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So, here is the question. Assume we have two circles. They lie in the two-dimensional space. The cartesian product of the sets representing two circles consists of 4-tuples, and thus, one may expect a geometric representation of that set in the fourth-dimensional space. Indeed, one may consider the cartesian product of a line (one-dimensional space) and a circle (two-dimensional space) as a cylindrical surface (one plus two, thus three-dimensional space). However, for two circles, one may come up with the torus representation, which seems intuitively reasonable. So, what makes two circles special in this case, and is there a general rule which allows us to determine whether there is a representation in a space of a lower dimension than the sum dimension?

P.S. I'm not sure about the tags. Sorry if they're misleading.

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    $\begingroup$ You should learn some basic topology in order to formulate more precise questions along these lines. I think what you are asking is for a necessary/sufficient conditions for the product of two planar sets to be topologically embeddable in $R^3$. $\endgroup$
    – Moishe Kohan
    Nov 20 at 15:31
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    $\begingroup$ The fact that both circles are 1 dimensional makes it plausible, but even that is not enough. Consider an X shape. The product of it with itself is not embeddable in 3 dimensions because of the local structure around the intersection point. $\endgroup$
    – Cheerful Parsnip
    Nov 20 at 15:35
  • $\begingroup$ @Moishe Kohan, ok, I will. I don't study topology or something related. I was just curious. Someone changed the tags to include general topology, but I hardly expected it would be so complicated. Thank you. $\endgroup$
    – Seeker
    Nov 20 at 15:42
  • $\begingroup$ @Cheerful Parsnip, very interesting! Thank you! $\endgroup$
    – Seeker
    Nov 20 at 15:42
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    $\begingroup$ Yes, it is actually complicated when stated properly. $\endgroup$
    – Moishe Kohan
    Nov 20 at 16:06

1 Answer 1

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From wikipedia:

If $N$ is a compact orientable $n$-dimensional manifold, then $N$ embeds in $\mathbb{R}^{ 2 n − 1}$.

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    $\begingroup$ Wow, it seems the question that arose from my curiosity is not so stupid :) $\endgroup$
    – Seeker
    Nov 20 at 15:36
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    $\begingroup$ It’s not a stupid question at all! It’s a great question. $\endgroup$
    – Joe
    Nov 20 at 16:16
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    $\begingroup$ I don't understand how this answers the question. $\endgroup$
    – Cheerful Parsnip
    Nov 20 at 20:41
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    $\begingroup$ @CheerfulParsnip The last sentence of the question asks for properties that guarantee an embedding into $\mathbb{R}^{2n-1}$. Whitney's theorem provides one. This answer quotes that theorem, with a reference. Proving it here is out of the question. $\endgroup$
    – Ethan Bolker
    Nov 20 at 20:51
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    $\begingroup$ @EthanBolker but the OP was explicitly asking about products (OP says "sum dimension"). For manifolds which are subsets of the plane, this really only applies to products of 1 -manifolds, which is nothing additional to what the OP already knows. $\endgroup$
    – Cheerful Parsnip
    Nov 21 at 4:53

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