Mixing
Schur-Weyl Duality (Example for k = 3)
Clash Royale CLAN TAG#URR8PPP
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I am an undergraduate student interested in representation theory.
I know that you can decompose the iterated tensor product $(C^n)^otimes k$ into the direct sum of irreducible $S_k times GL(V)$ representations $S^lambda otimes U^lambda$, where the length of $lambda$ is smaller equal n.
For the partitions $(k)$ and $(1^k)$ these are just the symmetric / skewsymmetric tensors.
However, I am interested in the third part (example for k = 3) $S^(21) otimes U^(21)$. I tried to compute both of them seperatly, but I'm stuck somewhere:
I think $U^(21)$ is the kernel of the canonical map $Alt^2(V) otimes V rightarrow Alt^3(V)$. How does $S^(21) otimes U^(21)$ look like though?
And how does this decompositions helps us in any way? I assume the general goal is to compute the multiplicities of irreducible representations?
Thank you in advance for advice!
representation-theory tensor-products tensor-decomposition weyl-group
asked Jul 31 at 15:23
Amjad
768613
add a comment |Â
up vote
2
down vote
favorite
I am an undergraduate student interested in representation theory.
I know that you can decompose the iterated tensor product $(C^n)^otimes k$ into the direct sum of irreducible $S_k times GL(V)$ representations $S^lambda otimes U^lambda$, where the length of $lambda$ is smaller equal n.
For the partitions $(k)$ and $(1^k)$ these are just the symmetric / skewsymmetric tensors.
However, I am interested in the third part (example for k = 3) $S^(21) otimes U^(21)$. I tried to compute both of them seperatly, but I'm stuck somewhere:
I think $U^(21)$ is the kernel of the canonical map $Alt^2(V) otimes V rightarrow Alt^3(V)$. How does $S^(21) otimes U^(21)$ look like though?
And how does this decompositions helps us in any way? I assume the general goal is to compute the multiplicities of irreducible representations?
Thank you in advance for advice!
representation-theory tensor-products tensor-decomposition weyl-group
asked Jul 31 at 15:23
Amjad
768613
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am an undergraduate student interested in representation theory.
I know that you can decompose the iterated tensor product $(C^n)^otimes k$ into the direct sum of irreducible $S_k times GL(V)$ representations $S^lambda otimes U^lambda$, where the length of $lambda$ is smaller equal n.
For the partitions $(k)$ and $(1^k)$ these are just the symmetric / skewsymmetric tensors.
However, I am interested in the third part (example for k = 3) $S^(21) otimes U^(21)$. I tried to compute both of them seperatly, but I'm stuck somewhere:
I think $U^(21)$ is the kernel of the canonical map $Alt^2(V) otimes V rightarrow Alt^3(V)$. How does $S^(21) otimes U^(21)$ look like though?
And how does this decompositions helps us in any way? I assume the general goal is to compute the multiplicities of irreducible representations?
Thank you in advance for advice!
representation-theory tensor-products tensor-decomposition weyl-group
asked Jul 31 at 15:23
Amjad
768613
I am an undergraduate student interested in representation theory.
I know that you can decompose the iterated tensor product $(C^n)^otimes k$ into the direct sum of irreducible $S_k times GL(V)$ representations $S^lambda otimes U^lambda$, where the length of $lambda$ is smaller equal n.
For the partitions $(k)$ and $(1^k)$ these are just the symmetric / skewsymmetric tensors.
However, I am interested in the third part (example for k = 3) $S^(21) otimes U^(21)$. I tried to compute both of them seperatly, but I'm stuck somewhere:
I think $U^(21)$ is the kernel of the canonical map $Alt^2(V) otimes V rightarrow Alt^3(V)$. How does $S^(21) otimes U^(21)$ look like though?
And how does this decompositions helps us in any way? I assume the general goal is to compute the multiplicities of irreducible representations?
Thank you in advance for advice!
representation-theory tensor-products tensor-decomposition weyl-group
asked Jul 31 at 15:23
Amjad
768613
asked Jul 31 at 15:23
Amjad
768613
asked Jul 31 at 15:23
Amjad
768613
asked Jul 31 at 15:23
Amjad
768613
768613
add a comment |Â
add a comment |Â
1 Answer
1
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up vote
2
down vote
Let $tau:V^otimes 3to V^otimes 3$ be defined by $tau(uotimes
votimes w)=votimes wotimes u$. Then the $S^(21)$ part of
$V^otimes 3$ is the kernel of $textrmid+tau+tau^2$.
answered Jul 31 at 15:36
Lord Shark the Unknown
84.4k950111
Thanks for your quick answer! If I understand you correctly, those are tensors of the form $sum_sigma in A_k v_sigma(1)...v_sigma(3)$. How does that fit/interact with $U^(21)$ though? Shouldn't I get just one subspace, let's say $H^(21)$?
– Amjad
Jul 31 at 15:45
No, they aren't. What you write down is the direct sum of the symmetric and antisymmetric tensors: the kernel of $1+tau+tau^2$ is a complement to that. @Amjad
– Lord Shark the Unknown
Jul 31 at 15:51
Okay, got it. And how does $S^(21) otimes U^(21)$ look like? Isn't $S^(21)$ already the whole part of the decomposition? And if it doesn't give me any new information, what is the purpose of computing $U^(21)? Sorry that I've got so many questions.
– Amjad
Jul 31 at 17:21
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Let $tau:V^otimes 3to V^otimes 3$ be defined by $tau(uotimes
votimes w)=votimes wotimes u$. Then the $S^(21)$ part of
$V^otimes 3$ is the kernel of $textrmid+tau+tau^2$.
answered Jul 31 at 15:36
Lord Shark the Unknown
84.4k950111
Thanks for your quick answer! If I understand you correctly, those are tensors of the form $sum_sigma in A_k v_sigma(1)...v_sigma(3)$. How does that fit/interact with $U^(21)$ though? Shouldn't I get just one subspace, let's say $H^(21)$?
– Amjad
Jul 31 at 15:45
No, they aren't. What you write down is the direct sum of the symmetric and antisymmetric tensors: the kernel of $1+tau+tau^2$ is a complement to that. @Amjad
– Lord Shark the Unknown
Jul 31 at 15:51
Okay, got it. And how does $S^(21) otimes U^(21)$ look like? Isn't $S^(21)$ already the whole part of the decomposition? And if it doesn't give me any new information, what is the purpose of computing $U^(21)? Sorry that I've got so many questions.
– Amjad
Jul 31 at 17:21
add a comment |Â
up vote
2
down vote
Let $tau:V^otimes 3to V^otimes 3$ be defined by $tau(uotimes
votimes w)=votimes wotimes u$. Then the $S^(21)$ part of
$V^otimes 3$ is the kernel of $textrmid+tau+tau^2$.
answered Jul 31 at 15:36
Lord Shark the Unknown
84.4k950111
Thanks for your quick answer! If I understand you correctly, those are tensors of the form $sum_sigma in A_k v_sigma(1)...v_sigma(3)$. How does that fit/interact with $U^(21)$ though? Shouldn't I get just one subspace, let's say $H^(21)$?
– Amjad
Jul 31 at 15:45
No, they aren't. What you write down is the direct sum of the symmetric and antisymmetric tensors: the kernel of $1+tau+tau^2$ is a complement to that. @Amjad
– Lord Shark the Unknown
Jul 31 at 15:51
Okay, got it. And how does $S^(21) otimes U^(21)$ look like? Isn't $S^(21)$ already the whole part of the decomposition? And if it doesn't give me any new information, what is the purpose of computing $U^(21)? Sorry that I've got so many questions.
– Amjad
Jul 31 at 17:21
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let $tau:V^otimes 3to V^otimes 3$ be defined by $tau(uotimes
votimes w)=votimes wotimes u$. Then the $S^(21)$ part of
$V^otimes 3$ is the kernel of $textrmid+tau+tau^2$.
answered Jul 31 at 15:36
Lord Shark the Unknown
84.4k950111
Let $tau:V^otimes 3to V^otimes 3$ be defined by $tau(uotimes
votimes w)=votimes wotimes u$. Then the $S^(21)$ part of
$V^otimes 3$ is the kernel of $textrmid+tau+tau^2$.
answered Jul 31 at 15:36
Lord Shark the Unknown
84.4k950111
answered Jul 31 at 15:36
Lord Shark the Unknown
84.4k950111
answered Jul 31 at 15:36
Lord Shark the Unknown
84.4k950111
answered Jul 31 at 15:36
Lord Shark the Unknown
84.4k950111
84.4k950111
Thanks for your quick answer! If I understand you correctly, those are tensors of the form $sum_sigma in A_k v_sigma(1)...v_sigma(3)$. How does that fit/interact with $U^(21)$ though? Shouldn't I get just one subspace, let's say $H^(21)$?
– Amjad
Jul 31 at 15:45
No, they aren't. What you write down is the direct sum of the symmetric and antisymmetric tensors: the kernel of $1+tau+tau^2$ is a complement to that. @Amjad
– Lord Shark the Unknown
Jul 31 at 15:51
Okay, got it. And how does $S^(21) otimes U^(21)$ look like? Isn't $S^(21)$ already the whole part of the decomposition? And if it doesn't give me any new information, what is the purpose of computing $U^(21)? Sorry that I've got so many questions.
– Amjad
Jul 31 at 17:21
add a comment |Â
Thanks for your quick answer! If I understand you correctly, those are tensors of the form $sum_sigma in A_k v_sigma(1)...v_sigma(3)$. How does that fit/interact with $U^(21)$ though? Shouldn't I get just one subspace, let's say $H^(21)$?
– Amjad
Jul 31 at 15:45
No, they aren't. What you write down is the direct sum of the symmetric and antisymmetric tensors: the kernel of $1+tau+tau^2$ is a complement to that. @Amjad
– Lord Shark the Unknown
Jul 31 at 15:51
Okay, got it. And how does $S^(21) otimes U^(21)$ look like? Isn't $S^(21)$ already the whole part of the decomposition? And if it doesn't give me any new information, what is the purpose of computing $U^(21)? Sorry that I've got so many questions.
– Amjad
Jul 31 at 17:21
Thanks for your quick answer! If I understand you correctly, those are tensors of the form $sum_sigma in A_k v_sigma(1)...v_sigma(3)$. How does that fit/interact with $U^(21)$ though? Shouldn't I get just one subspace, let's say $H^(21)$?
– Amjad
Jul 31 at 15:45
Thanks for your quick answer! If I understand you correctly, those are tensors of the form $sum_sigma in A_k v_sigma(1)...v_sigma(3)$. How does that fit/interact with $U^(21)$ though? Shouldn't I get just one subspace, let's say $H^(21)$?
– Amjad
Jul 31 at 15:45
No, they aren't. What you write down is the direct sum of the symmetric and antisymmetric tensors: the kernel of $1+tau+tau^2$ is a complement to that. @Amjad
– Lord Shark the Unknown
Jul 31 at 15:51
No, they aren't. What you write down is the direct sum of the symmetric and antisymmetric tensors: the kernel of $1+tau+tau^2$ is a complement to that. @Amjad
– Lord Shark the Unknown
Jul 31 at 15:51
Okay, got it. And how does $S^(21) otimes U^(21)$ look like? Isn't $S^(21)$ already the whole part of the decomposition? And if it doesn't give me any new information, what is the purpose of computing $U^(21)? Sorry that I've got so many questions.
– Amjad
Jul 31 at 17:21
Okay, got it. And how does $S^(21) otimes U^(21)$ look like? Isn't $S^(21)$ already the whole part of the decomposition? And if it doesn't give me any new information, what is the purpose of computing $U^(21)? Sorry that I've got so many questions.
– Amjad
Jul 31 at 17:21
add a comment |Â
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