# Embedding torus in space such that its 6-fold symmetry extends

The following question is Problem 1.1.2.c in Thurston's book "Three-dimensional geometry and topology". I have not managed to solve it despite quite a bit of effort.

One can obtain a 2-dimensional torus $T$ by identifying the sides of a hexagon in an appropriate way (see, for example, here). By rotating this hexagon, we can obtain an order $6$ self-map of $T$. The question is whether we can embed $T$ into either $\mathbb{R}^3$ or $S^3$ such that this self-map extends to an order $6$ self-map of the ambient space. My guess is that the answer is "no", at least for $\mathbb{R}^3$. I'm less sure about $S^3$.

Thanks!

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Oh, here's a far simpler strategy. If the torus embeds, it bounds a solid torus on one side. What automorphisms are their of order $6$ for a solid torus? –  Ryan Budney Jan 26 '12 at 0:10
You don't need a classification result for this problem. Hint: An automorphism of a solid torus has to preserve the meridional class of the torus (up to sign) -- the non-trivial cycle that bounds a disc in the solid torus. –  Ryan Budney Jan 26 '12 at 0:21
Wow, I'm a little embarrassed that I did not see that. Thanks Ryan! –  Lucy Jan 26 '12 at 0:28
Further discussion should take place on Meta: tea.mathoverflow.net/discussion/1294/… Please upvote this comment so it appears "above the fold" –  David White Jan 29 '12 at 17:17
This comment thread got a bit out of hand with "meta conversation." This kind of conversation is a good thing, but I think it's worth keeping MO itself on topic. So I deleted several of the meta comments. However, the full comment thread has been copied to meta.MO: tea.mathoverflow.net/discussion/1294/… –  Anton Geraschenko Jan 30 '12 at 2:31

The answer is no for both $\mathbb{R}^3$ and $S^3$. I'll give the details for $\mathbb{R}^3$; the other case is similar. Fix an embedding $T^2 \hookrightarrow \mathbb{R}^3$. The first step is to show that $T^2$ is the boundary of a closed regular neighborhood $N$ of a knot. This is a nontrivial fact; for an exposition, see for example this. The space $N$ is a solid torus, and thus up to homotopy there exists exactly one simple closed curve $\gamma$ in $T^2$ which bounds a disc in $N$. Any homeomorphism $\phi : \mathbb{R}^3 \rightarrow \mathbb{R}^3$ which preserves $T^2$ must take $\gamma$ to a curve on $T^2$ homotopic to $\gamma$. In other words, the restriction of $\phi$ to $T^2$ must fix a nonzero vector in $H_1(T^2;\mathbb{Z})$. But the automorphism in the question fixes no such vector.
You don't need to know that $N$ is a solid torus. It is enough to know that $N$ is a compact manifold with $T = \partial N$. Then by "one-half lives, one-half dies" (see Lemma 3.5 of Hatcher's 3-manifold notes) there is a unique cyclic subgroup $Z$ of $H_1(T)$ killed by the inclusion map of $T$ into $N$. So the homeomorphism fixes the subgroup $Z$ and thus has an eigenvalue of $\pm 1$, a contradiction. –  Sam Nead Jan 28 '12 at 9:20