It is easy to see that $T^2=S^1 \times S^1$ can be embedded in $\mathbb{R}^4$ but also there is an embedded in $\mathbb{R}^3$.

The question is $T^n = S^1 \times \ldots \times S^1$ can be embedded into $\mathbb{R}^{n+1}$, $n>2$.


If we have an embedding $T^k\subset\mathbb{R}^{k+1}$ then the subspace inclusion $\mathbb{R}^{k+1}\subset\mathbb{R}^{k+2}$ allows us to view $T^k$ as embedded in $\mathbb{R}^{k+2}$. The boundary of the tubular neighborhood of $T^k$ in $\mathbb{R}^{k+2}$ will then be a torus $T^{k+1}$ embedded in $\mathbb{R}^{k+2}$, and we proceed inductively.

  • $\begingroup$ I like your answer very much. But the tubular neighborhood defines an embedding of $T^{k+1}$ only after restricting it I guess. $\endgroup$ Dec 4 '14 at 7:45

You know that $S^1$ embeds in $\mathbb{R}^2$, but also $S^1\times\mathbb{R}$ embeds in $\mathbb{R}^2$ (consider the homeomorphism of the cylinder to $\mathbb{R}^2\setminus(0,0)$).

Hence $S^1\times\mathbb{R^2}$ embeds in $\mathbb{R}^3$, but since $S^1$ embeds in $\mathbb{R}^2$ you have that $S^1\times S^1$ embeds in $\mathbb{R}^3$.

You can then proceed by induction: assuming $T^{n-1}$ embeds in $\mathbb{R}^{n}$ (and also $T^{n-1}\times\mathbb{R}$ embeds in $\mathbb{R}^{n}$), you have that $T^{n-1}\times\mathbb{R}^2$ embeds in $\mathbb{R}^{n+1}$ and hence $T^{n}=T^{n-1}\times S^1$ embeds in $\mathbb{R}^{n+1}$ (and $T^{n}\times\mathbb{R}$ embeds in $\mathbb{R}^{n+1}$).


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