Mixing
Irreducible Dual Representation
$begingroup$
For a semisimple Lie Algebra $mathfrak{g}$ with Cartan Subalgebra $mathfrak{t}$, let $V(lambda)$ be the unique irreducible highest weight module with highest weight $lambda$.
I am asked to show that the dual representation $V(lambda)^*$ is irreducible, and to give a condition for $V(lambda)$ to be self dual.
For the first part, my thoughts are that if I can take a basis of $V(lambda)^*$ and show that the orbit of one of them under the action of $mathfrak{t}$ contains all of them, then maybe I'd be done. But perhaps for this I would actually have to show it for any general basis?
For the second part I have heard that the condition is whether or not $-1$ is in the Weyl group, but as my understanding of Lie Algebras is quite weak I'm not sure why the Weyl group is important here. I would appreciate any help that you might be able to offer, thank you!
representation-theory lie-algebras weyl-group
asked Jan 6 at 20:48
user366818user366818
948410
$endgroup$
add a comment |
$begingroup$
For a semisimple Lie Algebra $mathfrak{g}$ with Cartan Subalgebra $mathfrak{t}$, let $V(lambda)$ be the unique irreducible highest weight module with highest weight $lambda$.
I am asked to show that the dual representation $V(lambda)^*$ is irreducible, and to give a condition for $V(lambda)$ to be self dual.
For the first part, my thoughts are that if I can take a basis of $V(lambda)^*$ and show that the orbit of one of them under the action of $mathfrak{t}$ contains all of them, then maybe I'd be done. But perhaps for this I would actually have to show it for any general basis?
For the second part I have heard that the condition is whether or not $-1$ is in the Weyl group, but as my understanding of Lie Algebras is quite weak I'm not sure why the Weyl group is important here. I would appreciate any help that you might be able to offer, thank you!
representation-theory lie-algebras weyl-group
asked Jan 6 at 20:48
user366818user366818
948410
$endgroup$
2
$begingroup$
For the first one, you can use that there is a correspondence between submodules of one and quotients of the other. For the second part, note that the lowest weight vector is turned into a highest weight vector, so calculate the weight of it in the dual.
$endgroup$
– Tobias Kildetoft
Jan 7 at 7:43
$begingroup$
@TobiasKildetoft thank you for the response. I’m a little confused though, are you saying $forall U^* in V(lambda)^*, ; exists U in V(lambda) $ such that $U^* cong V(lambda)/U$ if so I can’t quite see why this is the case?
$endgroup$
– user366818
Jan 7 at 16:34
$begingroup$
Dualize the short exact sequence $Uto V(lambda)to V(lambda)/U$.
$endgroup$
– David Hill
Jan 8 at 17:26
add a comment |
$begingroup$
For a semisimple Lie Algebra $mathfrak{g}$ with Cartan Subalgebra $mathfrak{t}$, let $V(lambda)$ be the unique irreducible highest weight module with highest weight $lambda$.
I am asked to show that the dual representation $V(lambda)^*$ is irreducible, and to give a condition for $V(lambda)$ to be self dual.
For the first part, my thoughts are that if I can take a basis of $V(lambda)^*$ and show that the orbit of one of them under the action of $mathfrak{t}$ contains all of them, then maybe I'd be done. But perhaps for this I would actually have to show it for any general basis?
For the second part I have heard that the condition is whether or not $-1$ is in the Weyl group, but as my understanding of Lie Algebras is quite weak I'm not sure why the Weyl group is important here. I would appreciate any help that you might be able to offer, thank you!
representation-theory lie-algebras weyl-group
asked Jan 6 at 20:48
user366818user366818
948410
$endgroup$
For a semisimple Lie Algebra $mathfrak{g}$ with Cartan Subalgebra $mathfrak{t}$, let $V(lambda)$ be the unique irreducible highest weight module with highest weight $lambda$.
I am asked to show that the dual representation $V(lambda)^*$ is irreducible, and to give a condition for $V(lambda)$ to be self dual.
For the first part, my thoughts are that if I can take a basis of $V(lambda)^*$ and show that the orbit of one of them under the action of $mathfrak{t}$ contains all of them, then maybe I'd be done. But perhaps for this I would actually have to show it for any general basis?
For the second part I have heard that the condition is whether or not $-1$ is in the Weyl group, but as my understanding of Lie Algebras is quite weak I'm not sure why the Weyl group is important here. I would appreciate any help that you might be able to offer, thank you!
representation-theory lie-algebras weyl-group
representation-theory lie-algebras weyl-group
asked Jan 6 at 20:48
user366818user366818
948410
asked Jan 6 at 20:48
user366818user366818
948410
asked Jan 6 at 20:48
user366818user366818
948410
asked Jan 6 at 20:48
user366818user366818
948410
asked Jan 6 at 20:48
user366818user366818
948410
948410
2
$begingroup$
For the first one, you can use that there is a correspondence between submodules of one and quotients of the other. For the second part, note that the lowest weight vector is turned into a highest weight vector, so calculate the weight of it in the dual.
$endgroup$
– Tobias Kildetoft
Jan 7 at 7:43
$begingroup$
@TobiasKildetoft thank you for the response. I’m a little confused though, are you saying $forall U^* in V(lambda)^*, ; exists U in V(lambda) $ such that $U^* cong V(lambda)/U$ if so I can’t quite see why this is the case?
$endgroup$
– user366818
Jan 7 at 16:34
$begingroup$
Dualize the short exact sequence $Uto V(lambda)to V(lambda)/U$.
$endgroup$
– David Hill
Jan 8 at 17:26
add a comment |
2
$begingroup$
For the first one, you can use that there is a correspondence between submodules of one and quotients of the other. For the second part, note that the lowest weight vector is turned into a highest weight vector, so calculate the weight of it in the dual.
$endgroup$
– Tobias Kildetoft
Jan 7 at 7:43
$begingroup$
@TobiasKildetoft thank you for the response. I’m a little confused though, are you saying $forall U^* in V(lambda)^*, ; exists U in V(lambda) $ such that $U^* cong V(lambda)/U$ if so I can’t quite see why this is the case?
$endgroup$
– user366818
Jan 7 at 16:34
$begingroup$
Dualize the short exact sequence $Uto V(lambda)to V(lambda)/U$.
$endgroup$
– David Hill
Jan 8 at 17:26
2
2
$begingroup$
For the first one, you can use that there is a correspondence between submodules of one and quotients of the other. For the second part, note that the lowest weight vector is turned into a highest weight vector, so calculate the weight of it in the dual.
$endgroup$
– Tobias Kildetoft
Jan 7 at 7:43
$begingroup$
For the first one, you can use that there is a correspondence between submodules of one and quotients of the other. For the second part, note that the lowest weight vector is turned into a highest weight vector, so calculate the weight of it in the dual.
$endgroup$
– Tobias Kildetoft
Jan 7 at 7:43
$begingroup$
@TobiasKildetoft thank you for the response. I’m a little confused though, are you saying $forall U^* in V(lambda)^*, ; exists U in V(lambda) $ such that $U^* cong V(lambda)/U$ if so I can’t quite see why this is the case?
$endgroup$
– user366818
Jan 7 at 16:34
$begingroup$
@TobiasKildetoft thank you for the response. I’m a little confused though, are you saying $forall U^* in V(lambda)^*, ; exists U in V(lambda) $ such that $U^* cong V(lambda)/U$ if so I can’t quite see why this is the case?
$endgroup$
– user366818
Jan 7 at 16:34
$begingroup$
Dualize the short exact sequence $Uto V(lambda)to V(lambda)/U$.
$endgroup$
– David Hill
Jan 8 at 17:26
$begingroup$
Dualize the short exact sequence $Uto V(lambda)to V(lambda)/U$.
$endgroup$
– David Hill
Jan 8 at 17:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
To deduces irreducibility of $V(lambda)^*$, you don't really have to go to quotients. Given an invariant subspace $Wsubset V(lambda)^*$ consider its annihilator $U:={vin V(lambda):forallphiin W:phi(v)=0}$. This is clearly a linear subspace in $V(lambda)$ and a short computation shows that invariance of $W$ implies invariance of $U$. Knowing the $U=V(lambda)$ and $U={0}$ are the only possibilities for $U$, linear algebra shows that $W={0}$ or $W=V(lambda)^*$.
Concerning the highest weight of $V(lambda)^*$ you take a basis for $V(lambda)$ consisting of weight vectors and consider the dual basis of $V(lambda)^*$ to conclude that the weights of $V(lambda)^*$ are exactly the negatives of the weights of $V(lambda)$. In particular, the highest weight of $V(lambda)^*$ is $-mu$, where $mu$ is the lowest weight of $V(lambda)$. It can be shown that $mu=w_0(lambda)$, where $w_0$ is the so-called "longest element" in the Weyl group. (There are cases in which $w_0=-id$ and then any $V(lambda)$ is isomorphic to its dual, but in general it may happen that $w_0(lambda)=-lambda$ and hence $V(lambda)cong V(lambda)^*$ without $w_0$ being $-id$).
answered Jan 9 at 8:35
Andreas CapAndreas Cap
11.1k923
$endgroup$
$begingroup$
Thank you for this response! What do you mean by "invariance of $W$" here?
$endgroup$
– user366818
Jan 10 at 1:04
1
$begingroup$
What I mean is that if the action of any element of $mathfrak g$ sends $W$ to itself, then the action also sends $U$ to itself.
$endgroup$
– Andreas Cap
Jan 10 at 4:58
$begingroup$
I can't completely see why this was important because it seems like I can just use the fact $U$ takes only two values and get the result by Linear Algebra?
$endgroup$
– user366818
Jan 10 at 12:41
$begingroup$
Nevermind, I realised that the invariance shows that $U$ is a subrepresentation of $V(lambda)$ which is the important part.
$endgroup$
– user366818
Jan 10 at 12:46
1
$begingroup$
This is just linear algebra: For a proper subspace $Wsubset V(lambda)^*$ take a basis. Then the joint kernel of the basis elements coincides with the joint kernel of all elements of $W$. Clearly the joint kernel of the basis elements has at least dimension $dim(V(lambda))-dim(W)$ (and indeed this is the dimension of the joint kernel). Thus $Uneq{0}$ for $Wneq V(lambda)^*$.
$endgroup$
– Andreas Cap
Jan 10 at 13:25
|
show 1 more comment
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1 Answer
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1 Answer
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$begingroup$
To deduces irreducibility of $V(lambda)^*$, you don't really have to go to quotients. Given an invariant subspace $Wsubset V(lambda)^*$ consider its annihilator $U:={vin V(lambda):forallphiin W:phi(v)=0}$. This is clearly a linear subspace in $V(lambda)$ and a short computation shows that invariance of $W$ implies invariance of $U$. Knowing the $U=V(lambda)$ and $U={0}$ are the only possibilities for $U$, linear algebra shows that $W={0}$ or $W=V(lambda)^*$.
Concerning the highest weight of $V(lambda)^*$ you take a basis for $V(lambda)$ consisting of weight vectors and consider the dual basis of $V(lambda)^*$ to conclude that the weights of $V(lambda)^*$ are exactly the negatives of the weights of $V(lambda)$. In particular, the highest weight of $V(lambda)^*$ is $-mu$, where $mu$ is the lowest weight of $V(lambda)$. It can be shown that $mu=w_0(lambda)$, where $w_0$ is the so-called "longest element" in the Weyl group. (There are cases in which $w_0=-id$ and then any $V(lambda)$ is isomorphic to its dual, but in general it may happen that $w_0(lambda)=-lambda$ and hence $V(lambda)cong V(lambda)^*$ without $w_0$ being $-id$).
answered Jan 9 at 8:35
Andreas CapAndreas Cap
11.1k923
$endgroup$
$begingroup$
Thank you for this response! What do you mean by "invariance of $W$" here?
$endgroup$
– user366818
Jan 10 at 1:04
1
$begingroup$
What I mean is that if the action of any element of $mathfrak g$ sends $W$ to itself, then the action also sends $U$ to itself.
$endgroup$
– Andreas Cap
Jan 10 at 4:58
$begingroup$
I can't completely see why this was important because it seems like I can just use the fact $U$ takes only two values and get the result by Linear Algebra?
$endgroup$
– user366818
Jan 10 at 12:41
$begingroup$
Nevermind, I realised that the invariance shows that $U$ is a subrepresentation of $V(lambda)$ which is the important part.
$endgroup$
– user366818
Jan 10 at 12:46
1
$begingroup$
This is just linear algebra: For a proper subspace $Wsubset V(lambda)^*$ take a basis. Then the joint kernel of the basis elements coincides with the joint kernel of all elements of $W$. Clearly the joint kernel of the basis elements has at least dimension $dim(V(lambda))-dim(W)$ (and indeed this is the dimension of the joint kernel). Thus $Uneq{0}$ for $Wneq V(lambda)^*$.
$endgroup$
– Andreas Cap
Jan 10 at 13:25
|
show 1 more comment
$begingroup$
To deduces irreducibility of $V(lambda)^*$, you don't really have to go to quotients. Given an invariant subspace $Wsubset V(lambda)^*$ consider its annihilator $U:={vin V(lambda):forallphiin W:phi(v)=0}$. This is clearly a linear subspace in $V(lambda)$ and a short computation shows that invariance of $W$ implies invariance of $U$. Knowing the $U=V(lambda)$ and $U={0}$ are the only possibilities for $U$, linear algebra shows that $W={0}$ or $W=V(lambda)^*$.
Concerning the highest weight of $V(lambda)^*$ you take a basis for $V(lambda)$ consisting of weight vectors and consider the dual basis of $V(lambda)^*$ to conclude that the weights of $V(lambda)^*$ are exactly the negatives of the weights of $V(lambda)$. In particular, the highest weight of $V(lambda)^*$ is $-mu$, where $mu$ is the lowest weight of $V(lambda)$. It can be shown that $mu=w_0(lambda)$, where $w_0$ is the so-called "longest element" in the Weyl group. (There are cases in which $w_0=-id$ and then any $V(lambda)$ is isomorphic to its dual, but in general it may happen that $w_0(lambda)=-lambda$ and hence $V(lambda)cong V(lambda)^*$ without $w_0$ being $-id$).
answered Jan 9 at 8:35
Andreas CapAndreas Cap
11.1k923
$endgroup$
$begingroup$
Thank you for this response! What do you mean by "invariance of $W$" here?
$endgroup$
– user366818
Jan 10 at 1:04
1
$begingroup$
What I mean is that if the action of any element of $mathfrak g$ sends $W$ to itself, then the action also sends $U$ to itself.
$endgroup$
– Andreas Cap
Jan 10 at 4:58
$begingroup$
I can't completely see why this was important because it seems like I can just use the fact $U$ takes only two values and get the result by Linear Algebra?
$endgroup$
– user366818
Jan 10 at 12:41
$begingroup$
Nevermind, I realised that the invariance shows that $U$ is a subrepresentation of $V(lambda)$ which is the important part.
$endgroup$
– user366818
Jan 10 at 12:46
1
$begingroup$
This is just linear algebra: For a proper subspace $Wsubset V(lambda)^*$ take a basis. Then the joint kernel of the basis elements coincides with the joint kernel of all elements of $W$. Clearly the joint kernel of the basis elements has at least dimension $dim(V(lambda))-dim(W)$ (and indeed this is the dimension of the joint kernel). Thus $Uneq{0}$ for $Wneq V(lambda)^*$.
$endgroup$
– Andreas Cap
Jan 10 at 13:25
|
show 1 more comment
$begingroup$
To deduces irreducibility of $V(lambda)^*$, you don't really have to go to quotients. Given an invariant subspace $Wsubset V(lambda)^*$ consider its annihilator $U:={vin V(lambda):forallphiin W:phi(v)=0}$. This is clearly a linear subspace in $V(lambda)$ and a short computation shows that invariance of $W$ implies invariance of $U$. Knowing the $U=V(lambda)$ and $U={0}$ are the only possibilities for $U$, linear algebra shows that $W={0}$ or $W=V(lambda)^*$.
Concerning the highest weight of $V(lambda)^*$ you take a basis for $V(lambda)$ consisting of weight vectors and consider the dual basis of $V(lambda)^*$ to conclude that the weights of $V(lambda)^*$ are exactly the negatives of the weights of $V(lambda)$. In particular, the highest weight of $V(lambda)^*$ is $-mu$, where $mu$ is the lowest weight of $V(lambda)$. It can be shown that $mu=w_0(lambda)$, where $w_0$ is the so-called "longest element" in the Weyl group. (There are cases in which $w_0=-id$ and then any $V(lambda)$ is isomorphic to its dual, but in general it may happen that $w_0(lambda)=-lambda$ and hence $V(lambda)cong V(lambda)^*$ without $w_0$ being $-id$).
answered Jan 9 at 8:35
Andreas CapAndreas Cap
11.1k923
$endgroup$
To deduces irreducibility of $V(lambda)^*$, you don't really have to go to quotients. Given an invariant subspace $Wsubset V(lambda)^*$ consider its annihilator $U:={vin V(lambda):forallphiin W:phi(v)=0}$. This is clearly a linear subspace in $V(lambda)$ and a short computation shows that invariance of $W$ implies invariance of $U$. Knowing the $U=V(lambda)$ and $U={0}$ are the only possibilities for $U$, linear algebra shows that $W={0}$ or $W=V(lambda)^*$.
Concerning the highest weight of $V(lambda)^*$ you take a basis for $V(lambda)$ consisting of weight vectors and consider the dual basis of $V(lambda)^*$ to conclude that the weights of $V(lambda)^*$ are exactly the negatives of the weights of $V(lambda)$. In particular, the highest weight of $V(lambda)^*$ is $-mu$, where $mu$ is the lowest weight of $V(lambda)$. It can be shown that $mu=w_0(lambda)$, where $w_0$ is the so-called "longest element" in the Weyl group. (There are cases in which $w_0=-id$ and then any $V(lambda)$ is isomorphic to its dual, but in general it may happen that $w_0(lambda)=-lambda$ and hence $V(lambda)cong V(lambda)^*$ without $w_0$ being $-id$).
answered Jan 9 at 8:35
Andreas CapAndreas Cap
11.1k923
answered Jan 9 at 8:35
Andreas CapAndreas Cap
11.1k923
answered Jan 9 at 8:35
Andreas CapAndreas Cap
11.1k923
answered Jan 9 at 8:35
Andreas CapAndreas Cap
11.1k923
11.1k923
$begingroup$
Thank you for this response! What do you mean by "invariance of $W$" here?
$endgroup$
– user366818
Jan 10 at 1:04
1
$begingroup$
What I mean is that if the action of any element of $mathfrak g$ sends $W$ to itself, then the action also sends $U$ to itself.
$endgroup$
– Andreas Cap
Jan 10 at 4:58
$begingroup$
I can't completely see why this was important because it seems like I can just use the fact $U$ takes only two values and get the result by Linear Algebra?
$endgroup$
– user366818
Jan 10 at 12:41
$begingroup$
Nevermind, I realised that the invariance shows that $U$ is a subrepresentation of $V(lambda)$ which is the important part.
$endgroup$
– user366818
Jan 10 at 12:46
1
$begingroup$
This is just linear algebra: For a proper subspace $Wsubset V(lambda)^*$ take a basis. Then the joint kernel of the basis elements coincides with the joint kernel of all elements of $W$. Clearly the joint kernel of the basis elements has at least dimension $dim(V(lambda))-dim(W)$ (and indeed this is the dimension of the joint kernel). Thus $Uneq{0}$ for $Wneq V(lambda)^*$.
$endgroup$
– Andreas Cap
Jan 10 at 13:25
|
show 1 more comment
$begingroup$
Thank you for this response! What do you mean by "invariance of $W$" here?
$endgroup$
– user366818
Jan 10 at 1:04
1
$begingroup$
What I mean is that if the action of any element of $mathfrak g$ sends $W$ to itself, then the action also sends $U$ to itself.
$endgroup$
– Andreas Cap
Jan 10 at 4:58
$begingroup$
I can't completely see why this was important because it seems like I can just use the fact $U$ takes only two values and get the result by Linear Algebra?
$endgroup$
– user366818
Jan 10 at 12:41
$begingroup$
Nevermind, I realised that the invariance shows that $U$ is a subrepresentation of $V(lambda)$ which is the important part.
$endgroup$
– user366818
Jan 10 at 12:46
1
$begingroup$
This is just linear algebra: For a proper subspace $Wsubset V(lambda)^*$ take a basis. Then the joint kernel of the basis elements coincides with the joint kernel of all elements of $W$. Clearly the joint kernel of the basis elements has at least dimension $dim(V(lambda))-dim(W)$ (and indeed this is the dimension of the joint kernel). Thus $Uneq{0}$ for $Wneq V(lambda)^*$.
$endgroup$
– Andreas Cap
Jan 10 at 13:25
$begingroup$
Thank you for this response! What do you mean by "invariance of $W$" here?
$endgroup$
– user366818
Jan 10 at 1:04
$begingroup$
Thank you for this response! What do you mean by "invariance of $W$" here?
$endgroup$
– user366818
Jan 10 at 1:04
1
1
$begingroup$
What I mean is that if the action of any element of $mathfrak g$ sends $W$ to itself, then the action also sends $U$ to itself.
$endgroup$
– Andreas Cap
Jan 10 at 4:58
$begingroup$
What I mean is that if the action of any element of $mathfrak g$ sends $W$ to itself, then the action also sends $U$ to itself.
$endgroup$
– Andreas Cap
Jan 10 at 4:58
$begingroup$
I can't completely see why this was important because it seems like I can just use the fact $U$ takes only two values and get the result by Linear Algebra?
$endgroup$
– user366818
Jan 10 at 12:41
$begingroup$
I can't completely see why this was important because it seems like I can just use the fact $U$ takes only two values and get the result by Linear Algebra?
$endgroup$
– user366818
Jan 10 at 12:41
$begingroup$
Nevermind, I realised that the invariance shows that $U$ is a subrepresentation of $V(lambda)$ which is the important part.
$endgroup$
– user366818
Jan 10 at 12:46
$begingroup$
Nevermind, I realised that the invariance shows that $U$ is a subrepresentation of $V(lambda)$ which is the important part.
$endgroup$
– user366818
Jan 10 at 12:46
1
1
$begingroup$
This is just linear algebra: For a proper subspace $Wsubset V(lambda)^*$ take a basis. Then the joint kernel of the basis elements coincides with the joint kernel of all elements of $W$. Clearly the joint kernel of the basis elements has at least dimension $dim(V(lambda))-dim(W)$ (and indeed this is the dimension of the joint kernel). Thus $Uneq{0}$ for $Wneq V(lambda)^*$.
$endgroup$
– Andreas Cap
Jan 10 at 13:25
$begingroup$
This is just linear algebra: For a proper subspace $Wsubset V(lambda)^*$ take a basis. Then the joint kernel of the basis elements coincides with the joint kernel of all elements of $W$. Clearly the joint kernel of the basis elements has at least dimension $dim(V(lambda))-dim(W)$ (and indeed this is the dimension of the joint kernel). Thus $Uneq{0}$ for $Wneq V(lambda)^*$.
$endgroup$
– Andreas Cap
Jan 10 at 13:25
|
show 1 more comment
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For the first one, you can use that there is a correspondence between submodules of one and quotients of the other. For the second part, note that the lowest weight vector is turned into a highest weight vector, so calculate the weight of it in the dual.
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– Tobias Kildetoft
Jan 7 at 7:43
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@TobiasKildetoft thank you for the response. I’m a little confused though, are you saying $forall U^* in V(lambda)^*, ; exists U in V(lambda) $ such that $U^* cong V(lambda)/U$ if so I can’t quite see why this is the case?
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– user366818
Jan 7 at 16:34
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Dualize the short exact sequence $Uto V(lambda)to V(lambda)/U$.
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– David Hill
Jan 8 at 17:26