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 ?

  • 1
    $\begingroup$ The limit $q\to 1$ is just the regular representation of the symmetric group. $\endgroup$ Sep 4, 2012 at 11:06
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.