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For the uninitiated

Morse theory, as many other early alebraic-topology widgets, leads to a picture of smooth manifolds as being built up from 'cells', copies of $\mathbb{D}^n$ for varying $n$, 'glued' to each other by the usual topological tools; giving rise to (in some sense) a more natural picture of homology as 'coming from' cellular homology.


As an example, consider the torus $\mathbb{T}^2$: we begin with empty space, attach a 0-cell ($\mathbb{D}^0=$ a point), attach a 1-cell ($\mathbb{D}^1=$ a line) to your point (both ends of the line are attached to the point, creating a circle), attach another 1-cell (in the same way, to the same point, creating a sort of figure 8).

The hardest bit to visualise is next: attaching a 2-cell ($\mathbb{D}^2=$ a disk, which we will think of as its homeomorphism equivalent, a square). Begin by twisting your figure 8 so that one circle is in the xy-plane, the other in the xz-plane, now attach the top and bottom of your square (coloured red in picture below) to the xy circle (creating a 'curling round' tube) and the left and right edges (coloured blue below) of your square (now a tube) either side of the xz circle, completing the torus.

Torus cell decomposition


The above takes some thinking, but a little reading around shows that this is fairly easy to see. What makes it so easy is that the cells we are attaching are of adjacent dimensions, that is; we may easily identify the boundary of one with the entirety of another. Where it gets harder to visualise is when the dimensions of the cells we are attaching to one another differ by >1- the canonical example of this is the complex projective plane $\mathbb{CP}^2$, a 4-manifold built by attaching a disk to a point (making a sphere) and then attaching a 4-ball to that sphere.

The latter attaching map (wherein points are identified with their images), I know, may be thought of as the Hopf fibration $\partial \mathbb{D}^4=S^3 \to \mathbb{CP}^1=S^2 $, but I have no way of visualising this, particularly with regard to the interior of the 4-disk.

How does a 4 cell wrap around a 2 cell without producing a singularity of some kind? Is this analagous in some sense to Dehn surgery in which one uses a thickening? Is there a right way to think about this or can it only really be thought of 'intellectually'?

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What do you mean by «it can only be thought of 'intellectually'»? Are you using the word intellectually to mean incorrectly? –  Mariano Suárez-Alvarez Aug 4 '10 at 16:05
I guess he means as opposed to 'viscerally,' e.g. in a way that someone with no mathematical training would understand. –  Qiaochu Yuan Aug 4 '10 at 17:05
@Quiaochu- correct. @Mariano is right though; it was a stupid word to use. What I meant was sort of 'in terms of higher mathematical abstractions', but even that is too mealy and doesn't quite explain what I'm after- (CP^2 is a fairly high mathematical abstraction in itself after all). –  Tom Boardman Aug 4 '10 at 17:50
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5 Answers

1) Near a point of $S^2$ the picture looks like $\mathbb{C}$ sitting inside, naturally $\mathbb{C}^2$ (because $\mathbb{C}P^2\cong(\mathbb{C}^3\setminus{0})/\mathbb{C}^{\times}$ and $S^2=\mathbb{C}P^1$ is the image of a hyperplane in $\mathbb{C}^3$).

2) Think first about $S^2$: one attaches a disk to a point, contracting the circle on the disk's boundary. Now, to make $\mathbb{C}P^2$ one takes 4-disk and do pretty much the same but not for one circle but for all, well, fibers of the Hopf fibration at once — or, in other words, for all points of $S^2$ at once. So, since there were no singularities after gluing $S^2$, there will be no singularities here either.

(Not sure if it's an answer, but only hope it helps.)

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+1- not quite the whole way there, but certainly helpful! –  Tom Boardman Aug 4 '10 at 12:29
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Tom, I'm not sure I see how it's getting any harder in passing from a torus to a projective space. In your $\mathbb CP^2$ case, you have $\mathbb CP^1$ sitting inside of it, and the boundary of a regular (tubular) neighbourhood of the $\mathbb CP^1$ is $S^3$. And $D^4$ has $S^3$ as its boundary, so the attaching map is tautological. The normal bundle is the missing data and that's what your CW-decomposition is ignoring.

This is essentially what always happens. Perhaps the conceptual hump you're dealing with is that you're asking for CW-decompositions of manifolds. Morse functions generically only build homotopy-equivalences to CW-complexes, they do not put CW-structures on the manifold without some significant work. Moreover, CW-decompositions ignore some of the most essential properties of the manifold, like smooth structures.

If instead you work with handle decompositions, what I state in my first paragraph is basically a generality -- critical points amount to handle attachments and the gluing instructions are always given in a direct way from the flow lines of the Morse function's (suitably normalized) gradient. So the handle decomposition is on the given manifold -- unlike the CW-case where you only have a homotopy-equivalence to a CW-complex.

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Note that you're only attaching the boundary of the 4-cell. This is a map $S^3 \rightarrow S^2$. The resulting topological space has a completely intact copy of the interior of the 4-cell. This is similar to making $\mathbb{RP}^2$ by gluing a disk to a circle by the degree-2 map $\partial D^2 = S^1 \rightarrow S^1$ (given, say, in complex coordinates by $z\mapsto z^2$).

As for the issue of having high-dimensional spheres "wrap around" low-dimensional cells: up to homotopy, this can be phrased as a question about homotopy groups, that is, maps from spheres into some space $X$. (This doesn't address the question of singularities, but I think it is still helpful for trying to think about these gluing constructions.) These are groups $\pi_k(X)$ for all $k\geq 1$; when $k=1$, this just the fundamental group. In general, they are quite poorly understood. Even for the simplest possible CW complexes $X=S^n$ (except $n=1$), we don't know all the homotopy groups! This might sound like an easy problem, but it's really....REALLY...nontrivial. Now, when you make CW complexes, you're gluing on cells $e^k$ one by one; these are really just maps from $\partial e^k = S^{k-1}$ to the stuff you already had (the "$(k-1)$-skeleton"). People often only care about the homotopy types of these maps. In summary, while it's certainly a good idea to try and visualize these things, it's also good to be able to deal abstractly with gluing constructions.

The wikipedia article on homotopy groups of spheres is really interesting, and contains a sort of picture of the Hopf fibration. Basically, you want to think of it as a bunch of circles that somehow fill out $S^3$, which is just $\mathbb{R}^3$ with one extra point at $\infty$. Interestingly, any two of these circles are linked. So you can kind of think of it locally as the unit circle in the xy-plane, and then you can move that circle in 2 dimension's worth of directions, and in every one of those directions that you push the circle off itself you're going to get a new, nearby circle that's linked to the original one. (Of course, there is precisely an $S^2$'s worth of circles!) You can take the circle going through $\infty$ to be the z-axis. So by what I just said, all of these other circles need to wind once around the z-axis too.

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Concerning the Hopf fibration, the film "Dimensions" does a superb job of visualizing it (chapters 7-8 in the table of contents).

Concerning the visualization of the glueing, maybe it helps to visualize other, similar constructions to get at least some intuition. For instance, you could try to get a picture of how the torus looks like after every "meridian" (blue circle in your picture) has been squashed to a point.

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I think you understand attaching cells that have dimension gap better than you think! Consider any $S^n$ for $n>>0$, we can think of this as being the attachment of $D^n$ to a point which is 0-dimensional.

Also, I would recommend that you think of this attaching cells thing a bit more carefully. if you understand the hopf fibration then you do understand the attaching of the cell. The hopf fibration is not easy to picture (as far as I know). I would recommend thinking about the map in terms of what happens when you work with coordinates. I will be back later to add more... I promise.

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