Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $P$ be a parabolic subgroup of $GL(n)$ with Levi decomposition $P =MN$, where $N$ is the unipotent radical.

Let $\pi$ be an irreducible representation of $M(\mathbf{Z}_p)$ inflated to $P(\mathbf{Z}_p)$,

how does $ Ind_{P(\mathbf{Z}_p)}^{GL_n(\mathbf{Z}_p)} \pi$ decompose?

It would be sufficient for me to know the result in the simplest case, where $P$ is a Borel subgroup

share|improve this question
add comment

2 Answers

up vote 5 down vote accepted

As has been noted, the answer is known for $\mathrm{GL}_2$. For $n>2$ there are certain cases where $\rho:=\mathrm{Ind}_{P(\mathbb{Z}/p^r)}^{\mathrm{GL}_n(\mathbb{Z}/p^r)}\pi$ is irreducible (see for example the result of Hill referred to in the question Parabolic induction for GL(2,Z/pn)).

For $\mathrm{GL}_3$ there are two papers by Campbell and Nevins (http://arxiv.org/pdf/0710.3261v1.pdf and http://arxiv.org/pdf/0710.3263.pdf) which study the decomposition of representations of the form $\rho$. In particular the authors almost achieve a decomposition of $\mathrm{Ind}_{B(\mathcal{O}_r)}^{\mathrm{GL}_3(\mathcal{O}_r)}\mathbf{1}$, but it turns out there are irreducible consituents which depend on the residue characteristic $p$, and these are hard to pin down explicitly in a uniform way.

Contrary to what's stated in the question, the simplest case is not when $P$ is a Borel but when it is a maximal parabolic subgroup of $\mathrm{GL}_n$. In this case one can find an explicit set of representatives for $P(\mathcal{O}_r)\backslash \mathrm{GL}_n(\mathcal{O}_r)/P(\mathcal{O}_r)$, independent of $p$, and consequently a complete decomposition of $\mathrm{Ind}_{P(\mathcal{O}_r)}^{\mathrm{GL}_n(\mathcal{O}_r)}\mathbf{1}$. This was done by Hill in "On the nilpotent representations of $\mathrm{GL}_n(\mathcal{O})$", and is also studied in a more general context in Onn & Bader (http://arxiv.org/abs/math/0404408).

In general, the problem of decomposing $\mathrm{Ind}_{B(\mathcal{O}_r)}^{\mathrm{GL}_n(\mathcal{O}_r)}\mathbf{1}$ will in some way or another involve a description of $B(\mathcal{O}_r)\backslash \mathrm{GL}_n(\mathcal{O}_r)/B(\mathcal{O}_r)$, and this is claimed to be a wild problem in Onn, Prasad & Vaserstein (http://arxiv.org/pdf/math/0506094.pdf), although Prasad seems to express a small degree of reservation in Bruhat decomposition for G(R), R local ring or R=Z/p^r.

Looking at the results for $\mathrm{GL}_3$ it seems very likely that the problem of decomposing a general representation of the form $\rho$ is hopelessly complicated. As mentioned above there are partial results, and this is probably the most one can expect in general.

The "Iwahori decomposition plus the Bruhat decomposition over the residue field" approach alluded to in one of the comments seems unlikely to me to lead to a complete solution since it would reasonably be independent of $p$ while the decomposition problem in general probably depends on $p$ (because the spaces of double cosets do).

share|improve this answer
Let $I$ be the Iwahori, then $I//B(o)$ will not depend on the residue characteristic of $o$? –  plusepsilon.de Mar 17 '12 at 12:31
"Prasad seems to express a small degree of reservation"... how do you know? Thanks for another the very informative answer. Certainly, you noticed that the other questions have also been by me. Btw. do you know, what is know for the situation of SL(2)? Can we use here $Res_{SL_2(o)} Ind_{B(o)}^{GL_2(o)} \pi$ and mackey restriction induction formula? –  plusepsilon.de Mar 17 '12 at 12:42
Regarding the Iwahori, I was referring to the Iwahori decomposition as a way to get a coarse description of the double coset spaces. This is not enough in general. Describing the constituents of $\mathrm{Ind}_B^I\mathbf{1}$ is probably as difficult (i.e., untractable) as the original problem. –  A Stasinski Mar 17 '12 at 19:42
Regarding Prasad's comment, his use of the word 'seems' indicates some degree of uncertainty, although I am not sure this was his intention. The best thing would be to ask Prasad directly. –  A Stasinski Mar 17 '12 at 19:48
The wildness here comes from the wildness of the embedding problem for finite abelian groups (C. M. Ringel and M. Schmidmeier, Submodule categories of wild representation type, Journal of Pure and Applied Algebra 205/2 (2006), 412-422). –  Amritanshu Prasad May 2 '12 at 17:35
add comment

If $P$ is the Borel and $\pi$ is admissible, then $\pi$ is 1-dimensional and let $\Pi$ denote the induction. Let $K(r)$ be the principal congruence subgroup of elements in $GL_n(\mathbb Z_p)$ that are 1 modulo $p^r$. Then $\Pi$ is a direct limit of the representations $\Pi^{K(r)}$. Now it's easy to see that $\Pi^{K(r)} \cong Ind_{P(\mathbb Z/p^r)}^{GL_n(\mathbb Z/p^r)} \pi^{T \cap K(r)}$. It suffices to decompose these representations.

When $n = 2$ this was done by Casselman in his paper "The Restriction of a Representation of $GL_2(k)$ to $GL_2(o)$." (Math. Ann., 1973) Link: http://www.digizeitschriften.de/dms/img/?PPN=PPN235181684_0206&DMDID=dmdlog53 (see prop. 1 on p. 312). Basically the complement of each $\Pi^{K(r)}$ in $\Pi^{K(r+1)}$ is irreducible (once $r$ is large enough so that these spaces are nonzero). You can easily compute their dimensions, incidentally, because $P(\mathbb Z/p^r)\backslash GL_2(\mathbb Z/p^r) \cong \mathbb P^1(\mathbb Z/p^r)$, which has $p^{r-1}(p+1)$ elements.

Note that Casselman assumes that $\pi$ is of the form $\epsilon_0 \otimes 1$, but we can reduce to this case by twisting $\pi$ by $\chi \circ \det$ for characters $\chi$ of $\mathbb Z_p^\times$.

It's possible that this can be generalised to $GL_n$, I haven't tried to think about that.

EDIT: I would guess it's a wild problem in general to decompose this, even if $P$ is the Borel. Namely if you could decompose this representation, then you would certainly also know the $GL_n(\mathbb Z/p^r)$-endomorphism ring $\Pi^{K(r)}$. But by a standard Frobenius reciprocity argument (also used in Casselman) this is the space of functions $f: GL_n(\mathbb Z/p^r) \to \mathbb C$ such that $f(b k b') = \pi(b) f(k) \pi(b')$ for all $k \in GL_n(\mathbb Z/p^r)$ and $b, b' \in P(\mathbb Z/p^r)$. In particular you should understand the double coset space $P(\mathbb Z/p^r) \backslash GL_n(\mathbb Z/p^r) / P(\mathbb Z/p^r)$ (and this is exactly what's required when $\pi$ is trivial). But this may involve wild classification problems according to Bruhat decomposition for G(R), R local ring or R=Z/p^r. [You even need this for all $r$, so you'd need $P(\mathbb Z_p) \backslash GL_n(\mathbb Z_p) / P(\mathbb Z_p)$. Maybe this is known to be wild for $n$ bigger than some bound?]

share|improve this answer
Howe's 1973 Trans. AMS paper "On the principal series of $GL_n$ over $p$-adic fields." concerns the types of (unitary) principal series. These are roughly speaking the smallest irreducible constituents of the above representation restricted to $GL_n(\mathbb Z_p)$ (I'm not sure it's literally the smallest). Howe shows that his type occurs only once and determines the representation. –  fherzig Mar 16 '12 at 20:39
Thanks for this answer. I am familiar with the $N=2$ situation. Actually Silberger in "PGL(2) over the p adics" is more precise, but omits even characteristic. An analogue of the Bruhat decomposition would be certainly convenient, but I think that the Iwahori decomposition plus the Bruhat decomposition over the residue field should be enough to circumvent it via induction by steps. –  plusepsilon.de Mar 17 '12 at 8:51
Actually if you looked into the history of the question, you will see that I have described all this, before I decided to make the question a little bit more catchy because of zero reaction;) –  plusepsilon.de Mar 17 '12 at 8:54
I had not looked at the history... –  fherzig Mar 17 '12 at 13:16
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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