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I'm essentially reopening this old question of Ricky Demer which was never fully answered.

Essentially the original question: Suppose we have a topological field $F$ which is complete, Hausdorff, and non-discrete, and we put a Hausdorff topology on $F^n$ so as to make it a topological vector space over $F$; is is this topology necessarily the product topology (and hence complete, and hence closed in anything it embeds in)?

As the link shows, the answer is yes if $\tau$ comes from an absolute value on $F$, and it's easy to see the same argument works if it comes from a field ordering on $F$.

Note that the argument there shows that this question reduces to the following lemma:

Suppose we have a topological field $(F,\tau)$ which is Hausdorff and non-discrete, and we give $F$ a second topology $\tau'$, which is also Hausdorff, such that $(F,\tau')$ is a topological vector space over $(F,\tau)$. Does this force $\tau=\tau'$? What if we assume that $(F,\tau)$ is complete?

(I'm isolating completeness as a separate, possibly-unnecessary condition because the argument there only uses completeness in the reduction to the lemma, not in proving the lemma for valued fields.)

Related to this question in that one way to come up with a counterexample for both simultaneously would be to find a field with two (nondiscrete, Hausdorff) topologies with one strictly finer than the other.

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+1 $\hspace{.05 in}$ :-) $\hspace{.1 in}$ –  Ricky Demer Jul 24 '11 at 8:46
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4 Answers

Yes, it is "standard" that a finite-dimensional vector space over a complete, non-discrete, division algebra (!) has a unique topology compatible with a topological vector space structure (=Hausdorff, addition is continuous, scalar multiplication is continuous). This is probably proven (at least for real or complex scalars, but the proofs should generalize) in any source on general TVS's, e.g., section 5, my notes. Without completeness, or without discreteness, there are easy counter-examples.

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The proof there seems to be assuming that the topology comes from an absolute value. Again, I already know it's true in that case, the question is about general topological fields... –  Harry Altman Jul 24 '11 at 20:54
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For completeness, let's do one of paul's easy counterexamples. The field $\mathbb Q$ of rationals with the usual topology is not complete. (But comes from an absolute value and from a field ordering.) The two-dimensional vector space $\{a+b\sqrt{2}:a,b\in\mathbb Q\}$ with its usual topology (as a subset of the reals) is not the product topology.

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I think you need to read the question more carefully. I isolated completeness as possibly not necessary for the 1-dimensional case; I'm aware that it's necessary for higher dimensions... –  Harry Altman Jul 24 '11 at 20:48
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So, there is a property of topological division rings that is equivalent to finite dimensional tvs's have a unique topology, and the property is implied by normability, but I can't find any examples that are not normable.

From Nachbin, "On Strictly minimal topological division rings":

Theorem 7. Let K be a given topological division ring. In order that every finite-dimensional vector space over K should have only one admissible topology it is necessary and sufficient that K be strictly minimal and complete.

"Admissible" here means "vector space operations are continuous". As for "strictly minimal":

A topological division ring K is said to be minimal if its topology $T_K$ is a minimal element in the ordered set of all admissible topologies on K; that is, if there exists no admissible topology X on K such that $T\lt T_K$. The topological division ring is said to be strictly minimal if there exists no topology T on K admissible with respect to $T_K$ such that $T\lt T_K$; it amounts to the same to say that the only topology on K admissible with respect to $T_K$ is $T_K$ itself.

In particular, "strictly minimal" is equivalent to the $n=1$ case of the question (i.e. completeness is not necessary, which was known already in the normed case: see Proposition 2 of Section 2.2 of Chapter 1 of Bourbaki's TVS).

In Warner's "Topological Fields", "strictly minimal" is replaced with all tvs $E$ over $K$ being "straight" (p. 224), which means that for every nonzero $a\in E$, the map $\lambda\rightarrow \lambda a$ is a homeomorphism from K onto the one-dimensional subspace generated by a. Warner mentions that no examples of straight division rings are known other than "locally retrobounded" division rings, which is more general than normable (see p. 158), which implies to me that there are known examples of non-normed locally retrobounded (and thus, straight) division rings. But, I didn't see any actual examples in the book (though I only have access to the Google books version).

Nachbin apparently has a book, "Espacos vetoriais topologicos", published in 1948, where the notion of strictly minimal is introduced. Maybe there'd be an example in there...?

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OK, this is helpful. But what I'm looking for seems to be easier than this -- If I understand this correctly, I'm not looking for a field which is strictly minimal (equivalently straight) but not normable, I'm just looking for a field which isn't strictly minimal (equivalently straight). Perhaps I should get a copy of that book, maybe it'll be in one of the exercises missing from Google Books. (Actually for a field which is strictly minimal but not normable, any field which is orderable but not normable should work, IINM.) –  Harry Altman Jul 25 '11 at 2:20
    
I suppose really I want complete but not straight, rather. –  Harry Altman Jul 25 '11 at 2:27
    
Oh yeah, I forgot to add that I couldn't find any examples that were not straight/minimally strict, either! –  B R Jul 25 '11 at 2:32
    
Yes, SFAICT neither Warner nor Więsław (whose chapter on the subject seems to be just a copy of Nachbin) make any mention of whether or not there might exist non-discrete, non-straight fields. Annoying... –  Harry Altman Jul 25 '11 at 3:36
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up vote 1 down vote accepted

Well, now I feel silly -- on looking through Wieslaw again, I see he does give examples of non-discrete, non-straight fields, just not in that section. For instance, take two absolute values on the rationals; the topology they generate together still make the rationals a topological field, is not discrete, and is obviously finer than either of the ones you started with. Since it isn't even minimal, it can't be straight.

I'm not going to accept my own answer on this since this example presumably isn't complete, and I'd like a complete example. Which might well be in here too, if I keep looking... (Is the completion of this again a field? If so that should work, but I'd need to check if that's true.)

Edit: Nope, the completion of this isn't a field, so I'm still lacking for a complete example.

Edit again: OK, I'm now pretty sure Wieslaw gives one, so the answer is no. Wieslaw gives an example of a complete normed field which is not given by any absolute value (here "normed" means instead of |ab|=|a||b|, we only require |ab|≤|a||b| and |-a|=|a|). Furthermore, he shows given a power-multiplicative norm, you can find an absolute value that generates a coarser topology (so if a power-multiplicative norm wasn't equivalent to an absolute value, it isn't minimal). (Here power-multiplicative only means we require |an|=|an| for positive n, not for negative n.) And after a bit of staring at his example, I'm pretty sure it's power-multiplicative. So unless anyone can show that I've missed something, I'm going to consider this one closed.

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