# 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: