9
$\begingroup$

In the "Morse-Bott theory and equivariant cohomology" paper by D.M. Austin and P.J. Braam, the authors introduce the Morse-Bott-complex to calculate the de-Rham-cohomology of a compact manifold (using a Morse-Bott-function).

Where does this complex "come from", i.e. is it Austin and Braam's original work?

In the introduction, they state:

Floer's homology groups for homology 3-spheres asked for many generalizations to describe more general gluing laws governing Donaldson's polynomial invariants, both through incorporating other auxiliary information (see Fukaya [1] and Braam and Donaldson [2]) as well as by considering general 3-manifolds [3]. In the course of this work, new techniques were developed in these infinite dimensional cases relating to equivariant cohomology, cup products and various alternative approaches to problems of classical Morse theory. The purpose of this paper is to give a self-contained finite dimensional description of these new aspects.

Does this paragraph also refer to the Morse-Bott-complex or only to the section on "Equivariant cohomology and Morse-Bott theory"?

(The referenced works in the citation, as good as I could guess from the (preprint) references listed:

  1. Fukaya: Floer homology for oriented 3-manifolds
  2. Braam, Donaldson: Fukaya-Floer homology and gluing formulae for polynomial invariants
  3. Austin, Braam: Equivariant Floer theory and gluing Donaldson polynomials )
$\endgroup$
0

1 Answer 1

7
$\begingroup$

I do not know a precise answer to your question (and perhaps someone else does), but I'm fairly certain the idea can be traced back to the paper

Bott, Raoul, The stable homotopy of the classical groups. Ann. of Math. (2) 70 1959 313–337.

This is the paper where Bott proves his famous periodicity theorem, and the proof uses Morse-Bott theory, as it's now known. The relevant Sections are 3 and 4. In particular, I'm fairly certain that the spectral sequence of Corollary 4.2 is a special case of the "Morse-Bott complex" in the paper of Austin and Braam.

Disclaimer: I'm no historian, and I haven't thought about these things in some time!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.