Let $G$ be a reductive algebraic group defined over a non-Archimedean field $F$.
Let $k_F$ be its residue field, of characteristic $p$.
Assume $G$ is unramified over $F$, then it admits a hyperspecial compact subgroup $K$.

Let $V(\lambda)$ be the irreducible algebraic representation of $G(F)$ of highest weight $\lambda$ with respect to a maximal torus $T$ contained in a Borel subgroup $B$ (over $\bar{F}$).
Let $\mathcal O_\bar{F}$ be the ring of integers of $\bar{F}$.
Let $V_0(\lambda)$ be a lattice (an $\mathcal O_\bar{F}$-submodule) in $V(\lambda)$, which is stable under the action of $K$.
Consider $V_p(\lambda)=V_0(\lambda)\otimes \bar{k_F}$.

Let $U^+$ be the unipotent radical of the opposite Borel subgroup $B^+$. Let $N_0 = U^+\cap K$.


Is there a $K$-stable lattice $V_0(\lambda)$ with a good description of the invariants $V_p(\lambda)^{N_0}$ ?


Note that the $N_0$-action on $V_p(\lambda)$ factors through an action of the finite quotient $U^+(k_F)$.


If $G = SL_2$ over $\mathbb Q_p$, $B$ is the Borel of lower triangular matrices, and $\lambda=k$, we see that $V(\lambda)$ is isomorphic to $Sym^k(\bar{\mathbb Q_p}^2)$.

Let $K = SL_2(\mathbb Z_p)$, then $V_0(\lambda)=Sym^k(\bar{\mathbb Z_p}^2)$ is a $K$-stable lattice, and $V_p(\lambda)=Sym^k(\bar{\mathbb F_p}^2)$.

If we model $Sym^k(\bar{\mathbb Q_p}^2)$ as homogeneous polynomials in two variables $x,y$ of degree at most $k$, then we have a very nice description of $$ V_p(k)^{N_0} = Sym^k(\bar{\mathbb F_p}^2)^{N_0} $$

Indeed, if $f$ is $N_0$-invariant, then $$ f(x,ax+y) = f(x,y) \quad \forall a \in \mathbb F_p $$

By dehomogenization $T = \frac{y}{x}$, we may consider $f$ as a polynomial in one variable of degree at most $k$, and $$ f(T+z) = f(T) $$ for any $z\in \mathbb F_p$, hence $f(T)$ is of the form $f(T)=g(T^p-T)$.


In the framework of the $p$-adic Langlands program, one is concerned with finding separated $G$-stable lattices in irreducible locally algebraic representations of $G$ over $\bar{F}$, which are representations of the form $\sigma \otimes V(\lambda)$ where $\sigma$ is a smooth representation.

For example, when $G$ is split and $\sigma$ is an unramified principal series representation,$F=\mathbb Q_p$, $G$ has a connected center and the coxeter number of $G$ is $p$-small, E. Große-Klönne (http://arxiv.org/abs/1408.3369) shows that if $V_p(\lambda)$ is an irreducible representation of $\bar{k_F}[K]$, then one can construct a separated $G$-stable lattice in $\sigma \otimes V(\lambda) $ if the necessary conditions are fulfilled.

Looking carefully at the proof, one notices that it is caused by the fact that under these conditions, $V_p(\lambda)^{N_0}$ is one-dimensional.

However, the result is conjectured to be true also when $V_p(\lambda)$ is no longer irreducible. Previously, in the case of $GL_2(\mathbb Q_p)$, C. Breuil (http://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=139607&fileId=S1474748003000021) has succeeded to prove the conjecture also for some such cases ($p \le k < 2p-1 $).

It seems that in order to generalize his methods for other groups, one must simplify the algebraic structure and manipulations.

A significant simplification can be made by looking at the $N_0$-invariants (in fact, mod $p^n$ for any $n$), and this is the reason for my interest in this question.

  • $\begingroup$ Can you clarify your set-up (and maybe motivation) a little more? For example, the concrete study of linear algebraic groups apparently intended here involves fields (not rings) of definition. Also, when you have the group $N_0$ acting on a module in prime characteristic, I guess you mean the reduction mod $p$ of this group? In your last paragraph, it's usual to view the $k$th symmetric power as the space of homogeneous polynomials in two variables of degree $k$, so your formulation looks nonstandard. (And you might add a tag 'algebraic-groups'.) $\endgroup$ – Jim Humphreys Mar 6 '16 at 18:24
  • $\begingroup$ Thank you very much. You are right. I have tried to clarify the set-up and the motivation and explained the formulation in terms of the standard model for the symmetric power. Regarding $N_0$, I would like to keep it that way, even though it acts via its finite quotient. $\endgroup$ – assaferan Mar 8 '16 at 9:03

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.