On the depth r Bernstein projector

Abstract

In this paper we prove an explicit formula for the Bernstein projector to representations of depth $\le r$. As a consequence, we show that the depth zero Bernstein projector is supported on topologically unipotent elements and it is equal to the restriction of the character of the Steinberg representation. As another application, we deduce that the depth r Bernstein projector is stable. Moreover, for integral depths our proof is purely local.

Appendices

Appendix A. Properties of Moy–Prasad filtrations

In this section we provide proofs of some of the results, formulated in Sections 2 and 3. We are going to follow a standard strategy, first to pass to an unramified extension, thus reducing to a quasi-split case, then to pass to a Levi subgroup, thus reducing to a rank one case, and to finish by direct calculations. Though most of the results in this sections are well-known to specialists (see, for example, [30, Section 1]), we include details for completeness.

1.1 Set-up

Let $\mathbf {S}\subseteq \mathbf {G}$ be a maximal split torus, $\mathbf {M}:=\mathbf {Z}_\mathbf {G}(\mathbf {S})$ the corresponding minimal Levi subgroup of $\mathbf {G}$, set $\mathcal {A}:=\mathcal {A}_{\mathbf {S}}$, and let $\Phi (\mathcal {A})_{nd}\subseteq \Phi (\mathcal {A})$ be the set of non-divisible roots, that is, those $\alpha \in \Phi (\mathcal {A})$ such that $a/2\notin \Phi (\mathcal {A})$.

Lemma A.2

There exists a finite unramified extension $F'{/}F$ such that $\mathbf {G}':=\mathbf {G}_{F'}$ is quasi-split. Moreover, for every such extension, there exists a subtorus $\mathbf {S}'\supseteq \mathbf {S}$ of $\mathbf {G}$ defined over F such that $\mathbf {S}'_{F'}\subseteq \mathbf {G}'$ is a maximal split torus.

Proof

Assume first that $\mathbf {G}=\mathbf {GL}_{1}(D)$ for some finite-dimensional central division algebra D over F. In this case, both assertions are easy. Indeed, let $\dim _F D=d^2$, and let $F'{/}F$ be an unramified extension. Then $\mathbf {G}_{F'}$ is quasi-split if and only if $F'$ splits D. Moreover, this happens if and only if $F'\supseteq F^{(d)}$, where $F^{(d)}{/}F$ is an unramified extension of degree d. Furthermore, there exists an embedding $F^{(d)}\hookrightarrow D$ of F-algebras, whose image corresponds to a torus $\mathbf {S}'$ we are looking for.

Assume next that $\mathbf {G}=\mathbf {GL}_{1}(D)$ for some (not necessary central) finite-dimensional division algebra D over F. This case reduces to the first one, and is left to the reader.

Finally, the general case follows from the previous one. Indeed, $\mathbf {G}_{F'}$ is quasi-split if and only if $\mathbf {M}_{F'}$ is quasi-split, and if and only if the simply connected covering $\mathbf {M}_{F'}^{\mathrm{sc}}$ of $\mathbf {M}_{F'}$ is quasi-split. Thus we may replace $\mathbf {G}$ by $\mathbf {M}^{\mathrm{sc}}$, thus assuming that $\mathbf {G}$ is semisimple, simply-connected, and anisotropic. Next, decomposing $\mathbf {G}$ into simple factors, we may further assume that $\mathbf {G}$ is simple. Then $\mathbf {G}={\mathbf {SL}}_{\mathbf{1}}(D)$ for some finite-dimensional division algebra over F, and ${\mathbf {SL}}_{\mathbf{1}}$ denotes the kernel of the reduced norm (see [26, Thm 6.5. p. 285]). Since the assertion for ${\mathbf {SL}}_{\mathbf{1}}(D)$ follows from the assertion for $\mathbf {GL}_{1}(D)$, the proof is now complete. $\square$

1.2 Affine roots subgroups

(a) Choose a set of positive roots $\Phi (\mathcal {A})^+_{nd}\subseteq \Phi (\mathcal {A})_{nd}$, and a total order on $\Phi (\mathcal {A})_{nd}\cup \{0\}$ such that $\alpha >0$ if and only if $\alpha \in \Phi (\mathcal {A})^+_{nd}$. Set $\mathbf {U}_0:=\mathbf {M}$. Then the product map $\prod _{\alpha \in \Phi (\mathcal {A})_{nd}\cup \{0\}}\mathbf {U}_{\alpha }\rightarrow \mathbf {G}$ is an open embedding.

(b) For every $\alpha \in \Phi (\mathcal {A})$, $x\in \mathcal {A}$ and $r\in \mathbb R_{\ge 0}$, we denote by $\psi _{\alpha ,x,r}$ the smallest affine root $\psi \in \Psi (\mathcal {A})$ such that $\alpha _{\psi }=\alpha$ and $\psi (x)\ge r$. Set $U_{\alpha ,x,r}:=U_{\psi _{\alpha ,x,r}}\subseteq U_{\alpha }$ and $\mathfrak {u}_{\alpha ,x,r}:=\mathfrak {u}_{\psi _{\alpha ,x,r}}\subseteq \mathfrak {u}_{\alpha }$.

(c) We also set $U_{(\alpha ),x,r}:=U_{\alpha ,x,r}\cdot U_{2\alpha ,x,r}\subseteq U_{\alpha }$, if $2\alpha \in \Phi (\mathcal {A})$; $U_{(\alpha ),x,r}:=U_{\alpha ,x,r}$, if $2\alpha \notin \Phi (\mathcal {A})$; and $U_{0,x,r}:=M_r$.

1.3 The ${\mathbf {SL}}_{\mathbf{2}}$-case

Let $\mathbf {G}={\mathbf {SL}}_{\mathbf{2}}$, and let $\mathbf {S}\subseteq \mathbf {G}$ be the group of diagonal matrices. In this case, $\mathbf {G}$ and $\mathbf {S}$ have natural $\mathcal {O}$-structures, hence we have a natural identification $\mathcal {A}\overset{\thicksim }{\rightarrow }V_{\mathbf {G},\mathbf {S}}$ (see 2.9(a)), which identifies $\Phi (\mathcal {A})$ with $\pm \alpha$ and $\Psi (\mathcal {A})$ with $\pm \alpha +\mathbb Z$. Moreover, if the root subgroup $U_{\alpha }$ consists of matrices $\small {g_a=\left( \begin{matrix} 1 &{} a \\ 0&{} 1\end{matrix}\right) }$ with $a\in F$, then the affine root subgroup $U_{\alpha +n}\subseteq U_{\alpha }$ consists of $g_a\in U_{\alpha }$ with $\mathrm{val}_F(a)\ge n$.

1.4 The ${\mathbf {SU}}_{\mathbf{3}}$-case (compare [29, Ex. 1.15])

(a) Let K / F be a separable totally ramified quadratic extension, and let $\tau \in \mathrm{Gal}(K{/}F)$ be a non-trivial element. Let $\mathbf {G}={\mathbf {SU}}_{\mathbf{3}}$ be the special unitary group over F split over K, corresponding to the quadratic form $(\overline{x},\overline{y})\mapsto \sum _i x_i y^{\tau }_{3-i}$. Let $\mathbf {S}\subseteq \mathbf {G}$ the maximal torus, corresponding to diagonal matrices, and let $\alpha \in \Phi (\mathbf {G},\mathbf {S})$ be the non-divisible root such that $U_{\alpha }$ consists of upper triangular matrices. Then $U_{\alpha }$ consists of all elements of the form $\small {g_{a,b}= \left( \begin{matrix} 1 &{} -a &{} -b\\ 0&{} 1 &{} a^{\tau }\\ 0&{} 0&{} 1 \end{matrix}\right) , a,b\in K}$ such that $a a^{\tau }+b+b^{\tau }=0,$ while $U_{2\alpha }$ consists of all $g_{0,b}\in U_{\alpha }$.

(b) Set $\delta :=\max \{\mathrm{val}_K(b)|b+b^{\tau }+1=0\}$. Then $\delta \le 0$, and $\delta =0$ if and only if $p\ne 2$. For every $g_{a,b}\in U_{\alpha }$, we have $\mathrm{val}_K(b)\le 2\mathrm{val}_K(a)+\delta$, and for every $a\in K^{\times }$ there exists $g_{a,b}\in U_{\alpha }$ with $\mathrm{val}_K(b)= 2\mathrm{val}_K(a)+\delta$. On the other hand, as it was explained in [29, Ex. 1.15], for every $g_{0,b}\in U_{2\alpha }$, we have $\mathrm{val}_K(b)\in 2\mathbb Z+\delta +1$.

(c) Using the identification $\mathcal {A}\overset{\thicksim }{\rightarrow }V_{\mathbf {G},\mathbf {S}}$ corresponding to the standard $\mathcal {O}$-structure of $\mathbf {G}$ and $\mathbf {S}$ (see 2.9(a)), we identify the set of affine roots $\Psi (\mathcal {A})$ with the set

$$\begin{aligned} (\pm \alpha +\frac{1}{4}(2\mathbb Z+\delta ))\cup (\pm 2\alpha +\frac{1}{2}(2\mathbb Z+\delta +1)), \end{aligned}$$

where we divide by an extra 2, because our normalization uses valuation $\mathrm{val}_F=\frac{1}{2}\mathrm{val}_K$.

(d) In the notation of (c), for $\psi :=\alpha +\frac{1}{4}(2n+\delta )$, the subgroup $U_{\psi }$ consists of all $g_{a,b}\in U_{\alpha }$ such that $\mathrm{val}_K(b)\ge 2n+\delta$, while for $\psi :=2\alpha +\frac{1}{2}(2n+\delta +1)$ the subgroup $U_{\psi }$ consists of $g_{0,b}\in U_{\alpha }$ such that $\mathrm{val}_K(b)\ge 2n+\delta +1$.

(e) Using (d), for every $x\in \mathcal {A}$ and $r\in \mathbb R_{\ge 0}$, the subgroup $U_{\alpha ,x,r}$ consists of $g_{a,b}\in U_{\alpha }$ such that $\mathrm{val}_K(b)\ge 4r-4\alpha (x)$, while the subgroup $U_{2\alpha ,x,r}$ consists of $g_{0,b}\in U_{\alpha }$ such that $\mathrm{val}_K(b)\ge 2r-4\alpha (x)$. In particular, we have $U_{\alpha ,x,r}\cap U_{2\alpha ,x,r}=U_{2\alpha ,x,2r}$.

(f) We claim that an element $g_{a,b}\in U_{\alpha }$ belongs to $U_{(\alpha ),x,r}$ if and only if we have inequalities $\mathrm{val}_K(a)\ge 2r-2\alpha (x)-\frac{1}{2}\delta$ and $\mathrm{val}_K(b)\ge 2r-4\alpha (x)$.

By definition, $U_{(\alpha ),x,r}$ consists of elements of the form $g_{a,b'+b''}=g_{a,b'}\cdot g_{0,b''}$ such that $g_{a,b'}\in U_{\alpha ,x,r}$ and $g_{0,b''}\in U_{2\alpha ,x,r}$. In particular, we have $\mathrm{val}_K(b'')\ge 2r-4\alpha (x)$, and $\mathrm{val}_K(b')\ge 4r-4\alpha (x)$ (by (d)), hence $2\mathrm{val}_K(a)\ge \mathrm{val}_{K}(b')-\delta \ge 4r-4\alpha (x)-\delta$ (by (b)) and $\mathrm{val}_K(b'+b'')\ge \min \{\mathrm{val}_K(b'),\mathrm{val}_K(b'')\}\ge 2r-4\alpha (x)$.

Conversely, assume that an element $g_{a,b}\in U_{\alpha }$ satisfies $\mathrm{val}_K(a)\ge 2r-2\alpha (x)-\frac{1}{2}\delta$ and $\mathrm{val}_K(b)\ge 2r-4\alpha (x)$. Choose $g_{a,b'}\in U_{\alpha }$ with $\mathrm{val}_K(b')= 2\mathrm{val}_K(a)+\delta$, and set $b'':=b-b'$. Then $\mathrm{val}_K(b')\ge 4r-4\alpha (x)$ and $\mathrm{val}_K(b'')\ge \min \{\mathrm{val}_K(b),\mathrm{val}_K(b')\}\ge 2r-4\alpha (x)$. Thus $g_{a,b'}\in U_{\alpha ,x,r}$ and $g_{0,b''}\in U_{2\alpha ,x,r}$, hence $g_{a,b'+b''}\in U_{(\alpha ),x,r}$.

1.5 Levi subgroups

Let $\mathbf {L}\supseteq \mathbf {S}$ be a Levi subgroup of $\mathbf {G}$, and set $\mathcal {A}_{\mathbf {L}}:=\mathcal {A}_{\mathbf {L},\mathbf {S}}$.

(a) We have a natural projection $\mathrm{pr}_\mathbf {L}:\mathcal {A}\rightarrow \mathcal {A}_{\mathbf {L}}$ of affine spaces, compatible with the projection $V_{\mathbf {G},\mathbf {S}}\rightarrow V_{\mathbf {L},\mathbf {S}}$ of vector spaces (see [21, 1.10 and 1.11]).

(b) We have an inclusion $\Phi (\mathcal {A}_{\mathbf {L}})\subseteq \Phi (\mathcal {A})$, and every affine root $\psi \in \Psi (\mathcal {A})$ such that $\alpha _{\psi }\in \Phi (\mathcal {A}_{\mathbf {L}})$ induces an affine function $\psi _{\mathbf {L}}$ on $\mathcal {A}_{\mathbf {L}}$, which belongs to $\Psi (\mathcal {A}_{\mathbf {L}})$. Moreover, the correspondence $\psi \mapsto \psi _{\mathbf {L}}$ induces a bijection between the set of affine roots $\psi \in \Psi (\mathcal {A})$ such that $\alpha _{\psi }\in \Phi (\mathcal {A}_\mathbf {L})$ and the set $\Psi (\mathcal {A}_{\mathbf {L}})$.

(c) By definition, for every $\psi \in \Psi (\mathcal {A})$ such that $\alpha _{\psi }\in \Phi (\mathcal {A}_{\mathbf {L}})\subseteq \Phi (\mathcal {A})$, the affine root subgroup $U_{\psi }\subseteq U_{\alpha _{\psi }}$ equals $U_{\psi _{\mathbf {L}}}$.

(d) By (b) and (c), for every $\alpha \in \Phi (\mathcal {A}_{\mathbf {L}})\subseteq \Phi (\mathcal {A})$, $x\in \mathcal {A}$ and $r\in \mathbb R_{\ge 0}$, the affine root subgroup $U_{\alpha , x,r}\subseteq U_{\alpha }\subseteq G$ equals $U_{\alpha , \mathrm{pr}_{\mathbf {L}}(x),r}\subseteq U_{\alpha }\subseteq L$.

(e) For every $\alpha \in \Phi (\mathcal {A})\subseteq X^*(\mathbf {S})$, let $\mathbf {S}_{\alpha }$ be the connected component $(\mathrm{Ker}\alpha )^0$ and set $\mathbf {L}_{\alpha }:=\mathbf {Z}_\mathbf {G}(\mathbf {S}_{\alpha })$. Then $\mathbf {L}_{\alpha }$ is a Levi subgroup of $\mathbf {G}$ semisimple rank one, thus $\mathbf {L}^{\mathrm{sc}}_{\alpha }$ is isomorphic either to $\mathrm{R}_{F'{/}F}{\mathbf {SL}}_{\mathbf{2}}$ or to $\mathrm{R}_{F'{/}F}{\mathbf {SU}}_{\mathbf{3}}$ for some finite separable extension $F'{/}F$.

1.6 Weil restriction of scalars

Let $F'{/}F$ be a finite separable extension of ramification degree e, and set $\mathbf {G}':=\mathrm{R}_{F'{/}F}\mathbf {G}$. Then we have natural identifications $\mathbf {G}'(F)\cong \mathbf {G}(F')$ and $\mathcal {X}(\mathbf {G}')\cong \mathcal {X}(\mathbf {G})$. Moreover, since $\mathrm{val}_{F'}=e\mathrm{val}_{F}$, for every $x\in \mathcal {X}(\mathbf {G}')\cong \mathcal {X}(\mathbf {G})$ and $r\in \mathbb R_{\ge 0}$ the isomorphism $\mathbf {G}'(F)\cong \mathbf {G}(F')$ induces an isomorphism $G'_{x,r}\cong G_{x,er}$.

1.7 The unramified descent

(a) Let $F'{/}F$, $\mathbf {G}':=\mathbf {G}_{F'}$ and $\mathbf {S}'\subseteq \mathbf {S}$ be as in Lemma A.2. Let $\mathcal {A}'\subseteq \mathcal {X}(\mathbf {G}')$ be the apartment corresponding to $\mathbf {S}'_{F'}\subseteq \mathbf {G}'$, and set $\Gamma ':=\mathrm{Gal}(F'{/}F)$. Then $\mathcal {A}'$ is equipped with an action of $\Gamma '$, and we have a natural identification $\mathcal {A}\overset{\thicksim }{\rightarrow }\mathcal {A}'^{\Gamma '}$.

(b) Note that for $\alpha \in \Phi (\mathcal {A})$, the root group $\mathbf {U}'_{\alpha }:=(\mathbf {U}_{\alpha })_{F'}$ equals the product $\prod _{\alpha '}\mathbf {U}'_{\alpha '}$, where $\alpha '$ runs over the union of all $\alpha '\in \Phi (\mathcal {A}')$ such that $\alpha '|_{\mathcal {A}}=\alpha$ and all $\alpha '\in \Phi (\mathcal {A}')_{nd}$ such that $\alpha '|_{\mathcal {A}}=2\alpha$ (compare [7, 21.9]).

(c) Moreover, for every $x\in \mathcal {A}$ and $r\in \mathbb R_{\ge 0}$, the affine root subgroup $U_{\alpha ,x,r}\subseteq U_{\alpha }$ equals $U_{\alpha ,x,r}=(U'_{\alpha ,x,r})^{\Gamma '}$, where $U'_{\alpha ,x,r}\subseteq U'_{\alpha }$ is the product

$$\begin{aligned} \left( \prod _{\alpha '\in \Phi (\mathcal {A}'),\alpha '|_{\mathcal {A}}=\alpha } U'_{\alpha ',x,r}\right) \times \left( \prod _{\alpha '\in \Phi (\mathcal {A}')_{nd},\alpha '|_{\mathcal {A}}=2\alpha } U'_{\alpha ',x,2r}\right) , \end{aligned}$$

taken in every order (use, for example, [21, 10.19 and 11.5]).

(d) For every triple $(\alpha ,x,r)$ as in (b),(c) such that $2\alpha \in \Phi (\mathcal {A})$, we have the equality $U_{\alpha ,x,r}\cap U_{2\alpha ,x,r}=U_{2\alpha ,x,2r}$. Indeed, by (c), it suffices to show that $U'_{\alpha ,x,r}\cap U'_{2\alpha ,x,r}=U'_{2\alpha ,x,2r}$, which reduces to the equality $U'_{\alpha ',x,r}\cap U'_{2\alpha ',x,r}=U'_{2\alpha ',x,2r}$ for every $\alpha '\in \Phi (\mathcal {A}')$ such that $2\alpha '\in \Phi (\mathcal {A}')$. Enlarging $F'$, if necessary, we may assume that $\mathbf {G}'$ splits over a totally ramified extension. Using A.6(d),(e) and A.7, we reduce to the case $\mathbf {G}={\mathbf {SU}}_{\mathbf{3}}$, in which case the assertion was shown in A.5(e).

(e) We set $U'_{(\alpha ),x,r}:=U'_{\alpha ,x,r}\cdot U'_{2\alpha ,x,r}\subseteq U'_{\alpha }$, if $2\alpha \in \Phi (\mathcal {A})$, and $U'_{(\alpha ),x,r}:=U'_{\alpha ,x,r}$, otherwise. We claim that $U_{(\alpha ),x,r}=(U'_{(\alpha ),x,r})^{\Gamma '}$. If $2\alpha \notin \Phi (\mathcal {A})$, this follows from (c). If $2\alpha \in \Phi (\mathcal {A})$, we have to show that $(U'_{\alpha ,x,r}\cdot U'_{2\alpha ,x,r})^{\Gamma '}=(U'_{\alpha ,x,r})^{\Gamma '}\cdot (U'_{2\alpha ,x,r})^{\Gamma '}$. Since $U'_{\alpha ,x,r}\cap U'_{2\alpha ,x,r}=U'_{2\alpha ,x,2r}$ by (d), it suffices to show that $H^1(\Gamma ',U'_{2\alpha ,x,2r})=0$. Using Shapiro’s lemma, the assertion reduces to the vanishing of $H^1(\Gamma ',\mathcal {O}_{F'})$, which follows from the additive Hilbert 90 theorem.

(f) For every two triples $(\alpha ,x,r)$ and $(\alpha ,y,s)$ as in (b),(c) such that $2\alpha \in \Phi (\mathcal {A})$ we have $U_{(\alpha ),x,r}\cap U_{(\alpha ),y,s}=(U_{\alpha ,x,r}\cap U_{\alpha ,y,s})\cdot (U_{2\alpha ,x,r}\cap U_{2\alpha ,y,s})$.

Indeed, using (c)-(e) and arguing as in (e), we reduce the assertion to the corresponding equality of the $U'$’s. Then using A.6(d),(e) and A.7 we reduce to the case $\mathbf {G}={\mathbf {SU}}_{\mathbf{3}}$, in which case we finish by precisely the same arguments as A.5(f).

1.8 Applications

(a) Each $\mathfrak {u}_{\psi }\subseteq \mathfrak {u}_{\alpha }$ is an $\mathcal {O}$-lattice (see 2.3(c)). Indeed, by A.8(c), we reduce to the case when $\mathbf {G}$ is quasi-split and split over a totally ramified extension. Then using A.6(d),(e) and A.7, we reduce to the absolute rank one case, in which case the assertion follows from formulas of A.4 and A.5.

(b) For every $x\in \mathcal {A}$, $r\in \mathbb R_{\ge 0}$ and $n\in \mathbb N$, we have the equality $\varpi ^n\mathfrak {g}_{x,r}=\mathfrak {g}_{x,r+n}$. Again, this can be shown by the same strategy as in (a).

(c) For every $\psi \in \Psi (\mathcal {A})$ there exists a positive integer $n_{\psi }$ such that the set of $\psi '\in \Psi (\mathcal {A})$ with $\alpha _{\psi '}=\alpha _{\psi }$ equals $\psi +\frac{1}{n_{\psi }}\mathbb Z$ (see 2.7(a)). Again, we reduce to the absolute rank one case as in (a) and use the explicit formulas from A.4 and A.5.

Lemma A.10

(a) In the situation of Proposition 3.10, the subalgebra $\mathfrak {g}_{x,r}$ decomposes as a direct sum $\mathfrak {g}_{x,r}=\mathfrak {m}_r\oplus \prod _{\alpha \in \Phi (\mathcal {A})} \mathfrak {u}_{\alpha ,x,r}$.

(b) Assume in addition that either $r>0$ or x lies in a chamber of $[\mathcal {X}]$. For every order of $\Phi (\mathcal {A})_{nd}\cup \{0\}$ as in A.3(a), the product map $\prod _{\alpha \in \Phi (\mathcal {A})_{nd}\cup \{0\}} U_{(\alpha ),x,r}\rightarrow G_{x,r}$ is bijective.

Remark A.11

Actually, the map in (b) is bijective for every order of $\Phi (\mathcal {A})_{nd}\cup \{0\}$.

Proof

We show only (b), while the proof of (a) is similar but much easier.

Since $U_{(\alpha ),x,r}\subseteq U_{\alpha }$ for all $\alpha$, the injectivity follows from A.3(a). To show the surjectivity, assume first that $\mathbf {G}$ is quasi-split. In this case, the argument is standard (compare [27, 2.9]), and can be carried out as follows.

Let $Y\subseteq G_{x,r}$ be the image of the product map. Since Y is closed, and $\{G_{x,s}\}_{s\ge 0}$ form a basis of open neighbourhoods, it remains to show that $G_{x,r}\subseteq Y\cdot G_{x,s}$ for every $s\ge r$. For this it suffices to show that $Y\cdot G_{x,s}\subseteq Y\cdot G_{x,s^+}$ for every $s\ge r$. Since $G_{x,s}$ is generated by subgroups $U_{(\alpha ),x,s}$, it remains to show that $Y\cdot U_{(\alpha ),x,s}\subseteq Y\cdot G_{x,s^+}$.

If $s>0$, this follows from the inclusion $(G_{x,r},G_{x,s})\subseteq G_{x,r+s}$ (use [27, 2.4 and 2.7]). If $s=r=0$, and $\alpha =0$, this follows from the fact that $M_r$ normalizes each $U_{(\alpha ),x,r}$. If $\alpha \ne 0$, then $U_{(\alpha ),x,0}=U_{(\alpha ),x,s}$ for some $s>0$, because x belongs to a chamber, and the assertion is immediate.

For an arbitrary $\mathbf {G}$, let $F'{/}F$ and $\mathbf {G}'$ be as in 3.5(b). Note that the embedding $\mathcal {X}(\mathbf {G})\hookrightarrow \mathcal {X}(\mathbf {G}')$ maps chambers into chambers. Set $U'_{(0),x,r}:=M'_{x_{\mathbf {M}},r}$. As it was already shown, the assertion holds for $G'_{x,r}$ and $M'_{x_{\mathbf {M}},r}$. This implies that the product $\prod _{\alpha \in \Phi (\mathcal {A})_{nd}\cup \{0\}}U'_{(\alpha ),x,r}\rightarrow G'_{x,r}$ is bijective. Now the assertion follows from equalities $G_{x,r}=(G'_{x,r})^{\Gamma '}$, $M_{r}=(M'_{x_{\mathbf {M}},r})^{\Gamma '}$, which were our definitions, and $U_{(\alpha ),x,r}=(U'_{(\alpha ),x,r})^{\Gamma '}$ for all $\alpha \in \Phi (\mathcal {A})_{nd}$ (see A.8(d)). $\square$

Corollary A.12

Let (x, r) be as in Lemma A.10(b), $y\in \mathcal {A}$ and $s\in \mathbb R_{\ge 0}$. Then

(a) For every order of $\Phi (\mathcal {A})_{nd}\cup \{0\}$ as in A.3(a), the product map

$$\begin{aligned} \prod _{\alpha \in \Phi (\mathcal {A})_{nd}\cup \{0\}} (U_{(\alpha ),x,r}\cap U_{(\alpha ),y,s})\rightarrow G_{x,r}\cap G_{y,s} \end{aligned}$$

is bijective.

(b) The subgroup $G_{x,r}\cap G_{y,s}$ is generated by $M_{\max \{r,s\}}$ and the affine root subgroups $U_{\psi }$, where $\psi$ runs over all elements of $\Psi (\mathcal {A})$ such that $\psi (x)\ge r$ and $\psi (y)\ge s$.

Proof

(a) It follows from Lemma A.10 that the product map is injective and that every $g\in G_{x,r}\cap G_{y,s}$ uniquely decomposes as $g=\prod _{\alpha } g_{\alpha }$ such that $g_{\alpha }\in U_{(\alpha ),x,r}$. It remains to show that $g_{\alpha }\in U_{(\alpha ),y,s}$ for all $\alpha$.

If (y, s) also satisfies the assumption of Lemma A.10(b), the assertion follows from Lemma A.10 together with the observation that the product map $\prod _{\alpha }\mathbf {U}_{\alpha }\rightarrow \mathbf {G}$ is injective. Thus we may assume that $s=0$.

If $r=0$, then, by our assumption, x lies in a chamber of $[\mathcal {A}]$. Then every $y'\in [x,y)$, close enough to y, lies in a chamber $\sigma$ such that $y\in \mathrm{cl}(\sigma )$. Then $g\in G_x\cap G_y\subseteq G_{y'}$ (by Lemma 3.11), thus $g\in G_x\cap G_{y'}$. Thus, by the previous case, $g_{\alpha }\in U_{(\alpha ),y',0}\subseteq U_{(\alpha ),y,0}$.

Finally, if $r>0$, then there exists a point $x'\in \mathcal {A}$, lying in a chamber of $[\mathcal {A}]$ such that $G_{x,r}\subseteq G_{x'}$. Thus, $g\in G_{x'}\cap G_{y}$, hence $g_{\alpha }\in U_{(\alpha ),y,0}$ by the $r=0$ case.

(b) The assertion (b) follows from (a) and A.8(f). $\square$

A.13. Proof of Proposition 3.10. Lemma A.10 implies all the cases, except the case of $G_x$, which is not Iwahori. To show the remaining case (which is not used in this work), note that $G_x$ is generated by its Iwahori subgroups $G_y$, where y lies in a chamber $\sigma \subseteq \mathcal {A}$ such that $x\in \mathrm{cl}(\sigma )$. Since each $G_y$ is generated by $T_0$ and $U_{\psi }$ with $\psi (y)\ge 0$ by Lemma A.10(b), and inequality $\psi (y)\ge 0$ implies $\psi (x)\ge 0$, the assertion for $G_x$ follows as well.$\square$

A.14. Completion of the proof of Lemma 3.11. As indicated in 3.12, it remains to show that for every $x,y\in \mathcal {X}$ and $z\in [x,y]$, we have $G_{x,r}\cap G_{y,r}\subseteq G_{z,r}$ for $r>0$ and $\mathfrak {g}_{x,r}\cap \mathfrak {g}_{y,r}\subseteq \mathfrak {g}_{z,r}$ for $r\ge 0$. Choose an apartment $\mathcal {A}$ of $\mathcal {X}$ such that $x,y\in \mathcal {A}$.

By Corollary A.12(b), to show that $G_{x,r}\cap G_{y,r}\subseteq G_{z,r}$ for $r>0$ it suffices to show that for every $\psi \in \Psi (\mathcal {A})$ such that $\alpha (x)\ge r$ and $\alpha (y)\ge r$ we have $\alpha (z)\ge r$. But this follows from the assumption $z\in [x,y]$. The proof of the inclusion $\mathfrak {g}_{x,r}\cap \mathfrak {g}_{y,r}\subseteq \mathfrak {g}_{z,r}$ is similar, but easier.$\square$

A.15. Proof of Lemma 3.14. We show the assertion for $G_{x,r}$, while the assertion for $\mathfrak {g}_{x,r}$ is similar but easier. For $r=0$, the assertion follows from the 3.6(d).

Assume now that $r>0$. Enlarging $F^{\flat }$ if necessary, we can assume that $\mathbf {G}^{\flat }$ is split. Next, replacing F by a finite unramified extension and using 3.7(a), we can assume that $\mathbf {G}$ is quasi-split. Then, using Lemma A.10, it remains to show the corresponding equality for tori $T_r=T^0\cap T^{\flat }_{re}$, which was our definition, and a similar equality for each affine root subgroup $U_{(\alpha ),x,r}$.

Finally, using A.6(d), (e) and A.7, we reduce to the case of ${\mathbf {SL}}_{\mathbf{2}}$ and ${\mathbf {SU}}_{\mathbf{3}}$, which follow from formulas in A.4 and A.5(f), respectively.$\square$

Remark A.16

The formula of A.5(f) also implies that the conclusion of Lemma 3.14 is false, if $\mathbf {G}$ is ${\mathbf {SU}}_{\mathbf{3}}$, split over a wildly ramified quadratic extension.

Appendix B. Congruence subsets

Notation B.1

For every $r\in \mathbb R_{\ge 0}$, we set $G_r:=\cup _{x\in \mathcal {X}}G_{x,r}\subseteq G$ and $\mathfrak {g}_r:=\cup _{x\in \mathcal {X}}\mathfrak {g}_{x,r}\subseteq \mathfrak {g}$. By construction, both $G_r\subseteq G$ and $\mathfrak {g}_r\subseteq \mathfrak {g}$ are open and $\mathrm{Ad}G$-invariant. Moreover, we have $G_{r^+}=\cup _{s>r}G_s\subseteq G_r$ and $\mathfrak {g}_{r^+}=\cup _{s>r}\mathfrak {g}_s\subseteq \mathfrak {g}_r$.

1.1 Remark

(a) The set of $r\in \mathbb R_{\ge 0}$ such that $G_{r^+}\ne G_r$ (resp. $\mathfrak {g}_{r^+}\ne \mathfrak {g}_r$) is discrete. For example, this follows from the fact that any such r is optimal in the sense of [2, 2.3]. Alternatively, this can be seen as follows.

Choose any r such that $G_{r^+}\ne G_r$, and choose a chamber $\sigma \in [\mathcal {X}]$. Since all chambers are G-conjugate, there exists $x\in \mathrm{cl}(\sigma )$ such that $G_{x,r}\ne G_{x,r^+}$. Choose $k\in \mathbb Z$ such that $r\in (k,k+1]$. It thus suffices to show that the set of subgroups $\{G_{x,s}\}_{x\in \mathrm{cl}(\sigma ),s\in (k,k+1]}$ is finite.

Choose an apartment $\mathcal {A}\subseteq \mathcal {X}$ containing $\sigma$, and fix $x\in \mathrm{cl}(\sigma )$ and $s\in (k,k+1]$. Then the set $\{\psi \in \Psi (\mathcal {A})\,|\,\psi (x)\ge s\}$ contains the set $\{\psi \in \Psi (\mathcal {A})\,|\,\psi (\sigma )> k+1\}$ and is contained in the set $\{\psi \in \Psi (\mathcal {A})\,|\,\psi (\sigma )> k\}$. This implies the assertion.

(b) It can be shown that every r from (a) is rational. But even without this fact it follows from (a) that for every $r\in \mathbb R_{\ge 0}$, there exist $r',r''\in \mathbb Q_{\ge 0}$ such that $G_r=G_{r'}$ and $G_{r^+}=G_{r''}$, and similarly for $\mathfrak {g}$. Thus Lemma 8.5 follows from the following assertion.

Lemma B.3

For every $r\in \mathbb R_{\ge 0}$, the subsets $G_{r}\subseteq G$, $\mathfrak {g}_{r}\subseteq \mathfrak {g}$ and $\mathfrak {g}^*_{-r}\subseteq \mathfrak {g}^*$ are open, closed and stable.

Remark B.4

Under some mild restriction on the residual characteristic of F one can show a more precise result (with a simpler proof) asserting that $G_{r}$ (resp. $\mathfrak {g}_{r}$, resp. $\mathfrak {g}^*_{-r}$) is equal to the full preimage of a certain open and compact subset of the corresponding Chevalley space.

Proof

First we show that $G_0\subseteq G$ is closed. By 3.6, the subgroup $G^0\subseteq G$ is closed, and $G_0=\cup _{x\in \mathcal {X}}\mathrm{Stab}_{G^0}G(x)$. Then, by the Bruhat–Tits fixed point theorem, $G_0$ coincides with the set of all compact elements of $G^0$. But the set of all compact elements of G is closed. Indeed, choose a faithful representation $\rho :\mathbf {G}\hookrightarrow \mathbf {GL}_n$, and notice that $g\in G$ is compact if and only if $\det \rho (g)\in \mathcal {O}^{\times }$ and the characteristic polynomial of $\rho (g)$ has coefficients in $\mathcal {O}$.

Next we show that $G_r\subseteq G$ is closed for $r>0$. Since $G_0=\cup _{x\in \mathcal {X}} G_{x}\subseteq G$ is closed, and each $G_x$ is open and compact, it remains to show that for every $x\in \mathcal {X}$, the intersection $G_x\cap G_r$ is compact. By B.2(b), we may assume that $r\in \mathbb Q$, hence $r\in \frac{1}{m}\mathbb Z_{\ge 0}$ for some $m\in \mathbb N$. As in 6.7, for every $\Sigma \in \Theta _m$, we set $G_{\Sigma ,r}:=\cup _{\sigma \in \Sigma }G_{\sigma ,r}$. Then each $G_x\cap G_{\Sigma ,r}$ is compact, and it suffice to show that $G_x\cap G_r= G_x\cap G_{\Sigma ,r}$ for every $\Sigma \supseteq \Upsilon _{x,r}$. Equivalently, it suffices to show the equality of functions $1_{G_x}\cdot 1_{G_{\Sigma ,r}}=1_{G_x}\cdot 1_{G_{\Sigma ',r}}$ for every $\Sigma ',\Sigma \in \Theta _m$ such that $\Upsilon _{x,r}\subseteq \Sigma '\subseteq \Sigma$.

As in Lemma 6.8, we deduce from Lemma 3.11 that for every $\Sigma \in \Theta _m$ we have $1_{G_{\Sigma ,r}}=\sum _{\sigma \in \Sigma }(-1)^{\dim \sigma }1_{G_{\sigma ,r}}$. Thus we have to show that for every $\Sigma '\subseteq \Sigma$ as above, we have $\sum _{\sigma \in \Sigma \smallsetminus \Sigma '}(-1)^{\dim \sigma }(1_{G_x}\cdot 1_{G_{\sigma ,r}})=0$. Arguing as in Proposition 4.14(a), it remains to show that for every $\sigma ,\sigma '\in [\mathcal {X}_m]$ with $\sigma '\preceq \sigma$ and $\sigma \in \Gamma _r(\sigma ',x)$ we have the equality $1_{G_x}\cdot 1_{G_{\sigma ,r}}=1_{G_x}\cdot 1_{G_{\sigma ',r}}$. Equivalently, we have to show that $G_x\cap G_{\sigma ,r}=G_x\cap G_{\sigma ',r}$, that is, $G_x\cap G_{\sigma ',r}\subseteq G_{\sigma ,r}$.

Choose an apartment $\mathcal {A}\subseteq \mathcal {X}$, containing $\sigma ,x$. By Corollary A.12(b), the intersection $G_x\cap G_{\sigma ',r}$ is generated by $M_r$ and the affine root subgroups $U_{\psi }$, where $\psi \in \Psi (\mathcal {A})$ satisfies $\psi (x)\ge 0$ and $\psi (\sigma ')\ge r$. Thus we have to show that for every $\psi \in \Psi (\mathcal {A})$ such that $\psi (x)\ge 0$ and $\psi (\sigma ')\ge r$ we have $\psi (\sigma )\ge r$. Replacing $\psi$ with $r-\psi$ it suffices to show that for every $\psi \in \Psi _m(\mathcal {A})$ with $\psi (x)\le r$ and $\psi (\sigma ')\le 0$, we have $\psi (\sigma )\le 0$. But this is precisely the assumption $\sigma \in \Gamma _r(\sigma ',x)$.

This shows that every $G_r$ is closed. To show that $G_r$ is stable, we need to show that for every $\mathbf {G}(\overline{F})$-conjugate $g,g'\in G^{\mathrm{sr}}$ such that $g\in G_r$, we have $g'\in G_r$. In other words, we have to show that the subset $\mathcal {X}(g',r)\subseteq \mathcal {X}$ consisting of all $x\in \mathcal {X}$ such that $g'\in G_{x,r}$, is non-empty.

Since g and $g'$ are $\mathbf {G}(\overline{F})$-conjugate, and $F^{\mathrm{nr}}$ is of cohomological dimension one, we conclude that $g,g'$ are $\mathbf {G}(F^{\mathrm{nr}})$-conjugate, thus $\mathbf {G}(F^{\flat })$-conjugate for some finite unramified extension $F^{\flat }{/}F$. Set $\mathbf {G}^{\flat }:=\mathbf {G}_{F^{\flat }}$, $\mathcal {X}^{\flat }:=\mathcal {X}(\mathbf {G}^{\flat })$ and $\Gamma ^{\flat }:=\mathrm{Gal}(F^{\flat }{/}F)$. Then $g\in G_r\subseteq G^{\flat }_r$, hence $g'\in G\cap G^{\flat }_r$, because $G^{\flat }_r$ is $\mathrm{Ad}G^{\flat }$-invariant. Thus the subset $\mathcal {X}^{\flat }(g',r)\subseteq \mathcal {X}^{\flat }$ is non-empty. On the other hand, $\mathcal {X}^{\flat }(g',r)$ is $\Gamma ^{\flat }$-invariant, because $g'\in G$, and convex, by Lemma 3.11. Thus, by the Bruhat–Tits fixed point theorem, the set of fixed points $\mathcal {X}^{\flat }(g',r)^{\Gamma ^{\flat }}$ is non-empty. Since $\mathcal {X}^{\flat }(g',r)^{\Gamma ^{\flat }}$ equals $\mathcal {X}^{\flat }(g',r)\cap \mathcal {X}=\mathcal {X}(g',r)$ (by 2.5 and 3.7(a)), we are done.

The proof for $\mathfrak {g}_{r}$ is similar. Namely, for every $x\in \mathcal {X}$, we have $\mathfrak {g}=\cup _n \varpi ^{-n}\mathfrak {g}_x$. Thus to show that $\mathfrak {g}_{r}\subseteq \mathfrak {g}$ is closed, it remains to show that every $\mathfrak {g}_{r}\cap \varpi ^{-n}\mathfrak {g}_x$ is compact. Since $\varpi ^n\mathfrak {g}_r=\mathfrak {g}_{r+n}$ (see A.9(b)), it remains to show that the intersection $\mathfrak {g}_{r+n}\cap \mathfrak {g}_x$ is compact. This can be shown as in the group case.

Finally, the prove the result for $\mathfrak {g}^*_{-r}$ we can either mimic the proof for $\mathfrak {g}_{r}$, using the decomposition for $\mathfrak {g}^*_{x,-r}$, obtained from Lemma A.10(a) by duality, or to deduce it from a Lie algebra version of Proposition 4.14(a) by the Fourier transform. $\square$

B.5. Proof of Lemma 8.12. It suffices to show that $\pi$ induces bijections $\pi _x:G'_{x,r^+}\overset{\thicksim }{\rightarrow }G_{x,r^+}$ and $\pi _{x,y}:G'_{x,r^+}\cap G'_{y,r^+}\overset{\thicksim }{\rightarrow }G_{x,r^+}\cap G_{y,r^+}$ for $x,y\in \mathcal {X}(\mathbf {G})=\mathcal {X}(\mathbf {G}')$. Indeed, the surjectivity of $G'_{r^+}\overset{\thicksim }{\rightarrow }G_{r^+}$ follows from the surjectivity of the $\pi _x$’s, while injectivity follows from the surjectivity of the $\pi _{x,y}$’s and the injectivity of the $\pi _x$’s.

Replacing F by $F'$ as in A.8, we may assume that $\mathbf {G}$ and $\mathbf {G}'$ are quasi-split over F. Choose an apartment $\mathcal {A}\ni x,y$, corresponding to a maximal split torus $\mathbf {S}\subseteq \mathbf {G}$, and set $\mathbf {T}:=\mathbf {Z}_\mathbf {G}(\mathbf {S})$, and $\mathbf {T}':=\pi ^{-1}(\mathbf {T})\subseteq \mathbf {G}'$. Then $\mathbf {T}\subseteq \mathbf {G}$ is a maximal torus, and we have decompositions $G_{x,r^+}=T_{r^+}\times \prod _{\alpha }U_{(\alpha ),x,r^+}$ (by Lemma A.10) and $G_{x,r^+}\cap G_{y,r^+}=T_{r^+}\times \prod _{\alpha }(U_{(\alpha ),x,r^+}\cap U_{(\alpha ),y,r^+})$ (by Corollary A.12), and similarly for $\mathbf {G}'$.

Since $\pi$ induces isomorphisms between the $U_{\alpha }$’s, it remains to show that the induced map $T'_{r^+}\rightarrow T_{r^+}$ is an isomorphism. If $\mathbf {T}$ and $\mathbf {T}'$ are split, the assertion is easy. Namely, $\pi$ induces a morphism of $\mathbb F_p$-vector spaces $\pi _n:T'_n/T'_{n+1}\rightarrow T_n/T_{n+1}$ for every $n>0$. Hence each $\pi _n$ is an isomorphism, because the degree of $\pi$ is prime to p. Therefore $T'_{r^+}\rightarrow T_{r^+}$ is an isomorphism as well.

In general, let $F^{\flat }$ be the splitting field of $\mathbf {T}$ (and $\mathbf {T}'$), and let e be the ramification degree of $F^{\flat }{/}F$. Set $r^{\flat }:=er$, and $\Gamma ^{\flat }:=\mathrm{Gal}(F^{\flat }{/}F)$. Then $T_{r^+}=\mathrm{Ker}w_{\mathbf {T}}\cap \mathbf {T}(F^{\flat })^{\Gamma ^{\flat }}_{(r^{\flat })^+}$, where $w_\mathbf {T}$ is the Kottwitz homomorphism $\mathbf {T}(F^{\mathrm{nr}})\rightarrow X_*(\mathbf {T})_{\Gamma _{\mathrm{nr}}}$ (see 3.3(a)), and similarly for $\mathbf {T}'$. By the split case, $\pi$ induces an isomorphism $\mathbf {T}'(F^{\flat })^{\Gamma ^{\flat }}_{(r^{\flat })^+}\overset{\thicksim }{\rightarrow }\mathbf {T}(F^{\flat })^{\Gamma ^{\flat }}_{(r^{\flat })^+}$ of pro-p-groups. By the functoriality of the Kottwitz homomorphism, it remains to check that every element in the kernel of the homomorphism $X_*(\mathbf {T}')_{\Gamma _{\mathrm{nr}}}\rightarrow X_*(\mathbf {T})_{\Gamma _{\mathrm{nr}}}$ is torsion of prime to p order. Since this kernel is killed by $\deg \pi$, the proof is complete.$\square$

Appendix C. Quasi-logarithms

1.1 Quasi-logarithms

Let $\mathbf {G}$ be a reductive group over a field F.

(a) Following [17, 1.8], we call an $\mathrm{Ad}\mathbf {G}$-equivariant morphism of algebraic varieties $\mathcal {L}:\mathbf {G}\rightarrow \varvec{\mathfrak {g}}$ a quasi-logarithm, if $\mathcal {L}(1)=0$, and the induced map on tangent spaces $d\mathcal {L}_1:\mathfrak {g}=T_1(G)\rightarrow T_0(\mathfrak {g})=\mathfrak {g}$ is the identity map.

(b) Let $F^{\flat }{/}F$ be a field extension. Then a quasi-logarithm $\mathcal {L}:\mathbf {G}\rightarrow \varvec{\mathfrak {g}}$ induces a quasi-logarithm $\mathcal {L}_{F^{\flat }}:\mathbf {G}_{F^{\flat }}\rightarrow \varvec{\mathfrak {g}}_{F^{\flat }}$. Conversely, a quasi-logarithm $\mathcal {L}^{\flat }:\mathbf {G}_{F^{\flat }}\rightarrow \varvec{\mathfrak {g}}_{F^{\flat }}$ induces a quasi-logarithm $\mathrm{R}_{F^{\flat }{/}F}(\mathcal {L}^{\flat }):\mathrm{R}_{F^{\flat }{/}F}(\mathbf {G}_{F^{\flat }})\rightarrow \mathrm{R}_{F^{\flat }{/}F}(\varvec{\mathfrak {g}}_{F^{\flat }})=\varvec{\mathfrak {g}}\otimes _F F^{\flat }$.

(c) Since $\mathcal {L}$ is $\mathrm{Ad}\mathbf {G}$-equivariant, it induces a morphism $[\mathcal {L}]:\mathbf {c}_\mathbf {G}\rightarrow \mathbf {c}_{\varvec{\mathfrak {g}}}$ of the corresponding Chevalley spaces (compare [18, 5.2]).

1.2 Quasi-logarithms defined over $\mathcal {O}$

Let F be a local non-archimedean field of residual characteristic p.

(a) Assume that $\mathbf {G}$ is split over F. Then the Chevalley spaces $\mathbf {c}_\mathbf {G}$ and $\mathbf {c}_{\varvec{\mathfrak {g}}}$ have natural structures over $\mathcal {O}$. In this case, we say that a quasi-logarithm $\mathcal {L}:\mathbf {G}\rightarrow \varvec{\mathfrak {g}}$ is defined over $\mathcal {O}$, if the corresponding map $[\mathcal {L}]$ is defined over $\mathcal {O}$ (compare [18, 5.2]). Note that by [18, Lem 5.2.1] this notion is equivalent to the corresponding notion of [17, 1.8.8].

(b) For an arbitrary $\mathbf {G}$, we say that $\mathcal {L}:\mathbf {G}\rightarrow \varvec{\mathfrak {g}}$ is defined over $\mathcal {O}$, if $\mathcal {L}_{F^{\flat }}$ is defined over $\mathcal {O}_{F^{\flat }}$ for some or, equivalently, every splitting field $F^{\flat }$ of $\mathbf {G}$.

(c) Let $F^{\flat }{/}F$ be a finite Galois extension, and let $\mathcal {L}^{\flat }:\mathbf {G}_{F^{\flat }}\rightarrow \varvec{\mathfrak {g}}_{F^{\flat }}$ be a quasi-logarithm defined over $\mathcal {O}_{F^{\flat }}$. Then the quasi-logarithm $\mathrm{R}_{F^{\flat }{/}F}(\mathcal {L}^{\flat })$ (see C.1(b)) is also defined over $\mathcal {O}$.

(d) In the situation of (c), assume that $[F^{\flat }:F]$ is prime to p. Then the composition

$$\begin{aligned} \mathcal {L}:\mathbf {G}\hookrightarrow \mathrm{R}_{F^{\flat }{/}F}\mathbf {G}_{F^{\flat }}\overset{\mathrm{R}_{F^{\flat }{/}F}(\mathcal {L}^{\flat })}{\longrightarrow }\varvec{\mathfrak {g}}\otimes _F F^{\flat }\overset{\frac{1}{[F^{\flat }:F]}\mathrm{Tr}_{F^{\flat }{/}F}}{\longrightarrow }\varvec{\mathfrak {g}}\end{aligned}$$

is a quasi-logarithm defined over $\mathcal {O}$.

Lemma C.3

Assume that $\mathbf {G}$ is semisimple and simply connected and p is very good for $\mathbf {G}$ (see 8.10). Then $\mathbf {G}$ admits a quasi-logarithm defined over $\mathcal {O}$.

Proof

(compare [17, Lem 1.8.12]). Assume that $\mathbf {G}=\prod _i \mathrm{R}_{F_i{/}F}\mathbf {H}_i$ as in 8.10. By C.2(c), we can replace $\mathbf {G}$ by $\mathbf {H}_i$, thus assuming that $\mathbf {G}$ is absolutely simple. Using [17, Lem 1.8.9], we can replace $\mathbf {G}$ by its quasi-split inner form. Since p is good, $\mathbf {G}$ splits over a tamely ramified extension. Hence, using C.2(d), we may extend scalars to the splitting field of $\mathbf {G}$, thus assuming that $\mathbf {G}$ is split. In this case, the assertion was shown in [17, Lem 1.8.12], using the fact that $\mathbf {G}$ has a faithful representation, whose Killing form is non-degenerate over $\mathcal {O}$. Namely, one uses the standard representation, if $\mathbf {G}$ is classical, and the adjoint representation, if $\mathbf {G}$ is exceptional. $\square$

Lemma C.4

Let $\mathbf {G}$ be semisimple and simply connected, $p\ne 2$, and let $\mathcal {L}:\mathbf {G}\rightarrow \varvec{\mathfrak {g}}$ be a quasi-logarithm defined over $\mathcal {O}$. Then for every $x\in \mathcal {X}$ and $r\in \mathbb R_{\ge 0}$, $\mathcal {L}$ induces analytic isomorphisms $\mathcal {L}_{r}:G_{r^+}\overset{\thicksim }{\rightarrow }\mathfrak {g}_{r^+}$ and $\mathcal {L}_{x,r}:G_{x,r^+}\overset{\thicksim }{\rightarrow }\mathfrak {g}_{x,r^+}$.

Proof

Assume first that $\mathbf {G}$ is split. The assertion for $r=0$ was shown in [17, Prop 1.8.16]. Next we show that $\mathcal {L}$ induces an analytic isomorphism $\mathcal {L}_{x,r}:G_{x,r^+}\rightarrow \mathfrak {g}_{x,r^+}$ when $x\in \mathcal {X}$ is a hyperspecial vertex and $r=n\in \mathbb Z$. In this case, $G_{x,r^+}=G_{x,n+1}$ and $\mathfrak {g}_{x,r^+}=\mathfrak {g}_{x,n+1}$, so we have to show that $\mathcal {L}$ induces an analytic isomorphism $G_{x,n+1}\overset{\thicksim }{\rightarrow }\mathfrak {g}_{x,n+1}$. This is easy and it was shown in the course of the proof of [17, Prop 1.8.16]. We are going to deduce the general case from the particular case shown above.

Let $F^{\flat }{/}F$ be a finite Galois extension of ramification degree e, and set $r^{\flat }:=er$, $\Gamma ^{\flat }:=\mathrm{Gal}(F^{\flat }{/}F)$ and $\mathbf {G}^{\flat }:=\mathbf {G}_{F^{\flat }}$. Then $\mathcal {L}$ induces a quasi-logarithm $\mathcal {L}^{\flat }:=\mathcal {L}_{F^{\flat }}: \mathbf {G}^{\flat }\rightarrow \varvec{\mathfrak {g}}^{\flat }$, which is $\Gamma ^{\flat }$-equivariant and defined over $\mathcal {O}_{F^{\flat }}$. Moreover, since $\mathbf {G}$ is semisimple and simply connected, we have $G^0=G$ (see 3.6). Since $p\ne 2$, we have $G_{x,r^+}=(G^{\flat }_{x,(r^{\flat })^+})^{\Gamma ^{\flat }}$ and $\mathfrak {g}_{x,r^+}=(\mathfrak {g}^{\flat }_{x,(r^{\flat })^+})^{\Gamma ^{\flat }}$ (by Lemma 3.14).

Note that the assertion for $\mathcal {L}^{\flat }$ and $r^{\flat }$ implies that for $\mathcal {L}$ and r. Indeed, if $\mathcal {L}^{\flat }$ induces an isomorphism $\mathcal {L}^{\flat }_{x,r^{\flat }}$, then it is automatically $\Gamma ^{\flat }$-equivariant, thus induces an isomorphism $\mathcal {L}_{x,r}:=(\mathcal {L}^{\flat }_{x,r^{\flat }})^{\Gamma ^{\flat }}$ of Galois invariants. Therefore $\mathcal {L}$ induces a morphism $\mathcal {L}_{r}:G_{r^+}\rightarrow \mathfrak {g}_{r^+}$, which is surjective, because each $\mathcal {L}_{x,r}$ is surjective, and injective, because $\mathcal {L}^{\flat }_{r^{\flat }}$ is injective. Thus we can replace F by $F^{\flat }$, $\mathbf {G}$ by $\mathbf {G}^{\flat }$, and r by $r^{\flat }$.

Now the assertion is easy. Indeed, choosing $F^{\flat }$ to be a splitting field of $\mathbf {G}$, we can assume that $\mathbf {G}$ is split. Since $\mathcal {L}_{0}$ is injective, it is enough to show that $\mathcal {L}$ induces an isomorphism $\mathcal {L}_{x,r}$. Observe that both $G_{x,r^+}$ and $\mathfrak {g}_{x,r^+}$ do not change if we replace pair (x, r) by a close pair $(x',r')$. Thus we may assume that $r\in \frac{1}{m}\mathbb Z_{\ge 0}$ and x is a hyperspecial vertex of $[\mathcal {X}_m(\mathbf {G})]$ for some m.

Choose a finite extension $F^{\flat }$ of F of ramification degree m. Then $r^{\flat }=mr\in \mathbb N$, and x is a hyperspecial vertex of $[\mathcal {X}_m(\mathbf {G})]\subseteq [\mathcal {X}(\mathbf {G}^{\flat })]$. Hence the assertion for $\mathcal {L}^{\flat }_{x,r^{\flat }}$, shown in the first paragraph of the proof, implies the assertion for $\mathcal {L}_{x,r}$. $\square$