Subject: answer to a question from Brayton Gray
From: Natalia Castellana Vila
Date: Thu, 01 Mar 2007 10:41:53 +0100
>
> Subject: for your list
> From: Brayton Gray
> Date: Tue, 27 Feb 2007 15:46:52 -0600
>
> The sphere S^(2n+1) localized at an odd prime is an H space, but not a
loop space if n>1. It is conceivable that it's 2n+1 connected cover is a
loop space. Does anyone know anything about this?
>
Dear Brayton,
We've been thinking about your question. By a result of Sullivan,
the odd sphere localized at p, S^{2n+1}(p),
is a loop space if and only if n+1 divides p-1, so maybe the
question is about covers of localized spheres which are not loop
spaces?
It turns out that the covers of S^{2n+1}(p) are loop spaces if and
only if so is S^{2n+1}(p). The argument goes as follows (and the
connectivity of the cover doesn't play any role):
Let X denotes the 2n+1-connected cover of S^{2n+1}(p) and assume
it's a loop space. By using a result of Neisendorfer, [1], the
BZ/p-nullification P_{BZ/p}(X) is homotopy equivalent, up to
p-completion, to S^{2n+1}(p).
Since the nullification of a loop space is again a loop space, we see
that the p-completion of S^{2n+1}(p) must be a loop space. This implies
in turn by an arithmetic square argument that S^{2n+1}(p) itself is a
loop space, which happens iff n+1 divides p-1.
[1] Neisendorfer, Localization and connected covers of finite complexes.
Jérôme and Natàlia.
Dpt de Matemàtiques
Universitat Autònoma de Barcelona.