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.