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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
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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
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Wow, I'm a little embarrassed that I did not see that. Thanks Ryan! –  Lucy Jan 26 '12 at 0:28
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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
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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
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1 Answer

up vote 17 down vote accepted

Lurking here on MO, I've noticed that unanswered questions get bumped to the top periodically. Since this question was answered by Ryan Budney in the comments, I've decided to write his answer here (marked "community wiki" so I get no reputation points) to prevent this from happening.

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.

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Welcome to MO, Lucy. –  Ryan Budney Jan 28 '12 at 5:54
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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
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