 1 \end{pmatrix},$$ I get $Tr A^n \equiv 3+p 3 \binom{n}{3} \bmod p^2$ and $\det(A)^{p-1} \equiv 1-p \bmod{p^2}$. There is no particular reason that there has to be a solution to $3 \binom{n}{3} = -1$ modulo $p$ and, whenever there isn't, we lose.

I'll leave the previous answer standing below, as it explains why this works in the unramified case.

$\def\ZZ{\mathbb{Z}}$Yes, we can take $N = p^{n!}-1$.

Given $A \in GL_n(\ZZ/p^2)$, lift $A$ arbitrarily to $\tilde{A} \in GL_n(\ZZ_p)$. Let $\tilde{\lambda}_i$ be the eigenvalues of $\tilde{A}$; they live in the ring of integers of some finite extension of $\mathbb{Q}_p$. Let $\bar{A}$ be the reduction of $\tilde{A}$ modulo $p$ and $\bar{\lambda}_i$ be the reduction of $\tilde{\lambda_i}$ modulo $p$. Then $\bar{\lambda}_i$ is in $\mathbb{F}_{p^d}$ for some $d \leq n$. Moreover, since $\tilde{A}$ is invertible over $\ZZ_p$, we know that $\bar{\lambda}_i \neq 0$.

Let $\tau_i$ be the Teichmuller lift of $\bar{\lambda_i}$. So $\tau_i^N=1$. THIS IS THE BROKEN STEP: Write $\tilde{\lambda}_i = \tau_i (1+p \alpha_i)$. The left hand side of your equation is $$\sum_i \tau_i^N (1+p \alpha_i)^N -n = \sum_i (1+p \alpha_i)^N -n \equiv n + p N \sum_i \alpha_i -n \equiv -p \sum \alpha_i \bmod p^2.$$

On the right hand side, notice that $\prod \tau_i$ must be the Teichmuller lift of $\prod \bar{\lambda}_i = \det(\bar{A})$. Since $\det(\bar{A}) \in \mathbb{F}_p^{\ast}$, this shows that $\left( \prod \tau_i \right)^{p-1}=1$. So your right hand side is $$\prod \tau_i^{p-1} (1+p \alpha_i)^{p-1}-1 = \prod (1+p \alpha_i)^{p-1}-1 \equiv -p \sum \alpha_i \bmod p^2.$$

• ok --- wow ! is this something that's already known ? best, a. leverkuhn – A. Leverkuhn Dec 9 '13 at 15:45
• I hadn't seen it before, but that doesn't mean much. It seem to me like the kind of formula that is easier to prove than to conjecture. – David E Speyer Dec 9 '13 at 17:18
• could the problem could be circumvented if the matrix $g$ is assumed to be in $\text{SL}_n \big(\Bbb{Z} / p^2 \Bbb{Z} \big)$ or in the $n \times n$ orthogonal group over $\Bbb{Z} / p^2 \Bbb{Z}$ --- i.e. $\text{tr} \big( g^N \big) = 0 \, \text{mod} \, p^2$ ? – A. Leverkuhn Dec 10 '13 at 18:45
• I don't understand the i.e.: For $N$ sufficiently divisible, $Tr(g^N)$ should be $n \mod p$, not $0 \mod p$. In any case, I am pessimistic. – David E Speyer Dec 10 '13 at 19:06
• sorry --- $\text{tr}\big( g^N \big) - n = 0 \, \text{mod} \, p^2$. – A. Leverkuhn Dec 11 '13 at 1:46