Fundamental groups of the configuration spaces of all triangles and right triangles












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$begingroup$


This is a question from a past comprehensive exam:



Consider triangles in the plane, with vertices given by non-colinear points as usual. The space $T$ of all plane triangles can be given a natural quotient topology: $T$ is the quotient of an open subset of $mathbb{R}^6=(mathbb{R}^2)^3$ by the action of the symmetric group $S_3$ permuting the vertices of a triangle.
Let $R$ be the space of all right triangles in the plane. Let $i:Rto T$ be the inclusion map. Is the map $i_*:pi_1(R)to pi_1(T)$ surjective?



I do not know the answer and I only have a very vague picture of $T$ and $R$ as some fiber bundles. I don't really know where to start when faced with such problems. Any hint will be appreciated.










share|cite|improve this question









$endgroup$

















    1












    $begingroup$


    This is a question from a past comprehensive exam:



    Consider triangles in the plane, with vertices given by non-colinear points as usual. The space $T$ of all plane triangles can be given a natural quotient topology: $T$ is the quotient of an open subset of $mathbb{R}^6=(mathbb{R}^2)^3$ by the action of the symmetric group $S_3$ permuting the vertices of a triangle.
    Let $R$ be the space of all right triangles in the plane. Let $i:Rto T$ be the inclusion map. Is the map $i_*:pi_1(R)to pi_1(T)$ surjective?



    I do not know the answer and I only have a very vague picture of $T$ and $R$ as some fiber bundles. I don't really know where to start when faced with such problems. Any hint will be appreciated.










    share|cite|improve this question









    $endgroup$















      1












      1








      1


      2



      $begingroup$


      This is a question from a past comprehensive exam:



      Consider triangles in the plane, with vertices given by non-colinear points as usual. The space $T$ of all plane triangles can be given a natural quotient topology: $T$ is the quotient of an open subset of $mathbb{R}^6=(mathbb{R}^2)^3$ by the action of the symmetric group $S_3$ permuting the vertices of a triangle.
      Let $R$ be the space of all right triangles in the plane. Let $i:Rto T$ be the inclusion map. Is the map $i_*:pi_1(R)to pi_1(T)$ surjective?



      I do not know the answer and I only have a very vague picture of $T$ and $R$ as some fiber bundles. I don't really know where to start when faced with such problems. Any hint will be appreciated.










      share|cite|improve this question









      $endgroup$




      This is a question from a past comprehensive exam:



      Consider triangles in the plane, with vertices given by non-colinear points as usual. The space $T$ of all plane triangles can be given a natural quotient topology: $T$ is the quotient of an open subset of $mathbb{R}^6=(mathbb{R}^2)^3$ by the action of the symmetric group $S_3$ permuting the vertices of a triangle.
      Let $R$ be the space of all right triangles in the plane. Let $i:Rto T$ be the inclusion map. Is the map $i_*:pi_1(R)to pi_1(T)$ surjective?



      I do not know the answer and I only have a very vague picture of $T$ and $R$ as some fiber bundles. I don't really know where to start when faced with such problems. Any hint will be appreciated.







      algebraic-topology fundamental-groups configuration-space






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      asked Dec 17 '18 at 6:31









      user392347user392347

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          1 Answer
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          $begingroup$

          First, consider the map that sends a planar triangle to its circumcircle. A circle in the plane is determined by its center in $mathbb{R}^2$ and its radius in $mathbb{R}_{>0}$, so the space of circles is homeomorphic to $mathbb{R}^2 times mathbb{R}_{>0}$, which in any case is contractible. This define maps of fiber bundles



          begin{array}{ccccc}
          F_R & rightarrow & R & rightarrow & mathbb{R}^2 times mathbb{R}_{>0} simeq * \
          downarrow & & downarrow & & downarrow \
          F_T & rightarrow & T & rightarrow & mathbb{R}^2 times mathbb{R}_{>0} simeq *
          end{array}



          We are reduced to studying the map of fibers $F_R to F_T$. Let us identify these spaces more explicitly. First, $F_T = mathrm{UConf}_3(S^1)$ is just the unordered configuration space of three distinct points on a circle. On the other hand, the projection of a configuration in $F_R$ to the vertex with the right angle determines a fiber bundle $$S^1 setminus {*} to F_R to S^1,$$ which means $F_R xrightarrow{sim} S^1$ as $S^1 setminus {*}$ is contractible. Note that a generator of $pi_1 F_R cong mathbb{Z}$ is represented by a full revolution around the circle.



          Now, we have $$mathrm{UConf}_3(S^1) = mathrm{Conf}_3(S^1)/Sigma_3 simeq (S^1 sqcup S^1)/Sigma_3 cong S^1 / C_3.$$ Let me explain this a little. First, $mathrm{Conf}_3(S^1)$ has two path components corresponding to the clockwise and counterclockwise cyclic orderings of the points, and each component fibers over the circle with fiber a contractible simplex, so $mathrm{Conf}_3(S^1) simeq S^1 sqcup S^1$. There is a transposition in $Sigma_3$ that identifies the two cyclic orderings, and thus we are left with $mathrm{UConf}_3(S^1) simeq S^1 / C_3$. Here, a generator of $pi_1 mathrm{UConf}_3(S^1) cong mathbb{Z}$ is represented by a partial revolution of the circle.



          Combining these observations, we have
          begin{array}{ccc}
          pi_1 F_R & rightarrow & pi_1 F_T \
          downarrow & & downarrow \
          mathbb{Z} cong pi_1 S^1 & xrightarrow{3} & pi_1 (S^1/C_3) cong mathbb{Z}
          end{array}



          So it seems that the map induced on $pi_1$ by the inclusion $R to T$ is not surjective.






          share|cite|improve this answer









          $endgroup$













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            1 Answer
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            3












            $begingroup$

            First, consider the map that sends a planar triangle to its circumcircle. A circle in the plane is determined by its center in $mathbb{R}^2$ and its radius in $mathbb{R}_{>0}$, so the space of circles is homeomorphic to $mathbb{R}^2 times mathbb{R}_{>0}$, which in any case is contractible. This define maps of fiber bundles



            begin{array}{ccccc}
            F_R & rightarrow & R & rightarrow & mathbb{R}^2 times mathbb{R}_{>0} simeq * \
            downarrow & & downarrow & & downarrow \
            F_T & rightarrow & T & rightarrow & mathbb{R}^2 times mathbb{R}_{>0} simeq *
            end{array}



            We are reduced to studying the map of fibers $F_R to F_T$. Let us identify these spaces more explicitly. First, $F_T = mathrm{UConf}_3(S^1)$ is just the unordered configuration space of three distinct points on a circle. On the other hand, the projection of a configuration in $F_R$ to the vertex with the right angle determines a fiber bundle $$S^1 setminus {*} to F_R to S^1,$$ which means $F_R xrightarrow{sim} S^1$ as $S^1 setminus {*}$ is contractible. Note that a generator of $pi_1 F_R cong mathbb{Z}$ is represented by a full revolution around the circle.



            Now, we have $$mathrm{UConf}_3(S^1) = mathrm{Conf}_3(S^1)/Sigma_3 simeq (S^1 sqcup S^1)/Sigma_3 cong S^1 / C_3.$$ Let me explain this a little. First, $mathrm{Conf}_3(S^1)$ has two path components corresponding to the clockwise and counterclockwise cyclic orderings of the points, and each component fibers over the circle with fiber a contractible simplex, so $mathrm{Conf}_3(S^1) simeq S^1 sqcup S^1$. There is a transposition in $Sigma_3$ that identifies the two cyclic orderings, and thus we are left with $mathrm{UConf}_3(S^1) simeq S^1 / C_3$. Here, a generator of $pi_1 mathrm{UConf}_3(S^1) cong mathbb{Z}$ is represented by a partial revolution of the circle.



            Combining these observations, we have
            begin{array}{ccc}
            pi_1 F_R & rightarrow & pi_1 F_T \
            downarrow & & downarrow \
            mathbb{Z} cong pi_1 S^1 & xrightarrow{3} & pi_1 (S^1/C_3) cong mathbb{Z}
            end{array}



            So it seems that the map induced on $pi_1$ by the inclusion $R to T$ is not surjective.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              First, consider the map that sends a planar triangle to its circumcircle. A circle in the plane is determined by its center in $mathbb{R}^2$ and its radius in $mathbb{R}_{>0}$, so the space of circles is homeomorphic to $mathbb{R}^2 times mathbb{R}_{>0}$, which in any case is contractible. This define maps of fiber bundles



              begin{array}{ccccc}
              F_R & rightarrow & R & rightarrow & mathbb{R}^2 times mathbb{R}_{>0} simeq * \
              downarrow & & downarrow & & downarrow \
              F_T & rightarrow & T & rightarrow & mathbb{R}^2 times mathbb{R}_{>0} simeq *
              end{array}



              We are reduced to studying the map of fibers $F_R to F_T$. Let us identify these spaces more explicitly. First, $F_T = mathrm{UConf}_3(S^1)$ is just the unordered configuration space of three distinct points on a circle. On the other hand, the projection of a configuration in $F_R$ to the vertex with the right angle determines a fiber bundle $$S^1 setminus {*} to F_R to S^1,$$ which means $F_R xrightarrow{sim} S^1$ as $S^1 setminus {*}$ is contractible. Note that a generator of $pi_1 F_R cong mathbb{Z}$ is represented by a full revolution around the circle.



              Now, we have $$mathrm{UConf}_3(S^1) = mathrm{Conf}_3(S^1)/Sigma_3 simeq (S^1 sqcup S^1)/Sigma_3 cong S^1 / C_3.$$ Let me explain this a little. First, $mathrm{Conf}_3(S^1)$ has two path components corresponding to the clockwise and counterclockwise cyclic orderings of the points, and each component fibers over the circle with fiber a contractible simplex, so $mathrm{Conf}_3(S^1) simeq S^1 sqcup S^1$. There is a transposition in $Sigma_3$ that identifies the two cyclic orderings, and thus we are left with $mathrm{UConf}_3(S^1) simeq S^1 / C_3$. Here, a generator of $pi_1 mathrm{UConf}_3(S^1) cong mathbb{Z}$ is represented by a partial revolution of the circle.



              Combining these observations, we have
              begin{array}{ccc}
              pi_1 F_R & rightarrow & pi_1 F_T \
              downarrow & & downarrow \
              mathbb{Z} cong pi_1 S^1 & xrightarrow{3} & pi_1 (S^1/C_3) cong mathbb{Z}
              end{array}



              So it seems that the map induced on $pi_1$ by the inclusion $R to T$ is not surjective.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                First, consider the map that sends a planar triangle to its circumcircle. A circle in the plane is determined by its center in $mathbb{R}^2$ and its radius in $mathbb{R}_{>0}$, so the space of circles is homeomorphic to $mathbb{R}^2 times mathbb{R}_{>0}$, which in any case is contractible. This define maps of fiber bundles



                begin{array}{ccccc}
                F_R & rightarrow & R & rightarrow & mathbb{R}^2 times mathbb{R}_{>0} simeq * \
                downarrow & & downarrow & & downarrow \
                F_T & rightarrow & T & rightarrow & mathbb{R}^2 times mathbb{R}_{>0} simeq *
                end{array}



                We are reduced to studying the map of fibers $F_R to F_T$. Let us identify these spaces more explicitly. First, $F_T = mathrm{UConf}_3(S^1)$ is just the unordered configuration space of three distinct points on a circle. On the other hand, the projection of a configuration in $F_R$ to the vertex with the right angle determines a fiber bundle $$S^1 setminus {*} to F_R to S^1,$$ which means $F_R xrightarrow{sim} S^1$ as $S^1 setminus {*}$ is contractible. Note that a generator of $pi_1 F_R cong mathbb{Z}$ is represented by a full revolution around the circle.



                Now, we have $$mathrm{UConf}_3(S^1) = mathrm{Conf}_3(S^1)/Sigma_3 simeq (S^1 sqcup S^1)/Sigma_3 cong S^1 / C_3.$$ Let me explain this a little. First, $mathrm{Conf}_3(S^1)$ has two path components corresponding to the clockwise and counterclockwise cyclic orderings of the points, and each component fibers over the circle with fiber a contractible simplex, so $mathrm{Conf}_3(S^1) simeq S^1 sqcup S^1$. There is a transposition in $Sigma_3$ that identifies the two cyclic orderings, and thus we are left with $mathrm{UConf}_3(S^1) simeq S^1 / C_3$. Here, a generator of $pi_1 mathrm{UConf}_3(S^1) cong mathbb{Z}$ is represented by a partial revolution of the circle.



                Combining these observations, we have
                begin{array}{ccc}
                pi_1 F_R & rightarrow & pi_1 F_T \
                downarrow & & downarrow \
                mathbb{Z} cong pi_1 S^1 & xrightarrow{3} & pi_1 (S^1/C_3) cong mathbb{Z}
                end{array}



                So it seems that the map induced on $pi_1$ by the inclusion $R to T$ is not surjective.






                share|cite|improve this answer









                $endgroup$



                First, consider the map that sends a planar triangle to its circumcircle. A circle in the plane is determined by its center in $mathbb{R}^2$ and its radius in $mathbb{R}_{>0}$, so the space of circles is homeomorphic to $mathbb{R}^2 times mathbb{R}_{>0}$, which in any case is contractible. This define maps of fiber bundles



                begin{array}{ccccc}
                F_R & rightarrow & R & rightarrow & mathbb{R}^2 times mathbb{R}_{>0} simeq * \
                downarrow & & downarrow & & downarrow \
                F_T & rightarrow & T & rightarrow & mathbb{R}^2 times mathbb{R}_{>0} simeq *
                end{array}



                We are reduced to studying the map of fibers $F_R to F_T$. Let us identify these spaces more explicitly. First, $F_T = mathrm{UConf}_3(S^1)$ is just the unordered configuration space of three distinct points on a circle. On the other hand, the projection of a configuration in $F_R$ to the vertex with the right angle determines a fiber bundle $$S^1 setminus {*} to F_R to S^1,$$ which means $F_R xrightarrow{sim} S^1$ as $S^1 setminus {*}$ is contractible. Note that a generator of $pi_1 F_R cong mathbb{Z}$ is represented by a full revolution around the circle.



                Now, we have $$mathrm{UConf}_3(S^1) = mathrm{Conf}_3(S^1)/Sigma_3 simeq (S^1 sqcup S^1)/Sigma_3 cong S^1 / C_3.$$ Let me explain this a little. First, $mathrm{Conf}_3(S^1)$ has two path components corresponding to the clockwise and counterclockwise cyclic orderings of the points, and each component fibers over the circle with fiber a contractible simplex, so $mathrm{Conf}_3(S^1) simeq S^1 sqcup S^1$. There is a transposition in $Sigma_3$ that identifies the two cyclic orderings, and thus we are left with $mathrm{UConf}_3(S^1) simeq S^1 / C_3$. Here, a generator of $pi_1 mathrm{UConf}_3(S^1) cong mathbb{Z}$ is represented by a partial revolution of the circle.



                Combining these observations, we have
                begin{array}{ccc}
                pi_1 F_R & rightarrow & pi_1 F_T \
                downarrow & & downarrow \
                mathbb{Z} cong pi_1 S^1 & xrightarrow{3} & pi_1 (S^1/C_3) cong mathbb{Z}
                end{array}



                So it seems that the map induced on $pi_1$ by the inclusion $R to T$ is not surjective.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 17 '18 at 16:45









                JHFJHF

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