6
$\begingroup$

The answer to the question below is almost certainly known to the representation theorists; in fact, I'm pretty sure it can be extracted from Green's paper "The characters of the finite general linear groups" in Trans. Amer. Math. Soc. 80 (1955), 402–447. However, I am not a specialist in representation theory (I work in geometry), so I am having trouble extracting it from the literature.

This preamble brings me to my question. Fix some prime number $p$. Consider $1 \leq m \leq n$. Let $\vec{e}_1,\ldots,\vec{e}_n \in \mathbb{F}_p^n$ be the standard basis and let $\Gamma_{n,m}$ be the subgroup of $\text{GL}_n(\mathbb{F}_p)$ that stabilizes the vectors $\vec{e}_1,\ldots,\vec{e}_m$ point wise. Define $V_{n,m}$ to be the $\text{GL}_n(\mathbb{F}_p)$-representation obtained by inducing the $1$-dimensional trivial representation $\mathbb{C}$ of $\Gamma_{n,m}$ up to $\text{GL}_n(\mathbb{F}_p)$.

Question : What is the decomposition of $V_{n,m}$ into irreducible representations? Or at least how many terms appear in this decomposition (as a function of $n$ and $m$)? I have some reason (hope?) that the answer to this only depends on $m$ once $n$ is sufficiently large.

Finally, can anyone recommend an easy-to-read source for reading about the sort of representation theory that would go into answering this kind of question? Preferably one that is as concrete as possible (so one that e.g. doesn't assume that I am an expert in reductive groups).

EDIT : Some of the confusion in the comments comes from the fact that I changed the way I stated the question after I started writing it to be a little more "slick"; this made the question not obviously correspond to the title.

Here's the original formulation (which involved parabolic induction more directly). Define $\overline{\Gamma}_{n,m}$ to be the $\text{GL}_n(\mathbb{F}_p)$-stabilizer of the flag $$0 < \langle \vec{e}_1,\ldots,\vec{e}_m \rangle < \mathbb{F}_p^n.$$ Thus $\Gamma_{n,m} < \overline{\Gamma}_{n,m}$ and $\overline{\Gamma}_{n,m}$ is the parabolic subgroup which is the semidirect product of $\text{GL}_m(\mathbb{F}_p) \times \text{GL}_{n-m}(\mathbb{F}_p)$ and a unipotent subgroup $U_{n,m}$. If we induce the trivial $\Gamma_{n,m}$-representation $\mathbb{C}$ up to $\overline{\Gamma}_{n,m}$, we get a representation $\overline{V}_{n,m}$ of $\overline{\Gamma}_{n,m}$ which is the $\mathbb{C}$-group ring of $\text{GL}_{m}(\mathbb{F}_p)$ on which $\text{GL}_{n-m}(\mathbb{F}_p)$ and $U_{n,m}$ act trivially. The irreducible subrepresentations of $\overline{V}_{n,m}$ are thus exactly the irreducible representations of $\text{GL}_{m}(\mathbb{F}_p)$. This brings us to the following, which is similar to the question above (the first part of it would give an answer to the first question above).

Question : Let $M$ be an irreducible representation of $\text{GL}_{m}(\mathbb{F}_p)$. Regard $M$ as a representation of $\overline{\Gamma}_{n,m}$, and let $\tilde{M}$ be the result of inducing $M$ from $\overline{\Gamma}_{n,m}$ up to $\text{GL}_n(\mathbb{F}_p)$. How can we decompose $\tilde{M}$ into irreducibles? Can we at least get a bound on the number of irreducibles in $\tilde{M}$ as a function of $n$ and $m$ (which even better only depends on $m$ as long as $n$ is sufficiently large)?

$\endgroup$
  • 2
    $\begingroup$ So, if I've understood what you're saying correctly, $\Gamma_{n,m}$ is isomorphic to $(\mathbb{F}_p)^m \times \mathrm{GL}_{n-m}(p)$? In which case you're looking at a Levi subgroup of $\mathrm{GL}_n(p)$. Inducing from such subgroups is quite intractable on the whole. However if you first lifted the representation to a parabolic subgroup containing $\Gamma_{n,m}$ and then induce, then there is a very nice answer to your problem. $\endgroup$ – Jay Taylor Dec 24 '13 at 8:00
  • $\begingroup$ @Helen: Like Jay I am confused about your formulation. In the header you mention "parabolic" but your induction seems to be from a non-parabolic subgroup. Inducing the trivial character from a parabolic is organized systematically though difficult to work out explicitly in large dimensions (as is true of symmetric group representations). But inducing from a reductive subgroup is usually quite tricky. $\endgroup$ – Jim Humphreys Dec 24 '13 at 18:37
  • $\begingroup$ @JayTaylor : I edited the question to make it more like you suggest. $\endgroup$ – Helen M. Dec 24 '13 at 19:22
  • $\begingroup$ @JimHumphreys : I want to bring your attention to the edited version of the question, which is more in line with what you say the technology is good at doing. $\endgroup$ – Helen M. Dec 24 '13 at 19:22
2
$\begingroup$

If $m=1$ and $n>1$, then the decomposition is multiplicity-free and has $q$ irreducible representations. The way to see this is the following: The representation of $GL_n(\mathbf F_q)$ that you are looking at is actually the permutation representation on the space of $\mathbf C$-valued functions on the set $\mathbf F_q^{n*}$ of non-zero vectors (on which $GL_n(\mathbf F_q)$ acts transitively. Let us denote this space by $\mathbf C[\mathbf F_q^{n*}]$.

The algebra $End_{GL_n(\mathbf F_q)}\mathbf C[\mathbf F_q^{n*}]$ is parametrized by the space of functions $\mathbf F_q^{n*} \times \mathbf F_q^{n*}$ which are invariant under the diagonal action of $GL_n(\mathbf F_q)$ on this set (we call these relative positions of points). Given an orbit $O$, the action on $f\in \mathbf C[\mathbf F_q^{n*}]$ is given by $T_O f(x) = \sum_{(x,y)\in O} f(y)$.

A complete set relative positions is:

  1. $O_\lambda = \{(v, \lambda v)\mid \lambda\in \mathbf F_q^{n*}\}$ ($q-1$ relative positions),
  2. $O_* = \{(v,w)\mid \text{$v$ and $w$ are linearly independent}\}$.

Thus this endomorphism algebra is $q$-dimensional. Moreover, $T_{O_\lambda}$, as $\lambda$ runs over $\mathbf F_q^*$, along with the constant function $1$ form a basis for $End_{GL_n(\mathbf F_q)}\mathbf C[\mathbf F_q^{n*}]$.

We have $T_{O_\lambda}T_{O_\mu}(v, \theta v) = \#\{w\mid w = \lambda v,\; \mu w = \theta v\}$, which is $1$ or $0$ depending on whether or not $\lambda\mu=\theta$. Since this is independent of the order in which $\lambda$ and $\mu$ are taken, these elements commute. It is easy to see that $T_{O_*}$ commutes with all the other endomorphisms. It follows that the endomorphism algebra of this representation is commutative, and so the representation is multiplicity-free. The number of irreps occurring in it is therefore the dimension of the endomorphism algebra, which is $q$.

For larger values of $m$ I am not sure what the answer is, even whether the the number of representations depends on $q$ in a nice way or not, but my guess is that it will be a definite polynomial function of $q$.

This answers your last question in the affirmative for $m=1$: if $n\geq 2$, then the number of irreducibles in the decomposition does not depend on $n$.

In the general case, one can still compute the dimension of the endomorphism algebra (I am not sure whether or not it is commutative). This will give an upper bound on the number of irreducible representations in the decomposition of $V_{n,m}$.

Let $X_{m,n}$ denote the set of all $m$-tuples in $\mathbf F_q^{n*}$, (a superscript $*$ will always signify the subset of non-zero vectors) on which $GL_n(\mathbf F_q)$ acts diagonally. We are interested in the decomposition of the permutation representation $V_{n,m} =\mathbf C[X_{m,n}]$ of the group $GL_n(\mathbf F_q)$ into irreducibles. The dimension of the endomorphism algebra of this representation is the number of orbits for the action of $GL_n(\mathbf F_q)$ on pairs of $m$-tuples in $GL_n(\mathbf F_q)$, which is the same as the number of orbits of $2m$-tuples. In what follows, I will give a closed-form solution for the cardinality of $GL_n(\mathbf F_q)\backslash (\mathbf F_q^{n*})^k$ of $k$-tuples for each non-negative integer $k$. Then $\dim End_{GL_n(\mathbf F_q)} V_{n,m} = |C_{2m}|$ of the following computation. It will follow that this dimension does not depend on $n$ for $n\geq 2m$.

Let $C_k = GL_n(\mathbf F_q)\backslash (\mathbf F_q^{n*})^k$, and $C = \coprod_{k=0}^\infty C_k$. Give $C$ the structure of a rooted tree as follows: the root is the unique element of $C_1$. The orbit of $(v_1,\dotsc,v_k)$ in $C_k$ is connected by an edge to the orbit of $(v_1,\dotsc,v_{k+1})\in C_{k+1}$ for every $v_{k+1}\in \mathbf F_q^{n*}$ (we will say that the orbit of $(v_1,\dotsc,v_{k+1})$ is a child of the orbit of $(v_1,\dotsc,v_k)$). Let \begin{equation*} C_{k,r} = \{(v_1,\dotsc,v_k)\in C_k \mid v_1,\dotsc, v_k \text{ span an $r$-dimensional subspace}\}. \end{equation*} Now if $v_{k+1}$ and $v'_{k+1}$ are any two vectors in $\mathbf F_q^{n*}$, then $(v_1,\dotsc,v_k,v_{k+1})$ and $(v_1,\dotsc, v_k, v'_{k+1})$ lie in the same orbit in $C_{k+1}$ if and only if there exists an element of $GL_n(\mathbf F_q)$ which fixes each of $v_1,\dotsc, v_k$ and maps $v_{k+1}$ to $v'_{k+1}$.

Let $V$ denote the span of $v_1,\dotsc,v_k$. Any element of $GL_n(\mathbf F_q)$ which fixed each of $v_1,\dotsc,v_k$ fixes every vector in $V$. Thus if $v_{k+1}\in V^*$, then $(v_1,\dotsc,v_k,v_{k+1})$ and $(v_1,\dotsc, v_k, v'_{k+1})$ lie in the same orbit in $C_{k+1}$ if and only if $v_{k+1}= v'_{k+1}$. On the other hand, if $v_{k+1}\notin V$ then $(v_1,\dotsc,v_k,v_{k+1})$ and $(v_1,\dotsc, v_k, v'_{k+1})$ lie in the same orbit in $C_{k+1}$ if and only if $v'_{k+1}\notin V$.

It follows that an element of $C_{k,r}$ has $q^r -1$ children in $C_{k+1,r}$. Also it has one child in $C_{k+1,r+1}$ if $r<n$, and zero such children otherwise.

Thus, consider the matrix \begin{equation*} A_n = \begin{pmatrix} q-1 & 0 & 0 & \cdots & 0 & 0\\ 1 & q^2-1 & 0 &\cdots & 0 & 0\\ 0 & 1 & q^3-1& \cdots & 0 & 0\\ \vdots & \vdots &\vdots & \ddots & \vdots & \vdots\\ 0 & 0 & 0 & \cdots & q^{n-1}-1 & 0\\ 0 & 0 & 0 & \cdots & 1 & q^n - 1 \end{pmatrix} \end{equation*} If $\vec C_k$ is the column vector of length $n$ with entries $C_{k,1},\dotsc,C_{k,n}$, then we have \begin{equation*} \vec C_k = A_n \vec C_{k-1} \end{equation*} for $k = 1, 2, 3,\dotsc$. The vector $\vec C_1$ is the first coordinate vector $\vec e_1 = (1,0,\dotsc,0)$.

Finally, we have $|C_k| = \vec 1^t \vec C_k = \vec 1^t A_n^{k-1} \vec C_1$. A consequence of this explicit calculation is that $|C_k|$ is independent of $n$ for $k\leq n$. This fact is already apparent while computing the number of children.

$\endgroup$
  • 1
    $\begingroup$ I have some notes where this approach to analyzing permutation reps is explained in detail. I can send them to you if you send me an e-mail. $\endgroup$ – Amritanshu Prasad Dec 24 '13 at 7:15
  • $\begingroup$ Thanks! I'll send you an email later asking for those notes. $\endgroup$ – Helen M. Dec 24 '13 at 19:23
1
$\begingroup$

The edited question (and the answer to the first version) stilll leave me somewhat confused, so I'd suggest starting with concrete low-rank examples in order to focus better on the issues involved. (Sometimes the notation gets in the way, as in the switch from the $q$ in the header to the prime $p$; everything here can be done uniformly for any power $q$ of a fixed prime $p$.)

There is some useful older literature, starting with the thesis work which Steinberg did with Brauer in Toronto, which treats characters of $\mathrm{GL}_3(\mathbb{F}_q)$ and $\mathrm{GL}_4(\mathbb{F}_q)$: see his 1951 paper here. Soon afterwards J.A. Green worked out combinatorially the characters of all finite general linear groups, as noted in the question, though he needs a lot of careful notation involving partitions and the like: see his 1955 paper here.

For instance, in the case of $G=\mathrm{GL}_3(\mathbb{F}_q)$, there are (up to conjugacy) just two parabolic subgroups $P_1, P_2$ strictly between the standard Borel subgroup $B$ of upper triangular matrices and the full group $G$; these correspond to the two simple roots. The orders of all these groups are easily computed, along with the index of each parabolic in $G$. e.g., $$|G|=q^3(q-1)^3(q+1)(q^2+q+1),\: |P_i|=|B|(q+1) = q^3(q-1)^3(q+1)$$

Steinberg works out explicitly the degrees of all irreducible characters, which makes it easy to see how induction works for each type of character of $P_i$ arising from a Levi subgroup of type $\mathrm{GL}_2(\mathbb{F}_q)$ times the multiplicative group of $\mathbb{F}_q$. The only extra complication here is the fact that such a Levi subgroup has a family of characters of degree 1 which are trivial on the derived (=special linear) group.

In any case, the study of such characters doesn't depend on introducing geometric interpretations of the parabolic subgroups (though Steinberg does take advantage of some 2-transitive permutation characters in his paper). Concerning older literature, in his book on symmetric functions (either editin) Macdonald included a treatment of Green's work, which Springer reformulated in his lectures at IAS (published in Springer Lecture Notes 131). By 1976 the more sophisticated methods of Deligne-Lusztig had paved the way for determination of characters of all finite groups of Lie type, along lines Macdonald and Springer had conjectured. But general linear groups can be done using only combnatorial methods, due to some absence of "cuspidal" characters.

$\endgroup$

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.