Why does the real projective plane / Boy surface look like this?












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In geometry, Boy's surface is an immersion of the real projective plane in 3-dimensional space found by Werner Boy in 1901




My question is, you can see that the Boy surface is made up of three identical parts. But how does the number $3$ come up? I cannot see it in the definition of $mathbb{R}P^2$.



Boy surface










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    5












    $begingroup$



    In geometry, Boy's surface is an immersion of the real projective plane in 3-dimensional space found by Werner Boy in 1901




    My question is, you can see that the Boy surface is made up of three identical parts. But how does the number $3$ come up? I cannot see it in the definition of $mathbb{R}P^2$.



    Boy surface










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      5


      2



      $begingroup$



      In geometry, Boy's surface is an immersion of the real projective plane in 3-dimensional space found by Werner Boy in 1901




      My question is, you can see that the Boy surface is made up of three identical parts. But how does the number $3$ come up? I cannot see it in the definition of $mathbb{R}P^2$.



      Boy surface










      share|cite|improve this question









      $endgroup$





      In geometry, Boy's surface is an immersion of the real projective plane in 3-dimensional space found by Werner Boy in 1901




      My question is, you can see that the Boy surface is made up of three identical parts. But how does the number $3$ come up? I cannot see it in the definition of $mathbb{R}P^2$.



      Boy surface







      general-topology soft-question projective-geometry






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      asked 2 days ago









      JiuJiu

      496112




      496112






















          2 Answers
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          3












          $begingroup$

          There are probably many ways to answer this question, and I'm totally unqualified to do so but here is one thing: it is possible to construct versions of Boy's surface with 5-fold (and larger odd numbered) symmetry, (there are some lovely illustrations in Models of the Real Projective Plane: Computer Graphics of Steiner and Boy Surfaces
          - Francois Apery, you may have access to this at https://link.springer.com/content/pdf/bbm%3A978-3-322-89569-1%2F1.pdf)



          So the question could instead be why does the rotational symmetry of a Boy's type immersion have to be odd, it's probably something to do with the non-orientability of the surface.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thanks! Many interesting pictures in the link. I have a hard time imagining what a mobius strip with circular boundary looks like.
            $endgroup$
            – Jiu
            2 days ago










          • $begingroup$
            first link is partly missing, looking forward to seeing it! (see my profile image)
            $endgroup$
            – uhoh
            2 days ago










          • $begingroup$
            @uhuh thanks, fixed!
            $endgroup$
            – Alex J Best
            2 days ago



















          2












          $begingroup$

          3 occurs in the usual definition of $RP^2$ as the set of lines in $R^3$. That is, the quotient space of $R^3-0$ that identifies $xsim cx$ for all nonzero $xin R^3$ and nonzero real $c$. The homeomorphism $(x_1,x_2,x_3)to(x_2,x_3,x_1)$
          for example induces a threefold symmetry of $RP^2$.






          share|cite|improve this answer









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            2 Answers
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            2 Answers
            2






            active

            oldest

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            active

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            active

            oldest

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            3












            $begingroup$

            There are probably many ways to answer this question, and I'm totally unqualified to do so but here is one thing: it is possible to construct versions of Boy's surface with 5-fold (and larger odd numbered) symmetry, (there are some lovely illustrations in Models of the Real Projective Plane: Computer Graphics of Steiner and Boy Surfaces
            - Francois Apery, you may have access to this at https://link.springer.com/content/pdf/bbm%3A978-3-322-89569-1%2F1.pdf)



            So the question could instead be why does the rotational symmetry of a Boy's type immersion have to be odd, it's probably something to do with the non-orientability of the surface.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks! Many interesting pictures in the link. I have a hard time imagining what a mobius strip with circular boundary looks like.
              $endgroup$
              – Jiu
              2 days ago










            • $begingroup$
              first link is partly missing, looking forward to seeing it! (see my profile image)
              $endgroup$
              – uhoh
              2 days ago










            • $begingroup$
              @uhuh thanks, fixed!
              $endgroup$
              – Alex J Best
              2 days ago
















            3












            $begingroup$

            There are probably many ways to answer this question, and I'm totally unqualified to do so but here is one thing: it is possible to construct versions of Boy's surface with 5-fold (and larger odd numbered) symmetry, (there are some lovely illustrations in Models of the Real Projective Plane: Computer Graphics of Steiner and Boy Surfaces
            - Francois Apery, you may have access to this at https://link.springer.com/content/pdf/bbm%3A978-3-322-89569-1%2F1.pdf)



            So the question could instead be why does the rotational symmetry of a Boy's type immersion have to be odd, it's probably something to do with the non-orientability of the surface.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks! Many interesting pictures in the link. I have a hard time imagining what a mobius strip with circular boundary looks like.
              $endgroup$
              – Jiu
              2 days ago










            • $begingroup$
              first link is partly missing, looking forward to seeing it! (see my profile image)
              $endgroup$
              – uhoh
              2 days ago










            • $begingroup$
              @uhuh thanks, fixed!
              $endgroup$
              – Alex J Best
              2 days ago














            3












            3








            3





            $begingroup$

            There are probably many ways to answer this question, and I'm totally unqualified to do so but here is one thing: it is possible to construct versions of Boy's surface with 5-fold (and larger odd numbered) symmetry, (there are some lovely illustrations in Models of the Real Projective Plane: Computer Graphics of Steiner and Boy Surfaces
            - Francois Apery, you may have access to this at https://link.springer.com/content/pdf/bbm%3A978-3-322-89569-1%2F1.pdf)



            So the question could instead be why does the rotational symmetry of a Boy's type immersion have to be odd, it's probably something to do with the non-orientability of the surface.






            share|cite|improve this answer











            $endgroup$



            There are probably many ways to answer this question, and I'm totally unqualified to do so but here is one thing: it is possible to construct versions of Boy's surface with 5-fold (and larger odd numbered) symmetry, (there are some lovely illustrations in Models of the Real Projective Plane: Computer Graphics of Steiner and Boy Surfaces
            - Francois Apery, you may have access to this at https://link.springer.com/content/pdf/bbm%3A978-3-322-89569-1%2F1.pdf)



            So the question could instead be why does the rotational symmetry of a Boy's type immersion have to be odd, it's probably something to do with the non-orientability of the surface.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 days ago

























            answered 2 days ago









            Alex J BestAlex J Best

            2,08211225




            2,08211225












            • $begingroup$
              Thanks! Many interesting pictures in the link. I have a hard time imagining what a mobius strip with circular boundary looks like.
              $endgroup$
              – Jiu
              2 days ago










            • $begingroup$
              first link is partly missing, looking forward to seeing it! (see my profile image)
              $endgroup$
              – uhoh
              2 days ago










            • $begingroup$
              @uhuh thanks, fixed!
              $endgroup$
              – Alex J Best
              2 days ago


















            • $begingroup$
              Thanks! Many interesting pictures in the link. I have a hard time imagining what a mobius strip with circular boundary looks like.
              $endgroup$
              – Jiu
              2 days ago










            • $begingroup$
              first link is partly missing, looking forward to seeing it! (see my profile image)
              $endgroup$
              – uhoh
              2 days ago










            • $begingroup$
              @uhuh thanks, fixed!
              $endgroup$
              – Alex J Best
              2 days ago
















            $begingroup$
            Thanks! Many interesting pictures in the link. I have a hard time imagining what a mobius strip with circular boundary looks like.
            $endgroup$
            – Jiu
            2 days ago




            $begingroup$
            Thanks! Many interesting pictures in the link. I have a hard time imagining what a mobius strip with circular boundary looks like.
            $endgroup$
            – Jiu
            2 days ago












            $begingroup$
            first link is partly missing, looking forward to seeing it! (see my profile image)
            $endgroup$
            – uhoh
            2 days ago




            $begingroup$
            first link is partly missing, looking forward to seeing it! (see my profile image)
            $endgroup$
            – uhoh
            2 days ago












            $begingroup$
            @uhuh thanks, fixed!
            $endgroup$
            – Alex J Best
            2 days ago




            $begingroup$
            @uhuh thanks, fixed!
            $endgroup$
            – Alex J Best
            2 days ago











            2












            $begingroup$

            3 occurs in the usual definition of $RP^2$ as the set of lines in $R^3$. That is, the quotient space of $R^3-0$ that identifies $xsim cx$ for all nonzero $xin R^3$ and nonzero real $c$. The homeomorphism $(x_1,x_2,x_3)to(x_2,x_3,x_1)$
            for example induces a threefold symmetry of $RP^2$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              3 occurs in the usual definition of $RP^2$ as the set of lines in $R^3$. That is, the quotient space of $R^3-0$ that identifies $xsim cx$ for all nonzero $xin R^3$ and nonzero real $c$. The homeomorphism $(x_1,x_2,x_3)to(x_2,x_3,x_1)$
              for example induces a threefold symmetry of $RP^2$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                3 occurs in the usual definition of $RP^2$ as the set of lines in $R^3$. That is, the quotient space of $R^3-0$ that identifies $xsim cx$ for all nonzero $xin R^3$ and nonzero real $c$. The homeomorphism $(x_1,x_2,x_3)to(x_2,x_3,x_1)$
                for example induces a threefold symmetry of $RP^2$.






                share|cite|improve this answer









                $endgroup$



                3 occurs in the usual definition of $RP^2$ as the set of lines in $R^3$. That is, the quotient space of $R^3-0$ that identifies $xsim cx$ for all nonzero $xin R^3$ and nonzero real $c$. The homeomorphism $(x_1,x_2,x_3)to(x_2,x_3,x_1)$
                for example induces a threefold symmetry of $RP^2$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                Bob TerrellBob Terrell

                1,705710




                1,705710






























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