# 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.