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Defining a torus $T$ as $S^1 \times S^1$, it should follow that $T \subseteq \mathbb{R}^4$. But you can also think of a torus as a bagel, which means it's a subset of $\mathbb{R}^3$. Can anyone clarify this point?

user78720
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    Perhaps the issue would be clarified by asking the same question, but for a simpler space: Defining a one-point topological space $P$ as ${0}$, it should follow that $P\subseteq\mathbb{R}$. But you can also think of a point as being any given point in any topological space whatsoever. The resolution is that there is no "the one-point topological space"; any occurrence is just as good as any other. See the nLab article on "generalized the". – Zev Chonoles May 21 '13 at 04:50

3 Answers3

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These are two different embeddings of the same topological space, i.e. the surface of the bagel in $\mathbb R^3$ and the product of two circles in $\mathbb R^4$ are homeomorphic. Note that they are not isometric though; the torus embeded in $\mathbb R^4$ as a product of circles is flat, while the surface of the bagel has regions of positive curvature and of negative curvature.

Matt E
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Neither! A torus is an abstract topological space defined, as you say, to be the product $S^1 \times S^1$. Since, as it happens, $S^1 \subset \mathbb{R}^2$, we get a natural embedding of "the" torus into $\mathbb{R}^4$. However, there are of course other ways of representing this space; one can change coordinates, apply functions to it, and so on. And among these alterations are some that "project" the torus down into $\mathbb{R}^3$, where it resembles a bagel.

This is a little bit of an aberration, since not every two-dimensional manifold can be embedded in $\mathbb{R}^3$, the Klein bottle being a counterexample (also, this is related to your question in that it is more or less a twisted torus). And more generally, an $n$-dimensional manifold can be embedded in $\mathbb{R}^{2n}$ but not necessarily lower. So if you insist on thinking of a manifold as "being" a subset of some $\mathbb{R}^N$, that's the one to choose.

Ryan Reich
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  • A product $S^1\times S^1$ is not a subset of $\mathbb R^4$? – user42912 May 21 '13 at 06:28
  • It can be, but does not have to be. If you want to look at it that way, there is an obvious way of doing it. Think of it this way: a line is a subset of $\mathbb{R}^2$. So is the product of two lines, I.e. a plane, a subset of $\mathbb{R}^4$? Yes, but not necessarily. – Ryan Reich May 21 '13 at 12:15
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Neither. The torus is generally defined as an abstract topological space, independently of any "embedding" into another space. So in the strict sense, it is incorrect to say $T\subseteq \mathbb R^4$ or $T\subseteq \mathbb R^3$.

However, we often talk about abstract spaces like the torus as subsets of other spaces, commonly $\mathbb R^n$ for some $n$. How do we resolve this? The answer is something called an embedding. We say that a space $A$ can be embedded into a space $B$ if there is a continuous function $f:A\to B$ which maps $f$ injectively onto a subset of $B$, such that $f$ restricted to $B$ has a continuous inverse in the subspace topology$^1$. What you're noticing is that there are embeddings of the torus into $\mathbb R^3$ and $\mathbb R^4$.

  1. Note that this is for topological spaces. If we add additional structure to the space, then the embedding is required to preserve this structure in a suitable way. It is common for example to give the torus a manifold structure, and the embedding is then required to be an embedding of manifolds, which is somewhat involved to define. Depending on the additional structure imposed, the torus may fail to embedd into $\mathbb R^3$, as Matt points out.
Alex Becker
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  • How an abstract torus without any embeddings is defined? – user42912 May 21 '13 at 06:19
  • @user42912 As a set with a topology. I should mention that there are some subtle issues involved in calling the torus "the" torus, which I haven't gotten into. – Alex Becker May 21 '13 at 06:30
  • yes I know, but what kind of set and what kind of topology? – user42912 May 21 '13 at 06:32
  • @user42912 It can be whatever you want. Certain elements of $\mathbb R^4$ are notationally convenient. But generally when one says "the" torus one is referring to the equivalence class of spaces homeomorphic to whatever particular choice you make in defining it. – Alex Becker May 21 '13 at 07:14