Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Motivation

A topological vector space is a vector space over a (topological) field, K, that carries a topology such that addition and scalar multiplication are continuous maps, e.g., all normed vector spaces. Since $\mathbb{C} \approx \mathbb{R}^2$ as topological spaces (with their standard topologies), it's clear that the actual field cannot be detected by the topology alone. Also, cardinality considerations show that any topology on a finite field cannot be homeomorphic to a vector space over a field of characteristic $0$. But that still says nothing about whether there are some topologies, such that, for instance $\mathbb{Z}_p(t) \approx \mathbb{Q}$. I can't find any topological invariants that will distinguish just the fields (as vectors spaces of dimension 1). I suspect that if this problem is solvable, it will be just be topological group considerations, but I can't figure out how to do it.


So here are my two well-posed questions:

a) Suppose K and F are topological fields. If K and F are homeomorphic, then is it necessarily true that char K = char F?

b) Suppose K and F are topological fields. Further suppose that $V$ is a topological vector space over K and $W$ is a topological vector space over F. If $V$ and $W$ are homeomorphic, then is it necessarily true that char K = char F?

Obviously, an answer to b) implies an answer for a). I'm not sure how much of my thinking I should put here. But I do know that if char K = p, then we get a homeomorphism of $V$ onto itself by translation for each $v \in V$, which each have finite order p. These are mapped to self-homeomorphisms of $W$ which also have order p, but a priori, they don't "see" any of the algebraic structure of $W$, so I'm not sure this is actually an obstruction. Any thoughts (or references) to proofs or counterexamples would be greatly appreciated.

share|improve this question
    
Regarding a) - homeomorphic as spaces I presume. Ditto for b). –  David Roberts May 29 '11 at 3:27
    
Yes. Not via homeomorphic isomorphisms. I want to know if after forgetting the original algebraic structure, the characteristics are still detectable. –  Charlie Cunningham May 29 '11 at 3:38
2  
Do you allow the discrete topology for a topological field? –  Tom Goodwillie May 29 '11 at 4:02
    
I think the discrete and indiscrete topologies each always give homeomorphisms (assuming cardinalities are the same). But I'm interested in more "natural" topologies, although I'm not really sure how to make that precise, especially in the positive characteristic case. –  Charlie Cunningham May 29 '11 at 4:27
add comment

2 Answers

up vote 14 down vote accepted

I think all non-archimedean locally compact fields are homeomorphic: Their rings of integers are compact, metric and totally disconnected and hence are all homeomorphic (to the Cantor set). The same is true for the units in those rings. The field is then topologically the disjoint union of the ring and a countable number of copies of the units.

share|improve this answer
2  
One thing I sometimes like to point out in popular lectures is that such a ring of integers is also homeomorphic to the absolute Galois group of a number field. Which somewhat explains why $p$-adic representations are more informative than complex representations of Galois group. –  Minhyong Kim May 29 '11 at 7:52
1  
You need finite residue field to have compactness. –  Felipe Voloch May 29 '11 at 8:06
    
@Felipe: That's what I meant by "local" but I'll add it as a condition. –  Torsten Ekedahl May 29 '11 at 9:15
    
Thanks. I still need to think more about the structure of local fields. –  Charlie Cunningham May 31 '11 at 2:48
add comment

Torsten Ekedahl answered your first question in a nontrivial set of cases. As a concrete example, the characteristic two field $\mathbb{F}_2((t))$ is homeomorphic to the characteristic zero field $\mathbb{Q}_2$ by the map $$\sum_{k \gg -\infty} a_k t^k \mapsto \sum_{k \gg -\infty} a_k 2^k.$$ This also provides a counterexample to your second question in dimension greater than zero.

The answer to the question in the title is a bit more subtle. That is, there are topological vector spaces that only allow one characteristic. This is obvious for finite dimensional vector spaces over finite fields, but it's also true that there are no fields of positive characteristic that are homeomorphic to $\mathbb{R}$, because the additive structure would allow one to produce a finite group of automorphisms of the order topology of size greater than 2.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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