Question about $text{Hom}_{mathfrak{g}}(M,L)$












2














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







share|cite|improve this question





























    2














    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







    share|cite|improve this question



























      2












      2








      2


      1





      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







      share|cite|improve this question















      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 27 at 14:14

























      asked Nov 27 at 13:51









      James Cheung

      1234




      1234






















          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.






          share|cite|improve this answer























          • 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











          Your Answer





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          1 Answer
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          active

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          2














          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.






          share|cite|improve this answer























          • 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
















          2














          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.






          share|cite|improve this answer























          • 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














          2












          2








          2






          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.






          share|cite|improve this answer














          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 27 at 14:31

























          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


















          • 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


















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