# Why does the cartesian product of two two-dimensional figures lie in the three-dimensional space?

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?

• 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 $R^3$. Nov 20 at 15:31
• 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. Nov 20 at 15:35
• @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. Nov 20 at 15:42
• @Cheerful Parsnip, very interesting! Thank you! Nov 20 at 15:42
• Yes, it is actually complicated when stated properly. Nov 20 at 16:06

If $$N$$ is a compact orientable $$n$$-dimensional manifold, then $$N$$ embeds in $$\mathbb{R}^{ 2 n − 1}$$.
• @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. Nov 20 at 20:51