# 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.

# 2012-01-30

Let's discuss framings more carefully: A framing of an embedded $(k-1)$-sphere $K$ in a $(n-1)$-manifold $M$ is an isotopy class of trivializations of the normal bundle $\nu K \cong K \times \mathbb{R}^{n-k}$. (So $K$ needs a trivial normal bundle to begin with.) The difference between two *trivializations* is a map $K \to GL(n-k)$, but since we only care about trivializations up to isotopy, we only care about this map up to homotopy, so the difference between two *framings* is an element of $\pi_{k-1}(O(n-k))$. (Here we use that $GL(n-k)$ deformation retracts onto $O(n-k)$. Also, if $k-1 \geq 1$, we really mean $GL^+(n-k)$ and $SO(n-k)$.)

Thus the difference two framings of a knot in a $3$-manifold lies in $\pi_1(SO(2)) = \mathbb{Z}$. This explains the comment last time that the set of framings is a $\mathbb{Z}$-torsor. This is worth comparing to the case of $1$-knots (embedded $S^1$'s) in a $4$-manifold, in which case we have $\pi_1(SO(3)) = \mathbb{Z}/2\mathbb{Z}$, so there are two different ways to attach a $2$-handle to a $5$-manifold along a fixed embedding of $S^1$ in the $4$-manifold boundary.

Now there is a special case when the set of framings of a knot $K \subset M^3$ *can* be canonically indentified with $\mathbb{Z}$, when $[K] = 0 \in H_1(M;\mathbb{Z})$. (In particular, when $M=S^3$ this holds.) In this case $K = \partial \Sigma$ for some oriented, compact surface $\Sigma \subset M$, and we use this *Seifert surface* $\Sigma$ to define a preferred $0$-framing. (Note that algebraic topology just tells us that $K$ is the boundary of a singular $2$-chain, but not that the $2$-chain can be realized by a smooth surface. Seifert gave an explicit algorithm for constructing such a surface from a knot diagram when $M=S^3$, but there is also a general smooth topology argument using a map from $M \setminus K$ to $S^1$ and pulling back a regular value.) We arbitrarily orient $K$ and then orient $\Sigma$ so that the boundary orientation agrees with the given orientation on $K$. Then a framing of $K$ gives a parallel push-off $K'$ of $K$, with orientation coming from $K$. Make $K'$ transverse to $\Sigma$ and count intersections with sign ($+$ means $K'$ goes from the negative side of $\Sigma$ to the positive side and $-$ means the opposite). This associates an integer to the framing; this integer is otherwise known as the linking number $\mathop{lk}(K,K')$. (See Rolfsen, *Knots and Links*, for a beautiful discussion of ten different definitions of linking number.) Here are some pictures to illustrate this:

OK, now back to building $4$-manifolds. 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.

\[ M' = (M \setminus \phi(S^1 \times B^2)) \cup_\phi (B^2 \times S^1) \]

Here the gluing map is labelled $\phi$, by which I mean $\phi|_{S^1 \times S^1}$. How do we see this pictorially? Let $K = \phi(S^1 \times \{0\})$ and let $\nu = \phi(S^1 \times B^2)$, a knot and a tubular neighborhood in $M$. Let $\mu = \phi(\{p\} \times S^1)$, the *meridian curve*, and let $\lambda = \phi(S^1 \times \{p\})$, the *longitude curve*. (Here $p$ is just some arbitrary point in $S^1$.) The meridian $\mu$ is characterized up to isotopy by being homologically nontrivial in $\partial \nu$ but bounding a disk in $\nu$. On the other hand the longitude $\lambda$ depends on the framing of $K$, and is in fact exactly the parallel push-off we have been discussing above. ($\lambda$ is partially characterized by intersecting $\mu$ once, but that's not enough.) Then $M'$ can be described as the result of removing $\nu$ from $M$ and glueing back in a solid torus so that, after gluing it back in, $\lambda$ now bounds a disk (the $B^2$ in $B^2 \times S^1$) while $\mu$ does not. This is illustrated below:

**Basic examples (exercises)**

- $0$-framed surgery on the unknot in $S^3$ gives $S^1 \times S^2$. Thus if we build a $4$-manifold with a $0$-handle and a $2$-handle attached along the $0$-framed unknot, we cannot cap it off to a closed $4$-manifold by just attaching a $4$-handle.
- $\pm 1$-framed surgery on the unknot gives $S^3$. Thus we can build a closed $4$-manifold with one $0$-, one $2$- and one $4$-handle, with the $2$-handle attached along the $\pm 1$-framed unknot. In fact these $4$-manifolds are $\mathbb{CP}^2$ (for $+1$ framing) and $\overline{\mathbb{CP}^2}$ (for $-1$ framing).

I ended with an attempt to describe $S^2$'s inside $\mathbb{CP}^2$ and $\overline{\mathbb{CP}^2}$ with self-intersection $\pm 1$, but that's best left for the blog until the next post.

Here's the movie:

# 2012-01-27

(At the beginning of class, Eric Burgess presented a solution to the problem of seeing the $1$-parameter family of Morse functions associated to projecting a particular embedding of $S^1$ to various directions, see the video at the end of this post.)

**$4$-manifolds**: Now we extend the ideas from dimension $3$ to dimension $4$ to do our best to draw pictures of $4$-manifolds built from handles. We start with a single $0$-handle. It's boundary is $S^3 = \mathbb{R}^3 \cup \infty$, which we draw as $\mathbb{R}^3$, the ambient space in which the rest of the drawing will happen. This is just the background of our drawing - i.e. we don't really "draw" anything, we just imply it. We will assume no handles except the final $4$-handle will have attaching maps which hit $\infty$.

A $4$-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.

A $4$-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.

**Framings**: Consider an embedding $\phi : S^{k-1} \hookrightarrow M^{n-1}$. Under what conditions does this extend to an embedding $\Phi : S^{k-1} \times B^{n-k} \hookrightarrow M$? An obvious necessary condition is that the image $K$ of the embedding $\phi$ should have a trivial normal bundle, and this is a sufficient condition by the tubular neighborhood theorem. When $k-1=0$ we always have triviality. When $k-1=1$ we may not have triviality if $M$ is nonorientable (e.g. the core of a Mobius band). But if $M$ is orientable then $S^1$'s always have trivial normal bundles. Thing get more interesting when $k-1=1$ and $n-1 \geq 4$, but we will stay away from such high dimensions for now.

Now, assuming $K$ has trivial normal bundle $\nu$, there may be more than one trivialization of $\nu$ (isomorphism $\nu \to K \times B^{n-k}$). The standard proof of the tubular neighborhood theorem extends to show that isotopy classes of trivializations of $\nu$ are in one-to-one correspondence with isotopy classes of extensions $\Phi$ of $\phi$. A *framing* of $K$ is precisely an isotopy class of trivializations of the normal bundle $\nu$. We will discuss framings in full generality next time, but for now, consider two cases:

First, $n-1 = 0$ and $k-1=1$. So $K$ is a pair of points in a $1$-manifold. We claim that, up to isotopy, there are exactly four framings of $K$. These are illustrated below:

Next we show how each of these framings specifies a handle attachment:

Note that the two on the left are diffeomorphic, and the two on the right are diffeomorphic, and the distinction is orientability. This example generalizes to higher dimensional $0$-handles, with the upshot being that, if we agree that we are only working with orientable manifolds, we do not need to specify the framing of the attaching map of a $0$-handle and need simply draw the images the attaching maps (pairs of balls). Note that we could draw points rather than balls but that the use of balls allows us to distinguish different curves going over a single $1$-handle more easily than points would.

Next, consider $n-1=1$ and $k-1=2$. Now we are framing a simple closed curve in a surface, and a moment's thought shows that there are only two framings, but that handle attachment with either framing produces diffeomorphic manifolds. In fact, this phenomenon where two distinct framings produce the same manifold will persist in all dimensions, so we should immediately mod out by it. We leave this detail for the reader to sort out.

Finally for this post, 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.

Intuitively, 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.

Another useful way to draw a framed knot is to adopt the convention that we always use the *blackboard framing*, the framing where the pushoff is in the plane of the surface on which the surface is drawn. Here are some examples; note that, to acheive extra twists we introduce small kinks in $K$:

Here's the video: