Tensor product and invariant subspaces
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Let $V$ be a vector space of dimension $n$ over $mathbbC$.
Q.1 When thinking of $Votimes Votimes V $, if $V$ has basis $x_1,x_2,cdots,x_n$ then can we think $Votimes Votimes V$ as $mathbbC$-vector space of polynomials $sum_i,j a_ijx_ix_jx_k$ where $x_i$ and $x_j$ are non-commuting for $ineq j$.
If this is true, then consider the natural action of $S_3$ on $Votimes Votimes V$. This action makes $Votimes Votimes V$ an $mathbbC[S_n]$-module. Consider the element of the group algebra
$$alpha=(1)+(12)+(13)+(23)+(123)+(132).$$
The action of $alpha$ on $Votimes Votimes V$ leaves the subspace $Srm ym^3(V)$ invariant. Now I want to prove that converse is also true:
if an element $betain mathbbC[S_n]$ leaves $rm Sym^3(V)$ invariant, then $beta$ should some scalar multiple of $alpha$. How should I proceed for it?
I was thinking elements of $Votimes Votimes V$ as in Q.1, but I couldn't proceed for the question on $beta$
representation-theory symmetric-groups
edited Sep 1 at 5:24
Chickenmancer
3,057622
asked Sep 1 at 5:17
Beginner
3,45711123
add a comment |Â
up vote
2
down vote
favorite
Let $V$ be a vector space of dimension $n$ over $mathbbC$.
Q.1 When thinking of $Votimes Votimes V $, if $V$ has basis $x_1,x_2,cdots,x_n$ then can we think $Votimes Votimes V$ as $mathbbC$-vector space of polynomials $sum_i,j a_ijx_ix_jx_k$ where $x_i$ and $x_j$ are non-commuting for $ineq j$.
If this is true, then consider the natural action of $S_3$ on $Votimes Votimes V$. This action makes $Votimes Votimes V$ an $mathbbC[S_n]$-module. Consider the element of the group algebra
$$alpha=(1)+(12)+(13)+(23)+(123)+(132).$$
The action of $alpha$ on $Votimes Votimes V$ leaves the subspace $Srm ym^3(V)$ invariant. Now I want to prove that converse is also true:
if an element $betain mathbbC[S_n]$ leaves $rm Sym^3(V)$ invariant, then $beta$ should some scalar multiple of $alpha$. How should I proceed for it?
I was thinking elements of $Votimes Votimes V$ as in Q.1, but I couldn't proceed for the question on $beta$
representation-theory symmetric-groups
edited Sep 1 at 5:24
Chickenmancer
3,057622
asked Sep 1 at 5:17
Beginner
3,45711123
1
This is wrong, as @anon noticed. Perhaps you rather want to say "if an element $beta in mathbbCleft[S_nright]$ sends $V otimes V otimes V$ to $operatornameSym^3 V$, then $beta$ is a scalar multiple of $alpha$". This seems more likely to be true to me (but I am not sure and don't have the time to check).
– darij grinberg
Sep 1 at 23:58
@darijgrinberg That follows from the fact every element of the group algebra is expressible as a $Bbb C[G]$-linear combination of the isotypical projectors $e_V$ (of which the normalized $alpha$ is one). Probably a more elementary explanation for $S_3$ though.
– anon
Sep 2 at 0:01
Actually, it's almost correct. It's false if $dim V < 2$, but it's true if $dim V geq 3$.
– darij grinberg
Sep 2 at 0:05
@darij: Yes, this is actually the thing I was missing from my understanding. But thanks for the clear statement and repairing my doubt.
– Beginner
Sep 3 at 3:45
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $V$ be a vector space of dimension $n$ over $mathbbC$.
Q.1 When thinking of $Votimes Votimes V $, if $V$ has basis $x_1,x_2,cdots,x_n$ then can we think $Votimes Votimes V$ as $mathbbC$-vector space of polynomials $sum_i,j a_ijx_ix_jx_k$ where $x_i$ and $x_j$ are non-commuting for $ineq j$.
If this is true, then consider the natural action of $S_3$ on $Votimes Votimes V$. This action makes $Votimes Votimes V$ an $mathbbC[S_n]$-module. Consider the element of the group algebra
$$alpha=(1)+(12)+(13)+(23)+(123)+(132).$$
The action of $alpha$ on $Votimes Votimes V$ leaves the subspace $Srm ym^3(V)$ invariant. Now I want to prove that converse is also true:
if an element $betain mathbbC[S_n]$ leaves $rm Sym^3(V)$ invariant, then $beta$ should some scalar multiple of $alpha$. How should I proceed for it?
I was thinking elements of $Votimes Votimes V$ as in Q.1, but I couldn't proceed for the question on $beta$
representation-theory symmetric-groups
edited Sep 1 at 5:24
Chickenmancer
3,057622
asked Sep 1 at 5:17
Beginner
3,45711123
Let $V$ be a vector space of dimension $n$ over $mathbbC$.
Q.1 When thinking of $Votimes Votimes V $, if $V$ has basis $x_1,x_2,cdots,x_n$ then can we think $Votimes Votimes V$ as $mathbbC$-vector space of polynomials $sum_i,j a_ijx_ix_jx_k$ where $x_i$ and $x_j$ are non-commuting for $ineq j$.
If this is true, then consider the natural action of $S_3$ on $Votimes Votimes V$. This action makes $Votimes Votimes V$ an $mathbbC[S_n]$-module. Consider the element of the group algebra
$$alpha=(1)+(12)+(13)+(23)+(123)+(132).$$
The action of $alpha$ on $Votimes Votimes V$ leaves the subspace $Srm ym^3(V)$ invariant. Now I want to prove that converse is also true:
if an element $betain mathbbC[S_n]$ leaves $rm Sym^3(V)$ invariant, then $beta$ should some scalar multiple of $alpha$. How should I proceed for it?
I was thinking elements of $Votimes Votimes V$ as in Q.1, but I couldn't proceed for the question on $beta$
representation-theory symmetric-groups
representation-theory symmetric-groups
edited Sep 1 at 5:24
Chickenmancer
3,057622
asked Sep 1 at 5:17
Beginner
3,45711123
edited Sep 1 at 5:24
Chickenmancer
3,057622
asked Sep 1 at 5:17
Beginner
3,45711123
edited Sep 1 at 5:24
Chickenmancer
3,057622
edited Sep 1 at 5:24
Chickenmancer
3,057622
edited Sep 1 at 5:24
Chickenmancer
3,057622
3,057622
asked Sep 1 at 5:17
Beginner
3,45711123
asked Sep 1 at 5:17
Beginner
3,45711123
asked Sep 1 at 5:17
Beginner
3,45711123
3,45711123
1
This is wrong, as @anon noticed. Perhaps you rather want to say "if an element $beta in mathbbCleft[S_nright]$ sends $V otimes V otimes V$ to $operatornameSym^3 V$, then $beta$ is a scalar multiple of $alpha$". This seems more likely to be true to me (but I am not sure and don't have the time to check).
– darij grinberg
Sep 1 at 23:58
@darijgrinberg That follows from the fact every element of the group algebra is expressible as a $Bbb C[G]$-linear combination of the isotypical projectors $e_V$ (of which the normalized $alpha$ is one). Probably a more elementary explanation for $S_3$ though.
– anon
Sep 2 at 0:01
Actually, it's almost correct. It's false if $dim V < 2$, but it's true if $dim V geq 3$.
– darij grinberg
Sep 2 at 0:05
@darij: Yes, this is actually the thing I was missing from my understanding. But thanks for the clear statement and repairing my doubt.
– Beginner
Sep 3 at 3:45
add a comment |Â
1
This is wrong, as @anon noticed. Perhaps you rather want to say "if an element $beta in mathbbCleft[S_nright]$ sends $V otimes V otimes V$ to $operatornameSym^3 V$, then $beta$ is a scalar multiple of $alpha$". This seems more likely to be true to me (but I am not sure and don't have the time to check).
– darij grinberg
Sep 1 at 23:58
@darijgrinberg That follows from the fact every element of the group algebra is expressible as a $Bbb C[G]$-linear combination of the isotypical projectors $e_V$ (of which the normalized $alpha$ is one). Probably a more elementary explanation for $S_3$ though.
– anon
Sep 2 at 0:01
Actually, it's almost correct. It's false if $dim V < 2$, but it's true if $dim V geq 3$.
– darij grinberg
Sep 2 at 0:05
@darij: Yes, this is actually the thing I was missing from my understanding. But thanks for the clear statement and repairing my doubt.
– Beginner
Sep 3 at 3:45
1
1
This is wrong, as @anon noticed. Perhaps you rather want to say "if an element $beta in mathbbCleft[S_nright]$ sends $V otimes V otimes V$ to $operatornameSym^3 V$, then $beta$ is a scalar multiple of $alpha$". This seems more likely to be true to me (but I am not sure and don't have the time to check).
– darij grinberg
Sep 1 at 23:58
This is wrong, as @anon noticed. Perhaps you rather want to say "if an element $beta in mathbbCleft[S_nright]$ sends $V otimes V otimes V$ to $operatornameSym^3 V$, then $beta$ is a scalar multiple of $alpha$". This seems more likely to be true to me (but I am not sure and don't have the time to check).
– darij grinberg
Sep 1 at 23:58
@darijgrinberg That follows from the fact every element of the group algebra is expressible as a $Bbb C[G]$-linear combination of the isotypical projectors $e_V$ (of which the normalized $alpha$ is one). Probably a more elementary explanation for $S_3$ though.
– anon
Sep 2 at 0:01
@darijgrinberg That follows from the fact every element of the group algebra is expressible as a $Bbb C[G]$-linear combination of the isotypical projectors $e_V$ (of which the normalized $alpha$ is one). Probably a more elementary explanation for $S_3$ though.
– anon
Sep 2 at 0:01
Actually, it's almost correct. It's false if $dim V < 2$, but it's true if $dim V geq 3$.
– darij grinberg
Sep 2 at 0:05
Actually, it's almost correct. It's false if $dim V < 2$, but it's true if $dim V geq 3$.
– darij grinberg
Sep 2 at 0:05
@darij: Yes, this is actually the thing I was missing from my understanding. But thanks for the clear statement and repairing my doubt.
– Beginner
Sep 3 at 3:45
@darij: Yes, this is actually the thing I was missing from my understanding. But thanks for the clear statement and repairing my doubt.
– Beginner
Sep 3 at 3:45
add a comment |Â
1 Answer
1
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1
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$rm Sym^n(V)$ is already a subrepresentation of $V^otimes n$; it is a subspace stabilized by every element of the group algebra. As for elements $beta$ which fix each element $x$ of $rm Sym^n(V)$, observe
$$ beta x=left(sum_g c(g)gright)x=sum_g c(g)gx=sum_g c(g)x=left(sum_g c(g)right)x. $$
That is, $beta$ fixes nonzero symmetric tensors $x$ if and only if the sum of its coefficients is $1$. In that case, the sum of the coefficients of $beta-alpha$ is $0$, and so $beta$ fixes them if and only if it is in the coset $alpha+A$, where $A$ is the augmentation ideal of $Bbb C[G]$ and $alpha=|G|^-1sum_g g$ is the normalized sum of the group elements.
Note this argument works on any $Bbb C[G]$-module, where $V^otimes n$ is replaced with $V$, $S_n$ is replaced with $G$, and $rm Sym^n(V)$ is replaced by the invariant subspace $V^G$ (assuming it's nontrivial).
answered Sep 1 at 23:37
anon
71k5107210
Careful: You want $V^G neq 0$; otherwise, any element of $mathbbCleft[Gright]$ leaves any element of $V^G$ invariant.
– darij grinberg
Sep 1 at 23:57
Yeah, that's true.
– anon
Sep 2 at 0:00
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$rm Sym^n(V)$ is already a subrepresentation of $V^otimes n$; it is a subspace stabilized by every element of the group algebra. As for elements $beta$ which fix each element $x$ of $rm Sym^n(V)$, observe
$$ beta x=left(sum_g c(g)gright)x=sum_g c(g)gx=sum_g c(g)x=left(sum_g c(g)right)x. $$
That is, $beta$ fixes nonzero symmetric tensors $x$ if and only if the sum of its coefficients is $1$. In that case, the sum of the coefficients of $beta-alpha$ is $0$, and so $beta$ fixes them if and only if it is in the coset $alpha+A$, where $A$ is the augmentation ideal of $Bbb C[G]$ and $alpha=|G|^-1sum_g g$ is the normalized sum of the group elements.
Note this argument works on any $Bbb C[G]$-module, where $V^otimes n$ is replaced with $V$, $S_n$ is replaced with $G$, and $rm Sym^n(V)$ is replaced by the invariant subspace $V^G$ (assuming it's nontrivial).
answered Sep 1 at 23:37
anon
71k5107210
Careful: You want $V^G neq 0$; otherwise, any element of $mathbbCleft[Gright]$ leaves any element of $V^G$ invariant.
– darij grinberg
Sep 1 at 23:57
Yeah, that's true.
– anon
Sep 2 at 0:00
add a comment |Â
up vote
1
down vote
$rm Sym^n(V)$ is already a subrepresentation of $V^otimes n$; it is a subspace stabilized by every element of the group algebra. As for elements $beta$ which fix each element $x$ of $rm Sym^n(V)$, observe
$$ beta x=left(sum_g c(g)gright)x=sum_g c(g)gx=sum_g c(g)x=left(sum_g c(g)right)x. $$
That is, $beta$ fixes nonzero symmetric tensors $x$ if and only if the sum of its coefficients is $1$. In that case, the sum of the coefficients of $beta-alpha$ is $0$, and so $beta$ fixes them if and only if it is in the coset $alpha+A$, where $A$ is the augmentation ideal of $Bbb C[G]$ and $alpha=|G|^-1sum_g g$ is the normalized sum of the group elements.
Note this argument works on any $Bbb C[G]$-module, where $V^otimes n$ is replaced with $V$, $S_n$ is replaced with $G$, and $rm Sym^n(V)$ is replaced by the invariant subspace $V^G$ (assuming it's nontrivial).
answered Sep 1 at 23:37
anon
71k5107210
Careful: You want $V^G neq 0$; otherwise, any element of $mathbbCleft[Gright]$ leaves any element of $V^G$ invariant.
– darij grinberg
Sep 1 at 23:57
Yeah, that's true.
– anon
Sep 2 at 0:00
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$rm Sym^n(V)$ is already a subrepresentation of $V^otimes n$; it is a subspace stabilized by every element of the group algebra. As for elements $beta$ which fix each element $x$ of $rm Sym^n(V)$, observe
$$ beta x=left(sum_g c(g)gright)x=sum_g c(g)gx=sum_g c(g)x=left(sum_g c(g)right)x. $$
That is, $beta$ fixes nonzero symmetric tensors $x$ if and only if the sum of its coefficients is $1$. In that case, the sum of the coefficients of $beta-alpha$ is $0$, and so $beta$ fixes them if and only if it is in the coset $alpha+A$, where $A$ is the augmentation ideal of $Bbb C[G]$ and $alpha=|G|^-1sum_g g$ is the normalized sum of the group elements.
Note this argument works on any $Bbb C[G]$-module, where $V^otimes n$ is replaced with $V$, $S_n$ is replaced with $G$, and $rm Sym^n(V)$ is replaced by the invariant subspace $V^G$ (assuming it's nontrivial).
answered Sep 1 at 23:37
anon
71k5107210
$rm Sym^n(V)$ is already a subrepresentation of $V^otimes n$; it is a subspace stabilized by every element of the group algebra. As for elements $beta$ which fix each element $x$ of $rm Sym^n(V)$, observe
$$ beta x=left(sum_g c(g)gright)x=sum_g c(g)gx=sum_g c(g)x=left(sum_g c(g)right)x. $$
That is, $beta$ fixes nonzero symmetric tensors $x$ if and only if the sum of its coefficients is $1$. In that case, the sum of the coefficients of $beta-alpha$ is $0$, and so $beta$ fixes them if and only if it is in the coset $alpha+A$, where $A$ is the augmentation ideal of $Bbb C[G]$ and $alpha=|G|^-1sum_g g$ is the normalized sum of the group elements.
Note this argument works on any $Bbb C[G]$-module, where $V^otimes n$ is replaced with $V$, $S_n$ is replaced with $G$, and $rm Sym^n(V)$ is replaced by the invariant subspace $V^G$ (assuming it's nontrivial).
answered Sep 1 at 23:37
anon
71k5107210
edited Sep 1 at 23:59
answered Sep 1 at 23:37
anon
71k5107210
answered Sep 1 at 23:37
anon
71k5107210
answered Sep 1 at 23:37
anon
71k5107210
71k5107210
Careful: You want $V^G neq 0$; otherwise, any element of $mathbbCleft[Gright]$ leaves any element of $V^G$ invariant.
– darij grinberg
Sep 1 at 23:57
Yeah, that's true.
– anon
Sep 2 at 0:00
add a comment |Â
Careful: You want $V^G neq 0$; otherwise, any element of $mathbbCleft[Gright]$ leaves any element of $V^G$ invariant.
– darij grinberg
Sep 1 at 23:57
Yeah, that's true.
– anon
Sep 2 at 0:00
Careful: You want $V^G neq 0$; otherwise, any element of $mathbbCleft[Gright]$ leaves any element of $V^G$ invariant.
– darij grinberg
Sep 1 at 23:57
Careful: You want $V^G neq 0$; otherwise, any element of $mathbbCleft[Gright]$ leaves any element of $V^G$ invariant.
– darij grinberg
Sep 1 at 23:57
Yeah, that's true.
– anon
Sep 2 at 0:00
Yeah, that's true.
– anon
Sep 2 at 0:00
add a comment |Â
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1
This is wrong, as @anon noticed. Perhaps you rather want to say "if an element $beta in mathbbCleft[S_nright]$ sends $V otimes V otimes V$ to $operatornameSym^3 V$, then $beta$ is a scalar multiple of $alpha$". This seems more likely to be true to me (but I am not sure and don't have the time to check).
– darij grinberg
Sep 1 at 23:58
@darijgrinberg That follows from the fact every element of the group algebra is expressible as a $Bbb C[G]$-linear combination of the isotypical projectors $e_V$ (of which the normalized $alpha$ is one). Probably a more elementary explanation for $S_3$ though.
– anon
Sep 2 at 0:01
Actually, it's almost correct. It's false if $dim V < 2$, but it's true if $dim V geq 3$.
– darij grinberg
Sep 2 at 0:05
@darij: Yes, this is actually the thing I was missing from my understanding. But thanks for the clear statement and repairing my doubt.
– Beginner
Sep 3 at 3:45