This exercise is from Do Carmo, Riemannian Geometry.

Introduce a Riemannian metric on the torus $T^n$ in such a way that the natural projection $\pi:\mathbb{R}^n\to T^n$ given by $$\pi(x_1,...,x_n)=(e^{2\pi ix_1},...,e^{2\pi ix_n})$$ is a local isometry. Show that with this metric $T^n$ is isometric to the flat torus.*

*The flat torus is just $T^n=S^1\times ...\times S^1$ with the product Riemannian metric.

I don't know if this is correct, but in the Lee's book Introduction to Smooth Manifolds, there's something about proper actions that may be related:

The discrete group $\mathbb{Z}^n$ acts smoothly, freely and properly on $\mathbb{R}^n$ by translations. Then there is a unique smooth structure making the quotient map into a smooth covering map. To verify that this smooth structure is the same as the one we defined proviously (thinking of $T^n$ as the product manifold $S^1\times ...\times S^1$) we just check that the covering map $\pi:\mathbb{R}^n\to T^n$ [defined above] is a local diffeomorphism with the product smooth structure on $T^n$. Hence $\mathbb{R}^n/\mathbb{Z}^n$ is diffeomorphic to $T^n$.

So basically me question is if we can use the fact that $\mathbb{Z}^n$ acts on $\mathbb{R}^n$ to give $\mathbb{R}^n/\mathbb{Z}^n$ a Riemannian structure? Is there any bibliography that can be useful?

Thank you.

  • 1
    $\begingroup$ Yes, that works. $\endgroup$ – Qiaochu Yuan May 15 '16 at 2:57

An attempt of proof, please review carefully and tell me your opinion. Thanks and kind regards.

Recall that the $n$-torus $T^n$ is defined as the quotient $\Bbb{R}^n/G$ where $G$ is the group of integer translations in $\Bbb{R}^n$. It can be identified with the product of $n$ circles: $$ T^n = \underbrace{S^1 \times \ldots \times S^n}_{n \text{ times}}. $$ We can then define a natural projection $\pi: \Bbb{R}^n \longrightarrow T^n$ given by $$ \pi(x) = (e^{ix_1}, \ldots, e^{ix_n}), \quad x = (x_1, \ldots, x_n) \in \Bbb{R}^n. $$ For $u = (u_1, \ldots, u_n) \in T_x\Bbb{R}^n = \Bbb{R}^n$ we have $$ d\pi_x(u) = J(x)u = i(u_1 e^{i x_1}, \ldots, u_n e^{i x_n}). $$ where $J(x)$ is the Jacobian matrix of $\pi$ at $x$ given by $$ J(x) = \begin{bmatrix} ie^{ix_1} & 0 & \cdots & 0 \\ 0 & ie^{i x_2} & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & i e^{i x_n} \end{bmatrix}. $$

We can just define $$ \langle d\pi_x(u), d \pi_x(v) \rangle_{\pi(x)} = \langle u, v \rangle, \quad u, v \in T_x\Bbb{R}^n = \Bbb{R}^n $$ where $\langle \cdot, \cdot \rangle$ denotes the inner product of the Euclidean space. To get a local isometry, it suffices to restrict $\pi$ to a neighborhood of $x$ such that it is a diffeomorphism.

We now show that the identity map $i: \Bbb{R}^n/G \longrightarrow T^n$ is an isometry, that is, we show that the two metrics are the same. Let $u,v \in T_pT^n$. Then \begin{align*} \langle u, v \rangle_{(e^{i x_1}, \ldots, e^{i x_n})} & = \sum_1^n \langle d \pi_j (u), d \pi_j(v) \rangle_{e^{i x_j}} \\ & = \sum_1^n \langle u_j e^{i (x_j + \pi/2)}, v_j e^{i (x_j + \pi/2)} \rangle_{e^{i x_j}} \\ & = \sum_1^n u_j v_j \\ & = \langle u, v \rangle, \end{align*}
which completes the proof.


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