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







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    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







    share|cite|improve this question
























      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







      share|cite|improve this question













      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






      share|cite|improve this question













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      asked Nov 23 at 11:24









      Xu JiuCi

      102




      102






















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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.






          share|cite|improve this answer





















          • 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











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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.






          share|cite|improve this answer





















          • 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















          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.






          share|cite|improve this answer





















          • 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













          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.






          share|cite|improve this answer












          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 23 at 15:37









          Prototank

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          • 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




          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


















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