How can we construct a smooth structure of Möbius ring?
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In John Lee’s book, I was asked to find unique smooth structure on E to make the quotient map $q: mathbb{R}^{2}rightarrow E$ smooth . Where q is the quotient map to a equivalent class in $mathbb{R}^{2}:$
${(x,y)sim (x+n,(-1)^{n}y)}$
I want to use the natural embedding in $mathbb{R}^{3}$ and the projection map, but I don’t know how to proof the uniqueness, and furthermore, I want to find a more intrinsic way, what should I do?
differential-geometry differential-topology smooth-manifolds
asked Nov 23 at 11:24
Xu JiuCi
102
add a comment |
up vote
0
down vote
favorite
In John Lee’s book, I was asked to find unique smooth structure on E to make the quotient map $q: mathbb{R}^{2}rightarrow E$ smooth . Where q is the quotient map to a equivalent class in $mathbb{R}^{2}:$
${(x,y)sim (x+n,(-1)^{n}y)}$
I want to use the natural embedding in $mathbb{R}^{3}$ and the projection map, but I don’t know how to proof the uniqueness, and furthermore, I want to find a more intrinsic way, what should I do?
differential-geometry differential-topology smooth-manifolds
asked Nov 23 at 11:24
Xu JiuCi
102
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In John Lee’s book, I was asked to find unique smooth structure on E to make the quotient map $q: mathbb{R}^{2}rightarrow E$ smooth . Where q is the quotient map to a equivalent class in $mathbb{R}^{2}:$
${(x,y)sim (x+n,(-1)^{n}y)}$
I want to use the natural embedding in $mathbb{R}^{3}$ and the projection map, but I don’t know how to proof the uniqueness, and furthermore, I want to find a more intrinsic way, what should I do?
differential-geometry differential-topology smooth-manifolds
asked Nov 23 at 11:24
Xu JiuCi
102
In John Lee’s book, I was asked to find unique smooth structure on E to make the quotient map $q: mathbb{R}^{2}rightarrow E$ smooth . Where q is the quotient map to a equivalent class in $mathbb{R}^{2}:$
${(x,y)sim (x+n,(-1)^{n}y)}$
I want to use the natural embedding in $mathbb{R}^{3}$ and the projection map, but I don’t know how to proof the uniqueness, and furthermore, I want to find a more intrinsic way, what should I do?
differential-geometry differential-topology smooth-manifolds
differential-geometry differential-topology smooth-manifolds
asked Nov 23 at 11:24
Xu JiuCi
102
asked Nov 23 at 11:24
Xu JiuCi
102
asked Nov 23 at 11:24
Xu JiuCi
102
asked Nov 23 at 11:24
Xu JiuCi
102
asked Nov 23 at 11:24
Xu JiuCi
102
102
add a comment |
add a comment |
1 Answer
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votes
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1
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There is a unique differential structure on $E$ such that the quotient map $q:mathbb{R}^2to E$ is locally a diffeomorphism. So if you find any smooth structure which makes $q$ smooth, you're done. But this isn't so bad. The idea is that we should look at the universal cover and try to restrict to open sets where the quotient is giving us our charts.
Here's my attempt at this, with maybe some details skipped over: Let $U_1=lbrace [(x,y)]: 1/3<x<2/3rbrace$ and $U_2=lbrace [(x,y)]:1/2<x<3/2rbrace$ - I'm using $[(x,y)]$ to denote the equivalence class containing the tuple $(x,y)$. Let $overline{x}$ denote the map $mathbb{R}to [0,1)$ so that $x$ is sent to its value mod $1$. Then let $phi_1:U_1tomathbb{R}^2$ be given by $phi_1([(x,y)])=(overline{x},(-1)^{x-overline{x}}y)$ and $phi_2:U_2tomathbb{R}^2$ by $phi([x,y])=(overline{x}+1/2,(-1)^{x-(overline{x}+1/2)}y)$. It is not hard to check these are well-defined. Then note that $U_1cup U_2=E$ and $U_1cap U_2=lbrace [(x,y)]:1/3< x<1/2text{ or }1/2<x<2/3rbrace$. It should work out that $phi_2circphi_1^{-1}$ restricted to $phi_1(U_1cap U_2)$ has the formula $(x,y)mapsto (x+1/2,y)$ and $phi_1circphi_2^{-1}$ restricted to $phi_2(U_1cap U_2)$ has the formula
$(x,y)mapsto leftlbracebegin{array}[ll]
((x,y)&xin(5/6,1) \
(x-1,-y) &xin(1,7/6)end{array}right.$. So $lbrace phi_irbrace_{i=1,2}$ is a smooth atlas. Now we want to show that $q:mathbb{R}^2to E$ is smooth. Let $(x_0,y_0)inmathbb{R}^2$. Then $(x_0,y_0)$ is contained in an integral horizontal translation of $lbrace (x,y):1/3<x<2/3rbrace$ or an integral horizontal translation of $lbrace (x,y):1/2<x<3/2rbrace$. Suppose we are in the former case so that there is $ninmathbb{N}$ with $x-nin (1/3,2/3)$. Then choose an open ball $V$ around $(x_0,y_0)$ completely contained in the translated set. In this quotient this ball is an open ball in $U_1$ so we can post-compose with $phi_1$ and we get that $phicirc q:Vtomathbb{R}^2$ is given by $(x,y)mapsto (x-n,(-1)^ny)$. The other case is similar!
Like I said, there are some missing pieces but this is the general idea.
answered Nov 23 at 15:37
Prototank
1,015820
Thank you very much. I find myself not so good at construction. I believe that’s because I still don’t understand smooth structure that deep. Anyway, thank you for the answer!
– Xu JiuCi
Nov 25 at 0:23
add a comment |
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1 Answer
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1 Answer
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There is a unique differential structure on $E$ such that the quotient map $q:mathbb{R}^2to E$ is locally a diffeomorphism. So if you find any smooth structure which makes $q$ smooth, you're done. But this isn't so bad. The idea is that we should look at the universal cover and try to restrict to open sets where the quotient is giving us our charts.
Here's my attempt at this, with maybe some details skipped over: Let $U_1=lbrace [(x,y)]: 1/3<x<2/3rbrace$ and $U_2=lbrace [(x,y)]:1/2<x<3/2rbrace$ - I'm using $[(x,y)]$ to denote the equivalence class containing the tuple $(x,y)$. Let $overline{x}$ denote the map $mathbb{R}to [0,1)$ so that $x$ is sent to its value mod $1$. Then let $phi_1:U_1tomathbb{R}^2$ be given by $phi_1([(x,y)])=(overline{x},(-1)^{x-overline{x}}y)$ and $phi_2:U_2tomathbb{R}^2$ by $phi([x,y])=(overline{x}+1/2,(-1)^{x-(overline{x}+1/2)}y)$. It is not hard to check these are well-defined. Then note that $U_1cup U_2=E$ and $U_1cap U_2=lbrace [(x,y)]:1/3< x<1/2text{ or }1/2<x<2/3rbrace$. It should work out that $phi_2circphi_1^{-1}$ restricted to $phi_1(U_1cap U_2)$ has the formula $(x,y)mapsto (x+1/2,y)$ and $phi_1circphi_2^{-1}$ restricted to $phi_2(U_1cap U_2)$ has the formula
$(x,y)mapsto leftlbracebegin{array}[ll]
((x,y)&xin(5/6,1) \
(x-1,-y) &xin(1,7/6)end{array}right.$. So $lbrace phi_irbrace_{i=1,2}$ is a smooth atlas. Now we want to show that $q:mathbb{R}^2to E$ is smooth. Let $(x_0,y_0)inmathbb{R}^2$. Then $(x_0,y_0)$ is contained in an integral horizontal translation of $lbrace (x,y):1/3<x<2/3rbrace$ or an integral horizontal translation of $lbrace (x,y):1/2<x<3/2rbrace$. Suppose we are in the former case so that there is $ninmathbb{N}$ with $x-nin (1/3,2/3)$. Then choose an open ball $V$ around $(x_0,y_0)$ completely contained in the translated set. In this quotient this ball is an open ball in $U_1$ so we can post-compose with $phi_1$ and we get that $phicirc q:Vtomathbb{R}^2$ is given by $(x,y)mapsto (x-n,(-1)^ny)$. The other case is similar!
Like I said, there are some missing pieces but this is the general idea.
answered Nov 23 at 15:37
Prototank
1,015820
Thank you very much. I find myself not so good at construction. I believe that’s because I still don’t understand smooth structure that deep. Anyway, thank you for the answer!
– Xu JiuCi
Nov 25 at 0:23
add a comment |
up vote
1
down vote
accepted
There is a unique differential structure on $E$ such that the quotient map $q:mathbb{R}^2to E$ is locally a diffeomorphism. So if you find any smooth structure which makes $q$ smooth, you're done. But this isn't so bad. The idea is that we should look at the universal cover and try to restrict to open sets where the quotient is giving us our charts.
Here's my attempt at this, with maybe some details skipped over: Let $U_1=lbrace [(x,y)]: 1/3<x<2/3rbrace$ and $U_2=lbrace [(x,y)]:1/2<x<3/2rbrace$ - I'm using $[(x,y)]$ to denote the equivalence class containing the tuple $(x,y)$. Let $overline{x}$ denote the map $mathbb{R}to [0,1)$ so that $x$ is sent to its value mod $1$. Then let $phi_1:U_1tomathbb{R}^2$ be given by $phi_1([(x,y)])=(overline{x},(-1)^{x-overline{x}}y)$ and $phi_2:U_2tomathbb{R}^2$ by $phi([x,y])=(overline{x}+1/2,(-1)^{x-(overline{x}+1/2)}y)$. It is not hard to check these are well-defined. Then note that $U_1cup U_2=E$ and $U_1cap U_2=lbrace [(x,y)]:1/3< x<1/2text{ or }1/2<x<2/3rbrace$. It should work out that $phi_2circphi_1^{-1}$ restricted to $phi_1(U_1cap U_2)$ has the formula $(x,y)mapsto (x+1/2,y)$ and $phi_1circphi_2^{-1}$ restricted to $phi_2(U_1cap U_2)$ has the formula
$(x,y)mapsto leftlbracebegin{array}[ll]
((x,y)&xin(5/6,1) \
(x-1,-y) &xin(1,7/6)end{array}right.$. So $lbrace phi_irbrace_{i=1,2}$ is a smooth atlas. Now we want to show that $q:mathbb{R}^2to E$ is smooth. Let $(x_0,y_0)inmathbb{R}^2$. Then $(x_0,y_0)$ is contained in an integral horizontal translation of $lbrace (x,y):1/3<x<2/3rbrace$ or an integral horizontal translation of $lbrace (x,y):1/2<x<3/2rbrace$. Suppose we are in the former case so that there is $ninmathbb{N}$ with $x-nin (1/3,2/3)$. Then choose an open ball $V$ around $(x_0,y_0)$ completely contained in the translated set. In this quotient this ball is an open ball in $U_1$ so we can post-compose with $phi_1$ and we get that $phicirc q:Vtomathbb{R}^2$ is given by $(x,y)mapsto (x-n,(-1)^ny)$. The other case is similar!
Like I said, there are some missing pieces but this is the general idea.
answered Nov 23 at 15:37
Prototank
1,015820
Thank you very much. I find myself not so good at construction. I believe that’s because I still don’t understand smooth structure that deep. Anyway, thank you for the answer!
– Xu JiuCi
Nov 25 at 0:23
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
There is a unique differential structure on $E$ such that the quotient map $q:mathbb{R}^2to E$ is locally a diffeomorphism. So if you find any smooth structure which makes $q$ smooth, you're done. But this isn't so bad. The idea is that we should look at the universal cover and try to restrict to open sets where the quotient is giving us our charts.
Here's my attempt at this, with maybe some details skipped over: Let $U_1=lbrace [(x,y)]: 1/3<x<2/3rbrace$ and $U_2=lbrace [(x,y)]:1/2<x<3/2rbrace$ - I'm using $[(x,y)]$ to denote the equivalence class containing the tuple $(x,y)$. Let $overline{x}$ denote the map $mathbb{R}to [0,1)$ so that $x$ is sent to its value mod $1$. Then let $phi_1:U_1tomathbb{R}^2$ be given by $phi_1([(x,y)])=(overline{x},(-1)^{x-overline{x}}y)$ and $phi_2:U_2tomathbb{R}^2$ by $phi([x,y])=(overline{x}+1/2,(-1)^{x-(overline{x}+1/2)}y)$. It is not hard to check these are well-defined. Then note that $U_1cup U_2=E$ and $U_1cap U_2=lbrace [(x,y)]:1/3< x<1/2text{ or }1/2<x<2/3rbrace$. It should work out that $phi_2circphi_1^{-1}$ restricted to $phi_1(U_1cap U_2)$ has the formula $(x,y)mapsto (x+1/2,y)$ and $phi_1circphi_2^{-1}$ restricted to $phi_2(U_1cap U_2)$ has the formula
$(x,y)mapsto leftlbracebegin{array}[ll]
((x,y)&xin(5/6,1) \
(x-1,-y) &xin(1,7/6)end{array}right.$. So $lbrace phi_irbrace_{i=1,2}$ is a smooth atlas. Now we want to show that $q:mathbb{R}^2to E$ is smooth. Let $(x_0,y_0)inmathbb{R}^2$. Then $(x_0,y_0)$ is contained in an integral horizontal translation of $lbrace (x,y):1/3<x<2/3rbrace$ or an integral horizontal translation of $lbrace (x,y):1/2<x<3/2rbrace$. Suppose we are in the former case so that there is $ninmathbb{N}$ with $x-nin (1/3,2/3)$. Then choose an open ball $V$ around $(x_0,y_0)$ completely contained in the translated set. In this quotient this ball is an open ball in $U_1$ so we can post-compose with $phi_1$ and we get that $phicirc q:Vtomathbb{R}^2$ is given by $(x,y)mapsto (x-n,(-1)^ny)$. The other case is similar!
Like I said, there are some missing pieces but this is the general idea.
answered Nov 23 at 15:37
Prototank
1,015820
There is a unique differential structure on $E$ such that the quotient map $q:mathbb{R}^2to E$ is locally a diffeomorphism. So if you find any smooth structure which makes $q$ smooth, you're done. But this isn't so bad. The idea is that we should look at the universal cover and try to restrict to open sets where the quotient is giving us our charts.
Here's my attempt at this, with maybe some details skipped over: Let $U_1=lbrace [(x,y)]: 1/3<x<2/3rbrace$ and $U_2=lbrace [(x,y)]:1/2<x<3/2rbrace$ - I'm using $[(x,y)]$ to denote the equivalence class containing the tuple $(x,y)$. Let $overline{x}$ denote the map $mathbb{R}to [0,1)$ so that $x$ is sent to its value mod $1$. Then let $phi_1:U_1tomathbb{R}^2$ be given by $phi_1([(x,y)])=(overline{x},(-1)^{x-overline{x}}y)$ and $phi_2:U_2tomathbb{R}^2$ by $phi([x,y])=(overline{x}+1/2,(-1)^{x-(overline{x}+1/2)}y)$. It is not hard to check these are well-defined. Then note that $U_1cup U_2=E$ and $U_1cap U_2=lbrace [(x,y)]:1/3< x<1/2text{ or }1/2<x<2/3rbrace$. It should work out that $phi_2circphi_1^{-1}$ restricted to $phi_1(U_1cap U_2)$ has the formula $(x,y)mapsto (x+1/2,y)$ and $phi_1circphi_2^{-1}$ restricted to $phi_2(U_1cap U_2)$ has the formula
$(x,y)mapsto leftlbracebegin{array}[ll]
((x,y)&xin(5/6,1) \
(x-1,-y) &xin(1,7/6)end{array}right.$. So $lbrace phi_irbrace_{i=1,2}$ is a smooth atlas. Now we want to show that $q:mathbb{R}^2to E$ is smooth. Let $(x_0,y_0)inmathbb{R}^2$. Then $(x_0,y_0)$ is contained in an integral horizontal translation of $lbrace (x,y):1/3<x<2/3rbrace$ or an integral horizontal translation of $lbrace (x,y):1/2<x<3/2rbrace$. Suppose we are in the former case so that there is $ninmathbb{N}$ with $x-nin (1/3,2/3)$. Then choose an open ball $V$ around $(x_0,y_0)$ completely contained in the translated set. In this quotient this ball is an open ball in $U_1$ so we can post-compose with $phi_1$ and we get that $phicirc q:Vtomathbb{R}^2$ is given by $(x,y)mapsto (x-n,(-1)^ny)$. The other case is similar!
Like I said, there are some missing pieces but this is the general idea.
answered Nov 23 at 15:37
Prototank
1,015820
answered Nov 23 at 15:37
Prototank
1,015820
answered Nov 23 at 15:37
Prototank
1,015820
answered Nov 23 at 15:37
Prototank
1,015820
1,015820
Thank you very much. I find myself not so good at construction. I believe that’s because I still don’t understand smooth structure that deep. Anyway, thank you for the answer!
– Xu JiuCi
Nov 25 at 0:23
add a comment |
Thank you very much. I find myself not so good at construction. I believe that’s because I still don’t understand smooth structure that deep. Anyway, thank you for the answer!
– Xu JiuCi
Nov 25 at 0:23
Thank you very much. I find myself not so good at construction. I believe that’s because I still don’t understand smooth structure that deep. Anyway, thank you for the answer!
– Xu JiuCi
Nov 25 at 0:23
Thank you very much. I find myself not so good at construction. I believe that’s because I still don’t understand smooth structure that deep. Anyway, thank you for the answer!
– Xu JiuCi
Nov 25 at 0:23
add a comment |
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