9
$\begingroup$

Suppose $(M,g)$ is a Riemannian manifold equipped with a Morse function $f: M \rightarrow \mathbb R$. It's been shown that $f$ gives rise to a CW decomposition homeomorphic to $M$ under the generic assumption that $(f,g)$ satisfies Morse-Smale transversality. What additional conditions does one need to impose in this situation to ensure that the CW complex is regular? (Regular means the attaching maps are homeomorphisms.)

It is true that every CW complex is homotopy equivalent to a simplicial complex, and those are all regular, but I'm asking for regularity on the nose.

(Note: I originally asked the question on MSE.)

EDIT: I'm particularly interested in the case when Morse functions give rise to pseudo-regular CW complexes, by which I mean the incidence numbers of the cells all lie in $\{0,\pm 1\}$. By this definition, regular implies pseudo-regular, and pseudo-regular allows for a trivial boundary operator on the Morse-Smale complex.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
8
$\begingroup$

If the Morse function $f$ is perfect, then, for any choice of metric $g$, the attaching maps cannot be homeomorphism. Indeed if the Morse function was perfect, then the boundary operator of the associated Morse-Smale complex is trivial. If the attaching map was a homeomorphism, then the boundary operator cannot be zero. This can be seen from Thm. 4.5.1 of these notes.

In any case, the ''attaching map'' has a rather nasty description which makes verifying that it is a homeomorphism is nearly impossible.

I might be able to be of more help if you explained in more detail what would you do if you knew that the attaching maps are homeomorphisms.

Improve this answer
$\endgroup$
2
  • $\begingroup$ Thanks for the answer. I've updated the question to clarify the situation I'm interested in. $\endgroup$
    – Mike Catanzaro
    Commented Feb 24, 2016 at 15:06
  • 2
    $\begingroup$ $\mathbb{Z}$-perfect Morse functions are automatically pseudo-regular since all the incidence numbers are $0$. Functions with only even Morse indices are perfect. Moment maps of Hamiltonian $S^1$-actions are also perfect. The map $f: U(n)\to\mathbb{R}$ $T\mapsto {\rm Re}\;(AT)$, where $A$ is a symmetric $n\times n$matrix with with simple, positive spectrum is also perfect. $\endgroup$
    – Liviu Nicolaescu
    Commented Feb 24, 2016 at 15:27
4
$\begingroup$

As I mentioned earlier today, a manifold like $S^2 \times S^2$ with a standard Morse function (giving one 0-cell, two 2-cells and one 4-cell) can never be regular for silly reasons -- the attaching map from the 4-cell is a map $S^3 \to S^2 \vee S^2$ which can't be an embedding for dimension reasons.

This example shows there's no hope for anything generic satisfying your embedding condition. A minimal necessary condition would be for the k-skeleton to be locally k-dimensional, i.e. every point is in a closed k-cell. I'm not certain if that is a strong enough condition, but its a start.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .