4
$\begingroup$

Let $V$ and $W$ be complex irreducible representations of $GL_n(F_q)$ where $F_q$ is finite field. Is the decomposition of $V \otimes W$ into irreducible representations known?


PS

Same question: Decomposing tensor products of irreducible representations of reductive groups over a finite field

$\endgroup$
8
  • 2
    $\begingroup$ Something makes me feel Zelevinsky's "Representations of Finite Classical Groups" is the right source here. $\endgroup$ – darij grinberg Jul 28 '10 at 15:13
  • 5
    $\begingroup$ It has to be kept in mind here that the irreducibles themselves are not explicitly "known", but their characters were determined recursively in the classical 1955 Transactions AMS paper of J.A. Green. With this in mind, your problem is to work out the explicit irreducible character decomposition for the product of two explicitly given characters. Zelevinsky's 1981 Springer Lecture Notes No.869 offers an interesting approach to the representations in the spirit of classical Schur-Weyl theory, but without getting into the details of arbitrary tensor product decompositions. $\endgroup$ – Jim Humphreys Jul 28 '10 at 16:08
  • 3
    $\begingroup$ P.S. An elegant treatment of Green's ideas is given in Chapter IV of Symmetric Functions and Hall Polynomials by I.G. Macdonald (2nd ed., 1995, Oxford). As in Green's paper, the goal is an efficient description of irreducible characters using lots of combinatorics. But it remains technically challenging to compute an actual character table of any size, or to decompose tensor products, or to carry out branching rules. $\endgroup$ – Jim Humphreys Jul 28 '10 at 16:33
  • 1
    $\begingroup$ The Zelevinski approach uses parabolic induction to go from $GL_n(F_q)\times GL_m(F_q)$ to $GL_{n+m}(F_q)$. It would take some work to use this to answer the question. The tensor products of representations of symmetric groups are determined by Littlewood-Richardson coefficients but the relationship is complicated. $\endgroup$ – Bruce Westbury Jul 28 '10 at 20:12
  • 2
    $\begingroup$ @Pooja: For large enough $p$, the "patterns" of decomposition you get depend on the Weyl group but not essentially on $p$ even though the characters involve parameters depending on $p$. Look at GL$(2,p)$ or SL$(2,p)$ for $p$ large: "most" irreducible characters have degree $p-1$ or $p+1$ (roughly half of each), while a typical tensor product decomposes as a sum of roughly $p$ of these in limited patterns. Bookkeeping is tedious even in this simple case, of course, and hopeless in general. Each power$p^m$ expands this list of patterns in a way depending on $m$. $\endgroup$ – Jim Humphreys Jul 29 '10 at 13:05

Your Answer

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

Browse other questions tagged or ask your own question.