Converse of Schur's lemma for Lie algebras. [duplicate]
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Converse of Schur's Lemma in finite dimensional vector spaces
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The statement of Schur's lema for Lie algebras says that.
Let $(rho,mathcal{V})$ is a complex irreducible finite-dimensional representation of a Lie algebra $mathfrak{g}$. If $T$ conmmutes with $rho$, then exists $kinmathbb{C}$ such that $T = koperatorname{Id}_{mathcal{V}}$.
Is the reciprocal satisfied? That is, if the only operators that commute with the representation $rho$ are multiples of $operatorname{Id}_{mathcal{V}}$, then the representation is irreducible?
linear-algebra representation-theory lie-algebras
asked Jan 19 at 5:09
fer6268fer6268
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marked as duplicate by Dietrich Burde, Jyrki Lahtonen, José Carlos Santos linear-algebra
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Jan 19 at 12:16
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This question already has an answer here:
Converse of Schur's Lemma in finite dimensional vector spaces
1 answer
The statement of Schur's lema for Lie algebras says that.
Let $(rho,mathcal{V})$ is a complex irreducible finite-dimensional representation of a Lie algebra $mathfrak{g}$. If $T$ conmmutes with $rho$, then exists $kinmathbb{C}$ such that $T = koperatorname{Id}_{mathcal{V}}$.
Is the reciprocal satisfied? That is, if the only operators that commute with the representation $rho$ are multiples of $operatorname{Id}_{mathcal{V}}$, then the representation is irreducible?
linear-algebra representation-theory lie-algebras
asked Jan 19 at 5:09
fer6268fer6268
1294
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marked as duplicate by Dietrich Burde, Jyrki Lahtonen, José Carlos Santos linear-algebra
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Jan 19 at 12:16
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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– Dietrich Burde
Jan 19 at 9:55
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$begingroup$
This question already has an answer here:
Converse of Schur's Lemma in finite dimensional vector spaces
1 answer
The statement of Schur's lema for Lie algebras says that.
Let $(rho,mathcal{V})$ is a complex irreducible finite-dimensional representation of a Lie algebra $mathfrak{g}$. If $T$ conmmutes with $rho$, then exists $kinmathbb{C}$ such that $T = koperatorname{Id}_{mathcal{V}}$.
Is the reciprocal satisfied? That is, if the only operators that commute with the representation $rho$ are multiples of $operatorname{Id}_{mathcal{V}}$, then the representation is irreducible?
linear-algebra representation-theory lie-algebras
asked Jan 19 at 5:09
fer6268fer6268
1294
$endgroup$
This question already has an answer here:
Converse of Schur's Lemma in finite dimensional vector spaces
1 answer
The statement of Schur's lema for Lie algebras says that.
Let $(rho,mathcal{V})$ is a complex irreducible finite-dimensional representation of a Lie algebra $mathfrak{g}$. If $T$ conmmutes with $rho$, then exists $kinmathbb{C}$ such that $T = koperatorname{Id}_{mathcal{V}}$.
Is the reciprocal satisfied? That is, if the only operators that commute with the representation $rho$ are multiples of $operatorname{Id}_{mathcal{V}}$, then the representation is irreducible?
This question already has an answer here:
Converse of Schur's Lemma in finite dimensional vector spaces
1 answer
linear-algebra representation-theory lie-algebras
linear-algebra representation-theory lie-algebras
asked Jan 19 at 5:09
fer6268fer6268
1294
asked Jan 19 at 5:09
fer6268fer6268
1294
edited Jan 19 at 5:31
fer6268
asked Jan 19 at 5:09
fer6268fer6268
1294
asked Jan 19 at 5:09
fer6268fer6268
1294
asked Jan 19 at 5:09
fer6268fer6268
1294
1294
marked as duplicate by Dietrich Burde, Jyrki Lahtonen, José Carlos Santos linear-algebra
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Jan 19 at 12:16
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Jan 19 at 12:16
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
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See this question. There are many other duplicates at this site.
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– Dietrich Burde
Jan 19 at 9:55
add a comment |
1
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See this question. There are many other duplicates at this site.
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– Dietrich Burde
Jan 19 at 9:55
1
1
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See this question. There are many other duplicates at this site.
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– Dietrich Burde
Jan 19 at 9:55
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See this question. There are many other duplicates at this site.
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– Dietrich Burde
Jan 19 at 9:55
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No; consider the Lie algebra of 2 by 2 upper triangular matrices with the standard Lie bracket and the natural representation on $mathbb{C}^2$. It is easy to check that this is a counterexample.
However, it is true that $V$ must be indecomposable, i.e., there is no nontrivial decomposition as $V = V_1 oplus V_2$ where $V_1 ne V$ and $V_1 ne 0$. Otherwise the projection $pi_1: V rightarrow V_1$ commutes with $rho$, but is neither an isomorphism nor zero, so cannot be a multiple of the identity.
answered Jan 19 at 9:57
TedTed
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1 Answer
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$begingroup$
No; consider the Lie algebra of 2 by 2 upper triangular matrices with the standard Lie bracket and the natural representation on $mathbb{C}^2$. It is easy to check that this is a counterexample.
However, it is true that $V$ must be indecomposable, i.e., there is no nontrivial decomposition as $V = V_1 oplus V_2$ where $V_1 ne V$ and $V_1 ne 0$. Otherwise the projection $pi_1: V rightarrow V_1$ commutes with $rho$, but is neither an isomorphism nor zero, so cannot be a multiple of the identity.
answered Jan 19 at 9:57
TedTed
22.2k13361
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add a comment |
$begingroup$
No; consider the Lie algebra of 2 by 2 upper triangular matrices with the standard Lie bracket and the natural representation on $mathbb{C}^2$. It is easy to check that this is a counterexample.
However, it is true that $V$ must be indecomposable, i.e., there is no nontrivial decomposition as $V = V_1 oplus V_2$ where $V_1 ne V$ and $V_1 ne 0$. Otherwise the projection $pi_1: V rightarrow V_1$ commutes with $rho$, but is neither an isomorphism nor zero, so cannot be a multiple of the identity.
answered Jan 19 at 9:57
TedTed
22.2k13361
$endgroup$
add a comment |
$begingroup$
No; consider the Lie algebra of 2 by 2 upper triangular matrices with the standard Lie bracket and the natural representation on $mathbb{C}^2$. It is easy to check that this is a counterexample.
However, it is true that $V$ must be indecomposable, i.e., there is no nontrivial decomposition as $V = V_1 oplus V_2$ where $V_1 ne V$ and $V_1 ne 0$. Otherwise the projection $pi_1: V rightarrow V_1$ commutes with $rho$, but is neither an isomorphism nor zero, so cannot be a multiple of the identity.
answered Jan 19 at 9:57
TedTed
22.2k13361
$endgroup$
No; consider the Lie algebra of 2 by 2 upper triangular matrices with the standard Lie bracket and the natural representation on $mathbb{C}^2$. It is easy to check that this is a counterexample.
However, it is true that $V$ must be indecomposable, i.e., there is no nontrivial decomposition as $V = V_1 oplus V_2$ where $V_1 ne V$ and $V_1 ne 0$. Otherwise the projection $pi_1: V rightarrow V_1$ commutes with $rho$, but is neither an isomorphism nor zero, so cannot be a multiple of the identity.
answered Jan 19 at 9:57
TedTed
22.2k13361
answered Jan 19 at 9:57
TedTed
22.2k13361
answered Jan 19 at 9:57
TedTed
22.2k13361
answered Jan 19 at 9:57
TedTed
22.2k13361
22.2k13361
add a comment |
add a comment |
1
$begingroup$
See this question. There are many other duplicates at this site.
$endgroup$
– Dietrich Burde
Jan 19 at 9:55