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Let $A=d/dx$ and $B=d/dy$ be vector fields on $\mathbb{R}^2$, prove that they induce vector fields $X$ and $Y$ on the torus $T$ by $X(f)=A(f(\pi)), Y(f)=B(f(\pi))$ where $\pi$ is quotient map from $\mathbb{R}^2$ to torus.

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Welcome to math.SE. Here we answer questions for the people who might have them. Problem is, you have yet to propose one. What we like to see here is questions or very polite requests, together with perhaps some of your own thoughts on the topic. Once that is supplied, people will feel more inclined towards putting effort into solving someone else's problem. –  Arthur Oct 14 '12 at 22:10
    
On the definition(mathworld.wolfram.com/VectorField.html) it is written that vector field is a map from $R^n$ to $R^n$. By this definition we have that $X$ is vector field, nothing to prove? –  user44674 Oct 14 '12 at 22:25
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It doesn't say what a vector field on the torus is. Until you have that piece of information, showing whether or not anything is a vector field on the torus is impossible. So that's where I'd go. See how your source (be it internet or a book) would define a vector field on a torus, and see if $X$ and $Y$ can fit that role. –  Arthur Oct 14 '12 at 22:30
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Well, the torus can be defined as the quotient space $T=\Bbb R^2/\sim$, where $(u,v)\sim(u',v')$ iff $(u'-u),(v'-v)\in\Bbb Z$. Best is to draw on a squared paper.. Vectors of $A$ go right, $B$ goes up, with unit speed.

In particular, it is represented by the unit square $[0,1]^2$.

Practically, all we have to prove is that if points $U\sim V$ then $AU = AV$ and $BU=BV$. In this case, all smooth functions $f$ on $T$ can be 'lifted' to $\Bbb R^2$, i.e. can be viewed as a $\Bbb Z$-periodic function $f:\Bbb R^2\to\Bbb R$, that is, $fU\sim fV$ if $U\sim V$. And so..

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