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Let $K$ be a field equipped with an absolute value $\vert-\vert:K\rightarrow \mathbb{R}_{\geq 0}$ satisfying

  • $\vert k \vert = 0$ iff $k=0$
  • $\vert x+y \vert\leq\vert x\vert+\vert y\vert$
  • $\vert x\cdot y \vert = \vert x \vert\cdot\vert y\vert$
  • $K$ is complete with respect to the topology induced by the absolute value

(I think of $\mathbb{R,C}$ but also $\mathbb{Q}_p$ and $\mathbb{C}_p$)

Let $V$ be a $K$-vector space equipped with a norm $\Vert - \Vert: V\rightarrow \mathbb{R}_{\geq 0}$ satisfying

  • $\Vert v \Vert = 0$ iff $v=0$
  • $\Vert k\cdot v \Vert = \vert k \vert\cdot\Vert v\Vert$
  • $\Vert x+y \Vert \leq \Vert x \Vert + \Vert y \Vert$

Under which circumstances can we normalize any nonzero vector? More precisely, given $0\neq v \in V$ with norm $\Vert v \Vert = r \neq 1$, when does there exist some $k\in K$ such that $\vert k \vert = \frac{1}{r}$, ie. such that $\Vert kv \Vert = 1$?

One sufficient criterion is of cause that the absolute value $\vert-\vert:K\rightarrow\mathbb{R}_{\geq 0}$ is a surjective function (as is the case for $\mathbb{R}$ and $\mathbb{C}$).

In fact it is sufficient and necessary that the norm function $\Vert-\Vert:V\rightarrow\mathbb{R}_{\geq 0}$ factors through the image of the absolute value function. A priori I see no reason why this should always be the case. I am especially interested in the case, where $K$ is a local field in the sense of Wikipedia.

Thank you for your time.

Jonas Linssen
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  • And example where you can't: take $k=\mathbb{Q}$ with absolute value $|x|=0$ if $x=0$, $|x|=1$ if $x=1$. The topology induced by this absolute value is discrete, so $\mathbb{Q}$ is complete with respect to this norm. Now let the norm on $\mathbb{Q}^n$ be $\lVert (q_1,\ldots,q_n)\rVert = \sum |q_i|$ (the Hamming distance to $0$). Since the support of $x+y$ is contained in the union of the supports of $x$ and $y$, this will satisfy the conditions. But the only vectors that can be normalized are the ones that are already normal: the ones with exactly one nonzero entry. – Arturo Magidin Oct 19 '22 at 18:56
  • Thanks for this nice counterexample. Of cause I forgot that one has to also take into account the trivial norms. So I guess my question should simply be: can I normalize, when my base field is local? – Jonas Linssen Oct 19 '22 at 19:05
  • (Obviously it should be "and $|x|=1$ if $x\neq 0$", which you clearly understood, but I just wanted to note the typo since I can't edit the comment any more...) – Arturo Magidin Oct 19 '22 at 19:15
  • @TorstenSchoeneberg thank you for the comment, I would also accept this as an answer. – Jonas Linssen Oct 21 '22 at 13:42
  • $\mathbb{Q}_p$ is a normed vector space over $(\mathbb{Q},|\cdot|_p)$ if you choose $||\cdot||=\pi|\cdot|_p$. Here, no nonzero vector can be normalized. – Chilote Oct 21 '22 at 17:22

1 Answers1

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If $\Vert \cdot\Vert_1$ is any norm, then so is $\Vert \cdot \Vert_2 :=c \cdot \Vert \cdot \Vert_1$ for any $c \in (0,\infty)$, and these two norms are equivalent (cf. e.g. 1). So as soon as $\lvert K \rvert \subsetneq [0,\infty)$, which is always the case for a non-archimedean local field $(K, \lvert \cdot \rvert)$, you can enforce that some vectors cannot be scaled to $1$.

However, e.g. on finite dimensional vector spaces, all norms are equivalent if $K$ is complete as assumed. (Some standard proofs state this only for $\mathbb R$ and $\mathbb C$, and use local compactness, see e.g. 2, 3; for the general statement, look e.g. at Keith Conrad's Equivalence of Norms. In fact, just read that, it says everything I say here and more, better.). So one can always find a norm, equivalent to the given one, where every vector can be scaled to $1$. In fact, you can just choose any basis $b_1,...,b_n$, and with respect to this basis define the maximum norm $\Vert \kappa_1 b_1 + ... + \kappa_n b_n\Vert:=\max|\kappa_i|$. This way, any basis can play the role of a normal (unit) basis, without changing the topology.

Torsten Schoeneberg
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