A (not existing) self-homeomorphism of the figure eight knot complement

Question

I was recently looking at the figure eight knot complement $M$, as a once-punctured torus bundle over the surface with monodromy $ A=\begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} $ and its sister with monodromy $-A$.

I'm confused about the following: Suppose I take this bundle, with peripheral subgroup: $\lambda$ the boundary of the torus, and $\mu$ the meridian, which is vertical in my bundle picture. Now suppose I twist the top of the bundle in my picture by a full twist of $2\pi$ before gluing. I can't seem to reconcile the following three "facts":

1. Since I didn't actually change the monodromy I clearly get a self-homeomorphism of $M$.
2. This homeomorphism takes $\lambda$ to itself and $\mu$ to $\mu+\lambda$.
3. Such a homeomorphism doesn't exist as any homeomorphism will have to take $\mu$ to $\pm\mu$ by a theorem from ``knots are determined by their complements" by Gordon–Luecke.

So then, which one is wrong? Any help appreciated!

Answer 1

1. It's not so easy to understand "the top of my picture" since there is no picture. In particular, you'd need to describe the alleged self-homeomorphism of $M$ better. So I'll interpret your question as essentially asking if there's a self-homeomorphism of $M$ that extends the Dehn twist you describe. So 1 and 2 seem to be the same question.

2. Is incorrect (ie there is no such homeomorphism) because

3. is correct.

On the other hand, appealing to Gordon-Luecke is overkill, in my opinion. The assertion you're using is that only the trivial surgery (remove a neighborhood of the knot and put it right back where you found it) is $S^3$. If a homeomorphism as described in 2 existed then you'd be claiming that $+1$ surgery on the figure 8 knot is $S^3$. Thurston in his famous notes worked out all of the surgeries on the figure 8 knot; none of them is $S^3$ and $+1$ surgery yields a (non-trivial) Seifert fibered space. (Here's a reference: Brittenham, Mark; Wu, Ying-Qing, The classification of exceptional Dehn surgeries on 2-bridge knots. Comm. Anal. Geom. 9 (2001), no. 1, 97–113.) In any event, you could certainly find a representation of its fundamental group in a finite group by searching with GAP or some such program.

Perhaps a more direct route would be to observe that any homeomorphism as in 2 would have infinite order up to isotopy (since its restriction to the boundary has infinite order). But it's a consequence of Mostow rigidity (applied to the figure 8 knot complement) that any self-homeomorphism would have finite order up to isotopy.