Subject: Yuli's question From: Laurence Taylor Date: Fri, 24 Sep 2004 10:42:52 -0500 If no one else has finished off Yuli's question, here's a solution. Larry >> Are homology spheres stably parallelizable? There are certainly rational homology spheres which are not stably parallelizable - even smooth ones can have non-trivial Stiefel-Whitney classes. For integral ones John Klein showed they have fibre-homotopy-trivial stable tangent bundles so the smooth ones are stably parallelizable in dimensions not 4k. But even the topological ones are stably parallelizable here since \pi_{4k}(G/TOP) = Z detected by the L-class which also computes the signature which is 0. Since \pi_{4k}{BSO) = Z detected by the same L-class, the smooth ones are smoothly stably parallelizable. ________________________________________________________ Subject: Re: response and question From: browder@Math.Princeton.EDU Date: Fri, 24 Sep 2004 12:59:14 -0400 In case n = 4k the Pontryagin class is zero from the Index Theorem so the bundle is trivial. Bill ___________________________________________________________ Subject: Re: response and question From: Yuli Rudyak Date: Fri, 24 Sep 2004 13:08:09 -0400 (EDT) Thanking John, it seems I can prove the stable parallelizability now. In dimension 4k, the obstruction is the multiple of the $k$-th Pontryagin class, which, in turn is zero because the signarture is. By the way, the topological triviality of the stable topological bundle follows from the Cannon-Edwards double suspension theorem (double suspension of the homology sphere is homeomorphic to the sphere). ______________________________________________________________ Subject: Rudyak's question From: Tom Goodwillie Date: Fri, 24 Sep 2004 13:25:37 -0400 John Klein's argument can be completed in the remaining case as follows: M is a homology 4k-sphere, and John showed that its stable normal bundle is pulled back from a bundle on S^{4k}. To show that that bundle on the sphere is trivial it's enough to see that its kth Pontryagin class is zero. But that means evaluating the kth normal Pontryagin class of M on [M]. By Hirzebruch this number is proportional to the signature (since there are no lower Pontryagin classes around, since there is no cohomology below the top dimension). The signature must be zero (again because there is no homology below the top dimension). This argument fails in a couple of places if k=0, but that case can be left as an exercise. Tom Goodwillie