Subject: Re: Question for surgeons. A surgeon answers. From: Andrew Ranicki Date: Thu, 07 Jun 2007 06:57:24 +0100 Johannes Ebert wrote > > I want to see concrete counterexamples, i.e. homotopy > equivalences between smooth (!) manifolds M_0 \to M_1 which are not > covered by an isomorphism of the stable normal vector bundles. > Because I want to do calculations with them, I like to have "simple" > examples, i.e.: highly connected, small dimension, small Betti numbers, > known cohomology and computable rational homotopy type. Are there examples > where the normal bundles of M_0 and M_1 are distinguished by their > rational Pontryagin classes? It would certainly be very nice if one of the > manifolds is stably parallelizable. Is that possible? > Where shall I look in the literature? The topological surgery exact sequence for a product of spheres S^pxS^q with p>1,q>1,p+q>3 gives one homotopy equivalence h:M^{p+q}-->S^pxS^q for each (x,y) \in L_p(Z) \oplus L_q(Z) with L_*(Z)=\pi_*(G/TOP) the simply-connected surgery obstruction groups, and x (resp. y) the surgery obstruction of the p (resp. q)-dimensional normal map h\vert:h^{-1}(S^p)-->S^p (resp. h\vert:h^{-1}(S^q)-->S^q). The product S^pxS^q is stably parallelizable, so in particular the topological stable normal bundle is trivial. The topological stable normal bundle \nu_M is the pullback along h of the fibre homotopy trivial bundle over S^pxS^q classified by (x,y) \in L_p(Z)\oplus L_q(Z) \subset [S^pxS^q,G/TOP]. If (x,y) \neq (0,0) then \nu_M is not trivial, and if either p or q is 0(mod 4) then \nu_M has a nontrivial rational Pontryagin class. All this is in the topological category, but if p+q>4 and (x,y) \in im(\pi_p(G/O)\oplus \pi_q(G/O)) also in the smooth category. The original reference is S.P.Novikov "Homotopy equivalent smooth manifolds I." Izv. Akad. Nauk SSSR 28, 365-474 (1964) Conveniently, there is an English translation online http://www.mi.ras.ru/~snovikov/10.pdf For a more recent account see Example 20.4 of my 1992 CUP book "Algebraic L-theory and topological manifolds" http://www.maths.ed.ac.uk/~aar/books/topman.pdf Best, Andrew _________________________________________________________________ Subject: Re: Question for surgeons From: Laurence Taylor Date: Thu, 7 Jun 2007 10:46:11 -0400 An answer to Ebert's question The easiest example meeting all Ebert's requirements is S3 x S4. There are infinitely many homeomorphism types distinguished by the rational Pontryagin class in degree 4. All this follows from the surgery exact sequence of Sullivan & Wall. More concretely, let $H$ be the generator of K0(S4) = Z. Then x = 24H is fibre homotopically trivial and multiples of x are distinguished by their first Pontryagin class. With a bit more work one can find a four-dimensional vector bundle which is fibre homotopically trivial and stabilizes to x. The total space of the sphere bundle is an example homotopy equivalent to S3 x S4. Larry Taylor