Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I study surgery theory on 3-manifolds using the text book written by Gompf and Stipsicz.I can't understand \bf{slam-dunk} operation.

Let $K_{1}$ be the meridian of a knot $K_{2}$ in $S^3$, and $T$ is a tubular neighborhood of $K_{2}$.$K_{1}$ and $K_{2}$ has surgery coefficients $n \in \bf{N}$ and $r\in \bf{Q}\cup \{ \infty\}$ respectively.Then we surgery on $K_{2}$ removing $T$ and regluing $S^1 \times D^2$.

The book says that we can pull $K_{1}$ into $T$ and it intersects $\{ pt.\} \times D^2$ at only one point.I will be able to understand this facts because of integral surgery on $K_{2}$ if we can operate that,but I can't understand why we pull $K_{1}$ into $T$ forming $n\mu\pm\lambda$,where $\mu$ is the meridian of $T$ and $\lambda$ is the longitude of that.

Why can we this operation?

share|improve this question
1  
Crossposted to math.SE: math.stackexchange.com/q/265955/264 . In the future, please wait some time before posting your question in multiple fora, and when you do, provide links to the other posts - as you can imagine, it would be frustrating for someone to put time into answering your question here, only to see hear from you that you'd already gotten the solution elsewhere. –  Zev Chonoles Dec 28 '12 at 17:41
add comment

1 Answer

(Note: In an earlier version of this question, the asker referred to this very similar MO question. Apparently neither the explanation in Gompf and Stipsicz (page 163) nor the older MO answer is doing the trick, so I'll give a third explanation.)

We have a surgery diagram in a 3-manifold $M$ with two components $K_1$ and $K_2$. $K_2$ is arbitrary, but $K_1$ is a small linking circle for $K_2$; i.e. $K_1$ bounds a disk $D$ in $M$ which intersects $K_2$ exactly once. The surgery coefficient of $K_2$ is an integer $n$, and the surgery coefficient of $K_1$ is a rational number $r$.

Our goal is to show that if we delete $K_1$ from the surgery diagram and replace the coefficient of $K_2$ with $n-1/r$, the resulting surgery diagram describes the same 3-manifold as the original one.

Let $N_i$ be a regular neighborhood of $K_i$ in $M$. Let $S_i$ denote the surgery solid torus corresponding to $K_i$, so a meridian disk of $T_2$ in an $(n,1)$ curve on the boundary of $N_2$, and a meridian disk of $S_1$ is a $(p,q)$ (where $r=p/q$) on the boundary of $N_1$. Performing the surgery means replacing $N_i$ with $S_i$.

Let $A$ be a small annulus connecting the boundary of $N_1$ to the boundary of $N_2$. We can take $A$ to be a punctured version of the small disk $D$ referred to above. Let $R$ be a thickened neighborhood of $A$.

It is easy to see that $N_1 \cup R \cup N_2$ is a solid torus embedded in $M$. Furthermore, this solid torus is isotopic to $N_2$.

The key claim is that $S_1\cup R \cup S_2$ is also a solid torus. Once we have established this claim it follows that the result of the original surgery diagram can also be described by a single surgery on $K_2$. All that will be left to do is figure out the surgery coefficient.

To see that $S_1\cup R \cup S_2$ is a solid torus, notice that the meridian disk of $S_2$ intersects the annulus $A$ exactly once. (This is where the requirement that $n$ is an integer comes in.)

I'm running out of time, so I'll leave the calculation of the new surgery coefficient ($n-q/p$) as an exercise.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.