# Is there a symmetric embedding of the torus in $\mathbb{R}^3$?

The torus has the following symmetry exchanging the two circle coordinates: \begin{align} s : S^1 \times S^1 &\to S^1 \times S^1 \\ (x,y) &\mapsto (y,x) \end{align} This symmetry is not really visible with the usual embedding of the torus in $$\mathbb{R}^3$$. So this has inspired the following question: is there an embedding $$\varphi : S^1 \times S^1 \to \mathbb{R}^3$$ and an homeomorphism $$f : \mathbb{R}^3 \to \mathbb{R}^3$$ such that $$f \circ \varphi = \varphi \circ s$$?

In fact, I'm quite sure that it doesn't exist, but I don't know how to prove it. Studying the homology groups of $$\mathbb{R}^3 \setminus \phi(S^1 \times S^1)$$ for any embedding $$\phi$$, I could prove that it has two path-connected components, let say $$P_1$$ and $$P_2$$. Because $$S^1 \times S^1$$ is compact, at least one of the two path-connected components must be bounded, let say $$P_1$$, and we have that $$P_1 \cup \phi(S^1 \times S^1)$$ is compact, while $$P_2 \cup \phi(S^1 \times S^1)$$ is not, so there is no homeomorphism of $$\mathbb{R}^3$$ exchanging the two path-connected components, so if we could prove that any homeomorphism $$f$$ such that $$f \circ \phi = \phi \circ s$$ must exchange the two path-connected components, then we could conclude that it doesn't exist. Let $$x_0 \in S^1$$ be any point of the circle, and $$i_1 : S^1 \hookrightarrow S^1 \times \{x_0\}$$ and $$i_2 : S^1 \hookrightarrow \{x_0\} \times S^1$$ the two obvious embeddings of the circle in the torus $$S^1 \times S^1$$. I think that the map $$\phi \circ i_1$$ must be non-trivial in either $$P_1 \cup \phi(S^1 \times S^1)$$ or $$P_2 \cup \phi(S^1 \times S^1)$$, but not in both, and $$\phi \circ i_2$$ must be non-trivial in the other one. If this is true, we could conclude, but I don't know how to prove this.

• Since $s$ changes the orientation of the torus, such an homeomorphism would reverse it as well. Doesn't that imply the exchange of path-connected components? Feb 10, 2022 at 19:07
• For example, in the differentiable case, I can picture the $2$ usual tangent fields on the torus and see that an permuting the "coordinate circles" implies exchanging those two vectors. So, it changes the sign of the normal vector field (the cross product), and so every curve that goes "outwards the torus" is transformed in another one "going inwards". Feb 10, 2022 at 19:11
• @Compacto The path components are not homeomorphic. Feb 10, 2022 at 19:26
• I know, but @QuinnLesquimau wants to prove that such an homeomorphism, if it existed, would also be a homeomorphism between them. That way one could conclude that there's no such homeomorphism. Feb 10, 2022 at 19:28

If the embedding is locally the standard embedding of the $$2$$-disk in the $$3$$-disk then we can define orientations using local homology and follow ideas from the comments to the question. Since $$s$$ switches the orientation of the torus $$T = S^1 \times S^1$$, $$f$$ must switch the two components of $$\mathbb{R}^3 \setminus \varphi(T)$$. Since $$\varphi(T)$$ is compact and hence bounded, exactly one of these two components is bounded. But a subset of $$\mathbb{R}^3$$ is bounded iff its closure is compact, so this is a property preserved by homeomorphisms.
• What do you mean by orientation in a topological setting? Why does it follow that $f$ must switch the two components? 2 days ago