Let $p$ be an odd prime and $G$ a finitely generated pro-$p$ group. We say that $G$ is $p$-valued if there exits a map $\omega:G\to \mathbb{R}_{>0}\cup \{\infty\}$, which is called valuation, such that the following properties hold for all $g,h\in G$:
- $ \omega(g)>\frac{1}{p-1} $,
- $\omega(g)=\infty$ if and only if $g=1$,
- $ \omega(g^{-1}h)\geq \min\{\omega(g),\omega(h)\} $,
- $\omega(g^{-1}h^{-1}gh )\geq \omega(g)+\omega(h)$,
- $ \omega(g^{p})=\omega(g)+1 $.
Lazard uses the valuation $\omega$ to define a topology on $G$ by choosing the subgroup $G_{\nu}=\{g\in G~|~ \omega(g)\geq \nu\}$, with $\nu \in \mathbb{R}_{>0}$, as a fundamental system of neighborhoods of the identity. Then one can show that topology of $G$ coming from the valuation coincides with the topology of $G$ as a pro-$p$ group.
My question is the following:
Let $G$ be a $p$-valued pro-$p$ group. Can we always find a $p$-valutation $\omega'$ on $G$ such that $\omega'(g)>1$ for all $g\in G$?
Not if $p >2$. (Whereas for $p=2$, all $p$-valuations are of this kind, by definition.)
Consider the group $G:=\{ g \in GL_2(\mathbb Q_p): w(g-I_2) > \dfrac{1}{p-1} \}$, where $w: M_2(\mathbb Q_p) \rightarrow \mathbb Z$ is the additive version of the usual supremum norm, i.e.
$$w(\pmatrix{x_1&x_2\\x_3&x_4}) := \min(v_p(x_i)).$$
$G$ is a complete $p$-valued group with the $p$-valuation
$$\omega(g) := w(g-I_2)$$
according to Lazard, III. 3.2.6.
But for $p>2$, both $g:=\pmatrix{1&p\\0&1}$ and $h := \pmatrix{(p+1)^{-1}&0\\0&1}$ are in $G$, and $$g^{-1}h^{-1}gh = g^p,$$ so for any other $p$-valuation $\omega'$ such that $\omega '(h) >1$, we get a contradiction between conditions 4 and 5.