Morse and Cerf Theory
2012-02-01
Regarding the handle decompositions given last lecture for $\mathbb{CP}^2$ and $\overline{\mathbb{CP}^2}$, we observe that attaching a $2$-handle to a $0$-handle with framing $\pm 1$ to an unknot immediately yields an embedded sphere with self-intersection $\pm 1$. We see this, and a more general version, as follows (and illustrated at right in the half-dimensional version): If we attach a $2$-handle $B^2 \times B^2$ to a $4$-manifold $X$ along a null-homologous knot $K \subset \partial X$ with framing $n$, and $K = \partial \Sigma$ for some surface $\Sigma \subset \partial X$, then the interior of $\Sigma$ can be pushed in to the interior of $X$ so as to meet $\partial X$ transversely along $K$. Then gluing the core $B^2 \times \{0\}$ of the handle to $\Sigma$ we get a smooth closed surface $\Sigma' \subset X'$, where $X'$ is the result of attaching the handle to $X$.
The self-intersection of a surface in a $4$-manifold is the intersection of the surface with a nearby parallel copy of itself, counted with signs. We can push $\Sigma$ off itself in $X$ without self-intersections, and we can push $B^2 \times \{0\}$ off itself in $B^2 \times B^2$ without self-intersections, each restricting to a parallel push-off of $K$ in $\partial X$. The difference between these push-offs (literally their intersection number in the boundary of a tubular neighborhood of $K$) is exactly the framing $n$ of $K$ (yes, the signs do work out right). Thus the self-intersection of $\Sigma'$, denoted $\Sigma' \cdot \Sigma'$, is exactly equal to $n$. In the case of an unknot, $\Sigma$ is a disk, so $\Sigma'$ is a sphere.
Similarly, if one attaches two $2$-handles along nullhomologous knots, each one produces a closed surface, and the intersection number between the two surfaces is the linking number between the knots.
Ascending and Descending manifolds: We have been drawing handle diagrams as if all the handles were attached "at once", i.e. as if the corresponding critical points had the same value. But certainly in some instances this cannot be done; as in our examples where $2$-handles run over $1$-handles. To understand better when we can or cannot do this, we need to see larger handles than we have seen so far. In our argument that a Morse function on a manifold gives a decomposition of the manifold into handles, each handle is contained in a small neighborhood of the critical point, and "most" of the manifold is made up of products.
Instead, consider a gradient-like vector field $V$ for a Morse function $f$ and a critical point $p$. Then the ascending manifold for $p$, $A_p$, is the union of $\{p\}$ and all flow lines whose backward-time limit is $p$, while the descending manifold for $p$, $D_p$, is the union of $\{p\}$ and all flow lines whose forward-time limit is $p$. In local coordinates near $p$ in which $f = - x_1^2 - \ldots - x_k^2 + \ldots x_{k+1}^2 + \ldots x_n^2$, $D_p$ is the $x_1,\ldots,x_k$-space and $A_p$ is the $x_{k+1},\ldots,x_n$-space. Near every other point in $D_p$ (resp. $A_p$), forward (resp. backward) flow along $V$ for some time gives a local diffeomorphism into such a local coordinate chart around $p$, and hence we see that $D_p$ is a smooth $k$-dimensional submanifold and $A_p$ is a smooth $(n-k)$-dimensional submanifold. They intersect transversely at $p$. If $X$ is closed, so that there is no boundary to run into, they are diffeomorphic to $\mathbb{R}^k$ and $\mathbb{R}^{n-k}$.
Now consider a cobordism $X$ from $M_0$ to $M_1$ with Morse $f : X \to [0,1]$ and critical points $p_1, \ldots, p_m$, and suppose that all the ascending and descending manifolds of distinct critical values are disjoint. (Descending manifolds are always disjoing, and likewise ascending manifolds are always disjoint, but a point can be on the ascending manifold of one critical point and the descending manifold of another.) In this case, the attaching maps for the handles corresponding to the critical points all flow all the way down to $M_0$, and so we can see $X$ as built from $[0,1] \times M_0$ with all the handles attached simultaneously to $\{1\} \times M_0$. In fact, the handle $B^k \times B^{n-k}$ corresponding to critical point $p_i$ is precisely a small neighborhood of $D_{p_i} \cup A_{p_i}$, with $D_{p_i}$ being the core $B^k \times \{0\}$ and $A_{p_i}$ being the co-core $\{0\} \times B^{n-k}$.
In the next post we will investigate circumstances under which we can arrange for this disjointness to occur.
We know how to draw pictures of a $0$-handle with some $1$-handles, and now we can add the $2$-handles by drawing framed knots, some of which may go over the $1$-handles. To the right is a particularly nice example, called the Mazur manifold, involving one $0$-handle, one $1$-handle and one $2$-handle. It is contractible, roughly because the attaching circle for the $2$-handle has a clasp which, if undone, could slide off the $1$-handle leaving the $2$-handle only going once over the $1$-handle, and thus cancelling the $1$-handle (c.f. convertible roof). But it is not diffeomorphic to $B^4$, and it's boundary is a $3$-manifold with the same homology as $S^3$ (a homology sphere) but nontrivial $\pi_1$. In particular, we cannot complete this $4$-manifold to a closed $4$-manifold by attaching a $4$-handle; to do so we would need the boundary to be $S^3$. Thus we see that we need to think more carefully about surgery on $3$-manifolds in order to understand the boundaries of $4$-dimensional handlebodies and know whether we can cap them off to make closed $4$-manifolds.
$4$-dimensional $1$-handle is $B^1 \times B^3$ attached along $S^0 \times B^3$, a pair of balls. We draw these balls as small balls in $\mathbb{R}^3$, with dotted lines connecting the "feet" of each $1$-handle so we know which ball goes with which ball. Thus the illustration at right simply describes a $0$-handle with two $1$-handles attached. Note that the new boundary, after attaching these $1$-handles, is not $S^3$ because the interior of each ball is no longer in the boundary of the $4$-manifold, and each $S^0 \times B^3$ has been replaced with $B^1 \times S^2$. We should visualize each pair of balls as indicating "worm-holes" allowing us to tunnel from one region of space to another along an interval's worth of $S^2$'s. In particular, this new boundary here is exactly the connected sum of two copies of $S^1 \times S^2$. Thus we could not attach a $4$-handle at this point.
$4$-dimensional $2$-handle is $B^2 \times B^2$ attached along $S^1 \times B^2$, i.e. a solid torus. They can go over $1$-handles or not, as show in the diagram at right, involving two $1$-handles and two $2$-handles, one of which goes over one $1$-handle once and one of which does not go over any $1$-handles. Unfortunately just drawing the images of the attaching maps now no longer determines the attaching maps up to isotopy. To understand this better we now consider framings.
consider $n-1=1$ and $k-1=3$. Now we are framing a knot $K$ in a $3$-manifold $M$. A framing is a pair of linearly independent normal vectors to $K$ at each point along $K$ (varying smoothly along $K$ of course). To mod out by the issue mentioned in the preceding paragraph, we will assume that $K$ is oriented, $M$ is oriented, and that the tangent to $K$ followed by the two normal vectors is an oriented basis, so that, up to isotopy, we do not need to specify the second normal vector. Thus the framing is just given by a single nowhere zero vector field along $K$, normal to $K$. After identifying $\nu$ with a tubular neighborhood of $K$, this can be seen as a parallel copy of $K$ on the boundary of a tubular neighborhood, as shown at right.
this is characterized up to isotopy by the number of times it twists around $K$. I.e. somehow framings of $K$ can be identified with $\mathbb{Z}$. The problem is that, in general, there is no preferred $0$-framing. The best we can say in general is that the set of framings of $K$ is a $\mathbb{Z}$-torsor, or an affine space for $\mathbb{Z}$; it looks like $\mathbb{Z}$ but we don't know where $0$ is. Equivalently, given any framing, we can add or subtract $1$ to it in a consistent way to produce a new framing. We use the right-hand rule for the sign convention; an example of adding $+1$ to the previous drawing is shown at right.
