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