6
$\begingroup$

Wikipedia's article on ultrametric spaces seems to suggest that an ultrametic space can also be a normed vector space.

It seems to be impossible for an ultrametric to be induced by a vector space norm, because if $v$ is a nonzero vector, then $$d(0,2v)=\|2v\|=2\|v\|>\|v\|=\max(d(0,v),d(v,2v)) $$ contradicting $d$ being an ultrametric.

So what's going on there? Is there a concept of "normed vector space" that dispenses with the requirement that scalar multiplication scales the norm accordingly? (That would be more of a "normed abelian group", then).

$\endgroup$
7
$\begingroup$

The correct equality is $\|a v\|=|a|\|v\|$ for some absolute value $|\cdot|$ on the scalar field. Absolute values can be nonarchimedean and satisfy the ultrametric inequality. In such a case, $|k|\le 1$ for all $k\in\Bbb N$.

For instance, $\|(v_1,\cdots,v_n)\|:=\sqrt{|v_1|_p^2+\cdots+|v_n|_p^2}$ makes $\Bbb Q^n$ an ultrametric normed vector space over $\Bbb Q$, where $|\cdot|_p$ stands for the $p$-adic absolute value.

$\endgroup$
  • $\begingroup$ Ah, that makes sense. Thanks. $\endgroup$ – Henning Makholm Aug 10 '14 at 11:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.