8
$\begingroup$

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.

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

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.