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 $$\pi_1(S) \to \pi_1(S^3-K) \to Z$$ 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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    $\begingroup$ I assumed it was a deliberate joke. After all, these things aren't knots. Though it now seems to have been changed... $\endgroup$
    – HJRW
    Nov 14, 2009 at 19:56
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    $\begingroup$ 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. $\endgroup$
    – Tom Church
    Nov 14, 2009 at 22:28
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    $\begingroup$ This doesn't happen for knots in $S^3$ for homological reasons. $\endgroup$
    – Sam Nead
    Nov 14, 2009 at 22:38
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    $\begingroup$ What Sam said - homology implies that the fundamental group of S has even rank. But it is a good question, and it would be nice to have some simple criteria to show that an even-rank example is not geometric. The title was a deliberate joke, for the reason Henry gives, and I have changed it back. Greg, what gives?! $\endgroup$ Nov 15, 2009 at 0:27
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    $\begingroup$ @Theo Johnson-Freyd : The spelling of the title has been corrected before, and Danny changed it back. If Danny wants it this way, then I think we should respect his wishes, so I have reverted your edit. $\endgroup$ Jul 9, 2013 at 6:00

1 Answer 1


This question seems to be interesting even for the rank 2 free group. In this case, it is well-known (since the commutator is preserved up to conjugacy and inverse) that the outer automorphisms are the same as the mapping class group of a 2-punctured torus. The only possible candidates for outer automorphisms satisfying your condition are determined homologically as homology circles, and correspond to the trefoil and figure eight knot complements. But of the automorphisms corresponding to those outer autmorphisms, I'm not sure which satisfy your condition. I think this amounts to asking which curves in the figure 8 & trefoil knot complements normally generate the fundamental group. If one considers peripheral curves, then this is answered by the knot complement problem (I think the attribution of this for the figure 8 knot is Thurston, who classified the Dehn fillings on the figure 8 - I don't know who originally did this for the trefoil).

Another obervation is that if a curve normally generates, then there are finitely many conjugates that generate the fundamental group. Jason Callahan has ruled out the case of a non-peripheral curve in the figure 8 knot complement such that two conjugates generate the fundamental group. There's probably some other 3-manifold machinery that may be brought to bear on this question in the rank 2 case.


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