# 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: