Question: What is decomposition of the representation k[Flag(F_q)] as bimodule over GL_n(F_q) , Hecke(q) ?

(Let k=Complex numbers. Further question: is there any change for char k = p ? )

Remark: Hecke(q) is deformation of k[S_n] - which is semisimple, so there are no non-trivial deformations for generic q, I am not sure q=p^k is "generic", but I think it is true. So irreps of Hecke(q) are parametrized by Young diagramms of size "n". I guess the decomposition above is somewhat similar with Schur-Weyl duality so there should be some irreps of GL_n(F_q) parametrized by Young diagrams. Is there any independent description of these irreps ?

Notations and constructions:

F_q - finite field, Flag(F_q) - flag variety = GL_n(F_q) / Borel(F_q) , Hecke(q) - Hecke algebra.

GL_n(F_q) acts on Flag(F_q) in an obvious way - since any G acts on G/H.

To explain the action of Hecke(q) we need two facts:

1) For any G/H there is action of k[H\G/H] commuting with action of G see Florian Eisele answer here

2) k[ Borel\GL/Borel] is Hecke algebra. Some hints for this - recall Bruhat decomposition GL = BorelWeylBorel , so double coset as a set can be identified with Weyl group, however the convolution operation defines the Hecke algebra structure. It seems it is enough to check this for GL_3 only.

Further questions:

What about other semisimple algebraic groups ?

What should be limit q->1 ? In this limit both GL, Hecke goes to S_n.

Is there any relation with Schur-Weyl duality ?

What is more general context for this decomposition in view of Jim Humphreys remark:

for groups of Lie type there is a rich theory of what can be done if you induce up from the trivial character of a parabolic subgroup and then decompose the induced character using Hecke algebra methods ?

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    $\begingroup$ The limit $q\to 1$ is just the regular representation of the symmetric group. $\endgroup$ Sep 4, 2012 at 11:06
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    $\begingroup$ And yes: Plugging in a prime power for q is in the generic case and the Hecke algebra at this points is semisimple. The only places that can give non-semisimple specializations are roots of unity. $\endgroup$ Sep 4, 2012 at 14:58
  • $\begingroup$ @Johannes Hahn Is there simple way to see it ? $\endgroup$ Sep 4, 2012 at 19:19
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    $\begingroup$ Plugging in a prime power for $q$ puts you in the semisimple case when working over a field of characteristic zero because the Hecke algebra is the endomorphism algebra of $k[\text{flags}]$, which is (by Maschke's theorem) a semisimple representation. $\endgroup$ Sep 5, 2012 at 4:05

1 Answer 1


If you take the space $X_\lambda$ of flags of shape $\lambda$ (here $\lambda$ is a partition of $n$, and a flag of shape $\lambda$ is one where the $i$th subspace has dimension $\lambda_1+\dotsc+\lambda_i$), then there exists a family of irreducible representations of $V_\lambda$ of $GL_n(\mathbf F_q)$ such that

$k[X_\lambda] = V_\lambda \oplus (\oplus_{\mu>\lambda} V_\mu^{K_{\mu\lambda}})$

where $K_{\mu\lambda}$ is the number of SSYT of shape $\mu$ and type $\lambda$ (this is the analog of the Young rule). In partitcular, if you take full flags, then you get each $V_\lambda$ with multiplicity equal to the number of SYT of shape $\lambda$.

A more general statement than this is proved in Green's 1955 paper.

The above decomposition actually allows you to inductively define and compute the character of each $V_\lambda$. For example, one can exploit the fact that the cardinality of $X_\lambda$ is a $q$-analog of a multinomial coefficient to show that the dimension of $V_\lambda$ is a $q$-analog of the number of SYT of shape $\lambda$, i.e., a polynomial in $q$ whose value at $1$ is the number of SYT of shape $\lambda$.

Also, the set $V_\lambda^B$ of $B$-invariant vectors in $V_\lambda$ is a module for the Hecke algebra (whose dimension is the dimension of the corresponding symmetric group representation = the number of SYT's of shape $\lambda$). The answer to your question about decomposition as bi-modules is:

$k[\text{complete flags}] = \bigoplus_{\lambda} V_\lambda\otimes \tilde V_\lambda^B$,

where the tilde signifies taking contragredient.

  • $\begingroup$ @Amritanshu Prasad Thank you very much ! $\endgroup$ Sep 4, 2012 at 12:12

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