Prove that $f:\mathbb{R}^2\to T^2$ defined by $f(x,y)=(e^{2\pi i x},e^{2\pi iy})$ is covering map and also find covering tranformation group$=\{g:\mathbb{R}^2\to\mathbb{R}^2\mid g$ is diffeomorphism and $f\circ g=f\}$ for the first part f is surjective since $e^{2\pi i x}\in S$ and x can be any real numbers. So now i have to show that for any $ q\in T^2$ there is an open nbhood $U$ of $q$ satisfy the condition "For any connected component $W$ of $f^{-1}(U) ; W$ is an open set in $\mathbb{R}^2$ and $f\mid_W:W\to U$ is diffeomorphism" and what about the second part? I still have no idea how to do it.
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Hint: $f\circ g=f$ means that $\ g(x,y)-(x,y)\,\in\Bbb Z\times\Bbb Z$.
Since $g$, and therefore also $(x,y)\,\mapsto\,g(x,y)-(x,y)\ $ is continuous, and $\Bbb Z\times\Bbb Z$ is discrete, we have that it must be a constant $(a,b)\in\Bbb Z\times\Bbb Z$.
Berci
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could you give me any hints for the first part? or at least some theorems i should apply to find open nhood U and W? – user76608 Jun 10 '13 at 15:25
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Let $(x_0,y_0)$ be arbitrary point in $\Bbb R^2$ and consider its open neighborhood $(x_0-1/4,\ x_0+1/4)\times (y_0-1/4,\ y_0+1/4)$. – Berci Jun 10 '13 at 15:47
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Hint for part 2: consider maps $$g_1\colon\mathbb{R}^2\rightarrow\mathbb{R}^2:(x,y)\mapsto (x+1,y)$$ $$g_2\colon\mathbb{R}^2\rightarrow\mathbb{R}^2:(x,y)\mapsto (x,y+1).$$ Can you show that $f\circ g_i=f$?
Dan Rust
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yes,but according to berci's answer g(x,y)=(x+a,y+b) are also answers for part 2 for all integers a,b ,right? now what idea should i use for part 1? – user76608 Jun 10 '13 at 14:14
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Probably the easiest way to show that $f$ is a covering of the torus would be to identify the torus $T=S^1\times S^1$ with the square with identifications $I^2/\sim$ where $\sim$ identifies opposite edges. You can now easily lift open sets in $I^2/\sim$ back to $\mathbb{R}^2$ as $I^2$ is a fundamental region of the group action of $\mathbb{Z}^2$ on $\mathbb{R}^2$ given by translation. – Dan Rust Jun 10 '13 at 14:27
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I should add, my hint was meant to lead you to wondering if $g_1$ and $g_2$ generate the whole of the covering transformation group of $f$. – Dan Rust Jun 10 '13 at 14:39
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For the first part,I wanna show that there exists open nhood U of q and connected component W satisfy the condition. Could you hint me for the first part how to find them? what concepts or theorems should i use here? – user76608 Jun 10 '13 at 15:23
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Have you been given, or are you able to prove the following theorem? Given smooth covering spaces $f\colon M_1\rightarrow N_1$, $g\colon M_2\rightarrow N_2$, the map $f\times g\colon M_1\times M_2\rightarrow N_1\times N_2$ is a smooth covering space (here, $(f\times g)(x,y)=(f(x),g(y))$). (this question is relevant http://math.stackexchange.com/questions/24585/product-of-smooth-covering-maps-a-smooth-covering-map-j-lee-2-12) – Dan Rust Jun 10 '13 at 15:33