Is this geometric surface (flat torus) in $\mathbb R^3$ or not? I'm reading Barrett O'neill's Elementary Differential Geometry, and this example has me confused:

and the parametrization from Example 2.5 is:

lastly, Theorem 3.5 is:

On every compact surface $M \in \mathbb R^3$ there is a point at which
the Gaussian curvature $K$ is strictly positive.

What I understand is that we have that parametrization $x(u,v)$, which is mapping $(u,v)$ to 3 dimensions. However, we then make it a geometric surface by, instead of using the typical dot product for tangent vectors, using the metric tensor defined by $\langle x_u, x_u \rangle = 1$, etc. Due to the local isometry with $\mathbb R^2$, that means $T_0$ must have $K = 0$, which we typically couldn't have in $\mathbb R^3$ for a compact surface due to that theorem.
Here's my question: at the end he says that "$T_0$ can never be found in $\mathbb R^3$", but then what is the parametrization $x$ saying? that parametrization is explicitly a map $x: \mathbb R^2 \rightarrow \mathbb R^3$, so it seems like $T_0$ is in $\mathbb R^3$, even if it doesn't use the typical dot product of $\mathbb R^3$ as its inner product.
Which is it?
 A: The English is highly confusing in the text. The mapping he's giving is the parametrization of the usual torus in $\Bbb R^3$. The flat torus is $T_0$, which he describes by identifying the opposite edges of a square piece of paper. This torus $T_0$ can be parametrized as it sits in $\Bbb R^4$, as follows:
$$\mathbf x(u,v) = (\cos u,\sin u,\cos v,\sin v).$$
You can see that this is a product of circles, each in an $\Bbb R^2$.
A: Here, "to be found in $\mathbb{R}^3$" means that the torus is a sub-Riemannian manifold of $\mathbb{R}^3$. In particular, the metric in each point must be obtained by regarding tangent vectors on the torus as (tangent) vectors on $\mathbb{R}^3$ and computing computing their inner product w.r.t. the metric of $\mathbb{R}^3$. In more sophisticated terms, the metric on the embedded surface must equal the pullback of the ambient space's metric along the embedding. So as you pointed out yourself: "it doesn't use the typical dot product of $\mathbb{R}^3$ as its inner product", the latter is not the case for your torus.
So it is indeed a subset, but nor a sub-Riemannian manifold.
A: My interpretation is the following: suppose you were generating a model of the torus in space then while the point set would be given by the parametrization, the colour and lighting on a real torus would be given according to the metric pulled back from the ambient $\mathbb R^3$. So this flat torus would not appear as a real torus.
