slam-dunk or Kirby move

Question

I have difficulty understanding the slam-dunk. In slam-dunk you have two knots K1 and K2 with framings r (rational) and n (integer) such that K1 is the meridian of K2. Now we perform surgery on K2 first and pull K1 into the solid torus glued in during surgery on K2. Is this not the same as second Kirby move? and if yes, why does the framing of K2 change after this operation?

Answer 1

The point of the slam-dunk is that it transforms a surgery diagram with two link components $K_1,K_2$ (satisfying the conditions you've already pointed out) to a surgery diagram with just the link component $K_2$ but with a different surgery coefficient. I believe the "second Kirby move" is called a 2-handle slide in Gompf and Stipsicz's book which is where I've learned this material. I may be misunderstanding your question or more likely, missing something, but a 2-handle slide doesn't change the number of components in your surgery diagram, whereas the slam-dunk does, so I don't see how to interpret the slam-dunk as a 2-handle slide.

What follows is how I see that the framing changes and is really a rewrite of pp.163-164 in Gompf and Stipsicz's book.

Let $T$ be the solid torus glued in during surgery on $K_2$. Because the surgery coefficient of $K_2$ is integral, $K_1$ intersects the disk $\{\text{pt}\}\times D^2$ once, and so once you pull $K_1$ into $T$, it is isotopic to $S^1\times\{\text{pt}\}$ in $T$. This means that $T$ is a tubular neighborhood of $K_1$. Thus if we were to perform the surgery specified by the coefficient $r$ on $K_1$ now, we would again just cut out $T$ and glue it back in a second time.

What has happened is that the integral surgery coefficient of $K_2$ ensures that surgery on $K_1$ ends up working with the same solid torus $T$ -- in some sense $K_1$ has been moved into the same spot that $K_2$ was (though strictly speaking $K_1$ sits in a new manifold after surgery on $K_2$). This means that we could instead of doing two surgeries on $K_2$ then $K_1$ which end up involving the same $T$, just perform one on $K_2$, but with a different surgery coefficient.

To see what that surgery coefficient is, let's consider the case where $n=0$. There the first surgery on $K_2$ is simply a $\pi/2$ rotation of $H_1(\partial T)$. This changes the slope of surgery on $K_1$ from $r$ to $-1/r$. For general $n$, we have $n$ additional twists in this picture so we get $n-1/r$.

Answer 2

I've had trouble understanding the explanation in Gompf-Stipsicz for the surgery coefficient after slam-dunking. Here's another explanation. Let $K_2$ be a knot with integral surgery coefficient $n\in\mathbb{Z}$, and let $K_1$ be a meridian for $K_2$ with surgery coefficient $r=p/q\in \mathbb{Q}\cup \{\infty\}$. Let $T_i$, $i=1,2$ be closed tubular neighborhoods of $K_i$ where $T_1$ "hugs close" to $T_2$ like a ring around a finger. More precisely, arrange so that $$T_1\cap T_2 = (\partial T_1) \cap (\partial T_2) = A$$ for an annulus $A\cong S^1\times I$. Let $\gamma_i$ be a curve in $\partial T_i$ along which solid disks are attached when surgery is performed on $K_i$. In addition let $\mu,\lambda$ in $\partial T_2$ denote a meridian and 0-linking longitude of $K_2$, so that $$[\gamma_2 ]=[\lambda]+ n[\mu]\in H_1(\partial T_2).$$ We can arrange so that $\gamma_1\cap A$ consists of $p$ arcs going from one boundary component of $A$ to the other. Label these cyclically going around $K_2$ as $\alpha_1, \ldots,\alpha_p$. The portion of $\gamma_1$ outside of $A$ consists of $p$ arcs $\beta_1,\ldots,\beta_p$ where $\beta_i$ connects one endpoint of $\alpha_i$ to the opposite endpoint of $\alpha_{i+q}$ [subscripts interpretted modulo $p$]. We can nudge $\gamma_1$ slightly so that each of the arcs $\alpha_i$ travels along a closed curve parallel to $\gamma_2$ in $\partial T_2$.

We now perform surgery on $K_2$ by removing $\mathrm{int}\,T_2$ and gluing in a solid torus $T_2'$ so that curves parallel to $\gamma_2$ in $\partial T_2$ bound disks. As explained in Gompf-Stipsicz, the idea is to now drag $K_1$ into $T_2'$ until it becomes the core circle of $T_2'$. The issue is that the arcs $\alpha_i$ are in the way. We remedy this by pushing the arcs $\alpha_i$ through disks in $T_2'$ while keeping the endpoints of $\alpha_i$ fixed. Call the resulting arc $\alpha_i'$, so that $\alpha_i'$ is a curve parallel to $\gamma_2$ in $\partial T_2$ minus a small arc [where $\alpha_i$ was originally]. Let $\gamma_1'$ denote the curve obtained from $\gamma_1$ after this procedure. We are now free to drag $K_1$ inside $T_2'$. Since the surgery coefficient on $K_2$ was integral, $K_1$ can be isotoped to become a core circle of $T_2'$. Subsequently we can push the arcs $\beta_1,\ldots,\beta_p$ across $T_1$ [with endpoints fixed] so that $\gamma_1'$ lies on $\partial T_2=\partial T_2'$. Note $$[\gamma_1']= -p[\lambda]+(q-np)[\mu]\in H_1(\partial T_2).$$ [One can think of each arc $\alpha_i'$ contributing $-[\lambda]-n[\mu]$, and the arcs $\beta_1,\ldots,\beta_p$ altogether contributing to the $q[\mu]$ term.] We now perform surgery on $K_1$ by removing $\mathrm{int}\,T_2'$ and gluing in a solid torus so that curves parallel to $\gamma_1'$ in $\partial T_2$ bound disks. The effect of all of this starting from the original diagram is to erase $K_1$ and perform surgery on $K_2$ with coefficient $$-\frac{q-np}{p}=n-\frac{1}{r}.$$