A question on the distinguishable colorings of an object in relation to the number of fixed points in a...












1












$begingroup$


Note: the first half of this post is just for clarity , I'm sure most of you here can just skip to the part labelled (my actual question)



In class we were shown Burnsides counting theorem which states:



Let the finite group G act on the finite set $Omega$. Then $Omega/G$, the number of orbits of $G$ on $Omega$ is



$$Omega/G=frac{1}{|G|}sum_{gin G}|operatorname{fix}(g)| .tag{1}$$



This can be simplified (in terms of calculations) by grouping together elements in the same conjugacy class because they will have the same fixed points thus the equation (1) becomes



$$Omega/G=frac{1}{|G|}sum_{r=1}^m|K_r||operatorname{fix}(g_r)|. tag{2}$$



I also know that



$$operatorname{fix}(g)={xin Omega mid x^g=x}.$$



Take the following example: suppose we have a tetrahedron and we have 11 colors.



Then $|operatorname{fix}(1)|=11^4$ (i.e. all possible permutations).



The Question



I have read that in this case (of the tetrahedron) $$|operatorname{fix}(tau)|=|operatorname{fix}(omega)|=|operatorname{fix}(delta)|=11^2,$$



where





  • $tau$ is the rotation by $pi/3$ about a face,


  • $omega$ is the rotation by $-pi/3$ about a face, and


  • $delta$ is the rotation by $pi$ about an edge.


I'm a little confused at how this number is arrived at. Obviously 11 is the number of colors but I don't understand the significance of the exponent 2.






group-theory coloring







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Note: the first half of this post is just for clarity , I'm sure most of you here can just skip to the part labelled (my actual question)



    In class we were shown Burnsides counting theorem which states:



    Let the finite group G act on the finite set $Omega$. Then $Omega/G$, the number of orbits of $G$ on $Omega$ is



    $$Omega/G=frac{1}{|G|}sum_{gin G}|operatorname{fix}(g)| .tag{1}$$



    This can be simplified (in terms of calculations) by grouping together elements in the same conjugacy class because they will have the same fixed points thus the equation (1) becomes



    $$Omega/G=frac{1}{|G|}sum_{r=1}^m|K_r||operatorname{fix}(g_r)|. tag{2}$$



    I also know that



    $$operatorname{fix}(g)={xin Omega mid x^g=x}.$$



    Take the following example: suppose we have a tetrahedron and we have 11 colors.



    Then $|operatorname{fix}(1)|=11^4$ (i.e. all possible permutations).



    The Question



    I have read that in this case (of the tetrahedron) $$|operatorname{fix}(tau)|=|operatorname{fix}(omega)|=|operatorname{fix}(delta)|=11^2,$$



    where





    • $tau$ is the rotation by $pi/3$ about a face,


    • $omega$ is the rotation by $-pi/3$ about a face, and


    • $delta$ is the rotation by $pi$ about an edge.


    I'm a little confused at how this number is arrived at. Obviously 11 is the number of colors but I don't understand the significance of the exponent 2.






    group-theory coloring







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Note: the first half of this post is just for clarity , I'm sure most of you here can just skip to the part labelled (my actual question)



      In class we were shown Burnsides counting theorem which states:



      Let the finite group G act on the finite set $Omega$. Then $Omega/G$, the number of orbits of $G$ on $Omega$ is



      $$Omega/G=frac{1}{|G|}sum_{gin G}|operatorname{fix}(g)| .tag{1}$$



      This can be simplified (in terms of calculations) by grouping together elements in the same conjugacy class because they will have the same fixed points thus the equation (1) becomes



      $$Omega/G=frac{1}{|G|}sum_{r=1}^m|K_r||operatorname{fix}(g_r)|. tag{2}$$



      I also know that



      $$operatorname{fix}(g)={xin Omega mid x^g=x}.$$



      Take the following example: suppose we have a tetrahedron and we have 11 colors.



      Then $|operatorname{fix}(1)|=11^4$ (i.e. all possible permutations).



      The Question



      I have read that in this case (of the tetrahedron) $$|operatorname{fix}(tau)|=|operatorname{fix}(omega)|=|operatorname{fix}(delta)|=11^2,$$



      where





      • $tau$ is the rotation by $pi/3$ about a face,


      • $omega$ is the rotation by $-pi/3$ about a face, and


      • $delta$ is the rotation by $pi$ about an edge.


      I'm a little confused at how this number is arrived at. Obviously 11 is the number of colors but I don't understand the significance of the exponent 2.






      group-theory coloring







      share|cite|improve this question











      $endgroup$




      Note: the first half of this post is just for clarity , I'm sure most of you here can just skip to the part labelled (my actual question)



      In class we were shown Burnsides counting theorem which states:



      Let the finite group G act on the finite set $Omega$. Then $Omega/G$, the number of orbits of $G$ on $Omega$ is



      $$Omega/G=frac{1}{|G|}sum_{gin G}|operatorname{fix}(g)| .tag{1}$$



      This can be simplified (in terms of calculations) by grouping together elements in the same conjugacy class because they will have the same fixed points thus the equation (1) becomes



      $$Omega/G=frac{1}{|G|}sum_{r=1}^m|K_r||operatorname{fix}(g_r)|. tag{2}$$



      I also know that



      $$operatorname{fix}(g)={xin Omega mid x^g=x}.$$



      Take the following example: suppose we have a tetrahedron and we have 11 colors.



      Then $|operatorname{fix}(1)|=11^4$ (i.e. all possible permutations).



      The Question



      I have read that in this case (of the tetrahedron) $$|operatorname{fix}(tau)|=|operatorname{fix}(omega)|=|operatorname{fix}(delta)|=11^2,$$



      where





      • $tau$ is the rotation by $pi/3$ about a face,


      • $omega$ is the rotation by $-pi/3$ about a face, and


      • $delta$ is the rotation by $pi$ about an edge.


      I'm a little confused at how this number is arrived at. Obviously 11 is the number of colors but I don't understand the significance of the exponent 2.






      group-theory coloring




      group-theory coloring






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 7 at 19:15









      Amnotwhy

      283




      283










      asked Dec 8 '18 at 23:00









      can'tcauchycan'tcauchy

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          $begingroup$

          I'll assume you're coloring the faces; when you color the vertices the same thing happens.



          When you rotate about a face, then there are $11$ ways to color that face; the other three faces get permuted, so they have to be the same color if that coloring is fixed, and there are $11$ ways to pick that color. Altogether $11^2$ colorings fixed by the rotation.



          When you rotate about an edge, there are two pairs of faces that get swapped. If the rotation doesn't change the coloring, then each pair of faces must be the same color, so there are $11^2$ colorings ($11$ ways to color each pair).



          In general, when $Omega$ is a set of colorings of some ground set $S$, and $G$ acts on $Omega$ by acting on $S$, then each element of $g$ corresponds to a permutation $sigma_g$ of $S$. If a coloring is fixed by $g$, then the elements of each cycle of $sigma_g$ must all be the same color, so the number of such colorings is $r^k$ when there are $r$ colors and $sigma_g$ has $k$ cycles.






          share|cite|improve this answer









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            $begingroup$

            I'll assume you're coloring the faces; when you color the vertices the same thing happens.



            When you rotate about a face, then there are $11$ ways to color that face; the other three faces get permuted, so they have to be the same color if that coloring is fixed, and there are $11$ ways to pick that color. Altogether $11^2$ colorings fixed by the rotation.



            When you rotate about an edge, there are two pairs of faces that get swapped. If the rotation doesn't change the coloring, then each pair of faces must be the same color, so there are $11^2$ colorings ($11$ ways to color each pair).



            In general, when $Omega$ is a set of colorings of some ground set $S$, and $G$ acts on $Omega$ by acting on $S$, then each element of $g$ corresponds to a permutation $sigma_g$ of $S$. If a coloring is fixed by $g$, then the elements of each cycle of $sigma_g$ must all be the same color, so the number of such colorings is $r^k$ when there are $r$ colors and $sigma_g$ has $k$ cycles.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              I'll assume you're coloring the faces; when you color the vertices the same thing happens.



              When you rotate about a face, then there are $11$ ways to color that face; the other three faces get permuted, so they have to be the same color if that coloring is fixed, and there are $11$ ways to pick that color. Altogether $11^2$ colorings fixed by the rotation.



              When you rotate about an edge, there are two pairs of faces that get swapped. If the rotation doesn't change the coloring, then each pair of faces must be the same color, so there are $11^2$ colorings ($11$ ways to color each pair).



              In general, when $Omega$ is a set of colorings of some ground set $S$, and $G$ acts on $Omega$ by acting on $S$, then each element of $g$ corresponds to a permutation $sigma_g$ of $S$. If a coloring is fixed by $g$, then the elements of each cycle of $sigma_g$ must all be the same color, so the number of such colorings is $r^k$ when there are $r$ colors and $sigma_g$ has $k$ cycles.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                I'll assume you're coloring the faces; when you color the vertices the same thing happens.



                When you rotate about a face, then there are $11$ ways to color that face; the other three faces get permuted, so they have to be the same color if that coloring is fixed, and there are $11$ ways to pick that color. Altogether $11^2$ colorings fixed by the rotation.



                When you rotate about an edge, there are two pairs of faces that get swapped. If the rotation doesn't change the coloring, then each pair of faces must be the same color, so there are $11^2$ colorings ($11$ ways to color each pair).



                In general, when $Omega$ is a set of colorings of some ground set $S$, and $G$ acts on $Omega$ by acting on $S$, then each element of $g$ corresponds to a permutation $sigma_g$ of $S$. If a coloring is fixed by $g$, then the elements of each cycle of $sigma_g$ must all be the same color, so the number of such colorings is $r^k$ when there are $r$ colors and $sigma_g$ has $k$ cycles.






                share|cite|improve this answer









                $endgroup$



                I'll assume you're coloring the faces; when you color the vertices the same thing happens.



                When you rotate about a face, then there are $11$ ways to color that face; the other three faces get permuted, so they have to be the same color if that coloring is fixed, and there are $11$ ways to pick that color. Altogether $11^2$ colorings fixed by the rotation.



                When you rotate about an edge, there are two pairs of faces that get swapped. If the rotation doesn't change the coloring, then each pair of faces must be the same color, so there are $11^2$ colorings ($11$ ways to color each pair).



                In general, when $Omega$ is a set of colorings of some ground set $S$, and $G$ acts on $Omega$ by acting on $S$, then each element of $g$ corresponds to a permutation $sigma_g$ of $S$. If a coloring is fixed by $g$, then the elements of each cycle of $sigma_g$ must all be the same color, so the number of such colorings is $r^k$ when there are $r$ colors and $sigma_g$ has $k$ cycles.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 9 '18 at 0:34









                Misha LavrovMisha Lavrov

                45.7k656107




                45.7k656107






























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