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Let $S : \mathbb{R}^n → \mathbb{R}^n$ be linear invertible map, then $S$ projects to $\mathbb{T}^n$ diffeomorphism if and only if $S ∈ GL_n(\mathbb{Z})$.

I can't prove the right to left implication.

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That $S$ projects to a diffeo of $T^n$ means that $S$ is $\mathbb{Z}^n$-equivariant. – Neal Dec 4 '12 at 21:50
up vote 1 down vote accepted

I'm assuming that for you $\mathbb T^n = \mathbb R^n / \Gamma$, where $\Gamma$ is a full lattice in $\mathbb R^n$. In this case, the left to right implication is just the fact that if $S \in \text{GL}_n(\mathbb Z)$ then $S(\Gamma) \subset \Gamma$, hence universal property of the quotient concludes.

In fact, being the action of $\Gamma$ over $\mathbb R^n$ properly discontinuous, you can see $\mathbb T^n$ as a quotient in the category of differentiable manifolds. Hence your $S$ factors through $\mathbb T^n$ producing a map $\mathbb T^n \to \mathbb T^n$. In the same way you can factors its inverse, hence you get a diffeomorphism of the torus (you check that the two induced morphisms are inverse each other using again the universal property).

Edit: To do the converse, you don't even need the universal property. Assume in fact that $S$ is a linear invertible map inducing a diffeomorphism of the torus. Apparently I'm not allowed to draw commutative diagrams... So let $p \colon \mathbb R^n \to \mathbb T^n$ be the canonical projection and let $\overline{S} \colon \mathbb T^n \to \mathbb T^n$ the map induced by $S$. Then $p \circ S = \overline{S} \circ p$. This means that $p(S(\Gamma)) = \overline{S}(p(\Gamma)) = \overline{S}([\mathbf 0]) = [\mathbf 0]$, hence $S(\Gamma) \subset \Gamma$. Since $\Gamma = \mathbb Z^n$, and since the vectors $\mathbf e_i = (0,\ldots,1,\ldots,0)$ form a basis both for $\Gamma$ and for $\mathbb R^n$, you can write down the matrix of $S$ explicitly: in the $i$-th column you will have the coefficients of $S(\mathbf e_i)$ with respect to this basis; since $S(\Gamma) \subset \Gamma$ you see that $S(\mathbf e_i) \in \mathbb Z^n$, hence the coefficients must be integers! This implies $S \in M_n(\mathbb Z)$.

Now, your hypothesis is that $\overline{S}$ is a diffeomorphism of $\mathbb T^n$. Let $f \colon \mathbb T^n \to \mathbb T^n$ be its inverse. Since the action of $\mathbb Z^n$ over $\mathbb R^n$ is properly discontinous, then the canonical projection $p$ is a covering; since $\mathbb R^n$ is simply connected, this is the universal covering. Thus $f$ lifts (uniquely!) to a continuous map $T \colon \mathbb R^n \to \mathbb R^n$. Universal property of universal covering implies that $T$ is the inverse of $S$, hence $T \in \text{GL}_n(\mathbb R)$. The same reasoning of above shows $T \in M_n(\mathbb Z)$. Therefore $S, T \in \text{GL}_n(\mathbb Z)$.

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Thanks for the quick reply. I'm reading on wiki about the universal property of the quotient. However I really have trouble proving the converse statement. I already changed the question to correct this. – simon Dec 4 '12 at 22:14
Well, assume that $f \colon \mathbb T^n \to \mathbb T^n$ is a diffeomorphism induced by some linear invertible map $S$. Then, since $S$ factors through $\mathbb T^n$ it follows that $S(\Gamma) \subset \Gamma$. I'm assuming that for you $\Gamma = \mathbb Z^n$. In this case, the image of $\mathbf e_i$ must be of the form $\sum_{j = 1}^n a_j \mathbf e_j$ with $a_J \in \mathbb Z$, hence $S \in M_n(\mathbb Z)$, hence $S \in \text{GL}_n(\mathbb Z)$. – Mauro Porta Dec 4 '12 at 22:26
I don't understand: for example, why does a mapping $\left(\begin{matrix}2 & 0 \\ 0 & 1\end{matrix}\right)$ produce a diffeomorphism instead of a covering? – Alexei Averchenko Dec 4 '12 at 22:26
It's me to not understand. Your matrix is not in $\text{GL}_2(\mathbb Z)$, hence it doesn't produce a diffeomorphism of the torus. – Mauro Porta Dec 4 '12 at 22:28
Then I don't know. I'm a little bit sceptical because I don't see why a map $\mathbb T^n \to \mathbb T^n$ should be induced necessarily from a linear map in $\mathbb R^n$. This is true for a complex torus of dimension $1$ (i.e. endowed with complex structure), but I'm not sure about the general case. – Mauro Porta Dec 4 '12 at 22:49

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