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Every finite dimensional vector space over $\mathbb{R}$ or $\mathbb{C}$ has a unique topology that makes addition and scalar multiplication continuous. Is the same true of finite dimensional vector spaces over any locally compact field?

I assume that $F$ has a locally compact topology with respect to which addition, multiplication, and additive inverse are all continuous. If we also need to assume that multiplicative inverse is continuous on $F\setminus\{0\}$, that is fine. We can also assume the topology is metrizable if that helps.

Nik Weaver
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  • Maybe I'm misunderstanding something. Take the Zariski topology on $\Bbb{C}$, for instance. Scalar multiplication and addition are continuous with respect to the Zariski topology, but this topology is distinct from the usual Euclidean topology. – Alekos Robotis Oct 04 '23 at 01:56
  • Oh, I think I see. You are topologizing the field with a locally compact topology, so it rules out this type of issue. (Is that correct?) Anyway, I guess you are fixing the Euclidean topology for $\Bbb{R}$ or $\Bbb{C}$ to begin with? – Alekos Robotis Oct 04 '23 at 01:57
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    Yes, and I should have said "locally compact Hausdorff". Actually, I'm willing to assume it's metrizable; I'll add that to the question. – Nik Weaver Oct 04 '23 at 02:03
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    If $F$ is a local field and the vector space topology is Hausdorff, then yes. This can be found in Bourbaki's Topological Vector Spaces, Chapter 1, Section 2.3 "Vector subspaces of finite dimension", where $F$ is only assumed to be a complete non-discrete valued division ring. – Just a user Oct 04 '23 at 02:56
  • @Justauser Fantastic! I didn't know that the topology on any locally compact field comes from an absolute value, but Google tells me it does. (Do you know any way to prove this that doesn't use Haar measure? I want to use the uniqueness result in a book I'm writing, several chapters before I discuss Haar measure.) – Nik Weaver Oct 04 '23 at 03:41
  • @NikWeaver I took the liberty to write an answer mostly to collect various sources online to handle the potential novelties. I don't know how to establish the valuation without Haar measure... It feels hard to connect topology with real numbers, and Haar measure provides the bridge, also harmonic analysis on local fields is a fundamental tool in number theory. BTW, I enjoyed your Forcing for Mathematicians very much! Looking forward to the new book! – Just a user Oct 04 '23 at 04:31

1 Answers1

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The multiplicative inverse map is automatically continuous by $F$ being locally compact Hausdorff. (We cannot drop locally compactness, this is the difference between Banach field and nonarchimedean field in Kedlaya's paper.)

If $F$ is discrete, then definitely no. Take $F=\mathbb R$ with discrete topology, and $V=\mathbb R^n$ with the usual Euclidean topology. Otherwise (if $F$ is non-discrete), $F$ must be a local field, hence a complete non-discrete valued division ring. This can be found in e.g. Weil's Basic Number Theory, roughly the automorphism $x\mapsto ax$ of $F$ scales the Haar measure which can be used to define the valuation of $a\in F^{\times}$.

If the topology on $V$ is allowed to be non-Hausdorff, it can be the trivial topology. Otherwise, the answer is yes. This has been done at different levels of generality.

  • For local fields: Chapter 1, Section 2, Corollary 1 of Theorem 3 of Basic Number Theory.

  • For complete non-discrete valued division ring: Chapter 1, Section 2.3 "Vector subspaces of finite dimension" of Bourbaki's Topological Vector Spaces. Here is a sketch.

  • For complete non-discrete division algebra: MathOverFlow discussion. This can avoid establishing local fields can be valued.

Just a user
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