2
$\begingroup$

Let $G$ be a semi-simple algebraic group over $Q_p$ and $K$ in $G(Q_p)$ a maximal compact open subgroup. Let $\tilde{\pi}\colon \tilde{G}\rightarrow G$ be the simply connected cover. Then $\tilde{\pi}^{-1}(K)$ is a compact open subgroup of $\tilde{G}(Q_p)$. Is it necessarily maximal?

$\endgroup$
6
$\begingroup$

The answer is negative even for ${\rm{PGL}}_2$: the stabilizer of an edge in the building is a counterexample (with Iwahori preimage in ${\rm{SL}}_2(F)$). It likewise fails for ${\rm{PGL}}_n$ for every $n > 2$ by using stabilizers of chambers in the building.

In the work of Bruhat-Tits, analyzing maximal compact open subgroups of $G(F)$ (for connected semisimple $G$ over a non-archimedean local field $F$, say) is "easiest" when $G$ is split and simply connected. To understand more general split cases without the simply connected property involves additional work precisely to deal with the failure of surjectivity of $\widetilde{G}(F) \rightarrow G(F)$. To understand the origin of counterexamples with $G = {\rm{PGL}}_n$ (for $n > 1$), it is convenient to first carry out some general considerations (or jump ahead to the 2nd to last paragraph below if you are impatient).

Let $\pi: {\rm{SL}}_n \rightarrow {\rm{PGL}}_n$ be the natural central quotient map over $O_F$. The preimage in ${\rm{SL}}_n(F)$ of the (hyperspecial) maximal compact open subgroup $K_0 := {\rm{PGL}}_n(O_F)$ of ${\rm{PGL}}_n(F)$ is clearly $K'_0 := {\rm{SL}}_n(O_F)$, a (hyperspecial) maximal compact open subgroup in ${\rm{SL}}_n(F)$. I claim that $K_0$ is the unique maximal compact open subgroup of ${\rm{PGL}}_n(F)$ that contains $\pi(K'_0)$. To prove this directly, we briefly digress to recall an action construction.

The natural ${\rm{PGL}}_n$-action on itself via conjugation uniquely lifts to an action of ${\rm{PGL}}_n = {\rm{GL}}_n/\mathbf{G}_m$ on ${\rm{SL}}_n$ in an evident manner, namely via ${\rm{GL}}_n$-conjugation on ${\rm{SL}}_n$, say denoted $(g,x) \mapsto g.x$. Note that the action of $K'_0 = {\rm{PGL}}_n(O_F)$ on ${\rm{SL}}_n(F)$ preserves $K_0 = {\rm{SL}}_n(O_F)$. Thus, every ${\rm{PGL}}_n(F)$-conjugate of $K_0$ has preimage in ${\rm{SL}}_n(F)$ that is a maximal compact open subgroup, with such preimages given by $g.K'_0$ for $g \in {\rm{PGL}}_n(F)$.

Now consider a maximal compact open subgroup $K$ of ${\rm{PGL}}_n(F) = {\rm{GL}}_n(F)/F^{\times}$ containing $\pi(K_0)$. We will show that $K \subset {\rm{PGL}}_n(O_F) = K'_0$, so $K = K'_0$ as desired. For any $k \in K$, consider the unique lift of $k$-conjugation on ${\rm{PGL}}_n$ to an $F$-automorphism $f_k$ of ${\rm{SL}}_n$. Concretely, this is conjugation in ${\rm{GL}}_n(F)$ against a lift $\widetilde{k} \in {\rm{GL}}_n(F)$ of $k$.

Clearly $\pi(f_k(K'_0)) = k K_0 k^{-1} \subset K$, so $f_k(K'_0) \subset \pi^{-1}(K)$. But $\pi^{-1}(K)$ is a compact open subgroup of ${\rm{SL}}_n(F)$ which contains the maximal compact open subgroup $K'_0$ and hence is equal to $K'_0$, so $f_k(K'_0) \subset K'_0$. Hence, a lift $\widetilde{k} \in {\rm{GL}}_n(F)$ of $k$ normalizes $K'_0 = {\rm{SL}}_n(O_F)$; more precisely, $\widetilde{k} K'_0 \widetilde{k}^{-1} = K'_0$ by volume considerations (using that ${\rm{SL}}_n$ is "unimodular" as an algebraic group over $F$). In terms of the standard $O_F$-lattice $\Lambda = O_F^n$ inside $F^n$, this says that ${\rm{SL}}(\widetilde{k}.\Lambda) = {\rm{SL}}(\Lambda)$. But rather generally, a pair of $O_F$-lattices $\Lambda_1$ and $\Lambda_2$ in $F^n$ satisfies ${\rm{SL}}(\Lambda_1) = {\rm{SL}}(\Lambda_2)$ inside ${\rm{GL}}(F^n) = {\rm{GL}}_n(F)$ if and only if $\Lambda_1$ and $\Lambda_2$ are $F^{\times}$-multiples of each other, so $\widetilde{k} \in F^{\times} {\rm{GL}}(\Lambda) = F^{\times} {\rm{GL}}_n(O_F)$ and hence $k \in {\rm{PGL}}_n(O_F) = K_0$ as desired.

For $g \in {\rm{PGL}}_n(F)$, the ${\rm{SL}}_n(F)$-conjugacy class of $g.K'_0$ only depends on the image of $g$ in the set $$\pi({\rm{SL}}_n(F))\backslash {\rm{PGL}}_n(F)/K_0 = F^{\times}/(F^{\times})^nO_F^{\times}$$ (use det). This set has size $n$, yet (as a special case of Bruhat-Tits theory) ${\rm{SL}}_n(F)$ has $n$ distinct conjugacy classes of maximal compact open subgroups, so we have just built all of them. Hence, if $K$ is a maximal compact open subgroup of ${\rm{PGL}}_n(F)$ whose preimage $K'$ in ${\rm{SL}}_n(F)$ is a maximal compact open subgroup of ${\rm{SL}}_n(F)$ then necessarily $K' = g.K'_0$ for some $g \in {\rm{PGL}}_n(F)$. Thus, such $K$ would have to contain a conjugate of $\pi(K'_0)$. But we showed above that $\pi(K'_0)$ lies in a unique maximal compact open subgroup of ${\rm{PGL}}_n(F)$, namely $K_0$, so $K$ would have to be a conjugate of $K_0$!

The conclusion is that a maximal compact open subgroup $K$ of ${\rm{PGL}}_n(F)$ that is not conjugate to ${\rm{PGL}}_n(O_F)$ has preimage in ${\rm{SL}}_n(F)$ that is not maximal as a compact open subgroup. Any maximal compact open subgroup of the unimodular ${\rm{PGL}}_n(F)$ having volume (relative to a choice of Haar measure) distinct from that of ${\rm{PGL}}_n(O_F)$ is an example of such a $K$. Such $K$ exist for any $n \ge 2$, as stabilizers of a chamber in the building. But since the building can be most readily visualized and defined for $n = 2$, we now focus on describing an explicit counterexample for $n=2$.

Finally, consider ${\rm{PGL}}_2$. Let $K$ be the stabilizer of an edge in the building. Explicitly, this is described as follows. For the lattices $\Lambda = O_F^2$ and $\Lambda' = O_F \oplus \varpi O_F$ (with $\varpi$ a uniformizer of $F$) we choose some $g \in {\rm{GL}}_2(F)$ such that $g(\Lambda') = \Lambda$, so $g(\Lambda)$ corresponds to a line in the plane $\overline{\Lambda} := \varpi^{-1}\Lambda/\Lambda$ over the residue field. By surjectivity of ${\rm{PGL}}_2(\Lambda) \rightarrow {\rm{PGL}}_2(\overline{\Lambda})$, we can adjust the choice of $g$ so that also $g(\Lambda) = \varpi^{-1}O_F \oplus O_F = \varpi^{-1}\Lambda'$. Then $g$ swaps the homothety classes of the lattices $\Lambda$ and $\Lambda'$, so $g^2$ preserves these homothety classes; i.e., $g^2$ lies in the compact open subgroup $$\mathbf{K} := {\rm{PGL}}(\Lambda) \cap {\rm{PGL}}(\Lambda')$$ that is the intersection of the stabilizers of the endpoints of an edge of the building whereas $g \not\in \mathbf{K}$. Clearly $$K = \mathbf{K} \coprod g \mathbf{K} = N_{{\rm{PGL}}_2(F)}(\mathbf{K}).$$ This is exactly the stabilizer of the midpoint of an edge in the building, and that midpoint is its unique fixed point. Since any compact subgroup of ${\rm{PGL}}_2(F)$ must fix some point of the building, it follows that $K$ must be maximal as a compact open subgroup of ${\rm{PGL}}_2(F)$.

Either for volume reasons or fixed-point reasons with the building, $K$ is not conjugate to ${\rm{PGL}}_2(O_F)$, so its preimage in ${\rm{SL}}_2(F)$ is not maximal as a compact open subgroup of ${\rm{SL}}_2(F)$. Hence, by the general considerations above, $\pi^{-1}(K)$ cannot be maximal as a compact open subgroup of ${\rm{SL}}_2(F)$, and this can also be seen by direct calculation: by determinant reasons, $$\pi^{-1}(K) = \pi^{-1}(\mathbf{K}) = {\rm{SL}}_2(O_F) \cap {\rm{SL}}(O_F \oplus \varpi O_F)$$ is the Iwahori subgroup given by the preimage in ${\rm{SL}}_2(O_F)$ of the upper triangular Borel subgroup of ${\rm{SL}}_2(k)$ (with $k$ the residue field of $O_F$).

$\endgroup$
1
$\begingroup$

It seems to me this is a completely general fact. If $\tilde\pi^{-1}(K)$ is contained in a compact open subgroup $K'$, then $\tilde\pi(K')$ is a compact open subgroup containing $K$, thus equal to $K$; this implies $K'=\tilde\pi^{-1}(K)$.

$\endgroup$
  • 1
    $\begingroup$ I do not see why $\pi(K')$ should contain $K$, the map $\pi$ is not surjective on $Q_p$ points.... $\endgroup$ – user42721 Nov 23 '13 at 11:23

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.