The difference between the algebraic torus and the geometric torus

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?

The question can be formalized as finding a homeomorphism (or diffeomorphism) from the torus to itself that interchanges the two generators of its fundamental group. (These are represented concretely in the torus of revolution in $\mathbb R^3$ by one circle that goes around the big center hole, and another circle that goes around the "air space" inside the tire.)

If you're only interested in the torus as a topological space, it always has a self-homeomorphism that interchanges the two generators. Similarly, if you consider it as a smooth manifold, it has a self-diffeomorphism that does the same thing. This is most easily seen by representing the torus as $S^1\times S^1\subseteq\mathbb R^2\times\mathbb R^2 = \mathbb R^4$, where the self-diffeomorphism is given by $(x^1,x^2,x^3,x^4) \mapsto (x^3,x^4,x^1,x^2)$.

But if you want to consider it as a Riemannian manifold, then you're probably interested in finding an isometry that interchanges the generators. The diffeomorphism above is an isometry of the flat metric on $S^1\times S^1$ induced from the Euclidean metric on $\mathbb R^4$, so the two generators on that torus are geometrically indistinguishable from each other. But on the torus of revolution in $\mathbb R^3$ with its induced metric, there is no such isometry. One way to see this is to note that the big circle has a representative (the smallest inner circle) that passes only through points where the Gaussian curvature is negative. But every representative of the circle that goes around the "air space" has to pass through points where the Gaussian curvature is positive (the outer circle).

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.