Why is the exterior power $GL(bigwedge^kV)$ an irreducible representation of $GL(V)$?
$newcommand{GL}{operatorname{GL}}$
Let $V$ be a real $n$-dimensional vector space. For $1<k<n$ we have a natural representation of $GL(V)$ via the $k$ exterior power:
$rho:GL(V) to GL(bigwedge^kV)$, given by $rho(A)=bigwedge^k A$. I am trying to show $rho$ is an irreducible representation. Let $0neq W le bigwedge^kV$ be a subrepresentation. If we can show $W$ contains a non-zero decomposable element, we are done.
Indeed, suppose $W subsetneq bigwedge^kV$. Then, there exist a decomposable element $sigma=v_1 wedge dots wedge v_k neq 0$, such that $sigma notin W$. We assumed $W$ contains a non-zero decomposable element $sigma'=u_1 wedge dots wedge u_k neq 0$. Define a map $A in GL(V)$ by extending $u_i to v_i$. Then
$$rho(A) (sigma')=bigwedge^k A(u_1 wedge dots wedge u_k)=sigma notin W,$$
while $sigma' in W$, con
So, the question reduces to the following: Why does every non-zero subrepresentation contain a non-zero decomposable element?
I asked an even more naive question here-whether or not every subspace of dimension greater than $1$ contains a non-zero decomposable element?
group-theory representation-theory lie-groups exterior-algebra tensor-decomposition
asked 3 hours ago
Asaf Shachar
5,0843940
add a comment |
$newcommand{GL}{operatorname{GL}}$
Let $V$ be a real $n$-dimensional vector space. For $1<k<n$ we have a natural representation of $GL(V)$ via the $k$ exterior power:
$rho:GL(V) to GL(bigwedge^kV)$, given by $rho(A)=bigwedge^k A$. I am trying to show $rho$ is an irreducible representation. Let $0neq W le bigwedge^kV$ be a subrepresentation. If we can show $W$ contains a non-zero decomposable element, we are done.
Indeed, suppose $W subsetneq bigwedge^kV$. Then, there exist a decomposable element $sigma=v_1 wedge dots wedge v_k neq 0$, such that $sigma notin W$. We assumed $W$ contains a non-zero decomposable element $sigma'=u_1 wedge dots wedge u_k neq 0$. Define a map $A in GL(V)$ by extending $u_i to v_i$. Then
$$rho(A) (sigma')=bigwedge^k A(u_1 wedge dots wedge u_k)=sigma notin W,$$
while $sigma' in W$, con
So, the question reduces to the following: Why does every non-zero subrepresentation contain a non-zero decomposable element?
I asked an even more naive question here-whether or not every subspace of dimension greater than $1$ contains a non-zero decomposable element?
group-theory representation-theory lie-groups exterior-algebra tensor-decomposition
asked 3 hours ago
Asaf Shachar
5,0843940
A nice conceptual way to work with this is to first decompose the representation as a module for the group of diagonal matrices (into so-called weight spaces). Then note what happens to the weight of a vector in one of these subspaces when one acts by suitable upper triangular unipotent matrices.
– Tobias Kildetoft
3 hours ago
Thanks. Unfortunately, I really know barely nothing about the machinery of representation theory. Can you please elaborate on this or give me a reference? (I don't know what a weight of a vector is, and naive googling only found something in the context of representations of Lie algebras, not Lie groups).
– Asaf Shachar
3 hours ago
The definition is essentially the same. The representation decomposes as a sum of $1$-dimensional subspaces, and a vector in such a subspace will be acted on via a scalar. This scalar depends on the element acting, giving a linear character of the subgroup of diagonal matrices, and this character is what is called the weight of the vector. It may be a bit much to get into if none of this is familiar, but I would still advice you to try writing it up explicitly for $k=1$ when $dim(V) = 2$ to get a feel for what happens.
– Tobias Kildetoft
3 hours ago
add a comment |
$newcommand{GL}{operatorname{GL}}$
Let $V$ be a real $n$-dimensional vector space. For $1<k<n$ we have a natural representation of $GL(V)$ via the $k$ exterior power:
$rho:GL(V) to GL(bigwedge^kV)$, given by $rho(A)=bigwedge^k A$. I am trying to show $rho$ is an irreducible representation. Let $0neq W le bigwedge^kV$ be a subrepresentation. If we can show $W$ contains a non-zero decomposable element, we are done.
Indeed, suppose $W subsetneq bigwedge^kV$. Then, there exist a decomposable element $sigma=v_1 wedge dots wedge v_k neq 0$, such that $sigma notin W$. We assumed $W$ contains a non-zero decomposable element $sigma'=u_1 wedge dots wedge u_k neq 0$. Define a map $A in GL(V)$ by extending $u_i to v_i$. Then
$$rho(A) (sigma')=bigwedge^k A(u_1 wedge dots wedge u_k)=sigma notin W,$$
while $sigma' in W$, con
So, the question reduces to the following: Why does every non-zero subrepresentation contain a non-zero decomposable element?
I asked an even more naive question here-whether or not every subspace of dimension greater than $1$ contains a non-zero decomposable element?
group-theory representation-theory lie-groups exterior-algebra tensor-decomposition
asked 3 hours ago
Asaf Shachar
5,0843940
$newcommand{GL}{operatorname{GL}}$
Let $V$ be a real $n$-dimensional vector space. For $1<k<n$ we have a natural representation of $GL(V)$ via the $k$ exterior power:
$rho:GL(V) to GL(bigwedge^kV)$, given by $rho(A)=bigwedge^k A$. I am trying to show $rho$ is an irreducible representation. Let $0neq W le bigwedge^kV$ be a subrepresentation. If we can show $W$ contains a non-zero decomposable element, we are done.
Indeed, suppose $W subsetneq bigwedge^kV$. Then, there exist a decomposable element $sigma=v_1 wedge dots wedge v_k neq 0$, such that $sigma notin W$. We assumed $W$ contains a non-zero decomposable element $sigma'=u_1 wedge dots wedge u_k neq 0$. Define a map $A in GL(V)$ by extending $u_i to v_i$. Then
$$rho(A) (sigma')=bigwedge^k A(u_1 wedge dots wedge u_k)=sigma notin W,$$
while $sigma' in W$, con
So, the question reduces to the following: Why does every non-zero subrepresentation contain a non-zero decomposable element?
I asked an even more naive question here-whether or not every subspace of dimension greater than $1$ contains a non-zero decomposable element?
group-theory representation-theory lie-groups exterior-algebra tensor-decomposition
group-theory representation-theory lie-groups exterior-algebra tensor-decomposition
asked 3 hours ago
Asaf Shachar
5,0843940
asked 3 hours ago
Asaf Shachar
5,0843940
asked 3 hours ago
Asaf Shachar
5,0843940
asked 3 hours ago
Asaf Shachar
5,0843940
asked 3 hours ago
Asaf Shachar
5,0843940
5,0843940
A nice conceptual way to work with this is to first decompose the representation as a module for the group of diagonal matrices (into so-called weight spaces). Then note what happens to the weight of a vector in one of these subspaces when one acts by suitable upper triangular unipotent matrices.
– Tobias Kildetoft
3 hours ago
Thanks. Unfortunately, I really know barely nothing about the machinery of representation theory. Can you please elaborate on this or give me a reference? (I don't know what a weight of a vector is, and naive googling only found something in the context of representations of Lie algebras, not Lie groups).
– Asaf Shachar
3 hours ago
The definition is essentially the same. The representation decomposes as a sum of $1$-dimensional subspaces, and a vector in such a subspace will be acted on via a scalar. This scalar depends on the element acting, giving a linear character of the subgroup of diagonal matrices, and this character is what is called the weight of the vector. It may be a bit much to get into if none of this is familiar, but I would still advice you to try writing it up explicitly for $k=1$ when $dim(V) = 2$ to get a feel for what happens.
– Tobias Kildetoft
3 hours ago
add a comment |
A nice conceptual way to work with this is to first decompose the representation as a module for the group of diagonal matrices (into so-called weight spaces). Then note what happens to the weight of a vector in one of these subspaces when one acts by suitable upper triangular unipotent matrices.
– Tobias Kildetoft
3 hours ago
Thanks. Unfortunately, I really know barely nothing about the machinery of representation theory. Can you please elaborate on this or give me a reference? (I don't know what a weight of a vector is, and naive googling only found something in the context of representations of Lie algebras, not Lie groups).
– Asaf Shachar
3 hours ago
The definition is essentially the same. The representation decomposes as a sum of $1$-dimensional subspaces, and a vector in such a subspace will be acted on via a scalar. This scalar depends on the element acting, giving a linear character of the subgroup of diagonal matrices, and this character is what is called the weight of the vector. It may be a bit much to get into if none of this is familiar, but I would still advice you to try writing it up explicitly for $k=1$ when $dim(V) = 2$ to get a feel for what happens.
– Tobias Kildetoft
3 hours ago
A nice conceptual way to work with this is to first decompose the representation as a module for the group of diagonal matrices (into so-called weight spaces). Then note what happens to the weight of a vector in one of these subspaces when one acts by suitable upper triangular unipotent matrices.
– Tobias Kildetoft
3 hours ago
A nice conceptual way to work with this is to first decompose the representation as a module for the group of diagonal matrices (into so-called weight spaces). Then note what happens to the weight of a vector in one of these subspaces when one acts by suitable upper triangular unipotent matrices.
– Tobias Kildetoft
3 hours ago
Thanks. Unfortunately, I really know barely nothing about the machinery of representation theory. Can you please elaborate on this or give me a reference? (I don't know what a weight of a vector is, and naive googling only found something in the context of representations of Lie algebras, not Lie groups).
– Asaf Shachar
3 hours ago
Thanks. Unfortunately, I really know barely nothing about the machinery of representation theory. Can you please elaborate on this or give me a reference? (I don't know what a weight of a vector is, and naive googling only found something in the context of representations of Lie algebras, not Lie groups).
– Asaf Shachar
3 hours ago
The definition is essentially the same. The representation decomposes as a sum of $1$-dimensional subspaces, and a vector in such a subspace will be acted on via a scalar. This scalar depends on the element acting, giving a linear character of the subgroup of diagonal matrices, and this character is what is called the weight of the vector. It may be a bit much to get into if none of this is familiar, but I would still advice you to try writing it up explicitly for $k=1$ when $dim(V) = 2$ to get a feel for what happens.
– Tobias Kildetoft
3 hours ago
The definition is essentially the same. The representation decomposes as a sum of $1$-dimensional subspaces, and a vector in such a subspace will be acted on via a scalar. This scalar depends on the element acting, giving a linear character of the subgroup of diagonal matrices, and this character is what is called the weight of the vector. It may be a bit much to get into if none of this is familiar, but I would still advice you to try writing it up explicitly for $k=1$ when $dim(V) = 2$ to get a feel for what happens.
– Tobias Kildetoft
3 hours ago
add a comment |
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A nice conceptual way to work with this is to first decompose the representation as a module for the group of diagonal matrices (into so-called weight spaces). Then note what happens to the weight of a vector in one of these subspaces when one acts by suitable upper triangular unipotent matrices.
– Tobias Kildetoft
3 hours ago
Thanks. Unfortunately, I really know barely nothing about the machinery of representation theory. Can you please elaborate on this or give me a reference? (I don't know what a weight of a vector is, and naive googling only found something in the context of representations of Lie algebras, not Lie groups).
– Asaf Shachar
3 hours ago
The definition is essentially the same. The representation decomposes as a sum of $1$-dimensional subspaces, and a vector in such a subspace will be acted on via a scalar. This scalar depends on the element acting, giving a linear character of the subgroup of diagonal matrices, and this character is what is called the weight of the vector. It may be a bit much to get into if none of this is familiar, but I would still advice you to try writing it up explicitly for $k=1$ when $dim(V) = 2$ to get a feel for what happens.
– Tobias Kildetoft
3 hours ago