TITLE: The difference between the algebraic torus and the geometric torus
QUESTION [6 upvotes]: I know that the donut-shaped geometric object in $\mathbb{R}^3$ is homeomorphic to a square with identified opposite sites. However, while the latter has a clear symmetry between two dimensions, the former doesn't (as an idea to formalize this, I would think of the fact that you can put a string through the donut hole, and the string then identifies one direction of the two). 
Just as I was writing this question, I saw a similar question, Why is $S^1 \times S^1$ a Torus? ,
and a comment there says that representing the torus in $\mathbb{R}^4$ in $\mathbb{R}^3$ instead is what breaks symmetry. What I wonder is how this symmetry can be described in geometric terms? That is, which quantity tells us that one of the torus manifolds in $\mathbb{R}^3$ and $\mathbb{R}^4$ has some symmetry and the other doesn't? If you don't agree with my thoughts about symmetry, what other differences are there between the two?

REPLY [7 votes]: If you look at the torus of revolution in $\mathbb{R}^3$ and at $S^1 \times S^1$ using purely topological glasses, then they are indeed the same (homeomorphic). If, however, you look at them using "geometric" glasses, then they are different. By "geometric" glasses, I mean you are allowed to measure lengths, areas, angles, etc.
Given a submanifold $M$ of $\mathbb{R}^n$, one can use the inner product of $\mathbb{R}^n$ to endow $M$ with a Riemannian metric. The torus of revolution in $\mathbb{R}^3$ with the induced metric and the torus of revolution in $\mathbb{R}^4$ that comes from the embedding $S^1 \times S^1 \rightarrow \mathbb{R}^2 \times \mathbb{R}^2$ with the induced metric are not isometric - in fact, the torus in $\mathbb{R}^4$ is flat (and so is called the flat tori) while the torus of revolution has areas of positive, zero and negative curvature.
If you are not familiar with Riemannian metrics, you can think about this as follows: Given a submanifold $M$ of $\mathbb{R}^n$, one can use the inner product of $\mathbb{R}^n$ to measure the length of curves on $M$ (just by considering them as curves in $\mathbb{R}^n$ and calculating the usual length) and the angle between two curves on $M$. Using this observation, you can define an "intrinsic" distance function on $M$. Declare the distance between two points $P$ and $Q$ on $M$ to be the infimum of the lengths of all possible curves in $M$ connecting $P$ and $Q$. This gives you a distance function on $M$, turning it into a metric space. 
It turns out that by running this procedure for the two different embeddings of the tori (one in $\mathbb{R}^3$ and one in $\mathbb{R}^4$), you obtain two different, homeomorphic, diffeomorphic, but not isometric spaces. There isn't a distance preserving map between the two tori!
As for your comment about the symmetry, this can be formulated by saying that the group of symmetries (group of all distance preserving diffeomorphisms) of the flat tori is "larger" (has more degrees of freedom, has a larger dimension as a Lie group) then the group of symmetries of the tori of revolution. For the flat tori, identifying $\mathbb{C}$ with $\mathbb{R}^2$ and $S^1$ with $\{ e^{i\theta} \, | \, \theta \in \mathbb{R}\}$, given $\mu_1, \mu_2 \in \mathbb{R}$, the map
$$ (e^{i\theta_1}, e^{i\theta_2}) \mapsto (e^{i(\theta_1 + \mu_1)}, e^{i(\theta_2 + \mu_2)}) $$
is a distance preserving diffeomorphism of the flat torus. Using such a map, any point on the flat torus can be mapped to any other point and so all points "look" alike - we say that the flat torus is homogeneous. 
The group of symmetries of the torus of revolution is much smaller - if we consider only orientation preserving symmetries and the axis of revolution is the $z$ axis, then then we can rotate the torus around the $z$ axis by any angle or rotate the torus $\pi$ degrees clockwise or counterclockwise around any line on the $xy$ axis passing through the origin. In particular, you can't map any point to any other point using a distance-preserving symmetry.