Index-stable compact ${\varvec{p}}$-adic analytic groups

Abstract

A profinite group is index-stable if any two isomorphic open subgroups have the same index. Let p be a prime, and let G be a compact p-adic analytic group with associated $\mathbb {Q}_p$-Lie algebra $\mathcal {L}(G)$. We prove that G is index-stable whenever $\mathcal {L}(G)$ is semisimple. In particular, a just-infinite compact p-adic analytic group is index-stable if and only if it is not virtually abelian. Within the category of compact p-adic analytic groups, this gives a positive answer to a question of C. Reid. In the appendix, J-P. Serre proves that G is index-stable if and only if the determinant of any automorphism of $\mathcal {L}(G)$ has p-adic norm 1.

Appendix: A letter from Jean-Pierre Serre

Dear MM. Noseda and Snopce,

I have just seen your arXiv paper on “Index-stable compact p-adic analytic groups”. Your proof uses Lazard’s very nice results, but in fact these results are not necessary: p-adic integration is enough. Let me explain:

Let L be a finite-dimensional Lie algebra over $\mathbb {Q}_p$. Denote by |x| the p-adic norm of $\mathbb {Q}_p$. Consider the following property of L:

For every automorphism s of L, we have $|\mathrm {det}(s)| = 1$ (*).

This is true, for instance, if L is semisimple since $\mathrm {det}(s)=\pm 1$, which is the case you consider.

Assume property (*). Let $n = \dim L.$ Let u be a non-zero element of $\wedge ^nL^*$, where $L^*$ is the ${\mathbb {Q}}_p$-dual of L. Let G be a compact p-adic analytic group with Lie algebra L. Then u defines a right-invariant differential form $\omega _u$ on G of degree n. The corresponding measure $\mu _u = |\omega _u|$ is a non-zero right-invariant positive measure on G, hence is a Haar measure since G is compact. Property (*) implies that $\mu _u$ is invariant by every automorphism of L. Hence every isomorphism of G onto another group $G'$ carries $\mu _u$ (for G) into $\mu _u$ (for $G'$). This implies that, if $G, G'$ are open subgroups of some compact p-adic group $G''$, then they have the same index - as wanted.

Conversely, if (*) is not true for some s, and if $G''$ is compact with Lie algebra L, s defines a local automorphism of $G''$, and if G is a small enough open subgroup of $G''$, it is transformed by s into another open subgroup $G'$, and the ratio $(G'':G)/(G'':G')$ is equal to $|\mathrm {det} (s)|$, which is $\ne 1$.

Best wishes,

J-P. Serre