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I'm trying to prove the following formula. Suppose to have $p:\mathbb{R}^{d}\rightarrow\mathbb{T}^{d}$ the canonical projection of the real d- dimensional space in to the d-dimensional torus, and suppose to have a diffeomorphism (or more in genearl an homeomorphism) $\varphi:\mathbb{T}^{d}\longrightarrow\mathbb{T}^{d}$ that i can lift to a diffeomorphism $\tilde{\varphi}:\mathbb{R}^{d}\longrightarrow\mathbb{R}^{d}$ such that $p\circ\tilde{\varphi}=\varphi\circ p$ . It's true that i can write like $\tilde{\varphi}\left(x\right)=Ax+u\left(x\right)$ where $A\in Sl(\mathbb{Z},d)$ and $u\left(x\right)$ is periodic ?

I have proved that if A is a lift of a linear diffeomorphism of the torus than it lies in $Sl(\mathbb{Z},d)$. Any suggestion for the rest?

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A will have to be in $GL(\mathbb Z,d)$. –  Mariano Suárez-Alvarez Sep 25 '13 at 15:21
    
Yes, but because it's invertible and its inverse is the lift of the inverse, it lies in $Gl(Z,d)$ too. So its determinat must be +/-1 , right ? –  user52342 Sep 25 '13 at 15:33
    
But $SL(d,\mathbb Z)$ is the group of matrices of determinant equal to $1$, not $\pm1$. –  Mariano Suárez-Alvarez Sep 25 '13 at 16:16
    
Why do you need $A$ in the form for $\tilde \varphi$? It's simply true that any lift will be periodic. And of course, transforming such a lift by an element of $GL(d, \mathbb Z)$ will give an equivalent lift. But those are two independent statements. Now, if you demanded that the range of $\tilde \varphi$ should lie in $[0,1]^d$ that would be something else... –  Marek Sep 25 '13 at 16:38
    
Ok, you are right my fault... take it in $(+/-)Sl(Z,d)$. –  user52342 Sep 25 '13 at 16:39
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1 Answer

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If you have a continuous $\varphi \colon \mathbb{T}^d \to \mathbb{T}^d$ and a lift $\tilde{\varphi} \colon \mathbb{R}^d \to \mathbb{R}^d$ of $\varphi$, then for every $z \in \mathbb{Z}^d$, the function $\psi_z\colon \mathbb{R}^d \to \mathbb{R}^d; \: \psi_z(x) = \tilde{\varphi}(x+z) - \tilde{\varphi}(x)$ is continuous and $\mathbb{Z}^d$- valued, since $$p(\tilde{\varphi}(x+z)) = \varphi(p(x+z)) = \varphi(p(x)) = p(\tilde{\varphi}(x)),$$ hence constant. It is easy to see that $z \mapsto \psi_z(0) = \tilde{\varphi}(z) - \tilde{\varphi}(0)$ is $\mathbb{Z}$-linear, so we can write $\psi_z(0) = Az$ with a matrix $A \in M_d(\mathbb{Z})$. Then $u \colon \mathbb{R}^d \to \mathbb{R}^d;\; u(x) = \tilde{\varphi}(x) - Ax$ is $\mathbb{Z}^d$-periodic, since

$$u(x+z) = \tilde{\varphi}(x+z) - A(x+z) = \tilde{\varphi}(x) + \psi_z(x) - Ax - Az = u(x) + \bigl(\psi_z(x) - Az\bigr) = u(x).$$

If $\varphi$ is a homeomorphism, then $\tilde{\varphi}\bigl([0,1)^d\bigr)$ is a fundamental region of the lattice $\mathbb{Z}^d$, hence then $A \in GL_d(\mathbb{Z})$.

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Is the A a lift of a linear homeomorphism ? For this he as to stay in $Gl(Z,d)$? –  user52342 Sep 28 '13 at 14:22
    
$A$, or rather $x \mapsto Ax$ is the lift of a linear homeomorphism if and only if $A \in GL_d(\mathbb{Z})$. Thus in this construction, where you got $A$ from the lift of a homeomorphism, it is. And $u$ pushes down to a perturbation of the linear homeomorphism. –  Daniel Fischer Sep 28 '13 at 14:27
    
sorry but i cant see the linear Homeomorphism of the torus of which A is the lift... –  user52342 Sep 28 '13 at 14:45
    
$p\circ A$ is $\mathbb{Z}^d$-periodic, so $A$ induces $\alpha \colon \mathbb{T}^d \to \mathbb{T}^d$. $A^{-1}$ induces $\alpha^{-1}$, so $\alpha$ is a bijection of the torus. –  Daniel Fischer Sep 28 '13 at 16:00
    
but how you define $A^-1$ ? thank you.... –  user52342 Sep 28 '13 at 16:20
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