Talk:Contractibility of unit sphere in Hilbert space

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This article is so cool. —Preceding unsigned comment added by 195.176.0.55 (talk) 18:49, 9 May 2008 (UTC)

new article Cgwaldman 17:11, 16 October 2007 (UTC)

I'm not too happy with the most recent set of changes to this article.

All of the explanatory comments in the proofs were removed, making the article shorter and more streamlined, but harder to understand (for example, the fact that T maps S to its equator was removed). Is the goal to make articles terse, or to make them informative? Cgwaldman 20:35, 31 October 2007 (UTC)

hi, Cgwaldman. thanks for contributing this page. hope we see you around. let me respond to your comments. i don't think the current version of the article is terse. elaborating too much on trivial claims actually make the article more opaque.

consider, for instance, the second section. the key features of the proof are that the shift is an isometry and has no eigenvalues, therefore one has a curve from z to Tz that lies entirely on S. the argument in fact works with any powers of T, therefore any nonsurjective isometry. the pretty obvious fact that the unit sphere in Hilbert space can be mapped (linear) isometrically onto its own equator is not essential here. similarly, "there is no point z on the sphere S for which T(z) is the antipode − z of z" follows from that T has no point spectrum. ditto for "If z lies on S, then this path does not pass through the origin".

as for the last section, which particular basis one chooses for L²[0,1] is again irrelevant. the statement: "...two functions are considered to be identical if they agree on all but a countable set of points." is not quite right. to mention that measurable functions need not be continuous is distracting from the argument. also, it's misleading to say that the proof is less elementary and uses more functional analysis (in fact really no functional analysis is used) than the previous argument. Mct mht 02:14, 1 November 2007 (UTC)