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A character identity
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This is related to my question, but it concerns a specific point of the proof of Schur's Theorem.
Let $G$ be a finite group and $chi$ an irreducible character of $G$. Is it true that
$$forall gin G,qquadsum_hin Goverlinechi(h)chi(gh)=fracGchi(e),chi(g)quad?$$
This does not seem to be related to the orthogonality properties of the table of characters. This obviously true if $chi$ is a linear character, or if $g=e$. I checked it for $G=frak S_3$ and for one nonlinear character of $frak S_4$.
gr.group-theory rt.representation-theory finite-groups
edited 1 hour ago
YCor
26.4k380127
asked 1 hour ago
Denis Serre
28.7k791194
add a comment |
up vote
4
down vote
favorite
This is related to my question, but it concerns a specific point of the proof of Schur's Theorem.
Let $G$ be a finite group and $chi$ an irreducible character of $G$. Is it true that
$$forall gin G,qquadsum_hin Goverlinechi(h)chi(gh)=fracGchi(e),chi(g)quad?$$
This does not seem to be related to the orthogonality properties of the table of characters. This obviously true if $chi$ is a linear character, or if $g=e$. I checked it for $G=frak S_3$ and for one nonlinear character of $frak S_4$.
gr.group-theory rt.representation-theory finite-groups
edited 1 hour ago
YCor
26.4k380127
asked 1 hour ago
Denis Serre
28.7k791194
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
This is related to my question, but it concerns a specific point of the proof of Schur's Theorem.
Let $G$ be a finite group and $chi$ an irreducible character of $G$. Is it true that
$$forall gin G,qquadsum_hin Goverlinechi(h)chi(gh)=fracGchi(e),chi(g)quad?$$
This does not seem to be related to the orthogonality properties of the table of characters. This obviously true if $chi$ is a linear character, or if $g=e$. I checked it for $G=frak S_3$ and for one nonlinear character of $frak S_4$.
gr.group-theory rt.representation-theory finite-groups
edited 1 hour ago
YCor
26.4k380127
asked 1 hour ago
Denis Serre
28.7k791194
This is related to my question, but it concerns a specific point of the proof of Schur's Theorem.
Let $G$ be a finite group and $chi$ an irreducible character of $G$. Is it true that
$$forall gin G,qquadsum_hin Goverlinechi(h)chi(gh)=fracGchi(e),chi(g)quad?$$
This does not seem to be related to the orthogonality properties of the table of characters. This obviously true if $chi$ is a linear character, or if $g=e$. I checked it for $G=frak S_3$ and for one nonlinear character of $frak S_4$.
gr.group-theory rt.representation-theory finite-groups
gr.group-theory rt.representation-theory finite-groups
edited 1 hour ago
YCor
26.4k380127
asked 1 hour ago
Denis Serre
28.7k791194
edited 1 hour ago
YCor
26.4k380127
asked 1 hour ago
Denis Serre
28.7k791194
edited 1 hour ago
YCor
26.4k380127
edited 1 hour ago
YCor
26.4k380127
edited 1 hour ago
YCor
26.4k380127
26.4k380127
asked 1 hour ago
Denis Serre
28.7k791194
asked 1 hour ago
Denis Serre
28.7k791194
asked 1 hour ago
Denis Serre
28.7k791194
28.7k791194
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Yes. Let
$$ F(g) = sum_h in G overlinechi(h)chi(gh).$$
I claim that $F$ is a class function. Indeed,
$$F(s^-1 g s) = sum_h in G overlinechi(h)chi(s^-1gsh)=sum_h in G overlinechi(h)chi(g(shs^-1)) = sum_h' in G overlinechi(s^-1h's )chi(gh') = sum_h' in G overlinechi(h' )chi(gh') = F(g).$$
Thus, we can write $F$ as a combination of irreducible characters:
$$F(g) = sum_i a_chi_i chi_i(g).$$
Let $rho_chi_i$ be the representation with character $chi_i$. Then
$$a_chi_i = frac1G sum_g in G overlinechi_i(g)F(g) = frac1G sum_g,h overlinechi_i(g) overlinechi(h)chi(gh)= frac1G sum_h overlinechi(h) (mathrmTr( rho_chi(h) sum_g overlinechi_i(g)rho_chi(g) ). $$
Letting
$$A_i = sum_g overlinechi_i(g)rho_chi(g),$$
we have $$(*)a_chi_i = frac1G sum_h overlinechi(h) (mathrmTr( rho_chi(h) A_i).$$
I claim $A_i$ commutes with $rho_chi(h)$ for all $h$-s:
$$A_i rho_chi(h) =sum_g overlinechi_i(g)rho_chi(gh) =sum_g' overlinechi_i(g'h^-1)rho_chi(g')=sum_g' overlinechi_i(h^-1 g')rho_chi(g')=sum_g overlinechi_i(g)rho_chi(hg)= rho_chi(h)A_i.$$
Thus, by Schur's lemma, $A_i$ is a multiple of the identity:
$$A_i = c_i I$$
for some constant $c_i$, which can be extracted by taking traces:
$$c_i cdot dim(rho_i) = mathrmTr(A_i) = sum_g in Goverlinechi_i(g) chi(g) = 1_chi_i = chi |G|.$$
Thus, when $chi_i neq chi$ we have $A_i=0$ and so $a_chi_i=0$ in that case by $(*)$, and $F(g)$ must be proportional to $chi(g)$. To find the constant of proportionality, note that if $chi_i=chi$, we have from $(*)$ that
$$a_chi_i=frac1dim rho_chi_isum_h overlinechi_i(h) chi(h) = fracGchi(e),$$
as needed.
answered 27 mins ago
Ofir Gorodetsky
5,50212236
add a comment |
up vote
2
down vote
This is correct. It does not follow immediately from the orthogonality relations for group characters, but both they and the above formula follow from the orthogonality relations for individual matrix coefficients with respect to irreducible unitary representations which follow from Schur's Lemma (and were known to Schur). I won't give all details as they can be found in most texts ( and basically generalize to finite dimensional unitary representations of Lie groups), but the basic ideas are: Let $A$ and $B$ be finite dimensional unitary irreducible representations of a finite group $G$ and set $A(g) = [a_ij(g)],B(g) = [b_ij(g)]$ for each $g in G.$
For any (suitably sized) matrix $M$, set $T(A,B) = sum_g in G A(g^-1)MB(g).$ Then $A(h^-1)T(A,B)B(h) = T(A,B)$ for all $h in G,$ and it follows from Schur's Lemma that $T(A,B)$ is the zero matrix if $A,B$ are inequivalent, and that $T(A,B)$ is a scalar matrix if $A,B$ are equivalent.
It follows ( taking different choices for $M$ with exactly one non-zero entry $1$), that we obtain $sum_h in G a_ij(h)overlineb_kell(h))] = fracGrm dim Adelta_ikdelta_jell$ if $A,B$ are equivalent, or always $0$ if $A,B$ are inequivalent.
Now apply these finer relations in the case that $A = B$ affords character $chi$, and observe that $chi(h) = sum_ i= 1^chi(1) a_ii(h)$ while $chi(gh) = sum_i=1^chi(1) sum_k = 1^chi(1) a_ik(g)a_ki(h),$ and you get the formula you want.
Note also that your formula only depends on the character afforded by the representation, so only uses the equivalence type of the representation, so it is fine to choose a unitary representation.
answered 35 mins ago
Geoff Robinson
28.7k277107
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Yes. Let
$$ F(g) = sum_h in G overlinechi(h)chi(gh).$$
I claim that $F$ is a class function. Indeed,
$$F(s^-1 g s) = sum_h in G overlinechi(h)chi(s^-1gsh)=sum_h in G overlinechi(h)chi(g(shs^-1)) = sum_h' in G overlinechi(s^-1h's )chi(gh') = sum_h' in G overlinechi(h' )chi(gh') = F(g).$$
Thus, we can write $F$ as a combination of irreducible characters:
$$F(g) = sum_i a_chi_i chi_i(g).$$
Let $rho_chi_i$ be the representation with character $chi_i$. Then
$$a_chi_i = frac1G sum_g in G overlinechi_i(g)F(g) = frac1G sum_g,h overlinechi_i(g) overlinechi(h)chi(gh)= frac1G sum_h overlinechi(h) (mathrmTr( rho_chi(h) sum_g overlinechi_i(g)rho_chi(g) ). $$
Letting
$$A_i = sum_g overlinechi_i(g)rho_chi(g),$$
we have $$(*)a_chi_i = frac1G sum_h overlinechi(h) (mathrmTr( rho_chi(h) A_i).$$
I claim $A_i$ commutes with $rho_chi(h)$ for all $h$-s:
$$A_i rho_chi(h) =sum_g overlinechi_i(g)rho_chi(gh) =sum_g' overlinechi_i(g'h^-1)rho_chi(g')=sum_g' overlinechi_i(h^-1 g')rho_chi(g')=sum_g overlinechi_i(g)rho_chi(hg)= rho_chi(h)A_i.$$
Thus, by Schur's lemma, $A_i$ is a multiple of the identity:
$$A_i = c_i I$$
for some constant $c_i$, which can be extracted by taking traces:
$$c_i cdot dim(rho_i) = mathrmTr(A_i) = sum_g in Goverlinechi_i(g) chi(g) = 1_chi_i = chi |G|.$$
Thus, when $chi_i neq chi$ we have $A_i=0$ and so $a_chi_i=0$ in that case by $(*)$, and $F(g)$ must be proportional to $chi(g)$. To find the constant of proportionality, note that if $chi_i=chi$, we have from $(*)$ that
$$a_chi_i=frac1dim rho_chi_isum_h overlinechi_i(h) chi(h) = fracGchi(e),$$
as needed.
answered 27 mins ago
Ofir Gorodetsky
5,50212236
add a comment |
up vote
2
down vote
accepted
Yes. Let
$$ F(g) = sum_h in G overlinechi(h)chi(gh).$$
I claim that $F$ is a class function. Indeed,
$$F(s^-1 g s) = sum_h in G overlinechi(h)chi(s^-1gsh)=sum_h in G overlinechi(h)chi(g(shs^-1)) = sum_h' in G overlinechi(s^-1h's )chi(gh') = sum_h' in G overlinechi(h' )chi(gh') = F(g).$$
Thus, we can write $F$ as a combination of irreducible characters:
$$F(g) = sum_i a_chi_i chi_i(g).$$
Let $rho_chi_i$ be the representation with character $chi_i$. Then
$$a_chi_i = frac1G sum_g in G overlinechi_i(g)F(g) = frac1G sum_g,h overlinechi_i(g) overlinechi(h)chi(gh)= frac1G sum_h overlinechi(h) (mathrmTr( rho_chi(h) sum_g overlinechi_i(g)rho_chi(g) ). $$
Letting
$$A_i = sum_g overlinechi_i(g)rho_chi(g),$$
we have $$(*)a_chi_i = frac1G sum_h overlinechi(h) (mathrmTr( rho_chi(h) A_i).$$
I claim $A_i$ commutes with $rho_chi(h)$ for all $h$-s:
$$A_i rho_chi(h) =sum_g overlinechi_i(g)rho_chi(gh) =sum_g' overlinechi_i(g'h^-1)rho_chi(g')=sum_g' overlinechi_i(h^-1 g')rho_chi(g')=sum_g overlinechi_i(g)rho_chi(hg)= rho_chi(h)A_i.$$
Thus, by Schur's lemma, $A_i$ is a multiple of the identity:
$$A_i = c_i I$$
for some constant $c_i$, which can be extracted by taking traces:
$$c_i cdot dim(rho_i) = mathrmTr(A_i) = sum_g in Goverlinechi_i(g) chi(g) = 1_chi_i = chi |G|.$$
Thus, when $chi_i neq chi$ we have $A_i=0$ and so $a_chi_i=0$ in that case by $(*)$, and $F(g)$ must be proportional to $chi(g)$. To find the constant of proportionality, note that if $chi_i=chi$, we have from $(*)$ that
$$a_chi_i=frac1dim rho_chi_isum_h overlinechi_i(h) chi(h) = fracGchi(e),$$
as needed.
answered 27 mins ago
Ofir Gorodetsky
5,50212236
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Yes. Let
$$ F(g) = sum_h in G overlinechi(h)chi(gh).$$
I claim that $F$ is a class function. Indeed,
$$F(s^-1 g s) = sum_h in G overlinechi(h)chi(s^-1gsh)=sum_h in G overlinechi(h)chi(g(shs^-1)) = sum_h' in G overlinechi(s^-1h's )chi(gh') = sum_h' in G overlinechi(h' )chi(gh') = F(g).$$
Thus, we can write $F$ as a combination of irreducible characters:
$$F(g) = sum_i a_chi_i chi_i(g).$$
Let $rho_chi_i$ be the representation with character $chi_i$. Then
$$a_chi_i = frac1G sum_g in G overlinechi_i(g)F(g) = frac1G sum_g,h overlinechi_i(g) overlinechi(h)chi(gh)= frac1G sum_h overlinechi(h) (mathrmTr( rho_chi(h) sum_g overlinechi_i(g)rho_chi(g) ). $$
Letting
$$A_i = sum_g overlinechi_i(g)rho_chi(g),$$
we have $$(*)a_chi_i = frac1G sum_h overlinechi(h) (mathrmTr( rho_chi(h) A_i).$$
I claim $A_i$ commutes with $rho_chi(h)$ for all $h$-s:
$$A_i rho_chi(h) =sum_g overlinechi_i(g)rho_chi(gh) =sum_g' overlinechi_i(g'h^-1)rho_chi(g')=sum_g' overlinechi_i(h^-1 g')rho_chi(g')=sum_g overlinechi_i(g)rho_chi(hg)= rho_chi(h)A_i.$$
Thus, by Schur's lemma, $A_i$ is a multiple of the identity:
$$A_i = c_i I$$
for some constant $c_i$, which can be extracted by taking traces:
$$c_i cdot dim(rho_i) = mathrmTr(A_i) = sum_g in Goverlinechi_i(g) chi(g) = 1_chi_i = chi |G|.$$
Thus, when $chi_i neq chi$ we have $A_i=0$ and so $a_chi_i=0$ in that case by $(*)$, and $F(g)$ must be proportional to $chi(g)$. To find the constant of proportionality, note that if $chi_i=chi$, we have from $(*)$ that
$$a_chi_i=frac1dim rho_chi_isum_h overlinechi_i(h) chi(h) = fracGchi(e),$$
as needed.
answered 27 mins ago
Ofir Gorodetsky
5,50212236
Yes. Let
$$ F(g) = sum_h in G overlinechi(h)chi(gh).$$
I claim that $F$ is a class function. Indeed,
$$F(s^-1 g s) = sum_h in G overlinechi(h)chi(s^-1gsh)=sum_h in G overlinechi(h)chi(g(shs^-1)) = sum_h' in G overlinechi(s^-1h's )chi(gh') = sum_h' in G overlinechi(h' )chi(gh') = F(g).$$
Thus, we can write $F$ as a combination of irreducible characters:
$$F(g) = sum_i a_chi_i chi_i(g).$$
Let $rho_chi_i$ be the representation with character $chi_i$. Then
$$a_chi_i = frac1G sum_g in G overlinechi_i(g)F(g) = frac1G sum_g,h overlinechi_i(g) overlinechi(h)chi(gh)= frac1G sum_h overlinechi(h) (mathrmTr( rho_chi(h) sum_g overlinechi_i(g)rho_chi(g) ). $$
Letting
$$A_i = sum_g overlinechi_i(g)rho_chi(g),$$
we have $$(*)a_chi_i = frac1G sum_h overlinechi(h) (mathrmTr( rho_chi(h) A_i).$$
I claim $A_i$ commutes with $rho_chi(h)$ for all $h$-s:
$$A_i rho_chi(h) =sum_g overlinechi_i(g)rho_chi(gh) =sum_g' overlinechi_i(g'h^-1)rho_chi(g')=sum_g' overlinechi_i(h^-1 g')rho_chi(g')=sum_g overlinechi_i(g)rho_chi(hg)= rho_chi(h)A_i.$$
Thus, by Schur's lemma, $A_i$ is a multiple of the identity:
$$A_i = c_i I$$
for some constant $c_i$, which can be extracted by taking traces:
$$c_i cdot dim(rho_i) = mathrmTr(A_i) = sum_g in Goverlinechi_i(g) chi(g) = 1_chi_i = chi |G|.$$
Thus, when $chi_i neq chi$ we have $A_i=0$ and so $a_chi_i=0$ in that case by $(*)$, and $F(g)$ must be proportional to $chi(g)$. To find the constant of proportionality, note that if $chi_i=chi$, we have from $(*)$ that
$$a_chi_i=frac1dim rho_chi_isum_h overlinechi_i(h) chi(h) = fracGchi(e),$$
as needed.
answered 27 mins ago
Ofir Gorodetsky
5,50212236
edited 21 mins ago
answered 27 mins ago
Ofir Gorodetsky
5,50212236
answered 27 mins ago
Ofir Gorodetsky
5,50212236
answered 27 mins ago
Ofir Gorodetsky
5,50212236
5,50212236
add a comment |
add a comment |
up vote
2
down vote
This is correct. It does not follow immediately from the orthogonality relations for group characters, but both they and the above formula follow from the orthogonality relations for individual matrix coefficients with respect to irreducible unitary representations which follow from Schur's Lemma (and were known to Schur). I won't give all details as they can be found in most texts ( and basically generalize to finite dimensional unitary representations of Lie groups), but the basic ideas are: Let $A$ and $B$ be finite dimensional unitary irreducible representations of a finite group $G$ and set $A(g) = [a_ij(g)],B(g) = [b_ij(g)]$ for each $g in G.$
For any (suitably sized) matrix $M$, set $T(A,B) = sum_g in G A(g^-1)MB(g).$ Then $A(h^-1)T(A,B)B(h) = T(A,B)$ for all $h in G,$ and it follows from Schur's Lemma that $T(A,B)$ is the zero matrix if $A,B$ are inequivalent, and that $T(A,B)$ is a scalar matrix if $A,B$ are equivalent.
It follows ( taking different choices for $M$ with exactly one non-zero entry $1$), that we obtain $sum_h in G a_ij(h)overlineb_kell(h))] = fracGrm dim Adelta_ikdelta_jell$ if $A,B$ are equivalent, or always $0$ if $A,B$ are inequivalent.
Now apply these finer relations in the case that $A = B$ affords character $chi$, and observe that $chi(h) = sum_ i= 1^chi(1) a_ii(h)$ while $chi(gh) = sum_i=1^chi(1) sum_k = 1^chi(1) a_ik(g)a_ki(h),$ and you get the formula you want.
Note also that your formula only depends on the character afforded by the representation, so only uses the equivalence type of the representation, so it is fine to choose a unitary representation.
answered 35 mins ago
Geoff Robinson
28.7k277107
add a comment |
up vote
2
down vote
This is correct. It does not follow immediately from the orthogonality relations for group characters, but both they and the above formula follow from the orthogonality relations for individual matrix coefficients with respect to irreducible unitary representations which follow from Schur's Lemma (and were known to Schur). I won't give all details as they can be found in most texts ( and basically generalize to finite dimensional unitary representations of Lie groups), but the basic ideas are: Let $A$ and $B$ be finite dimensional unitary irreducible representations of a finite group $G$ and set $A(g) = [a_ij(g)],B(g) = [b_ij(g)]$ for each $g in G.$
For any (suitably sized) matrix $M$, set $T(A,B) = sum_g in G A(g^-1)MB(g).$ Then $A(h^-1)T(A,B)B(h) = T(A,B)$ for all $h in G,$ and it follows from Schur's Lemma that $T(A,B)$ is the zero matrix if $A,B$ are inequivalent, and that $T(A,B)$ is a scalar matrix if $A,B$ are equivalent.
It follows ( taking different choices for $M$ with exactly one non-zero entry $1$), that we obtain $sum_h in G a_ij(h)overlineb_kell(h))] = fracGrm dim Adelta_ikdelta_jell$ if $A,B$ are equivalent, or always $0$ if $A,B$ are inequivalent.
Now apply these finer relations in the case that $A = B$ affords character $chi$, and observe that $chi(h) = sum_ i= 1^chi(1) a_ii(h)$ while $chi(gh) = sum_i=1^chi(1) sum_k = 1^chi(1) a_ik(g)a_ki(h),$ and you get the formula you want.
Note also that your formula only depends on the character afforded by the representation, so only uses the equivalence type of the representation, so it is fine to choose a unitary representation.
answered 35 mins ago
Geoff Robinson
28.7k277107
add a comment |
up vote
2
down vote
up vote
2
down vote
This is correct. It does not follow immediately from the orthogonality relations for group characters, but both they and the above formula follow from the orthogonality relations for individual matrix coefficients with respect to irreducible unitary representations which follow from Schur's Lemma (and were known to Schur). I won't give all details as they can be found in most texts ( and basically generalize to finite dimensional unitary representations of Lie groups), but the basic ideas are: Let $A$ and $B$ be finite dimensional unitary irreducible representations of a finite group $G$ and set $A(g) = [a_ij(g)],B(g) = [b_ij(g)]$ for each $g in G.$
For any (suitably sized) matrix $M$, set $T(A,B) = sum_g in G A(g^-1)MB(g).$ Then $A(h^-1)T(A,B)B(h) = T(A,B)$ for all $h in G,$ and it follows from Schur's Lemma that $T(A,B)$ is the zero matrix if $A,B$ are inequivalent, and that $T(A,B)$ is a scalar matrix if $A,B$ are equivalent.
It follows ( taking different choices for $M$ with exactly one non-zero entry $1$), that we obtain $sum_h in G a_ij(h)overlineb_kell(h))] = fracGrm dim Adelta_ikdelta_jell$ if $A,B$ are equivalent, or always $0$ if $A,B$ are inequivalent.
Now apply these finer relations in the case that $A = B$ affords character $chi$, and observe that $chi(h) = sum_ i= 1^chi(1) a_ii(h)$ while $chi(gh) = sum_i=1^chi(1) sum_k = 1^chi(1) a_ik(g)a_ki(h),$ and you get the formula you want.
Note also that your formula only depends on the character afforded by the representation, so only uses the equivalence type of the representation, so it is fine to choose a unitary representation.
answered 35 mins ago
Geoff Robinson
28.7k277107
This is correct. It does not follow immediately from the orthogonality relations for group characters, but both they and the above formula follow from the orthogonality relations for individual matrix coefficients with respect to irreducible unitary representations which follow from Schur's Lemma (and were known to Schur). I won't give all details as they can be found in most texts ( and basically generalize to finite dimensional unitary representations of Lie groups), but the basic ideas are: Let $A$ and $B$ be finite dimensional unitary irreducible representations of a finite group $G$ and set $A(g) = [a_ij(g)],B(g) = [b_ij(g)]$ for each $g in G.$
For any (suitably sized) matrix $M$, set $T(A,B) = sum_g in G A(g^-1)MB(g).$ Then $A(h^-1)T(A,B)B(h) = T(A,B)$ for all $h in G,$ and it follows from Schur's Lemma that $T(A,B)$ is the zero matrix if $A,B$ are inequivalent, and that $T(A,B)$ is a scalar matrix if $A,B$ are equivalent.
It follows ( taking different choices for $M$ with exactly one non-zero entry $1$), that we obtain $sum_h in G a_ij(h)overlineb_kell(h))] = fracGrm dim Adelta_ikdelta_jell$ if $A,B$ are equivalent, or always $0$ if $A,B$ are inequivalent.
Now apply these finer relations in the case that $A = B$ affords character $chi$, and observe that $chi(h) = sum_ i= 1^chi(1) a_ii(h)$ while $chi(gh) = sum_i=1^chi(1) sum_k = 1^chi(1) a_ik(g)a_ki(h),$ and you get the formula you want.
Note also that your formula only depends on the character afforded by the representation, so only uses the equivalence type of the representation, so it is fine to choose a unitary representation.
answered 35 mins ago
Geoff Robinson
28.7k277107
answered 35 mins ago
Geoff Robinson
28.7k277107
answered 35 mins ago
Geoff Robinson
28.7k277107
answered 35 mins ago
Geoff Robinson
28.7k277107
28.7k277107
add a comment |
add a comment |
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