Mixing
Stabilizer in the definition of a Gelfand pair
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I am following the textbook Representation Theory of the Symmetric Groups, by Tullio C.-S., Fabio S., and Filippo T., and am confused at the definition of a Gelfand pair.
The definition is:
Let $G$ act transitively on a set $X$. Fix $x_0 in X$ and denote its stabilizer by $K={g in G : g x_0 = x_0}$. We can identify $X$ with the space of right cosets of $K$ in $G$ $(X = G/K)$. If the permutation representation $lambda$ of $G$ has a decomposition that is multiplicity-free, then $(G,K)$ is a Gelfand pair.
Here the permutation representation is the representation of $G$ on $L(X)$, the space of complex functions on $X$, where:
$$[lambda (g)f](x) = f(g^{-1}x)$$
I don't understand why $K$ is necessary in the definition. We can identify $X$ with $G/K$, but the action works on $X$ just the same. Doesn't this mean that all the stabilizers on $X$ form a Gelfand pair with $G$? What's the point of mentioning $K$ at all?
representation-theory symmetric-groups group-actions
asked Jan 28 at 21:37
Jason d'EonJason d'Eon
113
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add a comment |
$begingroup$
I am following the textbook Representation Theory of the Symmetric Groups, by Tullio C.-S., Fabio S., and Filippo T., and am confused at the definition of a Gelfand pair.
The definition is:
Let $G$ act transitively on a set $X$. Fix $x_0 in X$ and denote its stabilizer by $K={g in G : g x_0 = x_0}$. We can identify $X$ with the space of right cosets of $K$ in $G$ $(X = G/K)$. If the permutation representation $lambda$ of $G$ has a decomposition that is multiplicity-free, then $(G,K)$ is a Gelfand pair.
Here the permutation representation is the representation of $G$ on $L(X)$, the space of complex functions on $X$, where:
$$[lambda (g)f](x) = f(g^{-1}x)$$
I don't understand why $K$ is necessary in the definition. We can identify $X$ with $G/K$, but the action works on $X$ just the same. Doesn't this mean that all the stabilizers on $X$ form a Gelfand pair with $G$? What's the point of mentioning $K$ at all?
representation-theory symmetric-groups group-actions
asked Jan 28 at 21:37
Jason d'EonJason d'Eon
113
$endgroup$
$begingroup$
A better question is why mention $X$ at all?
$endgroup$
– David Hill
Jan 30 at 16:30
$begingroup$
$X$ is important, since $L(X)$ is the $FG$-module associated with $lambda$, so I think $lambda$ is dependent on $X$ and the action of $G$ on it.
$endgroup$
– Jason d'Eon
Jan 30 at 17:47
$begingroup$
No, $X$ is irrelevant. You can frame everything without referring to $X$ at all: Let $G$ act on the coset space $G/K$. We say that the pair $(G,K)$ is a Gelfand pair if the corresponding permutation representation is multiplicity free.
$endgroup$
– David Hill
Jan 30 at 18:35
$begingroup$
So by corresponding permutation representation, I should be thinking of it on the space $L(G/K)$?
$endgroup$
– Jason d'Eon
Jan 30 at 19:06
$begingroup$
Yes, exactly right.
$endgroup$
– David Hill
Jan 30 at 21:14
add a comment |
$begingroup$
I am following the textbook Representation Theory of the Symmetric Groups, by Tullio C.-S., Fabio S., and Filippo T., and am confused at the definition of a Gelfand pair.
The definition is:
Let $G$ act transitively on a set $X$. Fix $x_0 in X$ and denote its stabilizer by $K={g in G : g x_0 = x_0}$. We can identify $X$ with the space of right cosets of $K$ in $G$ $(X = G/K)$. If the permutation representation $lambda$ of $G$ has a decomposition that is multiplicity-free, then $(G,K)$ is a Gelfand pair.
Here the permutation representation is the representation of $G$ on $L(X)$, the space of complex functions on $X$, where:
$$[lambda (g)f](x) = f(g^{-1}x)$$
I don't understand why $K$ is necessary in the definition. We can identify $X$ with $G/K$, but the action works on $X$ just the same. Doesn't this mean that all the stabilizers on $X$ form a Gelfand pair with $G$? What's the point of mentioning $K$ at all?
representation-theory symmetric-groups group-actions
asked Jan 28 at 21:37
Jason d'EonJason d'Eon
113
$endgroup$
I am following the textbook Representation Theory of the Symmetric Groups, by Tullio C.-S., Fabio S., and Filippo T., and am confused at the definition of a Gelfand pair.
The definition is:
Let $G$ act transitively on a set $X$. Fix $x_0 in X$ and denote its stabilizer by $K={g in G : g x_0 = x_0}$. We can identify $X$ with the space of right cosets of $K$ in $G$ $(X = G/K)$. If the permutation representation $lambda$ of $G$ has a decomposition that is multiplicity-free, then $(G,K)$ is a Gelfand pair.
Here the permutation representation is the representation of $G$ on $L(X)$, the space of complex functions on $X$, where:
$$[lambda (g)f](x) = f(g^{-1}x)$$
I don't understand why $K$ is necessary in the definition. We can identify $X$ with $G/K$, but the action works on $X$ just the same. Doesn't this mean that all the stabilizers on $X$ form a Gelfand pair with $G$? What's the point of mentioning $K$ at all?
representation-theory symmetric-groups group-actions
representation-theory symmetric-groups group-actions
asked Jan 28 at 21:37
Jason d'EonJason d'Eon
113
asked Jan 28 at 21:37
Jason d'EonJason d'Eon
113
asked Jan 28 at 21:37
Jason d'EonJason d'Eon
113
asked Jan 28 at 21:37
Jason d'EonJason d'Eon
113
asked Jan 28 at 21:37
Jason d'EonJason d'Eon
113
113
$begingroup$
A better question is why mention $X$ at all?
$endgroup$
– David Hill
Jan 30 at 16:30
$begingroup$
$X$ is important, since $L(X)$ is the $FG$-module associated with $lambda$, so I think $lambda$ is dependent on $X$ and the action of $G$ on it.
$endgroup$
– Jason d'Eon
Jan 30 at 17:47
$begingroup$
No, $X$ is irrelevant. You can frame everything without referring to $X$ at all: Let $G$ act on the coset space $G/K$. We say that the pair $(G,K)$ is a Gelfand pair if the corresponding permutation representation is multiplicity free.
$endgroup$
– David Hill
Jan 30 at 18:35
$begingroup$
So by corresponding permutation representation, I should be thinking of it on the space $L(G/K)$?
$endgroup$
– Jason d'Eon
Jan 30 at 19:06
$begingroup$
Yes, exactly right.
$endgroup$
– David Hill
Jan 30 at 21:14
add a comment |
$begingroup$
A better question is why mention $X$ at all?
$endgroup$
– David Hill
Jan 30 at 16:30
$begingroup$
$X$ is important, since $L(X)$ is the $FG$-module associated with $lambda$, so I think $lambda$ is dependent on $X$ and the action of $G$ on it.
$endgroup$
– Jason d'Eon
Jan 30 at 17:47
$begingroup$
No, $X$ is irrelevant. You can frame everything without referring to $X$ at all: Let $G$ act on the coset space $G/K$. We say that the pair $(G,K)$ is a Gelfand pair if the corresponding permutation representation is multiplicity free.
$endgroup$
– David Hill
Jan 30 at 18:35
$begingroup$
So by corresponding permutation representation, I should be thinking of it on the space $L(G/K)$?
$endgroup$
– Jason d'Eon
Jan 30 at 19:06
$begingroup$
Yes, exactly right.
$endgroup$
– David Hill
Jan 30 at 21:14
$begingroup$
A better question is why mention $X$ at all?
$endgroup$
– David Hill
Jan 30 at 16:30
$begingroup$
A better question is why mention $X$ at all?
$endgroup$
– David Hill
Jan 30 at 16:30
$begingroup$
$X$ is important, since $L(X)$ is the $FG$-module associated with $lambda$, so I think $lambda$ is dependent on $X$ and the action of $G$ on it.
$endgroup$
– Jason d'Eon
Jan 30 at 17:47
$begingroup$
$X$ is important, since $L(X)$ is the $FG$-module associated with $lambda$, so I think $lambda$ is dependent on $X$ and the action of $G$ on it.
$endgroup$
– Jason d'Eon
Jan 30 at 17:47
$begingroup$
No, $X$ is irrelevant. You can frame everything without referring to $X$ at all: Let $G$ act on the coset space $G/K$. We say that the pair $(G,K)$ is a Gelfand pair if the corresponding permutation representation is multiplicity free.
$endgroup$
– David Hill
Jan 30 at 18:35
$begingroup$
No, $X$ is irrelevant. You can frame everything without referring to $X$ at all: Let $G$ act on the coset space $G/K$. We say that the pair $(G,K)$ is a Gelfand pair if the corresponding permutation representation is multiplicity free.
$endgroup$
– David Hill
Jan 30 at 18:35
$begingroup$
So by corresponding permutation representation, I should be thinking of it on the space $L(G/K)$?
$endgroup$
– Jason d'Eon
Jan 30 at 19:06
$begingroup$
So by corresponding permutation representation, I should be thinking of it on the space $L(G/K)$?
$endgroup$
– Jason d'Eon
Jan 30 at 19:06
$begingroup$
Yes, exactly right.
$endgroup$
– David Hill
Jan 30 at 21:14
$begingroup$
Yes, exactly right.
$endgroup$
– David Hill
Jan 30 at 21:14
add a comment |
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$begingroup$
A better question is why mention $X$ at all?
$endgroup$
– David Hill
Jan 30 at 16:30
$begingroup$
$X$ is important, since $L(X)$ is the $FG$-module associated with $lambda$, so I think $lambda$ is dependent on $X$ and the action of $G$ on it.
$endgroup$
– Jason d'Eon
Jan 30 at 17:47
$begingroup$
No, $X$ is irrelevant. You can frame everything without referring to $X$ at all: Let $G$ act on the coset space $G/K$. We say that the pair $(G,K)$ is a Gelfand pair if the corresponding permutation representation is multiplicity free.
$endgroup$
– David Hill
Jan 30 at 18:35
$begingroup$
So by corresponding permutation representation, I should be thinking of it on the space $L(G/K)$?
$endgroup$
– Jason d'Eon
Jan 30 at 19:06
$begingroup$
Yes, exactly right.
$endgroup$
– David Hill
Jan 30 at 21:14