6
$\begingroup$

Is it true that if $ E,F$ are two topological vector spaces (or say Banach spaces) over $\mathbb{R}$ such that they have nonempty open subsets $U\subset E, V\subset F$ which are homeomorphic, then the two vector spaces are isomorphic? If false, then what can we say if the two open subsets are $\mathcal{C}^1$-diffeomorphic?

$\endgroup$
  • $\begingroup$ I would guess that the first question is false, simply on the grounds that it is pretty trivially true for finite dimensional ones, so it's probably false for arbitrary vector spaces. $\endgroup$ – Simon Rose Sep 30 '15 at 9:49
  • $\begingroup$ My guess is that it is true in all spaces, since it is true for finite-dimentional ones. But I cannot give a proof or a disproof. $\endgroup$ – usr203050 Sep 30 '15 at 9:51
  • 1
    $\begingroup$ My guess for the second part is that the differential at some point (or may be at any point) can give an isomorphism between the two spaces. $\endgroup$ – usr203050 Sep 30 '15 at 10:06
  • 1
    $\begingroup$ Yes. The interesting question between those two extremes is the case of Lipschitz equivalence (for Banach spaces). Again much work has been done on this case, e.g. by Lindenstrauss et al. $\endgroup$ – shasta Sep 30 '15 at 10:19
9
$\begingroup$

This is false. All separable Banach spaces, for example, are homeomorphic.Indeed, there is a considerable body of work on when topological vector spaces are homeomorphic (see Bessaga and Pelczynski "Selected Topics in infinite-dimensional Topology" for starters).

$\endgroup$
  • $\begingroup$ Ok thank you. So it is far from evident in any case. $\endgroup$ – usr203050 Sep 30 '15 at 9:59
  • $\begingroup$ And what if the subsets are smoothly diffeomorphic? (The second question). $\endgroup$ – usr203050 Sep 30 '15 at 10:03
  • 1
    $\begingroup$ Isn't the diffeomorphic part answered by the comment of usr203050? $\endgroup$ – Jochen Wengenroth Sep 30 '15 at 10:17
  • 2
    $\begingroup$ If diffeomorphism is something that has differential as a bounded invertible operator between tangent spaces, then this very operator is linear isomorphism between spaces. If not, what concretely do you mean? $\endgroup$ – Fedor Petrov Sep 30 '15 at 10:17
  • $\begingroup$ That is what I mean! :-) Also, the comment of shasta gives something more. $\endgroup$ – usr203050 Sep 30 '15 at 10:29

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.