Timeline for Are all topological (finite-dim) real vector spaces homeomorphic to a coordinate space?
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
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| Dec 26, 2010 at 18:48 | vote | accept | Avi Steiner | ||
| Dec 26, 2010 at 2:58 | comment | added | Gerald Edgar | In fact true for any complete field in place of $\mathbb R$ ... even if not locally compact. | |
| Dec 26, 2010 at 2:51 | answer | added | algori | timeline score: 12 | |
| Dec 25, 2010 at 21:39 | comment | added | Avi Steiner | @Theo J: That makes sense. | |
| Dec 25, 2010 at 20:41 | comment | added | Theo Johnson-Freyd | As the Theo B said, if you do not insist that your vector spaces be Hausdorff, then the answer is trivially no: your favorite vector space with the indiscrete topology is such that all operations (addition, negation, multiplication) are continuous, but it is not homeomorphic to $\mathbb R^n$. | |
| Dec 25, 2010 at 15:01 | comment | added | KConrad | Another reference besides Weil's book is Alain Robert's A First Course in p-Adic Analysis, Appendix A.6 to Chapter II. His argument, like Weil's, uses Haar measure. It proceeds by induction on the dimension. | |
| Dec 25, 2010 at 11:53 | comment | added | Theo Buehler | @Pietro: I took that "isomorphic" means "as a linear space". I agree that it is not that difficult (essentially it boils down to compactness of the unit ball in standard $\mathbb{R}^{n}$). @Harry: No, uniform structures do not enter explicitly (but they lurk around, of course). Yes, uniform structures were invented by Weil, they arose in his investigations of topological groups in the late 30's. | |
| Dec 25, 2010 at 11:23 | comment | added | Pietro Majer | A strange question because, as remarked, isomorphism of TVS are in particular homeomorphism. Maybe you meant: if a real TVS is homeomorphic to $\mathbb{R}^n$ (as top. space), is it also isomorphic to it as TVS ? Yes, because a Hausdorff real TVS is finite dimensional iff is isomorphic to $\mathbb{R}^n$ for some $n$, iff it is locally compact; not trivial but not even difficult fact that you can find in any book on the subject. | |
| Dec 25, 2010 at 11:22 | comment | added | Harry Gindi | @Theo: Does he do it using uniform spaces (which were his invention, if I remember correctly)? | |
| Dec 25, 2010 at 8:45 | comment | added | Theo Buehler | If you insist that the vector space topology is Hausdorff (otherwise take a seminorm which is not a norm) then it is true that the dimension determines the homeomorphism type, but it is not trivial. It is easier if you require in addition local convexity. André Weil proves the general fact in one of the very first sections of his "Basic Number Theory". | |
| Dec 25, 2010 at 8:39 | comment | added | zroslav | Your isomorphism also gives you a homeomorphism | |
| Dec 25, 2010 at 7:55 | comment | added | Leandro | Since sum and scalar product have to be continuous, I guess that the space is in fact path connected, because of $\lambda\mapsto \lambda V+(1-\lambda)W$ is a path connecting any $V$ and $W\in V$. | |
| Dec 25, 2010 at 7:21 | history | asked | Avi Steiner | CC BY-SA 2.5 |