Examples of non-splittable norms

Question

Let $K$ be a complete non-archimedean field. A norm on a finite dimensional vector space $V$ is a function $| \cdot | : V \to \mathbf{R}$ which satisfies the usual norm properties (with the non-archimedean triangle inequality). If the valuation of $K$ is discrete, one can prove that the norm has a very simple form: there exists a basis $e_i$ of $V$ as a $K$-vector space such that $$ \left| \sum_i \lambda_i e_i \right| = \max |\lambda_i| |e_i| $$ (we can freely choose the $|e_i|$ of course). One call these norms splittable.

When $K$ is no longer discrete, there should be counterexamples, but I cannot find anywhere in the literature. I am wondering if anyone has written a counter-example down anywhere. Of course, any example must be at least two-dimensional.

For example if one considers $V^{\leqq 1} \subset V$ the $\mathcal{O}_K$-submodule consisting of elements of norm $\leqq 1$, then this is torsion-free hence flat over the valuation ring $\mathcal{O}_K$. If this is finitely generated it is therefore free and even one sees that the rank must agree with the dimension of the vector space. This then implies that the norm splits (and gives a proof in the discrete valuation case), so one sees that this phenomenon has to do with the non-noetherianness of the ring of integers.

Answer 1

All norms on finite-dimensional spaces over $K$ are splittable if, and only if, the field $K$ is maximally complete. You can find this result (and much more) in the paper by Boucksom–Eriksson “Spaces of norms, determinant of cohomology and Fekete points in non-Archimedean geometry” (Advances Math. 378, 2021, 107501), see Lemma 1.12.

Regarding the counterexample you are looking for, the argument is short enough so that I can give it here. Consider a field $K$ that is not maximally complete and a non-trivial immediate extension $L$ of $K$ (that is to say with the same value group and residue field as $K$). Let $a \in L \setminus K$. Then the space $V = K + K\cdot a$, with the norm induced by $L$, is a counterexample. Indeed, otherwise, you could find an orthonormal basis $(e_1,e_2)$ of $V$ (using $|L|=|K|$ to normalize). Since $\tilde L = \tilde K$, there exists $\alpha \in K$ such that $|e_1 - \alpha e_2| < 1$, and this is a contradiction.

Answer 2

The following (which was written before it was explicitly indicated that completeness was part of the hypothesis) is Example 11.1.14 of Kaletha and Prasad's book “Bruhat–Tits theory—a new approach”. (The numbering might not match the published version; I have an old draft.)

Assume that $K$ is not complete, and let $V$ be a $2$-dimensional $K$-vector subspace, containing $K$, of the completion $\widehat K$. Extend the valuation on $K$ canonically to a valuation $\omega$ on $\widehat K$, and then restrict it to a norm $\alpha$ on $V$. This restriction is not splitting. This follows from Lemma 11.1.13, which says that $\alpha$ is splitting if and only if a certain canonical map from $\smash{\widehat V}^\omega \mathrel{:=} V \otimes_K \widehat K$ to the completion $\smash{\widehat V}^\alpha$ of $V$ with respect to $\alpha$ is a bijection; but $\smash{\widehat V}^\alpha$ is just $\widehat K$, whereas $\smash{\widehat V}^\omega$ is a $2$-dimensional $\widehat K$-vector space.