Is $mathcal{U}(mathfrak{g})$ semisimple (as a module over itself)? (and related examples)












1












$begingroup$


Recall that an object in an abelian category is semisimple if it is a (finite) direct sum of simple objects. An abelian category is semisimple if every object is semisimple.



In studying representations of semisimple lie algebras, one often restricts to the subcategory called $mathcal{O}_{int}$ inside the category left modules for the universal enveloping algebra, $mathcal{U}(mathfrak{g})$. The main theorem about $mathcal{O}_{int}$ says it is semisimple (or, at least, that all the objects in $mathcal{O}_{int}$ are semisimple. Perhaps one needs to check several other things to see that $mathcal{O}_{int}$ is abelian as a subcategory).



I am new to this area and would like to motivate this restriction.



A ring $R$ is semisimple in the category of left $R$ modules iff the category of left $R$ modules is semisimple. Thus,




Is $mathcal{U}(mathfrak{g})$ semisimple as a left $mathcal{U}(mathfrak{g})$-module (for $mathfrak{g}$ a semisimple Lie algebra)?




As a bonus, are there other nice examples of $mathcal{U}(mathfrak{g})$-modules that aren't semisimple?






abstract-algebra representation-theory lie-algebras







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    No, $U(mathfrak{g})$ is not a semisimple $U(mathfrak{g})$-module. For your bonus questions, just consider Verma modules.
    $endgroup$
    – David Hill
    Jan 25 at 16:57






  • 2




    $begingroup$
    In categories of modules, there is no requirement that the sum be finite... any infinite sum of simple modules is also called semisimple. Maybe the case is different for the nomenclature of abelian categories, or not. I just thought I'd point out it isn't quite consistent with the normal usage for modules.
    $endgroup$
    – rschwieb
    Jan 25 at 17:39












  • $begingroup$
    @DavidHill How does one see they aren't semisimple?
    $endgroup$
    – AnonymousCoward
    Jan 26 at 7:44






  • 1




    $begingroup$
    A good example is to take the Verma module for sl2 with trivial head and show that the unique maximal submodule doesn't have a complement. That implies it isn't semisimple.
    $endgroup$
    – Matthew Towers
    Jan 26 at 13:16










  • $begingroup$
    Just to check: Is $mathcal{O}_{int}$ just the category of finite dimensional modules? (I can't think of any larger one which is semisimple and "natural" to define). I am mainly used to using such an $mathcal{O}$ for the BGG category $mathcal{O}$, with subscripts denoting various blocks (which are rarely semisimple).
    $endgroup$
    – Tobias Kildetoft
    Jan 27 at 6:43
















1












$begingroup$


Recall that an object in an abelian category is semisimple if it is a (finite) direct sum of simple objects. An abelian category is semisimple if every object is semisimple.



In studying representations of semisimple lie algebras, one often restricts to the subcategory called $mathcal{O}_{int}$ inside the category left modules for the universal enveloping algebra, $mathcal{U}(mathfrak{g})$. The main theorem about $mathcal{O}_{int}$ says it is semisimple (or, at least, that all the objects in $mathcal{O}_{int}$ are semisimple. Perhaps one needs to check several other things to see that $mathcal{O}_{int}$ is abelian as a subcategory).



I am new to this area and would like to motivate this restriction.



A ring $R$ is semisimple in the category of left $R$ modules iff the category of left $R$ modules is semisimple. Thus,




Is $mathcal{U}(mathfrak{g})$ semisimple as a left $mathcal{U}(mathfrak{g})$-module (for $mathfrak{g}$ a semisimple Lie algebra)?




As a bonus, are there other nice examples of $mathcal{U}(mathfrak{g})$-modules that aren't semisimple?






abstract-algebra representation-theory lie-algebras







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    No, $U(mathfrak{g})$ is not a semisimple $U(mathfrak{g})$-module. For your bonus questions, just consider Verma modules.
    $endgroup$
    – David Hill
    Jan 25 at 16:57






  • 2




    $begingroup$
    In categories of modules, there is no requirement that the sum be finite... any infinite sum of simple modules is also called semisimple. Maybe the case is different for the nomenclature of abelian categories, or not. I just thought I'd point out it isn't quite consistent with the normal usage for modules.
    $endgroup$
    – rschwieb
    Jan 25 at 17:39












  • $begingroup$
    @DavidHill How does one see they aren't semisimple?
    $endgroup$
    – AnonymousCoward
    Jan 26 at 7:44






  • 1




    $begingroup$
    A good example is to take the Verma module for sl2 with trivial head and show that the unique maximal submodule doesn't have a complement. That implies it isn't semisimple.
    $endgroup$
    – Matthew Towers
    Jan 26 at 13:16










  • $begingroup$
    Just to check: Is $mathcal{O}_{int}$ just the category of finite dimensional modules? (I can't think of any larger one which is semisimple and "natural" to define). I am mainly used to using such an $mathcal{O}$ for the BGG category $mathcal{O}$, with subscripts denoting various blocks (which are rarely semisimple).
    $endgroup$
    – Tobias Kildetoft
    Jan 27 at 6:43














1












1








1





$begingroup$


Recall that an object in an abelian category is semisimple if it is a (finite) direct sum of simple objects. An abelian category is semisimple if every object is semisimple.



In studying representations of semisimple lie algebras, one often restricts to the subcategory called $mathcal{O}_{int}$ inside the category left modules for the universal enveloping algebra, $mathcal{U}(mathfrak{g})$. The main theorem about $mathcal{O}_{int}$ says it is semisimple (or, at least, that all the objects in $mathcal{O}_{int}$ are semisimple. Perhaps one needs to check several other things to see that $mathcal{O}_{int}$ is abelian as a subcategory).



I am new to this area and would like to motivate this restriction.



A ring $R$ is semisimple in the category of left $R$ modules iff the category of left $R$ modules is semisimple. Thus,




Is $mathcal{U}(mathfrak{g})$ semisimple as a left $mathcal{U}(mathfrak{g})$-module (for $mathfrak{g}$ a semisimple Lie algebra)?




As a bonus, are there other nice examples of $mathcal{U}(mathfrak{g})$-modules that aren't semisimple?






abstract-algebra representation-theory lie-algebras







share|cite|improve this question











$endgroup$




Recall that an object in an abelian category is semisimple if it is a (finite) direct sum of simple objects. An abelian category is semisimple if every object is semisimple.



In studying representations of semisimple lie algebras, one often restricts to the subcategory called $mathcal{O}_{int}$ inside the category left modules for the universal enveloping algebra, $mathcal{U}(mathfrak{g})$. The main theorem about $mathcal{O}_{int}$ says it is semisimple (or, at least, that all the objects in $mathcal{O}_{int}$ are semisimple. Perhaps one needs to check several other things to see that $mathcal{O}_{int}$ is abelian as a subcategory).



I am new to this area and would like to motivate this restriction.



A ring $R$ is semisimple in the category of left $R$ modules iff the category of left $R$ modules is semisimple. Thus,




Is $mathcal{U}(mathfrak{g})$ semisimple as a left $mathcal{U}(mathfrak{g})$-module (for $mathfrak{g}$ a semisimple Lie algebra)?




As a bonus, are there other nice examples of $mathcal{U}(mathfrak{g})$-modules that aren't semisimple?






abstract-algebra representation-theory lie-algebras




abstract-algebra representation-theory lie-algebras






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 25 at 16:48







AnonymousCoward

















asked Jan 25 at 16:36









AnonymousCowardAnonymousCoward

3,1122435




3,1122435








  • 2




    $begingroup$
    No, $U(mathfrak{g})$ is not a semisimple $U(mathfrak{g})$-module. For your bonus questions, just consider Verma modules.
    $endgroup$
    – David Hill
    Jan 25 at 16:57






  • 2




    $begingroup$
    In categories of modules, there is no requirement that the sum be finite... any infinite sum of simple modules is also called semisimple. Maybe the case is different for the nomenclature of abelian categories, or not. I just thought I'd point out it isn't quite consistent with the normal usage for modules.
    $endgroup$
    – rschwieb
    Jan 25 at 17:39












  • $begingroup$
    @DavidHill How does one see they aren't semisimple?
    $endgroup$
    – AnonymousCoward
    Jan 26 at 7:44






  • 1




    $begingroup$
    A good example is to take the Verma module for sl2 with trivial head and show that the unique maximal submodule doesn't have a complement. That implies it isn't semisimple.
    $endgroup$
    – Matthew Towers
    Jan 26 at 13:16










  • $begingroup$
    Just to check: Is $mathcal{O}_{int}$ just the category of finite dimensional modules? (I can't think of any larger one which is semisimple and "natural" to define). I am mainly used to using such an $mathcal{O}$ for the BGG category $mathcal{O}$, with subscripts denoting various blocks (which are rarely semisimple).
    $endgroup$
    – Tobias Kildetoft
    Jan 27 at 6:43














  • 2




    $begingroup$
    No, $U(mathfrak{g})$ is not a semisimple $U(mathfrak{g})$-module. For your bonus questions, just consider Verma modules.
    $endgroup$
    – David Hill
    Jan 25 at 16:57






  • 2




    $begingroup$
    In categories of modules, there is no requirement that the sum be finite... any infinite sum of simple modules is also called semisimple. Maybe the case is different for the nomenclature of abelian categories, or not. I just thought I'd point out it isn't quite consistent with the normal usage for modules.
    $endgroup$
    – rschwieb
    Jan 25 at 17:39












  • $begingroup$
    @DavidHill How does one see they aren't semisimple?
    $endgroup$
    – AnonymousCoward
    Jan 26 at 7:44






  • 1




    $begingroup$
    A good example is to take the Verma module for sl2 with trivial head and show that the unique maximal submodule doesn't have a complement. That implies it isn't semisimple.
    $endgroup$
    – Matthew Towers
    Jan 26 at 13:16










  • $begingroup$
    Just to check: Is $mathcal{O}_{int}$ just the category of finite dimensional modules? (I can't think of any larger one which is semisimple and "natural" to define). I am mainly used to using such an $mathcal{O}$ for the BGG category $mathcal{O}$, with subscripts denoting various blocks (which are rarely semisimple).
    $endgroup$
    – Tobias Kildetoft
    Jan 27 at 6:43








2




2




$begingroup$
No, $U(mathfrak{g})$ is not a semisimple $U(mathfrak{g})$-module. For your bonus questions, just consider Verma modules.
$endgroup$
– David Hill
Jan 25 at 16:57




$begingroup$
No, $U(mathfrak{g})$ is not a semisimple $U(mathfrak{g})$-module. For your bonus questions, just consider Verma modules.
$endgroup$
– David Hill
Jan 25 at 16:57




2




2




$begingroup$
In categories of modules, there is no requirement that the sum be finite... any infinite sum of simple modules is also called semisimple. Maybe the case is different for the nomenclature of abelian categories, or not. I just thought I'd point out it isn't quite consistent with the normal usage for modules.
$endgroup$
– rschwieb
Jan 25 at 17:39






$begingroup$
In categories of modules, there is no requirement that the sum be finite... any infinite sum of simple modules is also called semisimple. Maybe the case is different for the nomenclature of abelian categories, or not. I just thought I'd point out it isn't quite consistent with the normal usage for modules.
$endgroup$
– rschwieb
Jan 25 at 17:39














$begingroup$
@DavidHill How does one see they aren't semisimple?
$endgroup$
– AnonymousCoward
Jan 26 at 7:44




$begingroup$
@DavidHill How does one see they aren't semisimple?
$endgroup$
– AnonymousCoward
Jan 26 at 7:44




1




1




$begingroup$
A good example is to take the Verma module for sl2 with trivial head and show that the unique maximal submodule doesn't have a complement. That implies it isn't semisimple.
$endgroup$
– Matthew Towers
Jan 26 at 13:16




$begingroup$
A good example is to take the Verma module for sl2 with trivial head and show that the unique maximal submodule doesn't have a complement. That implies it isn't semisimple.
$endgroup$
– Matthew Towers
Jan 26 at 13:16












$begingroup$
Just to check: Is $mathcal{O}_{int}$ just the category of finite dimensional modules? (I can't think of any larger one which is semisimple and "natural" to define). I am mainly used to using such an $mathcal{O}$ for the BGG category $mathcal{O}$, with subscripts denoting various blocks (which are rarely semisimple).
$endgroup$
– Tobias Kildetoft
Jan 27 at 6:43




$begingroup$
Just to check: Is $mathcal{O}_{int}$ just the category of finite dimensional modules? (I can't think of any larger one which is semisimple and "natural" to define). I am mainly used to using such an $mathcal{O}$ for the BGG category $mathcal{O}$, with subscripts denoting various blocks (which are rarely semisimple).
$endgroup$
– Tobias Kildetoft
Jan 27 at 6:43










1 Answer
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$begingroup$

If $U(mathfrak{g})$ were semi-simple as a $U(mathfrak{g})$-module, then the same would be true of any quotient. However, $U(mathfrak{g})/U(mathfrak{g})mathfrak{b}cong M(0)$, where $M(0)$ is the Verma module of highest weight $0$, $mathfrak{b}=mathfrak{h}oplusmathfrak{n}^+$ and $mathfrak{g}=mathfrak{n}^-oplusmathfrak{h}oplusmathfrak{n}^+$ is a triangular decomposition of $mathfrak{g}$.



It is well known, and easy to check that $M(0)$ is not semi-simple. Try it for $mathfrak{sl}_2$.






share|cite|improve this answer









$endgroup$













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    $begingroup$

    If $U(mathfrak{g})$ were semi-simple as a $U(mathfrak{g})$-module, then the same would be true of any quotient. However, $U(mathfrak{g})/U(mathfrak{g})mathfrak{b}cong M(0)$, where $M(0)$ is the Verma module of highest weight $0$, $mathfrak{b}=mathfrak{h}oplusmathfrak{n}^+$ and $mathfrak{g}=mathfrak{n}^-oplusmathfrak{h}oplusmathfrak{n}^+$ is a triangular decomposition of $mathfrak{g}$.



    It is well known, and easy to check that $M(0)$ is not semi-simple. Try it for $mathfrak{sl}_2$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      If $U(mathfrak{g})$ were semi-simple as a $U(mathfrak{g})$-module, then the same would be true of any quotient. However, $U(mathfrak{g})/U(mathfrak{g})mathfrak{b}cong M(0)$, where $M(0)$ is the Verma module of highest weight $0$, $mathfrak{b}=mathfrak{h}oplusmathfrak{n}^+$ and $mathfrak{g}=mathfrak{n}^-oplusmathfrak{h}oplusmathfrak{n}^+$ is a triangular decomposition of $mathfrak{g}$.



      It is well known, and easy to check that $M(0)$ is not semi-simple. Try it for $mathfrak{sl}_2$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        If $U(mathfrak{g})$ were semi-simple as a $U(mathfrak{g})$-module, then the same would be true of any quotient. However, $U(mathfrak{g})/U(mathfrak{g})mathfrak{b}cong M(0)$, where $M(0)$ is the Verma module of highest weight $0$, $mathfrak{b}=mathfrak{h}oplusmathfrak{n}^+$ and $mathfrak{g}=mathfrak{n}^-oplusmathfrak{h}oplusmathfrak{n}^+$ is a triangular decomposition of $mathfrak{g}$.



        It is well known, and easy to check that $M(0)$ is not semi-simple. Try it for $mathfrak{sl}_2$.






        share|cite|improve this answer









        $endgroup$



        If $U(mathfrak{g})$ were semi-simple as a $U(mathfrak{g})$-module, then the same would be true of any quotient. However, $U(mathfrak{g})/U(mathfrak{g})mathfrak{b}cong M(0)$, where $M(0)$ is the Verma module of highest weight $0$, $mathfrak{b}=mathfrak{h}oplusmathfrak{n}^+$ and $mathfrak{g}=mathfrak{n}^-oplusmathfrak{h}oplusmathfrak{n}^+$ is a triangular decomposition of $mathfrak{g}$.



        It is well known, and easy to check that $M(0)$ is not semi-simple. Try it for $mathfrak{sl}_2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 27 at 5:16









        David HillDavid Hill

        9,2261619




        9,2261619






























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