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](http://en.wikipedia.org/wiki/Hexagonal_tiling)). 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!



Embedding torus in space such that its 6-fold symmetry extends
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](http://math.iisc.ernet.in/~gadgil/expos/Tori.pdf). 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.


