Factoring out an element of a root subgroup to make a conjugation integral

affine-grassmanniangeometric-representation-theoryp-adic-groupsreductive-groupsrt.representation-theory

Fix a nonarchimedean local field $L/\mathbb{Q}_p$ with ring of integers $\mathcal{O}$, uniformizer $\varpi$, and residue field $k$. If I take a matrix
$$\begin{pmatrix} a & \varpi b \\ c & d \end{pmatrix} \in \mathrm{GL}_2(\mathcal{O})$$
then after conjugation by $\begin{pmatrix} \varpi^{-1} & 0 \\ 0 & \varpi \end{pmatrix}$, I get
$$\begin{pmatrix} a & \varpi^{-1} b \\ \varpi^2c & d \end{pmatrix}$$
For what I'm doing, it's problematic that there's an exponent of $-1$ before $b$, because this means that the conjugated matrix doesn't necessarily land in $\mathrm{GL}_2(\mathcal{O})$ anymore. But I can "fix this" by trying to "factor out the $\varpi^1$-term of $\varpi b$". More precisely, I can find a factorization
$$\begin{pmatrix} a & \varpi b \\ c & d \end{pmatrix} = \begin{pmatrix} 1 & \varpi[x] \\ 0 & 1 \end{pmatrix} \begin{pmatrix} a' & \varpi^2 b' \\ c' & d' \end{pmatrix}$$
such that $x \in k^\times$ (where $[-]: k^\times \to \mathcal{O}^\times$ is the usual multiplicative section) and such that
$$\begin{pmatrix} a' & \varpi b' \\ c' & d' \end{pmatrix} \in \mathrm{GL}_2(\mathcal{O}).$$
Specifically, you get that $x = [b_0]/[d_0]$, where I've written $b = \sum_{n=0}^\infty [b_n]\varpi^n$ and $d = \sum_{n=0}^\infty [d_n]\varpi^n$.

The problem is "fixed" because conjugating $\begin{pmatrix}a' & \varpi^2b' \\ c' & d' \end{pmatrix} \in \mathrm{GL}_2(\mathcal{O})$ by $\begin{pmatrix} \varpi^{-1} & 0 \\ 0 & \varpi \end{pmatrix}$ lands you in $\mathrm{GL}_2(\mathcal{O})$.

Question: is there a better conceptual way of thinking about what I just did? I want to do this for more general groups. In general, $G$ is a split reductive group, and I take a quasi-minuscule (but not minuscule!) cocharacter $\lambda$ (which for $\mathrm{GL}_2$ is the cocharacter $(1,-1)$, hence the conjugation above). I can define a parabolic subgroup $P_\lambda$ of $G(k)$ generated by the roots $\alpha$ of $G$ such that $\langle \alpha, \lambda \rangle \leq 0$.

Then given an element $g \in G(\mathcal{O})$ which reduces to $P_\lambda$ under the reduction map $G(\mathcal{O}) \to G(k)$, I want to factor it as
$$g = N_{\lambda^\vee}(\varpi x) \widetilde{g},$$
where $x \in \mathcal{O} \setminus \varpi \mathcal{O}$ and $\widetilde{g}$ satisfies
$$\varpi^{-\lambda} \widetilde{g} \varpi^\lambda \in G(\mathcal{O}).$$
By $N_{\lambda^\vee}: \mathcal{O} \to G(\mathcal{O})$ I mean the root subgroup for the root $\lambda^\vee$ (it's true that if $\lambda$ is quasi-minuscule, then $\lambda^\vee$ is a root). By $\varpi^\lambda$ I mean $\lambda(\varpi)$.

My thoughts: I want to use the fact that $G$ is generated by root subgroups and the torus, and then take $g$, write it as a product of elements of root subgroups, and then try to factor out the elements appearing in the root subgroup for the root $\lambda^\vee$. But I'm worried because root subgroups don’t necessarily commute with each other, so I fear I can't move these elements all the way to the left.

Best Answer

Elaborating on the comment above. Let $K$ be local field with ring of coefficients $O$, and residue field $k$. Let $S \hookrightarrow G(K)$ be a split maximal torus. The parabolic subgroup $P \hookrightarrow G(O)$, you defined can be described in two ways:

  1. defined as the fiber under the restriction of $G(O)\rightarrow G(k)$, of $P_\lambda(k) \rightarrow G(k)$, where $P_\lambda(k)$ is the subgroup of $G(k)$ generated by opposite Borel and $(\alpha, \lambda) \le 0$,
  2. As generated by affine functional of apartments, I proved example of $SL_2$ below .

In both cases, we have $$ (N^+(K) \cap P) \times Z(K)^\circ \times (N^-(K) \cap P) $$ where $N^+(K)$ is the unipotent radical of the minimal parabolic subgroup containing $S$ corresponding to $\Phi^+$.

Then we should understand $N^+ (K) \cap P$. Note then that in general we have an explicit isomorphism $$ \prod_{\alpha \in \Phi^{\pm, nd}} U_{\psi_\alpha} \simeq N^{\pm}(K) \cap P $$ over the indecomposable roots, and $\psi_\alpha$ is the smallest affine functional over the apartment of $S$ which has derivative $\alpha$ that is nonnegtive on the chamber that defines $P$.

Let us now do this explicitly in the case of $SL_2$.

  • The apartment here is $$A:=X_\bullet \otimes_\mathbb{Z} \mathbb{R} \simeq \mathbb{R}$$
  • The affine roots $\Phi:=\left \{ a+ j\,:\, a \in \Phi , j \in Z \right \}$ are all on the functionals on $A$. Note that the roots here $\Phi =\{ \pm 1 \}$.
  • For the affine root $\psi:= a+j$, we can introduce a function $U_{\psi}$ defines a root subgroup $U_a(\varpi^j O)$, an embedding $\varpi^j O$ in to the $a$-th root position.

Now $$N^+(K) \cap P \simeq U_{\psi_{\lambda^\vee}} \cdot \prod_{\Phi^{+,nd} \backslash \lambda^\vee } U_{\psi_\alpha}$$ I leave it as an exercise to argue that by factoring $U_{\psi_{\lambda^\vee}}$ out in the decomposition the remaining object lives in $P^{\ge \lambda}:= P \cap \varpi^\lambda G(O)\varpi^{-\lambda} $

footnote: Another (equivalent) definition: $P=P_x$ the maximal parabolic associated to a point $x \in A$, (the maximal ones are precisely those where $(y,x) \in Z$ for all $y \in \Phi$). It is the subgroup of $G(O)$ generated by $T(O)$ and $U_y(\varpi^{-\lfloor y,x \rfloor})$ for $y \in \Phi$. (The example you given is $x =2 \in A$. )

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