I am studying the flat torus $T^n=\mathbb{R}^n/\mathbb{Z}^n$. I am interested in the metric and the connection used. Unfortunately, in the books I am reading those things aren't defined. Does anyone knows this definition or a reference where I can find it? Thanks in advance for the help.
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3One natural metric is the one derived from $\mathbb R^n$: for $x, y \in \mathbb R^n$, $d(x + \mathbb Z^n, y + \mathbb Z^n) = \min{d_{\mathbb R^n}(x + x', y + y') \mid x', y' \in \mathbb Z^n}$. – Tunococ Jul 23 '13 at 08:36
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@Tunococ Isn't it more natural to define $d(x + \mathbb Z^n, y + \mathbb Z^n) = \min{d_{\mathbb R^n}(x, y + y') \mid y' \in \mathbb Z^n}$? If we think of the torus as the unit square with opposite sides identified, then to find the distance between $\overline{x}$ and $\overline{y}$ I start from $x$ and try to draw the shortest path to $y$, going through the sides if neccesary. So it feels like the $x$ should be fixed. – Blue Nov 11 '20 at 03:17
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Perhaps my definition doesn't actually give a metric; it's just the intuitive picture I have. – Blue Nov 11 '20 at 03:17
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You are right actually. Your definition of the same as mine. – Tunococ Nov 20 '20 at 13:12
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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.
Michael Albanese
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I found in Jost (Riemannian Geometry and geometric analysis, pages 3, 32) some deatils that may improve your answer. We can define the riemannian metric in a way such that $\pi: \mathbb{R}^n\to T^n$ is a local isometry: we choose an open subset $V\subset \mathbb{R}^n$ that doesn't contain two equivalent points, let $U=\pi(V)$, on the chart $(U,\pi^{-1}|_U)$ we define the metric as the one in $\pi^{-1}(U)$, this is well defined because we have that different points in different components of $\pi^{-1}(U)$ are obtained by translation. Do you think I may join your answer and this details? – John N. Jul 23 '13 at 09:10
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@JohnN. It's really as simple as Michael Albanese says. The action of $\mathbb{T}^N$ on itself by (left) translation canonically trivialises $T \mathbb{T}^n$, so that canonically, $T \mathbb{T}^n \cong \mathbb{T}^n \times \mathbb{R}^n$, which therefore admits the flat metric $g = dx^1 \otimes dx^1 + \cdots + dx^n \otimes dx^n$, whose Levi–Cività connection is the flat connection $d$... – Branimir Ćaćić Jul 23 '13 at 11:14
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2As often happens, John N., you're talking about the metric space structure and Michael was referring to the Riemannian metric. The latter gives the former and is more useful, since you want connection and curvature. – Ted Shifrin Jul 23 '13 at 11:14
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...which, If you're careful with your bookkeeping, you really can realise as being induced by the standard, $\mathbb{Z}^n$-invariant metric and connection on $\mathbb{R}^n$. – Branimir Ćaćić Jul 23 '13 at 11:15
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to Ted Shifrin: I was talking about the Riemannian metric. I should have written the complete name, sorry. – John N. Jul 23 '13 at 11:52
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To Branimir Ćaćić: I didn't understand your explaination, in particular: what is the action of (left) translation of $T^n$ on itself when $T^n=\mathbb{R}^n/\mathbb{Z}^n$? – John N. Jul 23 '13 at 12:05
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Since $\mathbb{Z}^n$ is a (trivially) normal subgroup of the abelian group $(\mathbb{R}^n,+)$, $\mathbb{T}^n = \mathbb{R}^n/\mathbb{Z}^n$ remains an abelian group under the induced group action. Indeed, $\mathbb{T}^n$ is a locally compact abelian Lie group. – Branimir Ćaćić Jul 27 '13 at 11:02
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I am slightly confused. Do we really need to show that the definition is well defined? I am probably mistaken, yet feel like your definition of the inverse is wrong. Rather I have seen it as $((\pi^{-1}))_q$, with the order of $$ and $^{-1}$ switched. In this way $((\pi^{-1})){\pi(p)}\circ(\pi)p=(id*)_p$. What am I missing? – plebmatician Apr 16 '19 at 08:01
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1@plebmatician: What does $\pi^{-1}$ mean given that $\pi$ is not invertible? – Michael Albanese Apr 16 '19 at 12:27
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Oh that is a good point. In the proof of showing that $(\pi_)p$ is an isomorphism we use that the smooth covering map is locally diffeomorphic around a neighbourhood $U$ of $p$, so that at $p$ we identify $(\pi)p$ with $((\pi\lvert_U)*)_p$ and then I guess $\pi^{-1}$ would mean $(\pi\lvert_U)^{-1}$. Does that make sense? – plebmatician Apr 16 '19 at 12:41
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1@plebmatician: Sure, but you still have to make a choice, namely $p \in \pi^{-1}(q)$. In order for the metric to be well-defined, it should not depend on the choice of $p$. – Michael Albanese Apr 16 '19 at 13:45