Induction, products, and tensor products

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$DeclareMathOperatorIndInd$Let $G, G'$ be groups, and $H'$ be a subgroup of finite index of $G'$. Let $(pi,V)$ be a representation of $G$, and $(sigma',W)$ a representation of $H'$. Consider the two representations

$$pi otimes Ind_H'^G' sigma'$$

$$Ind_G times H'^G times G' pi otimes sigma'$$

of $G times G'$. There is a $G times G'$-linear map from the first to the second: if $f: G' rightarrow W$ is an element of the underlying space of $Ind_H'^G' sigma'$ (so $f(h'g') = sigma'(h')f(g')$, and $G'$ acts on $f$ by right translation), and $v in V$, define

$$F: G times G' rightarrow V otimes W$$

$$F(g,g') = pi(g)v otimes f(g')$$

We see that $F$ lies in the underlying space of $Ind_G times H'^G times G' pi otimes sigma'$, and that $(v,f) mapsto F$ is $mathbb C$-bilinear, giving us a linear map of the underlying spaces

$$pi otimes Ind_H'^G' sigma' rightarrow Ind_G times H'^G times G' pi otimes sigma'$$

which is easily seen to be $G times G'$-linear. Is this map an isomorphism? It seems like it should be, but I haven't been able to prove it.

representation-theory

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asked Aug 2 at 15:20

D_S

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$DeclareMathOperatorIndInd$Let $G, G'$ be groups, and $H'$ be a subgroup of finite index of $G'$. Let $(pi,V)$ be a representation of $G$, and $(sigma',W)$ a representation of $H'$. Consider the two representations

$$pi otimes Ind_H'^G' sigma'$$

$$Ind_G times H'^G times G' pi otimes sigma'$$

of $G times G'$. There is a $G times G'$-linear map from the first to the second: if $f: G' rightarrow W$ is an element of the underlying space of $Ind_H'^G' sigma'$ (so $f(h'g') = sigma'(h')f(g')$, and $G'$ acts on $f$ by right translation), and $v in V$, define

$$F: G times G' rightarrow V otimes W$$

$$F(g,g') = pi(g)v otimes f(g')$$

We see that $F$ lies in the underlying space of $Ind_G times H'^G times G' pi otimes sigma'$, and that $(v,f) mapsto F$ is $mathbb C$-bilinear, giving us a linear map of the underlying spaces

$$pi otimes Ind_H'^G' sigma' rightarrow Ind_G times H'^G times G' pi otimes sigma'$$

which is easily seen to be $G times G'$-linear. Is this map an isomorphism? It seems like it should be, but I haven't been able to prove it.

representation-theory

share|cite|improve this question

asked Aug 2 at 15:20

D_S

12.7k51550

add a comment | 

up vote
1
down vote

favorite

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$DeclareMathOperatorIndInd$Let $G, G'$ be groups, and $H'$ be a subgroup of finite index of $G'$. Let $(pi,V)$ be a representation of $G$, and $(sigma',W)$ a representation of $H'$. Consider the two representations

$$pi otimes Ind_H'^G' sigma'$$

$$Ind_G times H'^G times G' pi otimes sigma'$$

of $G times G'$. There is a $G times G'$-linear map from the first to the second: if $f: G' rightarrow W$ is an element of the underlying space of $Ind_H'^G' sigma'$ (so $f(h'g') = sigma'(h')f(g')$, and $G'$ acts on $f$ by right translation), and $v in V$, define

$$F: G times G' rightarrow V otimes W$$

$$F(g,g') = pi(g)v otimes f(g')$$

We see that $F$ lies in the underlying space of $Ind_G times H'^G times G' pi otimes sigma'$, and that $(v,f) mapsto F$ is $mathbb C$-bilinear, giving us a linear map of the underlying spaces

$$pi otimes Ind_H'^G' sigma' rightarrow Ind_G times H'^G times G' pi otimes sigma'$$

which is easily seen to be $G times G'$-linear. Is this map an isomorphism? It seems like it should be, but I haven't been able to prove it.

representation-theory

share|cite|improve this question

asked Aug 2 at 15:20

D_S

12.7k51550

$DeclareMathOperatorIndInd$Let $G, G'$ be groups, and $H'$ be a subgroup of finite index of $G'$. Let $(pi,V)$ be a representation of $G$, and $(sigma',W)$ a representation of $H'$. Consider the two representations

$$pi otimes Ind_H'^G' sigma'$$

$$Ind_G times H'^G times G' pi otimes sigma'$$

of $G times G'$. There is a $G times G'$-linear map from the first to the second: if $f: G' rightarrow W$ is an element of the underlying space of $Ind_H'^G' sigma'$ (so $f(h'g') = sigma'(h')f(g')$, and $G'$ acts on $f$ by right translation), and $v in V$, define

$$F: G times G' rightarrow V otimes W$$

$$F(g,g') = pi(g)v otimes f(g')$$

We see that $F$ lies in the underlying space of $Ind_G times H'^G times G' pi otimes sigma'$, and that $(v,f) mapsto F$ is $mathbb C$-bilinear, giving us a linear map of the underlying spaces

$$pi otimes Ind_H'^G' sigma' rightarrow Ind_G times H'^G times G' pi otimes sigma'$$

which is easily seen to be $G times G'$-linear. Is this map an isomorphism? It seems like it should be, but I haven't been able to prove it.

representation-theory

share|cite|improve this question

asked Aug 2 at 15:20

D_S

12.7k51550

share|cite|improve this question

share|cite|improve this question

asked Aug 2 at 15:20

D_S

12.7k51550

asked Aug 2 at 15:20

D_S

12.7k51550

asked Aug 2 at 15:20

D_S

12.7k51550

12.7k51550

add a comment | 

add a comment | 

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You're aiming to show that
$$ hom_G times H'(kGotimes kG', Votimes W) cong hom_G(kG,V) otimes hom _H'(kG',W). $$
You know that there's an injection from the right hand side to the left, and you want to show that it is onto, or equivalently that every $Gtimes H'$-hom from $kG otimes kG'$ to $V otimes W$ can be written as a sum of homs of the form $alpha otimes beta$ where $alpha in hom_G(kG,V)$ and $beta in hom_H'(kG',W)$.

To do this we'll make use of the fact that $kG'$ is free of finite rank over $kH$. The result is true for rank one:
$$ hom_Gtimes H'(kG otimes kH',Votimes W) cong hom_G(kG,V)otimes hom_H'(kG,W)$$
because both sides identify with $Votimes W$ in a canonical way.

Let $g_1,ldots, g_n$ be a set of right coset reps for $H'$ in $G'$. Then since $kGotimes kG' cong bigoplus_i kG otimes kHg_i$:
$$hom_Gtimes H'(kG otimes kG', Votimes W) = bigoplus_i hom_Gtimes H' (kG otimes kHg_i,Votimes W) \ cong bigoplus_i hom_G(kG,V)otimes hom_H'(kHg_i,W) cong hom_G(kG,V) otimes hom_H'(kG',W)$$
These isomorphisms (going from right to left) tell you that $hom_Gtimes G'(kGotimes kG' ,Votimes W)$ is spanned by tensor products $alphaotimesbeta$ as above. You can even see how it goes wrong when $H'$ doesn't have finite index: then the first $bigoplus$ would have to be a $prod$.

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answered Aug 3 at 13:42

m_t_

6,85122244

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

You're aiming to show that
$$ hom_G times H'(kGotimes kG', Votimes W) cong hom_G(kG,V) otimes hom _H'(kG',W). $$
You know that there's an injection from the right hand side to the left, and you want to show that it is onto, or equivalently that every $Gtimes H'$-hom from $kG otimes kG'$ to $V otimes W$ can be written as a sum of homs of the form $alpha otimes beta$ where $alpha in hom_G(kG,V)$ and $beta in hom_H'(kG',W)$.

To do this we'll make use of the fact that $kG'$ is free of finite rank over $kH$. The result is true for rank one:
$$ hom_Gtimes H'(kG otimes kH',Votimes W) cong hom_G(kG,V)otimes hom_H'(kG,W)$$
because both sides identify with $Votimes W$ in a canonical way.

Let $g_1,ldots, g_n$ be a set of right coset reps for $H'$ in $G'$. Then since $kGotimes kG' cong bigoplus_i kG otimes kHg_i$:
$$hom_Gtimes H'(kG otimes kG', Votimes W) = bigoplus_i hom_Gtimes H' (kG otimes kHg_i,Votimes W) \ cong bigoplus_i hom_G(kG,V)otimes hom_H'(kHg_i,W) cong hom_G(kG,V) otimes hom_H'(kG',W)$$
These isomorphisms (going from right to left) tell you that $hom_Gtimes G'(kGotimes kG' ,Votimes W)$ is spanned by tensor products $alphaotimesbeta$ as above. You can even see how it goes wrong when $H'$ doesn't have finite index: then the first $bigoplus$ would have to be a $prod$.

share|cite|improve this answer

answered Aug 3 at 13:42

m_t_

6,85122244

add a comment | 

up vote
1
down vote



accepted

You're aiming to show that
$$ hom_G times H'(kGotimes kG', Votimes W) cong hom_G(kG,V) otimes hom _H'(kG',W). $$
You know that there's an injection from the right hand side to the left, and you want to show that it is onto, or equivalently that every $Gtimes H'$-hom from $kG otimes kG'$ to $V otimes W$ can be written as a sum of homs of the form $alpha otimes beta$ where $alpha in hom_G(kG,V)$ and $beta in hom_H'(kG',W)$.

To do this we'll make use of the fact that $kG'$ is free of finite rank over $kH$. The result is true for rank one:
$$ hom_Gtimes H'(kG otimes kH',Votimes W) cong hom_G(kG,V)otimes hom_H'(kG,W)$$
because both sides identify with $Votimes W$ in a canonical way.

Let $g_1,ldots, g_n$ be a set of right coset reps for $H'$ in $G'$. Then since $kGotimes kG' cong bigoplus_i kG otimes kHg_i$:
$$hom_Gtimes H'(kG otimes kG', Votimes W) = bigoplus_i hom_Gtimes H' (kG otimes kHg_i,Votimes W) \ cong bigoplus_i hom_G(kG,V)otimes hom_H'(kHg_i,W) cong hom_G(kG,V) otimes hom_H'(kG',W)$$
These isomorphisms (going from right to left) tell you that $hom_Gtimes G'(kGotimes kG' ,Votimes W)$ is spanned by tensor products $alphaotimesbeta$ as above. You can even see how it goes wrong when $H'$ doesn't have finite index: then the first $bigoplus$ would have to be a $prod$.

share|cite|improve this answer

answered Aug 3 at 13:42

m_t_

6,85122244

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

You're aiming to show that
$$ hom_G times H'(kGotimes kG', Votimes W) cong hom_G(kG,V) otimes hom _H'(kG',W). $$
You know that there's an injection from the right hand side to the left, and you want to show that it is onto, or equivalently that every $Gtimes H'$-hom from $kG otimes kG'$ to $V otimes W$ can be written as a sum of homs of the form $alpha otimes beta$ where $alpha in hom_G(kG,V)$ and $beta in hom_H'(kG',W)$.

To do this we'll make use of the fact that $kG'$ is free of finite rank over $kH$. The result is true for rank one:
$$ hom_Gtimes H'(kG otimes kH',Votimes W) cong hom_G(kG,V)otimes hom_H'(kG,W)$$
because both sides identify with $Votimes W$ in a canonical way.

Let $g_1,ldots, g_n$ be a set of right coset reps for $H'$ in $G'$. Then since $kGotimes kG' cong bigoplus_i kG otimes kHg_i$:
$$hom_Gtimes H'(kG otimes kG', Votimes W) = bigoplus_i hom_Gtimes H' (kG otimes kHg_i,Votimes W) \ cong bigoplus_i hom_G(kG,V)otimes hom_H'(kHg_i,W) cong hom_G(kG,V) otimes hom_H'(kG',W)$$
These isomorphisms (going from right to left) tell you that $hom_Gtimes G'(kGotimes kG' ,Votimes W)$ is spanned by tensor products $alphaotimesbeta$ as above. You can even see how it goes wrong when $H'$ doesn't have finite index: then the first $bigoplus$ would have to be a $prod$.

share|cite|improve this answer

answered Aug 3 at 13:42

m_t_

6,85122244

You're aiming to show that
$$ hom_G times H'(kGotimes kG', Votimes W) cong hom_G(kG,V) otimes hom _H'(kG',W). $$
You know that there's an injection from the right hand side to the left, and you want to show that it is onto, or equivalently that every $Gtimes H'$-hom from $kG otimes kG'$ to $V otimes W$ can be written as a sum of homs of the form $alpha otimes beta$ where $alpha in hom_G(kG,V)$ and $beta in hom_H'(kG',W)$.

To do this we'll make use of the fact that $kG'$ is free of finite rank over $kH$. The result is true for rank one:
$$ hom_Gtimes H'(kG otimes kH',Votimes W) cong hom_G(kG,V)otimes hom_H'(kG,W)$$
because both sides identify with $Votimes W$ in a canonical way.

Let $g_1,ldots, g_n$ be a set of right coset reps for $H'$ in $G'$. Then since $kGotimes kG' cong bigoplus_i kG otimes kHg_i$:
$$hom_Gtimes H'(kG otimes kG', Votimes W) = bigoplus_i hom_Gtimes H' (kG otimes kHg_i,Votimes W) \ cong bigoplus_i hom_G(kG,V)otimes hom_H'(kHg_i,W) cong hom_G(kG,V) otimes hom_H'(kG',W)$$
These isomorphisms (going from right to left) tell you that $hom_Gtimes G'(kGotimes kG' ,Votimes W)$ is spanned by tensor products $alphaotimesbeta$ as above. You can even see how it goes wrong when $H'$ doesn't have finite index: then the first $bigoplus$ would have to be a $prod$.

share|cite|improve this answer

answered Aug 3 at 13:42

m_t_

6,85122244

share|cite|improve this answer

share|cite|improve this answer

answered Aug 3 at 13:42

m_t_

6,85122244

answered Aug 3 at 13:42

m_t_

6,85122244

answered Aug 3 at 13:42

m_t_

6,85122244

6,85122244

add a comment | 

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