Distance between two points on the Clifford torus How can I obtain the distance between two points $\mathbf{x}=(x_1,x_2,x_3,x_4)$ and $\mathbf{y}=(y_1,y_2,y_3,y_4)$ that belong to the $2$-torus $\mathbb{S}^1\times \mathbb{S}^1$? This is, I want to measure the distance (I do not require the geodesic) of $\mathbf{x}$ to $\mathbf{y}$ along the manifold
$$\mathbb{S}^1\times\mathbb{S}^1=\big\{(x_1,x_2,x_3,x_4)\in\mathbb{R}^4:x_1^2+x_2^2=1,x_3^2+x_4^2=1\big\}.$$
I guess that there should be a way of doing this considering an isomorphism(?) via the parametrization $\mathbf{x}(\theta,\phi)=(\cos\theta,\sin\theta,\cos\phi,\sin\phi)$ with $\theta,\phi\in[-\pi,\pi)$ and then measuring the distances in $[-\pi,\pi)^2$. However, I do not know whether this could be possible nor how to prove that the parametrization is an isomorphism, since I will need to be able to compute distances in $\mathbb{S}^1\times\mathbb{S}^1$ for doing that.
Any ideas on this or in another direction will be appreciated, as well as references.
 A: Harald's answer is correct (though the wording of the justification is not quite: we need that the parametrization is Euclidean, not just flatness of the image). 
Your initial thought of how to approach it is just about right, except you want to measure the distances on the flat torus $\mathbb T^2 = (\mathbb R / 2 \pi \mathbb Z)^2$ rather than in $[-\pi,\pi)^2$, since doing the latter would give large distances for points on either side of $\pi$, ignoring the "wrap-around".
What you need to show is that your parametrization $\mathbf x : \mathbb T^2 \to \mathbb R^4$ is an isometric immersion. To do this, we just need to compute the (pullback of the) Euclidean metric in the coordinates from $\mathbb T^2$. Starting with 
$$dx^i = \frac{\partial x^i}{\partial \theta} d\theta + \frac{\partial x^i}{\partial \phi} d\phi,$$
compute from your parametrization
$$\frac{\partial\mathbf{x}}{\partial\theta}=\left(-\sin\theta,\cos\theta,0,0\right),\ \frac{\partial\mathbf{x}}{\partial\phi}=\left(0,0,-\sin\phi,\cos\phi\right);$$
so 
$$dx^{1}=-\sin\theta d\theta,\ dx^{2}=\cos\theta d\theta,\ dx^{3}=-\sin\phi d\phi,\ dx^{4}=\cos\phi d\phi.$$
Thus we have $$\delta_{ij}dx^{i}dx^{j}=\sin^{2}\theta d\theta^{2}+\cos^{2}\theta d\theta^{2}+\sin^{2}\phi d\phi^{2}+\cos^{2}\phi d\phi^{2}=d\theta^{2}+d\phi^{2},$$
which is the standard metric on the flat torus $\mathbb T^2$. Thus these spaces are isomorphic as Riemannian manifolds and thus as metric spaces, so you can take the shortest Euclidean distance $\sqrt{\Delta \theta^2 + \Delta \phi^2}$ as suggested by Harald.
