Timeline for Nice S¹-action implies existence of unconditional basis?
Current License: CC BY-SA 3.0
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| Feb 18, 2018 at 18:54 | comment | added | Jochen Glueck | Maybe it is worthwhile noting that, though the answer to your question is "no" in general (as you explained in your post below), one can still show that the eigenvectors of the action are total in $V$. This is a consequence of the Jacbos-de Leeuw-Glicksberg decomposition (even without the assumption that all eigenspaces be finite dimensional). | |
| Feb 18, 2018 at 13:57 | comment | added | Andrew Stacey | Section 5 of arxiv.org/abs/math/0612096 contains a discussion of the various possibilities for an action of $S^1$ on a LCTVS. It doesn't discuss bases, but might be useful background for $S^1$-actions. | |
| Feb 18, 2018 at 13:02 | comment | added | YCor | I didn't say they're defined only for separable spaces, I said that in the Wikipedia page they're defined only for separable spaces. I didn't realize that the assumption implies separability, that's indeed correct. | |
| Feb 18, 2018 at 13:01 | answer | added | André Henriques | timeline score: 3 | |
| Feb 18, 2018 at 12:55 | comment | added | André Henriques | @YCor. Yes, I was assuming that the action is by linear maps (I've edited the question to make that explicit). The fact that the action is by bounded operators follows from the continuity of the map $S^1\times V\to V$. Also, my assumption that the eigenspaces of $S^1$ are finite dimensional implies that the vector space is separable. (I have to say that I don't agree your the statement that unconditional bases are only defined for separable spaces, but that's irrelevant for the present question.) | |
| Feb 18, 2018 at 12:49 | history | edited | André Henriques | CC BY-SA 3.0 |
added 7 characters in body
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| Feb 18, 2018 at 10:53 | comment | added | YCor | You seem to mean implicitly that the action is by linear maps. Continuity then implies that the action is by bounded operators. Possibly you have more implicit assumptions. Could you be more specific? Second, unconditional bases are defined only in separable spaces in en.wikipedia.org/wiki/Schauder_basis; is $V$ assumed separable? | |
| Feb 18, 2018 at 10:24 | history | asked | André Henriques | CC BY-SA 3.0 |