Here we focus on examples of handles. Recall that an $n$-dimensional $k$-handle is $H = H^n_k = B^k \times B^{n-k}$ with $\partial H$ divided into two regions, the attaching region $S^{k-1} \times B^{n-k}$ and the free region $B^k \times S^{n-k-1}$, and a handle is attached to a pre-existing cobordism $X$ from $M_0$ to $M_1$ via an embedding of the attaching region into $M_0$, producing a new cobordsim $X'$ from $M_0$ to $M_1'$, containing $X$. Note that we have not yet discussed carefully how $M_1'$ is obtained from $M_1$, but when we do discuss this, the general method will be known as surgery.

First we note that all of this even makes sense when $k=0$ or $k=n$, with the convention that $B^0$ is a point and $S^{-1} = \emptyset$.1Dimensional0and1Handles For example, $S^1$ is built with a $0$-handle $B^0 \times B^1$, attached along an embedding of $\emptyset$, i.e. not attached to anything, following by a $1$-handle $B^1 \times B^0$ attached along an embedding of $S^0 \times B^0$, as in the figure to the right. This picture obviously generalizes to $S^n$ built with a $0$-handles and a $n$-handle.

Our main results thus far concern Morse functions with zero or one critical points, but these immediately imply the following general result:

Corollary: Every cobordism decomposes into a sequence of products and handles. In particular, every closed $n$-manifold can be built starting with a $0$-handle, then attaching some number of other handles of index $0 \leq k \leq n$, and then capping off with a $n$-handle.

(In fact we can always arrange that, if the manifold is connected, we only need one $0$-handle and one $n$-handle, but this fact is not entirely trivial and we will try to prove it carefully later.)

So here is a sequence of examples:

Dimension $1$: $S^1$ again, but with more handles; notice the different possible ways you might cancel pairs of $0$- and $1$-handles:


Dimension $2$: Here is a standard picture of a torus decomposed into a $0$-handle, two $1$-handles and  a $2$-handle, with products in between. In terms of this decomposition into elementary cobordisms, the second $1$-handle is attached to the top of the product above the first $1$-handle. However, we can let the attaching map for the second $1$-handle flow down along the gradient vector field through the product and past the first $1$-handle (as long as we are not unlucky and don't get sucked into the first $1$-handle's critical point - this is again an issue to be discussed more carefully soon), and then see both $1$-handle attached simultaneously to the boundary of the $0$-handle.


Dimension $3$: Now it gets interesting. First, we must abandon hope of embedding the $3$-manifold in $\mathbb{R}^3$ and seeing the Morse function as the height function. So instead we will just draw some handle decompositions and, perhaps, some level sets of the Morse functions. First recall the three kinds of handles in dimension $3$:


We can put these together as follows, for a simple example:


Note that, after attaching the $2$-handle, we have a ball again, so we might as well not have attached the $1$- and the $2$-handle at all. I.e. these two handles can be cancelled in a way that will be made precise in due time. (This is the convertible roof, see end of Lecture 5 video for the hand gestures,) This picture is hard to look at so we can flatten it and draw only the images of the attaching maps in the boundary of the $0$-handle (identifying $S^2$ with $\mathbb{R}^2 \cup \infty$) as follows:


Note that we could have many $1$-handles, so some labelling of the feet is appropriate, and note that we only need to draw the core of the attaching map of each $2$-handle (the image of $S^1 \times \{0\} \subset S^1 \times B^1$) to specify the isotopy class of the attaching map. In fact, the same could be said for $1$-handles (assuming everything is oriented) but it is visually convenient to draw the whole disk. So here is another example:


So this has three $1$-handles, labelled $A$, $B$ and $C$, and three $2$-handles, labelled $a$, $b$ and $c$, and, of course, a $0$-handle that is the "background" to this picture and a $3$-handle that caps everything off. First, to see that you can cap it off with a $3$-handle you need to verify that the boundary is $S^2$.

Exercise: show that this manifold is $S^1 \times S^2$.

I'll end here, although in the lecture I then discussed Heegaard splittings. I'll write that up next time. Here's the video: