Simple formula for the dimension of weight spaces of Verma module?irreducible highest weight modulesProperties of the longest element in a Weyl groupHow to use Weyl dimension formula?Weight spaces of Verma modulesInducing highest weight modulesDimension of a weight space which is of weight $0$.How to find the multiplicity of weight in a Verma module?Composition series for Verma modules.finding high weight vector in Verma moduleThe relation between Weyl character formula and Frobenius characteristic map

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Simple formula for the dimension of weight spaces of Verma module?


irreducible highest weight modulesProperties of the longest element in a Weyl groupHow to use Weyl dimension formula?Weight spaces of Verma modulesInducing highest weight modulesDimension of a weight space which is of weight $0$.How to find the multiplicity of weight in a Verma module?Composition series for Verma modules.finding high weight vector in Verma moduleThe relation between Weyl character formula and Frobenius characteristic map













0












$begingroup$


Let $mathfrakg$ be a simple Lie algebra (e.g. $mathfraksl_n$), and let $M_lambda$ be the Verma module with highest weight $lambda$.
Is there a simple formula for $dim (M_lambda)_mu$, where $(M_lambda)_mu$ denotes the weight space with weight $mu$? Of course Weyl character formula gives me the generating function, but I am looking for a closed form formula for $dim (M_lambda)_mu$.




Edit : Apparently this is called the Kostant partition function.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Which formulation of Weyl's character formula do you mean? I recall it usually being given directly, rather than via a generating function.
    $endgroup$
    – Tobias Kildetoft
    Mar 18 at 17:01










  • $begingroup$
    @TobiasKildetoft For instance, if you expand $1/Delta(z)$ into a power series where $Delta(z)$ is a Weyl denominator, the coefficients of this series are the certain dimensions that I want.
    $endgroup$
    – Henry
    Mar 19 at 19:35










  • $begingroup$
    As you write in your edit, it's the Kostant partition function: the number of ways to write $lambda-mu$ as a non-negative integer linear combination of positive roots. You won't get a better formula than that in general.
    $endgroup$
    – Stephen
    Mar 20 at 22:54










  • $begingroup$
    @Stephen But isn't Kostant partition function piecewise polynomial? If an explicit form of the piecewise polynomial is known for, say, $A_n$, then that would be helpful.
    $endgroup$
    – Henry
    Mar 20 at 23:18















0












$begingroup$


Let $mathfrakg$ be a simple Lie algebra (e.g. $mathfraksl_n$), and let $M_lambda$ be the Verma module with highest weight $lambda$.
Is there a simple formula for $dim (M_lambda)_mu$, where $(M_lambda)_mu$ denotes the weight space with weight $mu$? Of course Weyl character formula gives me the generating function, but I am looking for a closed form formula for $dim (M_lambda)_mu$.




Edit : Apparently this is called the Kostant partition function.










share|cite|improve this question











$endgroup$











  • $begingroup$
    Which formulation of Weyl's character formula do you mean? I recall it usually being given directly, rather than via a generating function.
    $endgroup$
    – Tobias Kildetoft
    Mar 18 at 17:01










  • $begingroup$
    @TobiasKildetoft For instance, if you expand $1/Delta(z)$ into a power series where $Delta(z)$ is a Weyl denominator, the coefficients of this series are the certain dimensions that I want.
    $endgroup$
    – Henry
    Mar 19 at 19:35










  • $begingroup$
    As you write in your edit, it's the Kostant partition function: the number of ways to write $lambda-mu$ as a non-negative integer linear combination of positive roots. You won't get a better formula than that in general.
    $endgroup$
    – Stephen
    Mar 20 at 22:54










  • $begingroup$
    @Stephen But isn't Kostant partition function piecewise polynomial? If an explicit form of the piecewise polynomial is known for, say, $A_n$, then that would be helpful.
    $endgroup$
    – Henry
    Mar 20 at 23:18













0












0








0





$begingroup$


Let $mathfrakg$ be a simple Lie algebra (e.g. $mathfraksl_n$), and let $M_lambda$ be the Verma module with highest weight $lambda$.
Is there a simple formula for $dim (M_lambda)_mu$, where $(M_lambda)_mu$ denotes the weight space with weight $mu$? Of course Weyl character formula gives me the generating function, but I am looking for a closed form formula for $dim (M_lambda)_mu$.




Edit : Apparently this is called the Kostant partition function.










share|cite|improve this question











$endgroup$




Let $mathfrakg$ be a simple Lie algebra (e.g. $mathfraksl_n$), and let $M_lambda$ be the Verma module with highest weight $lambda$.
Is there a simple formula for $dim (M_lambda)_mu$, where $(M_lambda)_mu$ denotes the weight space with weight $mu$? Of course Weyl character formula gives me the generating function, but I am looking for a closed form formula for $dim (M_lambda)_mu$.




Edit : Apparently this is called the Kostant partition function.







representation-theory lie-algebras verma-modules






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 21 at 6:21









Saad

20.2k92352




20.2k92352










asked Mar 17 at 22:26









HenryHenry

1,731726




1,731726











  • $begingroup$
    Which formulation of Weyl's character formula do you mean? I recall it usually being given directly, rather than via a generating function.
    $endgroup$
    – Tobias Kildetoft
    Mar 18 at 17:01










  • $begingroup$
    @TobiasKildetoft For instance, if you expand $1/Delta(z)$ into a power series where $Delta(z)$ is a Weyl denominator, the coefficients of this series are the certain dimensions that I want.
    $endgroup$
    – Henry
    Mar 19 at 19:35










  • $begingroup$
    As you write in your edit, it's the Kostant partition function: the number of ways to write $lambda-mu$ as a non-negative integer linear combination of positive roots. You won't get a better formula than that in general.
    $endgroup$
    – Stephen
    Mar 20 at 22:54










  • $begingroup$
    @Stephen But isn't Kostant partition function piecewise polynomial? If an explicit form of the piecewise polynomial is known for, say, $A_n$, then that would be helpful.
    $endgroup$
    – Henry
    Mar 20 at 23:18
















  • $begingroup$
    Which formulation of Weyl's character formula do you mean? I recall it usually being given directly, rather than via a generating function.
    $endgroup$
    – Tobias Kildetoft
    Mar 18 at 17:01










  • $begingroup$
    @TobiasKildetoft For instance, if you expand $1/Delta(z)$ into a power series where $Delta(z)$ is a Weyl denominator, the coefficients of this series are the certain dimensions that I want.
    $endgroup$
    – Henry
    Mar 19 at 19:35










  • $begingroup$
    As you write in your edit, it's the Kostant partition function: the number of ways to write $lambda-mu$ as a non-negative integer linear combination of positive roots. You won't get a better formula than that in general.
    $endgroup$
    – Stephen
    Mar 20 at 22:54










  • $begingroup$
    @Stephen But isn't Kostant partition function piecewise polynomial? If an explicit form of the piecewise polynomial is known for, say, $A_n$, then that would be helpful.
    $endgroup$
    – Henry
    Mar 20 at 23:18















$begingroup$
Which formulation of Weyl's character formula do you mean? I recall it usually being given directly, rather than via a generating function.
$endgroup$
– Tobias Kildetoft
Mar 18 at 17:01




$begingroup$
Which formulation of Weyl's character formula do you mean? I recall it usually being given directly, rather than via a generating function.
$endgroup$
– Tobias Kildetoft
Mar 18 at 17:01












$begingroup$
@TobiasKildetoft For instance, if you expand $1/Delta(z)$ into a power series where $Delta(z)$ is a Weyl denominator, the coefficients of this series are the certain dimensions that I want.
$endgroup$
– Henry
Mar 19 at 19:35




$begingroup$
@TobiasKildetoft For instance, if you expand $1/Delta(z)$ into a power series where $Delta(z)$ is a Weyl denominator, the coefficients of this series are the certain dimensions that I want.
$endgroup$
– Henry
Mar 19 at 19:35












$begingroup$
As you write in your edit, it's the Kostant partition function: the number of ways to write $lambda-mu$ as a non-negative integer linear combination of positive roots. You won't get a better formula than that in general.
$endgroup$
– Stephen
Mar 20 at 22:54




$begingroup$
As you write in your edit, it's the Kostant partition function: the number of ways to write $lambda-mu$ as a non-negative integer linear combination of positive roots. You won't get a better formula than that in general.
$endgroup$
– Stephen
Mar 20 at 22:54












$begingroup$
@Stephen But isn't Kostant partition function piecewise polynomial? If an explicit form of the piecewise polynomial is known for, say, $A_n$, then that would be helpful.
$endgroup$
– Henry
Mar 20 at 23:18




$begingroup$
@Stephen But isn't Kostant partition function piecewise polynomial? If an explicit form of the piecewise polynomial is known for, say, $A_n$, then that would be helpful.
$endgroup$
– Henry
Mar 20 at 23:18










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