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
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).
$\mathrm{Ind}_B^I\mathbf{1}$ is probably as difficult (i.e., untractable) as the original problem.
$\endgroup$
Commented
Mar 17, 2012 at 19:42
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?]
