$p$-adic analytic pro-$p$ group satisfies a pro-$p$ identity?

Question

Let $p$ be a prime. Let $w$ be an element of a free pro-$p$ group $F_r$ of finite rank $r\geq 2$. Then we say that a pro-$p$ group $G$ satisfies the pro-$p$ identity $w$ if for every homomorphism $ f:F_r\to G$ we have $f(w)=1$. So, is it known that any $p$-adic analytic pro-$p$ group $G$ satisfies some (nontrivial) pro-$p$ identity $w$? Any relevant literature would be appreciated.

Answer 1

The question is closely related to that of the linearity (via continuous representations) of the free pro-$p$ group $F_r$ aver pro-$p$ rings. Using the universal property of the $\mathbb{Z}_p$-algebra $\Lambda_p$ of formal power series in $rn^2$ commuting variables $x_{ij}^{k}$, with $1\leq i,j\leq n$ and $1\leq k\leq r$, the latter question can be reduced to examining just one pro-$p$ group, say $H_p(n)$.

Firstly, $\Lambda_p$ can be viewed as a pro-$p$ ring with the topology defined by the family of ideals $B_{l,m}$, where $B_{l,m}$ is formed by all the power series $f=\sum a_w w$, with $v_p(a_w)\geq l$ for $w$ of degree $\leq m-1$. Next we can consider the congruence subgroup $\Gamma\leq \mathrm{GL}_n(\Lambda_p)$ defined by the ideal of formal power series with zero constant terms; thus $\Gamma$ is a pro-$p$ group which contains the matrices $1+X_k$, $k=1,\ldots,r$, where $X_k=(x_{ij}^{k})$. We define $H_p(n)$ to be the closed subgroup generated by all the matrices $1+X_k$. This group satisfies the following two interesting properties:

1. For any pro-$p$ ring $R$, and every family of square matrices $f_k=(f_{ij}^{(k)})$, $k=1,\ldots,r$, with $f_{ij}^{(k)}$ lying in the Jacobson radical of $R$ (to ensure convergence), there exists a (continuous) group morphism from $H_p(n)$ onto the pro-$p$ group generated by the matrices $1+f_k$. This is readily seen since the map $x_{ij}^{(k)}\mapsto f_{ij}^{(k)}$ extends to a (continuous) $\mathbb{Z}_p$-algebra morphism from $\Lambda_p$ to $R$, which in turns extends to the corresponding matrix rings, and induces then the desired group morphism.

2. $H_p(n)$ is free iff $F_k$ embeds (continuously) in $\mathrm{GL}_n(R)$ for some pro-$p$ (or, more generally, profinite) ring $R$. The direct implication is obvious (take $R=\Lambda_p$); for the converse, one can reduce the problem to the case where $F_k$ lies in the congruence subgroup defined by the Jacobson radical of $R$, and then use (1) to see that $H_p(n)$ maps onto $F_r$, and so $H_p(n)\cong F_r$, as desired.

The result (2) is due to Zubkov (see https://link.springer.com/article/10.1007/BF00969315).

Returning to your question, note that an analytic pro-$p$ group $G$ embeds in $\mathrm{GL}_n(\mathbb{Z}_p)$ for some $n$, and that $G$ satisfies a pro-$p$ identity iff the intersection $G_1$ of $G$ with the congruence subgroup $\mathrm{GL}_n^1(\mathbb{Z}_p)$ (defined by the ideal $p\mathbb{Z}_p$) does (this is obvious since $G/G_1$ is finite). So we may assume without loss of generality that $G\leq \mathrm{GL}_n^1(\mathbb{Z}_p)$. Thus all is reduced to checking pro-$p$-identities in $\mathrm{GL}_n^1(\mathbb{Z}_p)$. By virtue of (1), this would be the case if $H_p(n)$ does, i.e., $H_p(n)$ is not free (by virtue of (1), again).

Zubkov proved that this is the case for $n=2$ and $p>2$. Recently, the latter was completed to $p=2$ by Ben-Ezra and Zelmanov. According to Zelmanov, for $n$ fixed, $H_p(n)$ satisfies a pro-$p$ identity provided that $p$ is greater than some integer $g(n)$. The proof is sketched in the expository paper "Infinite Algebras and Pro-$p$ Groups", Progress in Mathematics, Vol. 248, 403–-413. As far as I'm aware, the full proof was never published. Further, still, one has to remove the restriction on $p$. Thus your question is still open in full generality.