Subject: Responce on Pontryagn classes of $HP^n$. Date: Wed, 11 Apr 2001 05:09:18 +0200 (CEST) From: Yuli Rudyak To: Don Davis On Tue, 3 Apr 2001, Don Davis wrote: > Four postings here: A new question about projective spaces, > and three comments on texts...........DMD > ______________________________________ > Subject: faked quaternion spaces > Date: Tue, 03 Apr 2001 15:32:03 -0400 (EDT) > From: wilking@hans.math.upenn.edu > > Given a simply connected differentiable manifold $M$ > with the integral cohomology ring of HP^n. > Is M homotopically equivalent to HP^n > if all (or some) of its Pontrjagin classes > are the same? > > Conversely, if M is homotopically equivalent to HP^n > are some (or all) of the Pontrjagin classes the same? There are manifolds which are homotopy equivalent to $HP^n$ and have different Pontryagin classes. Actually, it is true for every simply connected manifold with suitable cohomology. I give the argument for PL manifolds. Consider the surgery exact sequence for $HP^n$. The group [HP^n, G/PL]\otimes Q =\prod H^{4k}(HP^n) is large, and so the kernel of the map s:[HP^n, G/PL] \to Z is large. Every element of this kernel gives us a space homotopy equivalent to $HP^n$. Now, the result follows since $H^*(G/PL;Q)=Q[p_1, ..., p_k, ...]$ where $p_i$ are the corresponding universal Pontryagin classes (and these Pontryagin classes yield the homotopy equivalencc G/PL = \prod K(Q,4k)) Yuli Rudyak