# Regular CW complex arising from a Morse decomposition

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.

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.
• $\mathbb{Z}$-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. – Liviu Nicolaescu Feb 24 '16 at 15:27
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.