Question about $text{Hom}_{mathfrak{g}}(M,L)$
Let $mathfrak{g}$ be a complex semisimple Lie algebra.
Suppose $M,N,L$ are $mathfrak{g}$-modules, $N$ is a $mathfrak{g}$-submodule of $M$.
Does this implies $text{Hom}_{mathfrak{g}}(N,L)le text{Hom}_{mathfrak{g}}(M,L)$ as a vector subspace?
If yes, how to prove it?
modules
asked Nov 27 at 13:51
James Cheung
1234
add a comment |
Let $mathfrak{g}$ be a complex semisimple Lie algebra.
Suppose $M,N,L$ are $mathfrak{g}$-modules, $N$ is a $mathfrak{g}$-submodule of $M$.
Does this implies $text{Hom}_{mathfrak{g}}(N,L)le text{Hom}_{mathfrak{g}}(M,L)$ as a vector subspace?
If yes, how to prove it?
modules
asked Nov 27 at 13:51
James Cheung
1234
add a comment |
Let $mathfrak{g}$ be a complex semisimple Lie algebra.
Suppose $M,N,L$ are $mathfrak{g}$-modules, $N$ is a $mathfrak{g}$-submodule of $M$.
Does this implies $text{Hom}_{mathfrak{g}}(N,L)le text{Hom}_{mathfrak{g}}(M,L)$ as a vector subspace?
If yes, how to prove it?
modules
asked Nov 27 at 13:51
James Cheung
1234
Let $mathfrak{g}$ be a complex semisimple Lie algebra.
Suppose $M,N,L$ are $mathfrak{g}$-modules, $N$ is a $mathfrak{g}$-submodule of $M$.
Does this implies $text{Hom}_{mathfrak{g}}(N,L)le text{Hom}_{mathfrak{g}}(M,L)$ as a vector subspace?
If yes, how to prove it?
modules
modules
asked Nov 27 at 13:51
James Cheung
1234
asked Nov 27 at 13:51
James Cheung
1234
edited Nov 27 at 14:14
asked Nov 27 at 13:51
James Cheung
1234
asked Nov 27 at 13:51
James Cheung
1234
asked Nov 27 at 13:51
James Cheung
1234
1234
add a comment |
add a comment |
1 Answer
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Firstly, since the functor Hom is contravariant in the first variable, what you expect is a map going the other way. For a general abelian category, the only situation in which you expect to have an inclusion $$mathrm{Hom}(A,B) subseteq mathrm{Hom}(C,B)$$ is when $A$ is a quotient of $C$. For a semi-simple Lie algebra $mathfrak{g}$ over $mathbf{C}$, the category of finite-dimensional modules is semi-simple, so there are (in general, many) ways to represent a given submodule $N subseteq M$ as a quotient. Choosing one of them gives you the desired inclusion (which is not canonical in general!).
However, the category of all $mathfrak{g}$-modules is far from semi-simple, so there is no such inclusion in the generality you are asking for. For instance, if $mathfrak{g}=mathfrak{sl}_2(mathbf{C})$ and you take $N=L=mathbf{C}$ to be the trivial representation and $M$ to be the co-Verma module containing it as a submodule, then
$$mathrm{Hom}_mathfrak{g}(M,mathbf{C})=0,$$ so there can be no inclusion as in your question.
answered Nov 27 at 14:12
Stephen
10.5k12237
But how to represent a given submodule $Nsubseteq M$ as a quotient of $M$ in the semisimple Lie algebra case?
– James Cheung
Nov 27 at 14:21
@JamesCheung Are your representations finite dimensional $mathbf{C}$-vector spaces? If so, you must choose a complementary submodule. There are many ways to do so, in general. It's impossible for me to specify one of them canonically, as I noted in my answer. One procedure would be to start with a positive definite Hermitian form on $M$ and to average it over the special unitary group to get an invariant positive definite form, which you could then use by taking orthogonal complements.
– Stephen
Nov 27 at 14:23
Thank you for your detailed answer.
– James Cheung
Nov 27 at 15:50
add a comment |
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Firstly, since the functor Hom is contravariant in the first variable, what you expect is a map going the other way. For a general abelian category, the only situation in which you expect to have an inclusion $$mathrm{Hom}(A,B) subseteq mathrm{Hom}(C,B)$$ is when $A$ is a quotient of $C$. For a semi-simple Lie algebra $mathfrak{g}$ over $mathbf{C}$, the category of finite-dimensional modules is semi-simple, so there are (in general, many) ways to represent a given submodule $N subseteq M$ as a quotient. Choosing one of them gives you the desired inclusion (which is not canonical in general!).
However, the category of all $mathfrak{g}$-modules is far from semi-simple, so there is no such inclusion in the generality you are asking for. For instance, if $mathfrak{g}=mathfrak{sl}_2(mathbf{C})$ and you take $N=L=mathbf{C}$ to be the trivial representation and $M$ to be the co-Verma module containing it as a submodule, then
$$mathrm{Hom}_mathfrak{g}(M,mathbf{C})=0,$$ so there can be no inclusion as in your question.
answered Nov 27 at 14:12
Stephen
10.5k12237
But how to represent a given submodule $Nsubseteq M$ as a quotient of $M$ in the semisimple Lie algebra case?
– James Cheung
Nov 27 at 14:21
@JamesCheung Are your representations finite dimensional $mathbf{C}$-vector spaces? If so, you must choose a complementary submodule. There are many ways to do so, in general. It's impossible for me to specify one of them canonically, as I noted in my answer. One procedure would be to start with a positive definite Hermitian form on $M$ and to average it over the special unitary group to get an invariant positive definite form, which you could then use by taking orthogonal complements.
– Stephen
Nov 27 at 14:23
Thank you for your detailed answer.
– James Cheung
Nov 27 at 15:50
add a comment |
Firstly, since the functor Hom is contravariant in the first variable, what you expect is a map going the other way. For a general abelian category, the only situation in which you expect to have an inclusion $$mathrm{Hom}(A,B) subseteq mathrm{Hom}(C,B)$$ is when $A$ is a quotient of $C$. For a semi-simple Lie algebra $mathfrak{g}$ over $mathbf{C}$, the category of finite-dimensional modules is semi-simple, so there are (in general, many) ways to represent a given submodule $N subseteq M$ as a quotient. Choosing one of them gives you the desired inclusion (which is not canonical in general!).
However, the category of all $mathfrak{g}$-modules is far from semi-simple, so there is no such inclusion in the generality you are asking for. For instance, if $mathfrak{g}=mathfrak{sl}_2(mathbf{C})$ and you take $N=L=mathbf{C}$ to be the trivial representation and $M$ to be the co-Verma module containing it as a submodule, then
$$mathrm{Hom}_mathfrak{g}(M,mathbf{C})=0,$$ so there can be no inclusion as in your question.
answered Nov 27 at 14:12
Stephen
10.5k12237
But how to represent a given submodule $Nsubseteq M$ as a quotient of $M$ in the semisimple Lie algebra case?
– James Cheung
Nov 27 at 14:21
@JamesCheung Are your representations finite dimensional $mathbf{C}$-vector spaces? If so, you must choose a complementary submodule. There are many ways to do so, in general. It's impossible for me to specify one of them canonically, as I noted in my answer. One procedure would be to start with a positive definite Hermitian form on $M$ and to average it over the special unitary group to get an invariant positive definite form, which you could then use by taking orthogonal complements.
– Stephen
Nov 27 at 14:23
Thank you for your detailed answer.
– James Cheung
Nov 27 at 15:50
add a comment |
Firstly, since the functor Hom is contravariant in the first variable, what you expect is a map going the other way. For a general abelian category, the only situation in which you expect to have an inclusion $$mathrm{Hom}(A,B) subseteq mathrm{Hom}(C,B)$$ is when $A$ is a quotient of $C$. For a semi-simple Lie algebra $mathfrak{g}$ over $mathbf{C}$, the category of finite-dimensional modules is semi-simple, so there are (in general, many) ways to represent a given submodule $N subseteq M$ as a quotient. Choosing one of them gives you the desired inclusion (which is not canonical in general!).
However, the category of all $mathfrak{g}$-modules is far from semi-simple, so there is no such inclusion in the generality you are asking for. For instance, if $mathfrak{g}=mathfrak{sl}_2(mathbf{C})$ and you take $N=L=mathbf{C}$ to be the trivial representation and $M$ to be the co-Verma module containing it as a submodule, then
$$mathrm{Hom}_mathfrak{g}(M,mathbf{C})=0,$$ so there can be no inclusion as in your question.
answered Nov 27 at 14:12
Stephen
10.5k12237
Firstly, since the functor Hom is contravariant in the first variable, what you expect is a map going the other way. For a general abelian category, the only situation in which you expect to have an inclusion $$mathrm{Hom}(A,B) subseteq mathrm{Hom}(C,B)$$ is when $A$ is a quotient of $C$. For a semi-simple Lie algebra $mathfrak{g}$ over $mathbf{C}$, the category of finite-dimensional modules is semi-simple, so there are (in general, many) ways to represent a given submodule $N subseteq M$ as a quotient. Choosing one of them gives you the desired inclusion (which is not canonical in general!).
However, the category of all $mathfrak{g}$-modules is far from semi-simple, so there is no such inclusion in the generality you are asking for. For instance, if $mathfrak{g}=mathfrak{sl}_2(mathbf{C})$ and you take $N=L=mathbf{C}$ to be the trivial representation and $M$ to be the co-Verma module containing it as a submodule, then
$$mathrm{Hom}_mathfrak{g}(M,mathbf{C})=0,$$ so there can be no inclusion as in your question.
answered Nov 27 at 14:12
Stephen
10.5k12237
edited Nov 27 at 14:31
answered Nov 27 at 14:12
Stephen
10.5k12237
answered Nov 27 at 14:12
Stephen
10.5k12237
answered Nov 27 at 14:12
Stephen
10.5k12237
10.5k12237
But how to represent a given submodule $Nsubseteq M$ as a quotient of $M$ in the semisimple Lie algebra case?
– James Cheung
Nov 27 at 14:21
@JamesCheung Are your representations finite dimensional $mathbf{C}$-vector spaces? If so, you must choose a complementary submodule. There are many ways to do so, in general. It's impossible for me to specify one of them canonically, as I noted in my answer. One procedure would be to start with a positive definite Hermitian form on $M$ and to average it over the special unitary group to get an invariant positive definite form, which you could then use by taking orthogonal complements.
– Stephen
Nov 27 at 14:23
Thank you for your detailed answer.
– James Cheung
Nov 27 at 15:50
add a comment |
But how to represent a given submodule $Nsubseteq M$ as a quotient of $M$ in the semisimple Lie algebra case?
– James Cheung
Nov 27 at 14:21
@JamesCheung Are your representations finite dimensional $mathbf{C}$-vector spaces? If so, you must choose a complementary submodule. There are many ways to do so, in general. It's impossible for me to specify one of them canonically, as I noted in my answer. One procedure would be to start with a positive definite Hermitian form on $M$ and to average it over the special unitary group to get an invariant positive definite form, which you could then use by taking orthogonal complements.
– Stephen
Nov 27 at 14:23
Thank you for your detailed answer.
– James Cheung
Nov 27 at 15:50
But how to represent a given submodule $Nsubseteq M$ as a quotient of $M$ in the semisimple Lie algebra case?
– James Cheung
Nov 27 at 14:21
But how to represent a given submodule $Nsubseteq M$ as a quotient of $M$ in the semisimple Lie algebra case?
– James Cheung
Nov 27 at 14:21
@JamesCheung Are your representations finite dimensional $mathbf{C}$-vector spaces? If so, you must choose a complementary submodule. There are many ways to do so, in general. It's impossible for me to specify one of them canonically, as I noted in my answer. One procedure would be to start with a positive definite Hermitian form on $M$ and to average it over the special unitary group to get an invariant positive definite form, which you could then use by taking orthogonal complements.
– Stephen
Nov 27 at 14:23
@JamesCheung Are your representations finite dimensional $mathbf{C}$-vector spaces? If so, you must choose a complementary submodule. There are many ways to do so, in general. It's impossible for me to specify one of them canonically, as I noted in my answer. One procedure would be to start with a positive definite Hermitian form on $M$ and to average it over the special unitary group to get an invariant positive definite form, which you could then use by taking orthogonal complements.
– Stephen
Nov 27 at 14:23
Thank you for your detailed answer.
– James Cheung
Nov 27 at 15:50
Thank you for your detailed answer.
– James Cheung
Nov 27 at 15:50
add a comment |
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