This problem was given to me by a friend:

Prove that $\Pi_{i=1}^m \mathbb{S}^{n_i}$ can be smoothly embedded in a Euclidean space of dimension $1+\sum_{i=1}^m n_i$.

The solution is apparently fairly simple, but I am having trouble getting a start on this problem. Any help?

  • $\begingroup$ Induction maybe? $\endgroup$
    – tomasz
    Jun 21, 2012 at 17:54
  • $\begingroup$ I don't know but I would try suitable generalized spherical coordinates. Much the same way we do the well known embedding of the common torus $\mathbb{S}^1\times \mathbb{S}^1$ as a surface in $\mathbb{R}^3$. $\endgroup$ Jun 21, 2012 at 17:56

2 Answers 2

  • Note first that $\mathbb{R}\times\mathbb{S}^n$ smoothly embeds in $\mathbb{R}^{n+1}$ for each $n$, via $(t,\textbf{p})\mapsto e^t\textbf{p}$.
  • Taking the Cartesian product with $\mathbb{R}^{m-1}$, we find that $\mathbb{R}^m\times\mathbb{S}^n$ smoothly embeds in $\mathbb{R}^{m}\times\mathbb{R}^n$ for each $m$ and $n$.
  • By induction, it follows that $\mathbb{R}\times\prod_{i=1}^m \mathbb{S}^{n_i}$ smoothly embeds in a Euclidean space of dimension $1+\sum_{i=1}^m n_i$.

The desired statement follows.

  • $\begingroup$ Very nice!! (and other characters) $\endgroup$ Feb 19, 2013 at 19:37
  • $\begingroup$ can you please elaborate the induction part? how to go about it? $\endgroup$
    – RagingBull
    Apr 19, 2019 at 16:51

EDIT: I cannot delete this post as it's been already accepted :(

The proof is carried out by induction on $m.$

$m=1$ is trivial by choosing coordinates $(x^{(1)}_0,x^{(1)}_1,...,x^{(1)}_{n_1})$ where $\sum_{j=0}^{n_1} (x_j^{(1)})^2=1$, so let $m=2,$ then similarly $S^{n_1} \times S^{n_2}$ is embedded in $\mathbb{R}^{n_1+n_2+2}$ which also lies in the hypersurface $H$ with equation $\sum_{j=0}^{n_1} (x_j^{(1)})^2+\sum_{j=0}^{n_2} (x_j^{(2)})^2=2.$ In fact, $H$ is diffeomorphic to $S^{n_1+n_2+1}.$ Now, the embedding is missing at least a point, for example $p=(0,...,0,1,1) \in H,$ so by stereographic projection $S^{n_1+n_2+1} \setminus \{p\}$ is diffeomorphic to $\mathbb{R}^{n_1+n_2+1}.$

Suppose that the assertion holds for $<m,$ then $(S^{n_1}\times...\times S^{n_{m-1}}) \times S^{n_m}$ is embedded diffeomorphically in $\mathbb{R}^{n_1+...+n_{m-1}+1} \times \mathbb{R}^{n_m+1}\cong \mathbb{R}^{n_1+...+n_{m}+2}$ hence following the same argument you can reduce the dimension by $1,$ so the result.

  • 2
    $\begingroup$ Would you mind explaining the 'hence following the same argument'? Suppose you know that $S^{n_1}\times S^{n_2}$ embeds into $\mathbb{R}^{n_1+n_2+1}$, then I can see that $(S^{n_1}\times S^{n_2})\times S^{n_3}$ embeds into $\mathbb{R}^{n_1+n_2+n_3+2}$, but how do you reduce the dimension? You need an analogue for $H$, but the embedding $S^{n_1}\times S^{n_2}\hookrightarrow\mathbb{R}^{n_1+n_2+1}$ does not lie in some scaled version of $S^{n_1+n_2+1}$, right? So what do you take for $H$? $\endgroup$ Dec 15, 2012 at 18:26
  • 1
    $\begingroup$ I did not see how you can use this argument inductively. It works only on the first step it seems. $\endgroup$
    – Marra
    Sep 11, 2013 at 19:40
  • $\begingroup$ Dear Daan and Gustavo, seems there's no way to use the induction! urgh... $\endgroup$ Sep 12, 2013 at 19:43

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