## Fibered nots (non-geometric HNN extensions of free groups normally generated by the monodromy)

A fibered knot is a knot $K$ in the $3$-sphere whose complement is a surface bundle over a circle. If $S$ is the fiber, the fundamental group of $S$ is free (of even rank), and the fundamental group of the complement is an HNN extension where the $Z$ is generated by the meridian of the knot.

Since surgery on $K$ recovers the $3$-sphere, the group $\pi_1(S^3-K)$ has the interesting property that it is normally generated by (the conjugacy class of) the meridian.

What I didn't realize until recently is that there are many examples of "non-geometric" automorphisms $\phi$ of free groups $F$ for which the associated HNN extension $F \to G \to Z$ is normally generated by the conjugacy class of the monodromy. One simple example is the case $F = \langle a,b,c \rangle$ and $\phi$ is the automorphism $a \to c^{-1}abac, b \to bac, c\to bc$.

Is there any systematic way of generating such examples? Is there a classification? One reason to be interested is that such examples can be used to construct smooth $4$-manifolds which are topologically $S^4$ but not obviously diffeomorphically $S^4$.

Edit: a link to the construction is http://lamington.wordpress.com/2009/11/09/4-spheres-from-fibered-knots/ (this explains the construction in the case of a fibered knot, but the group-theoretic condition is the only important ingredient).

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Should "generating element" be "stable letter"? Please explain the construction of the 4-manifold or give a link. Thank you. – Sam Nead Nov 11 2009 at 2:55
I was scratching my head about "fibered nots"... then I read the question and found out it was a typo in the title. – Ilya Nikokoshev Nov 14 2009 at 19:44
I assumed it was a deliberate joke. After all, these things aren't knots. Though it now seems to have been changed... – HW Nov 14 2009 at 19:56
How do you see that the fundamental group of S will have even rank? It seems that S could have multiple cusps (in particular, an even number of cusps) transitively permuted by the monodromy. – Tom Church Nov 14 2009 at 22:28
This doesn't happen for knots in $S^3$ for homological reasons. – Sam Nead Nov 14 2009 at 22:38