Let $M$ be a compact finite-dimensional smooth manifold. I have a question about the relationship between the statements that a Morse function induces a handle decomposition for $M$, and that it induces a CW decomposition for $M$.

A Morse function induces a handle decomposition

Denote by $X(M;f;s)$ the manifold $M$ with an $s$--handle attached by $f\colon\,(\partial D^s)\times D^{n-s}\to M$.

Theorem: Let f be a $C^\infty$ function on $M$ with no critical points on $f^{-1}[-\epsilon,\epsilon]$ except $k$ nondegenerate ones on $f^{-1}(0)$, all of index $s$. Then $f^{-1}[-\infty,\epsilon]$ is diffeomorphic to $X(f^{-1}[-\infty,-\epsilon];f_1,\ldots,f_k;s)$ (for suitable fi).

Historical note: This was stated by Smale in 1961, with proof outline. Milnor's Morse Theory, Theorem 3.2 states and proves a weaker, homotopy version of the theorem, where there is only one handle in play. I asked about a proof of this theorem in this MO question, and it turned out that the first complete proof appeared in Palais, simplified lated by Fukui [Math. Sem. Notes Kobe Univ. 3 (1975), no. 1, paper no. X, pp. 1-4]. There's an alternative proof given in Appendix C to Madsen-Tornehave.

Discussion: Roughly, the theorem states that passing a critical point of a Morse function corresponds to attaching a handle. Thus, a Morse function induces a handle decomposition for $M$.

A Morse function induces a CW decomposition

Let $f$ be a Morse function on $M$. Choosing a complete Riemannian metric on $M$ determines a stratification of $M$ into cells $D(p)$ (the unstable (descending) manifold for a critical point $p$ of $f$) in which two points lie in the same stratum if they are on the same unstable manifold. Each $D(p)$ is homeomorphic to an open cell, but the closure $\overline{D(p)}$ can be complicated.

Theorem: The union of compactified unstable manifolds $\bigcup \overline{D(p)}$ gives a CW decomposition of $M$ that is homeomorphic to $M$.

Historical note: A nice discussion of this theorem may be found in Bott's excellent Morse Theorem Indomitable, page 104. Milnor's Morse Theory derives a homotopy version of this statement (Theorem 3.5) from the homotopy version of the statement that a Morse function induces a handle decomposition (Theorem 3.2). The theorem seems to have been first proven by Kalmbach, and was recently strengthened to give the explicit characteristic maps by Lizhen Qin (understanding his papers is the motivation for my question).

The two statements given above look to me as though they should be very similar, especially since the homotopy version of the second follows directly from the homotopy version for the first in Milnor's book. But briefly searching through the literature makes it seem that they are virtually independant- papers proving one aren't even cited in papers proving the other, and the proofs look to me to be entirely unconnected. I don't understand why, probably because I'm having difficulty breaking free from the intuitive picture of the proof in Milnor's book, which works fine up to homotopy.

Question: Can you give an example, or intuition, for a case in which one of the above theorems is difficult but the other is easy? Is there an example for a compact finite-dimensional manifold with a Morse function such that the handlebody decomposition can be read straight off the Morse function, but the reading off the CW decomposition takes substantial extra work? Or the converse?
Stated differently, where does the "up to homotopy proof" on page 23 of Milnor conceptually collapse when we are working up to diffeomorphism instead of up to homotopy?
  • $\begingroup$ In the 2nd highlighted theorem you quote, shouldn't "that is diffeomorphic to $M$' be erased? Stating that it's a CW-decomposition of $M$ should be enough. CW-complexes don't have smooth structures so it's not clear what the latter part means. $\endgroup$ Jan 25, 2012 at 11:28
  • $\begingroup$ I suppose I view the CW-structure on $M$ to be the less natural thing. To construct an explicit CW-decomposition on $M$ you either have to do something like modify the flow-lines for the gradient of $f$, or do some kind of blow-up procedure to turn the (naturally discontinuous) cellular attaching maps into continuous maps. Neither option is particularly pleasant. The handle decomposition is rather natural provided $f$ has some modest restrictions on it. But to answer your question it depends on what "reading off" means. There's substantially less information in a CW-decomposition so... $\endgroup$ Jan 25, 2012 at 11:38
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    $\begingroup$ Note that a handle decomposition can be turned around, in a way that corresponds to replacing $f$ by $-f$. But the same is not true for a CW structure up to homotopy. $\endgroup$ Jan 25, 2012 at 14:05
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    $\begingroup$ The second Theorem is true for generic Riemannian metric only. The standard 2-torus in $R^3$, height function and the standard induced metric is a counterexamlpe (the boundary of upper 1-cell does not belong to the 0-cell). So in general we have only decomposition into disjoint union of balls, not CW-structure. $\endgroup$
    – Petya
    Jan 25, 2012 at 20:13
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    $\begingroup$ @GustavoGranja here's an updated link for that paper: arxiv.org/abs/1102.2838, Qin's An application of topological equivalence to Morse theory (published as doi.org/10.1007/s11784-020-00843-z) $\endgroup$
    – David Roberts
    Sep 24, 2021 at 1:00

2 Answers 2


The second of the theorems you quoted is considerably harder to prove. The gist of the proof is as follows. Consider the closure $\overline{D(p)}$ of $D(p)$ in $M$. Then Lizhen Qin proves that it admits a resolution in the sense of semi-algebraic geometry. More precisely he constructs a compact space $\widehat{D(p)}$ and a continuous surjective map $\pi: \widehat{D(p)}\to\overline{D(p)}$ with the following properties.

$\bullet$ The space $\widehat{D(p)}$ is homeomorphic to a closed ball of dimension equal to the Morse index of $p$.

$\bullet$ The restriction of $\pi$ to the interior of $\widehat{D(p)}$ induces a homeomorphism onto $D(p)$.

The theorem requires that the gradient flow satisfy the Morse-Smale transversality condition wheras no such requirement is needed for the handle decomposition theorem. Moreover, the result is very sensitive to the behavior of the gradient flow near the critical points. In such a region the flow is a linear flow given by a symmetric matrix, the Hessian of $f$ at that particular point. If the eigenvalues are $\pm 1$ things are fine. For different eigenvalues things can go horribly wrong.

In Chap. 8 of my paper Tame flows I show that under appropriate conditions on the eigenvalues of the Hessians at the critical points the Morse-Smale condition is equivalent to the requirement that the stratification by unstable manifolds be a Whitney regular stratification. Moreover I give examples and pictures describing how the Whitney regularity is destroyed if the spectra of the Hessians do not satisfy those constraints.

Another very good reference for these topics is Burghelea-Friedlander-Kappeler survey arXiv: 1101.0778. Burghelea has an alternate and much simpler argument for Lizhen Qin's result, and the paper arXiv: 1101.0778 is much more readable than Qin's.

In the shameless-plug department, I ought to mention the recent 2nd edition of my book An Invitation to Morse Theory. In Chapter 4 I discuss at length these issues without the tameness assumption.

  • $\begingroup$ Liviu, are you sure that Burghelea et. al. produce the CW decomposition in the reference you cited? I could not find it mentioned there. In fact, I did a search of the file for the term "CW" and it is not to be found there. Also, my impression was that Burghelea and company only handle the case of a metric which is standard near the critical point set. $\endgroup$
    – John Klein
    Jan 25, 2012 at 16:57
  • $\begingroup$ It's not in their paper, I agree. In that paper they produce the compact space $\widehat{D}(p)$ and the projection $\pi$ so that the preimage of $D(p)$ in $\widehat{D(p)}$ is open dense. Their construction of $\widehat{D(p)}$ is identical to Qin's but the presentation is cleaner. As Qin mentions in his paper these facts alone imply immediately (via some nontrivial results in topology) that $\widehat{D}(p)$ is topological disk. Qin gives an alternate proof that avoids these topological results. I find this part of his proof hard to diggest. $\endgroup$ Jan 25, 2012 at 17:07
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    $\begingroup$ Thanks Liviu. I presume that you mean "some nontrivial results in topology" refers to the $h$-cobordism theorem (or maybe the Poincare conjecture). Lizhen wanted to avoid using such a big weapon, so he found an alternative proof--a proof which uses more elementary tools. $\endgroup$
    – John Klein
    Jan 25, 2012 at 21:20
  • $\begingroup$ Yep. That's what I meant. $\endgroup$ Jan 26, 2012 at 9:39

Another reference I would like to mention is

Sharko, V.V. Functions on manifolds, Translations of Mathematical Monographs, Volume 131. American Mathematical Society, Providence, RI (1993). Algebraic and topological aspects, Translated from the Russian by V. V. Minachin.

He really does use some aspects of the crossed complex related to the handlebody decomposition, rather than the CW-filtration. We have suggested this as a line of possible development in our book "Nonabelian algebraic topology: filtered spaces, crossed complexes, cubical homotopy groupoids" (EMS Tract 15, 2011) since we work there with filtered spaces, and you get such from a Morse function on a manifold. I feel there is more to do there, using for example the tensor product technology explained in our book. This tensor product does reflect the usual cell decomposition of the product $E^m \times E^n$, $m,n \geqslant 0$, where $E^m, E^n$ have cell decompositions with 3 cells.

  • $\begingroup$ This sounds very interesting! $\endgroup$ Jan 28, 2012 at 13:37

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