How orbits of $G$-sets and these characters are related












2












$begingroup$


I've been learning about induced representations recently and I've come across something which I'm very confused about;



For any $G$-set $X$, the number of orbits is equal to $(1_G, chi_{mathbb{C}[X]})$ where $1_G$ denotes the character of the trivial representation and $chi_{mathbb{C}[X]}$ denotes the character of the permutation representation defined below;



$mathbb{C}[X]$ is the natural permutation representation with basis ${{e_x : x in X}}$ and the action of $g$ being $ge_{x} = e_{gx}$.



I have no idea why this is true. I know that $Ind_{H}^{G} mathbb{C} simeq mathbb{C}[G/H]$ and the explanation I've seen goes like this;



If $X$ is transitive, then $(1_G, chi_{mathbb{C}[X]}) = 1$, which can be shown through Frobenius Reciprocity. If $X$ isn't transitive, then $X = bigcup_{i=1}^{n} X_i$ where the $X_i$ are transitive. Then, $(1_G, chi_{mathbb{C}[X]}) = (1_G, chi_{mathbb{C}[X_1] oplus cdots oplus mathbb{C}[X_n]})= (1_G, chi_{mathbb{C}[X_1]}) + ... + (1_G, chi_{mathbb{C}[X_n]}) = n$.



I mean, I understand every line of the proof and can follow it, but why does the value $(1_G, chi_{mathbb{C}[X]})$ give you the number of orbits of a $G$-set? What does this value have to do with orbits of $G$-sets?










share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I've been learning about induced representations recently and I've come across something which I'm very confused about;



    For any $G$-set $X$, the number of orbits is equal to $(1_G, chi_{mathbb{C}[X]})$ where $1_G$ denotes the character of the trivial representation and $chi_{mathbb{C}[X]}$ denotes the character of the permutation representation defined below;



    $mathbb{C}[X]$ is the natural permutation representation with basis ${{e_x : x in X}}$ and the action of $g$ being $ge_{x} = e_{gx}$.



    I have no idea why this is true. I know that $Ind_{H}^{G} mathbb{C} simeq mathbb{C}[G/H]$ and the explanation I've seen goes like this;



    If $X$ is transitive, then $(1_G, chi_{mathbb{C}[X]}) = 1$, which can be shown through Frobenius Reciprocity. If $X$ isn't transitive, then $X = bigcup_{i=1}^{n} X_i$ where the $X_i$ are transitive. Then, $(1_G, chi_{mathbb{C}[X]}) = (1_G, chi_{mathbb{C}[X_1] oplus cdots oplus mathbb{C}[X_n]})= (1_G, chi_{mathbb{C}[X_1]}) + ... + (1_G, chi_{mathbb{C}[X_n]}) = n$.



    I mean, I understand every line of the proof and can follow it, but why does the value $(1_G, chi_{mathbb{C}[X]})$ give you the number of orbits of a $G$-set? What does this value have to do with orbits of $G$-sets?










    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      I've been learning about induced representations recently and I've come across something which I'm very confused about;



      For any $G$-set $X$, the number of orbits is equal to $(1_G, chi_{mathbb{C}[X]})$ where $1_G$ denotes the character of the trivial representation and $chi_{mathbb{C}[X]}$ denotes the character of the permutation representation defined below;



      $mathbb{C}[X]$ is the natural permutation representation with basis ${{e_x : x in X}}$ and the action of $g$ being $ge_{x} = e_{gx}$.



      I have no idea why this is true. I know that $Ind_{H}^{G} mathbb{C} simeq mathbb{C}[G/H]$ and the explanation I've seen goes like this;



      If $X$ is transitive, then $(1_G, chi_{mathbb{C}[X]}) = 1$, which can be shown through Frobenius Reciprocity. If $X$ isn't transitive, then $X = bigcup_{i=1}^{n} X_i$ where the $X_i$ are transitive. Then, $(1_G, chi_{mathbb{C}[X]}) = (1_G, chi_{mathbb{C}[X_1] oplus cdots oplus mathbb{C}[X_n]})= (1_G, chi_{mathbb{C}[X_1]}) + ... + (1_G, chi_{mathbb{C}[X_n]}) = n$.



      I mean, I understand every line of the proof and can follow it, but why does the value $(1_G, chi_{mathbb{C}[X]})$ give you the number of orbits of a $G$-set? What does this value have to do with orbits of $G$-sets?










      share|cite|improve this question











      $endgroup$




      I've been learning about induced representations recently and I've come across something which I'm very confused about;



      For any $G$-set $X$, the number of orbits is equal to $(1_G, chi_{mathbb{C}[X]})$ where $1_G$ denotes the character of the trivial representation and $chi_{mathbb{C}[X]}$ denotes the character of the permutation representation defined below;



      $mathbb{C}[X]$ is the natural permutation representation with basis ${{e_x : x in X}}$ and the action of $g$ being $ge_{x} = e_{gx}$.



      I have no idea why this is true. I know that $Ind_{H}^{G} mathbb{C} simeq mathbb{C}[G/H]$ and the explanation I've seen goes like this;



      If $X$ is transitive, then $(1_G, chi_{mathbb{C}[X]}) = 1$, which can be shown through Frobenius Reciprocity. If $X$ isn't transitive, then $X = bigcup_{i=1}^{n} X_i$ where the $X_i$ are transitive. Then, $(1_G, chi_{mathbb{C}[X]}) = (1_G, chi_{mathbb{C}[X_1] oplus cdots oplus mathbb{C}[X_n]})= (1_G, chi_{mathbb{C}[X_1]}) + ... + (1_G, chi_{mathbb{C}[X_n]}) = n$.



      I mean, I understand every line of the proof and can follow it, but why does the value $(1_G, chi_{mathbb{C}[X]})$ give you the number of orbits of a $G$-set? What does this value have to do with orbits of $G$-sets?







      abstract-algebra group-theory representation-theory group-actions group-rings






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      edited Jan 8 at 22:05









      Eric Wofsey

      187k14215344




      187k14215344










      asked Jan 8 at 19:16









      the manthe man

      773715




      773715






















          2 Answers
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          $begingroup$

          The inner product $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of intertwining maps from the trivial representation $mathbb{C}$ to $mathbb{C}[X]$. A linear map $T:mathbb{C}tomathbb{C}[X]$ is determined by $T(1)$, and is intertwining iff $T(1)$ is fixed by every element of $G$. So, $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of vectors in $mathbb{C}[X]$ which are fixed by every element of $G$.



          Now, when is an element $sum c_xe_xinmathbb{C}[X]$ fixed by every element of $G$? Exactly when the coefficients $c_x$ are constant on each orbit (since the action of $G$ just permutes these coefficients transitively within each orbit). Letting $X_1,dots,X_n$ be the orbits, this means that the vectors $v_i=sum_{xin X_i} e_x$ are a basis for the space of vectors fixed by $G$, and in particular that space has dimension $n$.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            I really like Eric Wofsey's answer (+1), which was posted as I was writing this. However, this is a different solution, so I'll post it too.



            My solution follows from the orbit-stabilizer theorem.



            Let's take a closer look at what $newcommandCC{mathbb{C}}chi_{CC[X]}$ actually is.
            $$chi_{CC[X]}(g) = newcommandtr{operatorname{tr}}tr(pi_g) = sum_{xin X} langle e_x,ge_xrangle = sum_{xin X} [gx=x], $$
            i.e. $chi_{CC[X]}(g)$ counts the number of elements of $X$ fixed by $G$.



            Then
            $$(1_G,chi_{CC[X]}) = frac{1}{|G|}sum_{gin G} chi_{CC[X]}(g)
            =frac{1}{|G|}sum_{gin G}sum_{xin X} [gx=x],$$

            i.e. the inner product is the average number of elements of $X$ fixed.
            Rearranging the sums, we have
            $$(1_G,chi_{CC[X]})=frac{1}{|G|}sum_{xin X} sum_{gin G}[gx=x] = frac{1}{|G|} sum_{xin X} |operatorname{Stab}(x)|.$$
            Now pull the cardinality of the group in, and recognize the term in the summation as the index of the stabilizer
            $$(1_G,chi_{CC[X]})=sum_{xin X} frac{|newcommandStab{operatorname{Stab}}Stab(x)|}{|G|}=sum_{xin X}frac{1}{[G:Stab(x)]},$$
            but the index of the stabilizer equals the size of the orbit, so we have
            $$(1_G,chi_{CC[X]})=sum_{xin X} frac{1}{|Gx|} =sum_{text{orbits}} 1 =#{text{orbits}} $$



            A note on notation



            For a statement $P$, I use the notation $$[P]:=begin{cases} 1 & text{ $P$ is true} \ 0 & text{otherwise}.end{cases}$$



            Note



            Also this result is essentially just Burnside's lemma (modulo my comment on the character counting the number of fixed points).






            share|cite|improve this answer











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              2 Answers
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              $begingroup$

              The inner product $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of intertwining maps from the trivial representation $mathbb{C}$ to $mathbb{C}[X]$. A linear map $T:mathbb{C}tomathbb{C}[X]$ is determined by $T(1)$, and is intertwining iff $T(1)$ is fixed by every element of $G$. So, $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of vectors in $mathbb{C}[X]$ which are fixed by every element of $G$.



              Now, when is an element $sum c_xe_xinmathbb{C}[X]$ fixed by every element of $G$? Exactly when the coefficients $c_x$ are constant on each orbit (since the action of $G$ just permutes these coefficients transitively within each orbit). Letting $X_1,dots,X_n$ be the orbits, this means that the vectors $v_i=sum_{xin X_i} e_x$ are a basis for the space of vectors fixed by $G$, and in particular that space has dimension $n$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                The inner product $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of intertwining maps from the trivial representation $mathbb{C}$ to $mathbb{C}[X]$. A linear map $T:mathbb{C}tomathbb{C}[X]$ is determined by $T(1)$, and is intertwining iff $T(1)$ is fixed by every element of $G$. So, $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of vectors in $mathbb{C}[X]$ which are fixed by every element of $G$.



                Now, when is an element $sum c_xe_xinmathbb{C}[X]$ fixed by every element of $G$? Exactly when the coefficients $c_x$ are constant on each orbit (since the action of $G$ just permutes these coefficients transitively within each orbit). Letting $X_1,dots,X_n$ be the orbits, this means that the vectors $v_i=sum_{xin X_i} e_x$ are a basis for the space of vectors fixed by $G$, and in particular that space has dimension $n$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  The inner product $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of intertwining maps from the trivial representation $mathbb{C}$ to $mathbb{C}[X]$. A linear map $T:mathbb{C}tomathbb{C}[X]$ is determined by $T(1)$, and is intertwining iff $T(1)$ is fixed by every element of $G$. So, $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of vectors in $mathbb{C}[X]$ which are fixed by every element of $G$.



                  Now, when is an element $sum c_xe_xinmathbb{C}[X]$ fixed by every element of $G$? Exactly when the coefficients $c_x$ are constant on each orbit (since the action of $G$ just permutes these coefficients transitively within each orbit). Letting $X_1,dots,X_n$ be the orbits, this means that the vectors $v_i=sum_{xin X_i} e_x$ are a basis for the space of vectors fixed by $G$, and in particular that space has dimension $n$.






                  share|cite|improve this answer









                  $endgroup$



                  The inner product $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of intertwining maps from the trivial representation $mathbb{C}$ to $mathbb{C}[X]$. A linear map $T:mathbb{C}tomathbb{C}[X]$ is determined by $T(1)$, and is intertwining iff $T(1)$ is fixed by every element of $G$. So, $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of vectors in $mathbb{C}[X]$ which are fixed by every element of $G$.



                  Now, when is an element $sum c_xe_xinmathbb{C}[X]$ fixed by every element of $G$? Exactly when the coefficients $c_x$ are constant on each orbit (since the action of $G$ just permutes these coefficients transitively within each orbit). Letting $X_1,dots,X_n$ be the orbits, this means that the vectors $v_i=sum_{xin X_i} e_x$ are a basis for the space of vectors fixed by $G$, and in particular that space has dimension $n$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 8 at 22:01









                  Eric WofseyEric Wofsey

                  187k14215344




                  187k14215344























                      2












                      $begingroup$

                      I really like Eric Wofsey's answer (+1), which was posted as I was writing this. However, this is a different solution, so I'll post it too.



                      My solution follows from the orbit-stabilizer theorem.



                      Let's take a closer look at what $newcommandCC{mathbb{C}}chi_{CC[X]}$ actually is.
                      $$chi_{CC[X]}(g) = newcommandtr{operatorname{tr}}tr(pi_g) = sum_{xin X} langle e_x,ge_xrangle = sum_{xin X} [gx=x], $$
                      i.e. $chi_{CC[X]}(g)$ counts the number of elements of $X$ fixed by $G$.



                      Then
                      $$(1_G,chi_{CC[X]}) = frac{1}{|G|}sum_{gin G} chi_{CC[X]}(g)
                      =frac{1}{|G|}sum_{gin G}sum_{xin X} [gx=x],$$

                      i.e. the inner product is the average number of elements of $X$ fixed.
                      Rearranging the sums, we have
                      $$(1_G,chi_{CC[X]})=frac{1}{|G|}sum_{xin X} sum_{gin G}[gx=x] = frac{1}{|G|} sum_{xin X} |operatorname{Stab}(x)|.$$
                      Now pull the cardinality of the group in, and recognize the term in the summation as the index of the stabilizer
                      $$(1_G,chi_{CC[X]})=sum_{xin X} frac{|newcommandStab{operatorname{Stab}}Stab(x)|}{|G|}=sum_{xin X}frac{1}{[G:Stab(x)]},$$
                      but the index of the stabilizer equals the size of the orbit, so we have
                      $$(1_G,chi_{CC[X]})=sum_{xin X} frac{1}{|Gx|} =sum_{text{orbits}} 1 =#{text{orbits}} $$



                      A note on notation



                      For a statement $P$, I use the notation $$[P]:=begin{cases} 1 & text{ $P$ is true} \ 0 & text{otherwise}.end{cases}$$



                      Note



                      Also this result is essentially just Burnside's lemma (modulo my comment on the character counting the number of fixed points).






                      share|cite|improve this answer











                      $endgroup$


















                        2












                        $begingroup$

                        I really like Eric Wofsey's answer (+1), which was posted as I was writing this. However, this is a different solution, so I'll post it too.



                        My solution follows from the orbit-stabilizer theorem.



                        Let's take a closer look at what $newcommandCC{mathbb{C}}chi_{CC[X]}$ actually is.
                        $$chi_{CC[X]}(g) = newcommandtr{operatorname{tr}}tr(pi_g) = sum_{xin X} langle e_x,ge_xrangle = sum_{xin X} [gx=x], $$
                        i.e. $chi_{CC[X]}(g)$ counts the number of elements of $X$ fixed by $G$.



                        Then
                        $$(1_G,chi_{CC[X]}) = frac{1}{|G|}sum_{gin G} chi_{CC[X]}(g)
                        =frac{1}{|G|}sum_{gin G}sum_{xin X} [gx=x],$$

                        i.e. the inner product is the average number of elements of $X$ fixed.
                        Rearranging the sums, we have
                        $$(1_G,chi_{CC[X]})=frac{1}{|G|}sum_{xin X} sum_{gin G}[gx=x] = frac{1}{|G|} sum_{xin X} |operatorname{Stab}(x)|.$$
                        Now pull the cardinality of the group in, and recognize the term in the summation as the index of the stabilizer
                        $$(1_G,chi_{CC[X]})=sum_{xin X} frac{|newcommandStab{operatorname{Stab}}Stab(x)|}{|G|}=sum_{xin X}frac{1}{[G:Stab(x)]},$$
                        but the index of the stabilizer equals the size of the orbit, so we have
                        $$(1_G,chi_{CC[X]})=sum_{xin X} frac{1}{|Gx|} =sum_{text{orbits}} 1 =#{text{orbits}} $$



                        A note on notation



                        For a statement $P$, I use the notation $$[P]:=begin{cases} 1 & text{ $P$ is true} \ 0 & text{otherwise}.end{cases}$$



                        Note



                        Also this result is essentially just Burnside's lemma (modulo my comment on the character counting the number of fixed points).






                        share|cite|improve this answer











                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          I really like Eric Wofsey's answer (+1), which was posted as I was writing this. However, this is a different solution, so I'll post it too.



                          My solution follows from the orbit-stabilizer theorem.



                          Let's take a closer look at what $newcommandCC{mathbb{C}}chi_{CC[X]}$ actually is.
                          $$chi_{CC[X]}(g) = newcommandtr{operatorname{tr}}tr(pi_g) = sum_{xin X} langle e_x,ge_xrangle = sum_{xin X} [gx=x], $$
                          i.e. $chi_{CC[X]}(g)$ counts the number of elements of $X$ fixed by $G$.



                          Then
                          $$(1_G,chi_{CC[X]}) = frac{1}{|G|}sum_{gin G} chi_{CC[X]}(g)
                          =frac{1}{|G|}sum_{gin G}sum_{xin X} [gx=x],$$

                          i.e. the inner product is the average number of elements of $X$ fixed.
                          Rearranging the sums, we have
                          $$(1_G,chi_{CC[X]})=frac{1}{|G|}sum_{xin X} sum_{gin G}[gx=x] = frac{1}{|G|} sum_{xin X} |operatorname{Stab}(x)|.$$
                          Now pull the cardinality of the group in, and recognize the term in the summation as the index of the stabilizer
                          $$(1_G,chi_{CC[X]})=sum_{xin X} frac{|newcommandStab{operatorname{Stab}}Stab(x)|}{|G|}=sum_{xin X}frac{1}{[G:Stab(x)]},$$
                          but the index of the stabilizer equals the size of the orbit, so we have
                          $$(1_G,chi_{CC[X]})=sum_{xin X} frac{1}{|Gx|} =sum_{text{orbits}} 1 =#{text{orbits}} $$



                          A note on notation



                          For a statement $P$, I use the notation $$[P]:=begin{cases} 1 & text{ $P$ is true} \ 0 & text{otherwise}.end{cases}$$



                          Note



                          Also this result is essentially just Burnside's lemma (modulo my comment on the character counting the number of fixed points).






                          share|cite|improve this answer











                          $endgroup$



                          I really like Eric Wofsey's answer (+1), which was posted as I was writing this. However, this is a different solution, so I'll post it too.



                          My solution follows from the orbit-stabilizer theorem.



                          Let's take a closer look at what $newcommandCC{mathbb{C}}chi_{CC[X]}$ actually is.
                          $$chi_{CC[X]}(g) = newcommandtr{operatorname{tr}}tr(pi_g) = sum_{xin X} langle e_x,ge_xrangle = sum_{xin X} [gx=x], $$
                          i.e. $chi_{CC[X]}(g)$ counts the number of elements of $X$ fixed by $G$.



                          Then
                          $$(1_G,chi_{CC[X]}) = frac{1}{|G|}sum_{gin G} chi_{CC[X]}(g)
                          =frac{1}{|G|}sum_{gin G}sum_{xin X} [gx=x],$$

                          i.e. the inner product is the average number of elements of $X$ fixed.
                          Rearranging the sums, we have
                          $$(1_G,chi_{CC[X]})=frac{1}{|G|}sum_{xin X} sum_{gin G}[gx=x] = frac{1}{|G|} sum_{xin X} |operatorname{Stab}(x)|.$$
                          Now pull the cardinality of the group in, and recognize the term in the summation as the index of the stabilizer
                          $$(1_G,chi_{CC[X]})=sum_{xin X} frac{|newcommandStab{operatorname{Stab}}Stab(x)|}{|G|}=sum_{xin X}frac{1}{[G:Stab(x)]},$$
                          but the index of the stabilizer equals the size of the orbit, so we have
                          $$(1_G,chi_{CC[X]})=sum_{xin X} frac{1}{|Gx|} =sum_{text{orbits}} 1 =#{text{orbits}} $$



                          A note on notation



                          For a statement $P$, I use the notation $$[P]:=begin{cases} 1 & text{ $P$ is true} \ 0 & text{otherwise}.end{cases}$$



                          Note



                          Also this result is essentially just Burnside's lemma (modulo my comment on the character counting the number of fixed points).







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 8 at 22:19

























                          answered Jan 8 at 22:13









                          jgonjgon

                          14.6k22042




                          14.6k22042






























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