Metric of the flat torus

Solution 1:

The standard metric on $\mathbb{R}^n$, namely $g = \sum_{i=1}^ndx^i\otimes dx^i$, is invariant under translations, i.e. for $f: \mathbb{R}^n \to \mathbb{R}^n$ of the form $f(x) = x + a$ where $a = (a^1, \dots, a^n) \in \mathbb{R}^n$, we have

\begin{align*} f^*g &= f^*\left(\sum_{i=1}^ndx^i\otimes dx^i\right)\\ &= \sum_{i=1}^n f^*dx^i\otimes f^*dx^i\\ &= \sum_{i=1}^nd(x^i\circ f)\otimes d(x^i\circ f)\\ &= \sum_{i=1}^nd(x^i + a^i)\otimes d(x^i+a^i)\\ &= \sum_{i=1}^ndx^i\otimes dx^i\\ &= g. \end{align*}

In particular, $g$ is invariant under translations by elements of the lattice $\mathbb{Z}^n$.

As $\pi : \mathbb{R}^n \to T^n$ is a smooth covering map, it is a local diffeomorphism. So given $q \in T^n$ and $p \in \pi^{-1}(q)$, $(\pi_*)_p : T_p\mathbb{R}^n \to T_qT^n$ is an isomorphism. Now for $v, w \in T_qT^n$, define

$$\hat{g}_q(v, w) := g_p((\pi_*)_p^{-1}(v), (\pi_*)_p^{-1}(w)).$$

To see this definition is well-defined (i.e. independent of the choice of $p$), note that if $p' \in \pi^{-1}(q)$, there is $a \in \mathbb{Z}^n$ such that $p' = p + a$. Letting $f : \mathbb{R}^n \to \mathbb{R}^n$ denote translation by $a$, we have $f(p) = p'$ and $\pi\circ f = \pi$, so $(\pi_*)_p = ((\pi\circ f)_*)_p = (\pi_*)_{f(p)}\circ(f_*)_p = (\pi_*)_{p'}\circ(f_*)_p$. Therefore,

\begin{align*} g_{p'}((\pi_*)_{p'}^{-1}(v), (\pi_*)_{p'}^{-1}(w)) &= g_{f(p)}((f_*)_p\circ(f_*)_p^{-1}\circ(\pi_*)_{p'}^{-1}(v), (f_*)_p\circ(f_*)_p^{-1}\circ(\pi_*)_{p'}^{-1}(w))\\ &= g_{f(p)}((f_*)_p\circ((\pi_*)_{p'}\circ(f_*)_p)^{-1}(v), (f_*)_p\circ((\pi_*)_{p'}\circ(f_*)_p)^{-1}(w))\\ &= g_{f(p)}((f_*)_p\circ(\pi_*)_p^{-1}(v), (f_*)_p\circ(\pi_*)_p^{-1}(w))\\ &= (f^*g)_p((\pi_*)_p^{-1}(v), (\pi_*)_p^{-1}(w))\\ &= g_p((\pi_*)_p^{-1}(v), (\pi_*)_p^{-1}(w))\\ &= \hat{g}_q(v, w). \end{align*}

So $\hat{g}$ is a well-defined Riemannian metric on $T^n$. By construction, $\pi^*\hat{g} = g$; that is, $\pi$ is a local isometry.

The exact same argument can be used to show that if $\pi : M \to N$ is a smooth covering map, and $g$ is a Riemannian metric on $M$ which is invariant under the deck transformations of $\pi$, then it descends to a Riemannian metric $\hat{g}$ on $N$, and $\pi^*\hat{g} = g$; that is, $\pi$ is a local isometry.