An incompressible surface

a torus being compressedHave a look at the this torus, embedded in a standard way in S^3.  The shaded disk D is inside the torus with its boundary on the torus itself.  Consider \partial D, the boundary of D.  It represents a nontrivial loop in the torus, but since it bounds a disk in S^3, the loop is trivial in S^3.  Such a disk is called a compressing disk for the torus, and we can compress the torus by cutting along \partial D and gluing in two copies of D as shown in the second picture.  This is also called surgering along \partial D.  In this case, we obtain a sphere by compressing along D.


Notice that when we compress along D we decrease the genus of our surface by one.  With that in mind, consider the Seifert surface F for the knot K we looked at previously.  Are there any compressing disks for F?  Suppose there was a compressing disk D and \partial D didn’t separate F into two components .  Then we could compress F along D, obtaining a new surface F'.  Since we didn’t touch anything near the boundary of F, F' is still a Seifert surface for K.  But F is a punctured torus, so F' must be a disk, and the only knot bounding a disk is the unknot.

On the other hand, if there was a compressing disk D such that \partial D did separate F into two components, then one of these components would have to be an annulus.   As above, we would get a disk bounding the knot K.  (Why?  Working out these details is a fun exercise using Euler characteristic arguments.)

So if you believe that the trefoil K is knotted, then the surface F must be incompressible!


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