To test Morse theory, one day I tried to compute the homology of Lens spaces. You can build a lens space out of $S^3 = \{ |z|^2 + |w|^2 = 1: z, w \in \mathbb{C} \}$ using the quotient by $z \mapsto e^{2\pi i /p}z, w \mapsto e^{2\pi i q/p}w$. The quotient by this action L(p,q) is a lens space.

With Morse theory estimates the Betti numbers of a manifold by counting the critical points of a specified index. For any real-valued function $f: L(p,q) \to \mathbb{R}$, we look for critical points where $\nabla f(p_0) = 0$. Expanding around p0 $f(p) = f(p_0) + (p - p_0)^T A (p - p_0) + O(|p - p_0|^2)$ for some matrix $A$, the Hessian. If the eigenvalues of A are all real and nonzero and the critical values of f are all different then $f$ is called "Morse". The Morse theory says L(p,q) is homotopic to a CW-complex with a cell complex of dimension k for each critical point of index k.

The Morse function I chose is $h = r \cos p\theta$ where $r,\theta$ come from $z = r e^{i \theta}$. Also let $w = \rho e^{i \phi}$. This function is well-defined on the lens space L(p,q) and its critical points are (+/-1,0,0,0) and its images under the deck transformation. To see this, notice the gradient in this coordinate system is $$\nabla = \left( \frac{\partial }{\partial r}, \frac{1}{r}\frac{\partial }{\partial \theta},, \frac{\partial }{\partial \rho}, \frac{1}{\rho} \frac{\partial }{\partial \phi}\right) $$ and you look for points where the gradient is normal to S3. However, $H_1[L(p,q)]= \mathbb{Z}/p$ suggesting we should have a saddle point of index 1.

I am told that Morse theory only gives you the singular $\mathbb{R}$ homology. Did I find all the critical points correctly? Is there a way to get the $\mathbb{Z}$ homology using Morse homology or other piece of differential topology?

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    $\begingroup$ Doesn't your function have a whole critical circle when $z=0$? It does not appear to be Morse. There are Morse functions on the 3-dimensional lens spaces with 4 critical points: min, max and two saddles. Those appear to be the most natural to me. By your above comment about the CW-complex equivalent to the space, everything works in $\mathbb Z$ homology. $\endgroup$ – Ryan Budney Jun 28 '10 at 23:01
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    $\begingroup$ Your function is not a smooth function. $\endgroup$ – Petya Jun 28 '10 at 23:18
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    $\begingroup$ I suggest you to consider the restriction of your function $h$ to a line $w=const$. Is it a smooth function everywhere? $\endgroup$ – Petya Jun 29 '10 at 1:37
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    $\begingroup$ Answer on my question (I asked it in order to help you) is "no". Function $h$ is not smooth аt $z=0$. Consider the map $p: z \mapsto (z,\sqrt{1-|z|^2})$. It is a smooth embedding of a neighborhood of 0 in $\mathbb C$ to $S^3$. If function $h$ is smooth then $h\circ p$ must be smooth. That is obviously wrong. $\endgroup$ – Petya Jun 29 '10 at 2:22
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    $\begingroup$ 1. Again: your function $h$ is prepared from the function $r\ cos p\theta$ on $\mathbb C={\mathbb R}^2$. This function is not smooth at the origin. Why? That function is homogeneous of degree $1$ function. Only linear functions on ${\mathbb R}^N$ are homogeneous of degree $1$ functions which are smooth at the origin. Your function is definitely not linear. 2. (r,\theta) are not smooth coordinates on $\mathbb C$. $\endgroup$ – Petya Jun 29 '10 at 10:46

Is there a way to get the $\mathbb{Z }$ homology via Morse Homology? Sure, indeed, historically, the underlying ideas and constructions by means of flow lines were one of the first attempt to build concretely the homology of a manifold. Actually, it seems to me that this construction was a little abandoned for a while, especially by functional analysts, because the alternative Morse theory by means of sublevel sets, popularized e.g. by Richard Palais in the '60, was somehow simpler, as it did not use transversality arguments. Then the Morse complex and homology construction became suddenly in fashion again after Andreas Floer's work, in which it became a spectacular tool to build new invariants aimed to prove the Arnold conjecture.

In any case, yes, you can certainly perform the Morse complex construction with $\mathbb{Z}$ coefficients for a $C^2$ functional $f$, even on a possibly infinite dimensional Hilbert manifold $M$, and prove that the resulting homology is isomorphic to the singular, provided the functional enjoys a list of properties (some of them are given for free in the case of compact manifolds).

Specifically, it should be bounded from below; have nondegenerate critical points (i.e., with invertible Hessian operator) with finite Morse indices; it should have complete sublevel sets, and lastly, it should satisfy the Palais-Smale compactness condition (PS). Up to a $C^{\infty}$-small perturbation of the metric, as an application of the Sard-Smale theorem, you may even assume that the negative gradient flow of $f$ has the property that for any pair of critical points $x$ and $y$, whose Morse indices $m(x)$ and $m(y)$ differ at most by 2 (that's a technical detail), the unstable manifold $W^u(x)$ of $x$ and the stable manifold $W^s(y)$ of $y$ meet transversally.

As a consequence, these intersections are flow-invariant, embedded submanifolds of dimension equal to $m(x)$ - $m(y)$ (thus empty if $m(x)$ $\leq$ $ m(y)$). And orientable in a canonical way, starting from arbitrarily choosen orientations of all unstable manifolds of critical points (the unstable manifold of a critical point is an embedded submanifold diffeomorphic with its tangent space, whose dimension coincides with the Morse index of the critical point). Note: these intersections are in any case orientable, even if $M$ is not. Moreover, if $m(x)-m(y)\leq1$, they meet any regular level set of $f$ in a compact set (thus finite).

As you probably know, this allows to define $C_k(f)$ as the free abelian group generated by the (possibly infinite) set $\mathrm{crit}_k(f)$ of all critical points of index $k$; moreover a boundary operator $$\partial:C_k(f)\to C_{k-1}(f)$$

is well-defined by linear extension starting from the generators $\mathrm{crit}_k(f)$. It just sends $x\in \mathrm{crit}_k(f)$ to a certain algebraic sum of the endpoints $y$ of all flow lines emanating from $x$ at time $-\infty$, and converging to critical points $y$ of index $m(x)-1$ at time $+\infty$. The mentioned compactness property ensures that there are finitely many of these flow lines; and the sign in front of any $y$ in the sum is chosen to be + or - according whether the orientation of $W^u(x)\cap W^s(y)$ coincides or not with the one induced by the gradient flow. Then one proves that this way one has really defined a true boundary, that is, the relation $\partial^2=0$ hold. Finally, one proves that the corresponding complex produces the singular homology. And this is the Morse homology with $\mathbb{Z}$ coefficients. (it is not completely clear to me whether your query was on the general side, or particularly aimed to your specific function. I hope recalling the general facts may be in any case of some aid.)

Warning. Actually this is one of the way pepole tell the story; it involves not too many technicalities (if you do it well), but everything looks like a kind of magic. In fact, the whole story should start by a cellular decomposition induced by the functional; this puts the isomorphism of the homologies into a more general and natural fact about cellular homologies (if you want to understand Morse homology, read chapter 5 of Albrecht Dold's Lectures on Algebraic Topology, on cellular spaces and cellular complexes); and then the nice picture of the Morse complex with points and lines becomes rather a representation (of the cellular complex) than a smart construction, which is even closer to the historical developement of these ideas by giants like Morse, Bott, Thom, Smale.

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    $\begingroup$ "Moreover, they meet any regular level set of f in a compact set." That in general is not true if $m(x)-m(y)>1$. $\endgroup$ – Petya Jun 29 '10 at 1:33
  • $\begingroup$ @ Pietro Majer - What do you think might be going wrong with my argument then? $\endgroup$ – john mangual Jun 29 '10 at 5:01
  • $\begingroup$ Thanks Petya, you are of course right; fixed now. It escaped to me and you immediately noticed it. Good! $\endgroup$ – Pietro Majer Jun 29 '10 at 6:34
  • $\begingroup$ @ Pietro - Your answer might be easier to read if it's split into paragraphs. $\endgroup$ – john mangual Jun 29 '10 at 9:16
  • $\begingroup$ You're right, done. As I'll get some free time I'll think to the other question of yours. $\endgroup$ – Pietro Majer Jun 29 '10 at 9:49

I tried a computation, I hope it's correct. I considered the map h = Re $z^p$ + Re $w^p$ on the sphere, and I found four sets of critical points: (1) ($\mu_{1}^{+} / \sqrt{2}$, $\mu_{2}^{+} / \sqrt{2}$), (2) ($\mu_{1}^{-}/ \sqrt{2}$, $\mu_{2}^{-} / \sqrt{2}$), (3) ($\mu_{1}^{+}$, 0) and (0, $\mu_{2}^{+}$), (4) ($\mu_{1}^{-}$, 0) and (0, $\mu_{2}^{-}$), where $\mu_{j}^+$ denotes a p-root of 1 and $\mu_{j}^-$ denotes a p-root of -1.

The first set gives maxima (where h = 2), the second set gives minima (h=-2) and the other two give index 1 and 2 points (I think). The first and second set contribute to p points in the lens space, the other two contribute 2 points each. Perhaps doing Morse homology as Pietro suggested counting connecting flow lines gives you the correct result.... but I have not tried.

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  • $\begingroup$ So, on the lens space, you get 4p critical points total? $\endgroup$ – john mangual Jun 30 '10 at 21:50
  • $\begingroup$ There are $2p + 4$ of them... I have not checked that they are all non-degenerate. I suspect they are not... the set (3) and (4) seem too few to give the correct homology. $\endgroup$ – Diego Matessi Jul 1 '10 at 7:03

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