Proof By Proposition 4.18 of [Reference Li and Mihatsch11],Footnote ² the quotient ${\Gamma }\backslash \mathcal {I}(g)$ is a proper scheme over $\operatorname {Spec} O_{\breve F}$ with empty generic fiber. Since the maximal reduced subscheme $\mathcal {N}_{\mathrm {red}}$ is $0$ -dimensional, cf. (2.12), this scheme has to be artinian.

Proof The set $\mathcal {L}(x/{\pi })$ is empty if $x/{\pi }$ is not integral.

Proof of Theorem 2.11 (1)

First recall the statement: Let ${\gamma }\in G^{\prime }_{\mathrm {rs}}$ be regular semi-simple. The statement we would like to prove only depends on the orbit of ${\gamma }$ , so we may assume without loss of generality that $\gamma = \gamma (x)$ for some $x\in GL_n(F)$ . Set $\widetilde x = x/{\pi }$ . The assumption is

$$ \begin{align*}\mathrm{char}(\widetilde x; T) = {\pi}^{-n}\mathrm{Inv}({\gamma}(x), {\pi}T) \in T^n + {\pi} O_F[T],\end{align*} $$

which implies that $\widetilde {x}$ is topologically nilpotent. In particular, $\widetilde {\gamma } := \gamma (\widetilde {x})$ lies in $GL_{2n}(F)$ and is regular semi-simple. It satisfies $\mathrm {Inv}(\widetilde {{\gamma }}; T) = {\pi }^{-n}\mathrm {Inv}({\gamma }; {\pi }T)$ and our task is to show that

(3.7) $$ \begin{align}\operatorname{Orb}({\gamma},f^\prime_{1/2},s) =\operatorname{Orb}(\widetilde{{\gamma}},1_{GL_{2n}(O_F)},s).\end{align} $$

To this end, we compare the two quantities ${\Omega }({\gamma }, {\Lambda }_+ \oplus \Lambda _-, s)$ and ${\Omega }(\widetilde {\gamma }, {\Lambda }_+ \oplus \Lambda _-, s)$ for $(\Lambda _+, \Lambda _-) \in \mathcal {L}(\widetilde x)$ . Choose $h_1$ and $h_2$ in $H'$ such that $h_1O^{2n}_F = {\Lambda }_+ \oplus \Lambda _-$ and $h_2O^{2n}_F = \gamma ({\Lambda }_+ \oplus \Lambda _-)$ . Write $h^{-1}_1 = (\begin {smallmatrix} a & \\ & b \end {smallmatrix})$ and $h_2 = (\begin {smallmatrix} c & \\ & d \end {smallmatrix})$ . We obtain from Definition 3.3 that

(3.8) $$ \begin{align}\begin{aligned} {\Omega}(\widetilde{{\gamma}},\, {\Lambda}_+ \oplus \Lambda_-,\, s) & = \Omega\left(\begin{pmatrix} ac & axd/{\pi} \\ bc & bd \end{pmatrix},\, s\right)\\ & = \Omega\left(\begin{pmatrix} ac & axd \\ bc & bd \end{pmatrix},\, s\right)q^{-ns} = {\Omega}({\gamma},\, {\Lambda}_+ \oplus \Lambda_-,\, s)q^{-ns}. \end{aligned}\end{align} $$

Substituting this into (3.6) for $\widetilde {\gamma }$ yields (3.5) for $\gamma $ which proves (3.7) as desired.

Proof As x was assumed to be integral over $O_F$ , the ring $O_F[x]$ is of the form $O_F[T]/(P(T))$ for some monic polynomial $P(T) \in O_F[T]$ . We consider the reduction $\mkern 1.5mu\overline {\mkern -1.5muP\mkern -1.5mu}\mkern 1.5mu$ of P modulo $\pi $ . It will factor as $\mkern 1.5mu\overline {\mkern -1.5muP\mkern -1.5mu}\mkern 1.5mu = T^m f(T)$ with $f(0) \neq 0$ . By Hensel’s lemma, this factorization lifts to a factorization of P which defines the desired factorization $R_0\times R_1$ .

Proof Every $O_F[\widetilde x]$ -lattice is a direct sum of an $R_0$ -lattice and an $R_1$ -lattice. It is easily seen that this defines a bijection $\mathcal {L}(\widetilde x)\overset {\sim }{\to } \mathcal {L}(x_0/{\pi }) \times \mathcal {L}(x_1/{\pi })$ . Furthermore, the definition of ${\Omega }({\gamma }, {\Lambda }_+ \oplus \Lambda _-, s)$ is multiplicative in such direct sums. The desired identity then follows from (3.5).

Proof We again use (3.5) to express the left hand side. A general identity (see Lemma 3.22 of [Reference Li and Mihatsch11]) states that the transfer factor of a lattice ${\Lambda } = {\Lambda }_+\oplus {\Lambda }_-\in \mathcal {L}_0({\gamma })$ is given by

(3.13) $$ \begin{align} \qquad\quad {\Omega}({\gamma}, {\Lambda}_+ \oplus \Lambda_-, s) = (-1)^{[(\gamma\Lambda)_-:z\Lambda_+] + [(\gamma\Lambda)_-:\Lambda_-]}q^{([(\gamma\Lambda)_+:z\Lambda_-] + [(\gamma\Lambda)_-:z\Lambda_+])s},\end{align} $$

where $z = z_{\gamma }$ and where $[{\Lambda }_1:{\Lambda }_2]$ denotes the length of ${\Lambda }_1/{\Lambda }_2$ . In the situation of (3.5), we apply this formula to the element ${\gamma } = {\gamma }(x)$ and a lattice ${\Lambda } = {\Lambda }_+ \oplus {\Lambda }_-$ with $({\Lambda }_+, {\Lambda }_-) \in \mathcal {L}(\widetilde x)$ . We have that $z = (\begin {smallmatrix} 0 & x \\ 1 & 0 \end {smallmatrix})$ and that ${\gamma }\Lambda = {\Lambda }$ because x is topologically nilpotent under the assumption $\det (\widetilde x) \in O_F^\times $ . The assumption moreover implies that any $\mathcal {L}(\widetilde x)$ is the set of lattice pairs $({\Lambda }_+, {\Lambda }_-)$ that satisfy

$$ \begin{align*}\widetilde x {\Lambda}_+ = {\Lambda}_- = {\Lambda}_+.\end{align*} $$

We see that $z\Lambda _+ = \Lambda _-$ and $z\Lambda _- = \pi \Lambda _+$ for all such $({\Lambda }_+, {\Lambda }_-)$ . Substituting this in (3.13) gives ${\Omega }({\gamma }(x), {\Lambda }_+ \oplus \Lambda _-, s) = q^{ns}$ for all $({\Lambda }_+, {\Lambda }_-) \in \mathcal {L}(\widetilde x)$ and hence

(3.14) $$ \begin{align}\operatorname{Orb}({\gamma},f^\prime_{1/2},s) = q^{ns}q^{-ns}\sum_{\Lambda = (\Lambda_+,\Lambda_-) \in \mathcal{L}(\widetilde x)} [O_{L_x}^{\times}:R^{\times}_{\Lambda}]. \end{align} $$

This is precisely the combinatorial description of $\operatorname {Orb}(\widetilde {x},1_{GL_n(O_F)})$ and the proof is complete.

Proof The assumption that $x\sigma (x){\pi }^{\varepsilon }=(x\varpi )^2$ is topologically nilpotent implies that the determinant $\det (g(x))$ lies in $O_F^\times $ . (Here and in the following, the determinant is meant in the sense of the reduced norm $GL_n(B)\to F^\times $ .) Hence, the condition $h_1^{-1}g h_2 \in GL_n(O_B)$ holds if and only if $h_1^{-1}g h_2 \in M_n(O_B)$ . Since $h_1^{-1}g h_2 = h_1^{-1} h_2(1 + h_2^{-1}x\varpi h_2)$ and since $h_2^{-1}x\varpi h_2$ is topologically nilpotent by assumption, this is equivalent to $h_1^{-1} h_2 \in GL_n(O_E)$ and $h_2^{-1} x {\sigma }(h_2) \in M_n(O_E)$ . Given a pair $(h_1, h_2)$ with $h_1^{-1}h_2 \in GL_n(O_E)$ , consider the lattice $\Lambda = h_1 O_E^n= h_2 O_E^n$ . Then $h_2^{-1} x {\sigma }(h_2)$ lies in $M_n(O_E)$ if and only if $x\sigma ({\Lambda }) \subseteq \Lambda $ . Rewriting the definition of $\operatorname {Orb}(g, 1_{GL_n(O_B)})$ in this way gives (3.17).

Proof Since $\operatorname {Orb}(g, 1_{GL_n(O_B)})$ only depends on the orbit of g and since every regular semi-simple orbit contains a representative of the form $g(x)$ we may assume without loss of generality that $g = g(x)$ for some $x\in GL_n(E)$ . In this case $z^2_g/\pi = x\sigma (x)$ . Moreover, $GL_n(O_B) = GL_n(O_E) + M_n(O_E)\varpi $ because B is the division algebra. So $h_1^{-1} h_2 + h_1^{-1} x {\sigma }(h_2) \varpi \in GL_n(O_B)$ with $h_i \in GL_n(E)$ can only hold if $h_1^{-1} h_2 \in GL_n(O_E)$ and $h_2^{-1} x\sigma (h_2) \in M_n(O_E)$ . The second condition implies that $x\sigma (x)$ is integral over $O_F$ because it is conjugate to $h_2^{-1}x\sigma (h_2) {\sigma }(h_2^{-1}x\sigma (h_2))$ . This was to be shown.

Proof of Theorem 2.11 (2)

Let $g\in G_{1/2, \mathrm {rs}}$ be such that ${\pi }^{-n}\mathrm {Inv}(g; {\pi }T) \in T^n + {\pi }O_F[T]$ . Denote by $\varpi _{{\lambda }}\in B_{\lambda }$ the element fixed in (3.15). Without loss of generality we may assume that $g = 1 + x \varpi _{1/2}$ for some $x\in GL_n(E)$ . The assumption on g is then equivalent to $x\sigma (x)$ being topologically nilpotent. The element $\widetilde g = 1 + x \varpi _{0}$ hence lies in $G_{0, \mathrm {rs}}$ and satisfies $\mathrm {Inv}(\widetilde g;T) = {\pi }^{-n}\mathrm {Inv}(g; {\pi }T)$ . Lemma 3.9 applies to both g and $\widetilde g$ , and yields

(3.18) $$ \begin{align} \operatorname{Orb}(g, 1_{GL_n(O_B)}) = \operatorname{Orb}(\widetilde g, 1_{GL_{2n}(O_F)}), \end{align} $$

as desired.

Proof Every $O_F[x\sigma (x)]$ -lattice is a direct sum of an $R_0$ -lattice and an $R_1$ -lattice. This defines a bijection $\mathcal {L}^\sigma (x)\overset {\sim }{\to } \mathcal {L}^{\sigma }(x_0)\times \mathcal {L}^{\sigma }(x_1)$ . The desired identity now follows from Lemma 3.9 which can be applied to all three orbital integrals because $x\sigma (x)$ , $x_0{\sigma }(x_0)$ and $x_1{\sigma }(x_1)$ are all integral by assumption.

Proof The assumption that $x\sigma (x)$ is integrally invertible implies that $\mathcal {L}^{\sigma }(x)$ is the set of $O_E$ -lattices ${\Lambda }\subseteq E^n$ such that $x\sigma ({\Lambda }) = {\Lambda }$ . In this case, the lattice counting expression in Lemma 3.9 equals $\operatorname {Orb}^\sigma (x,1_{GL_n(O_E)})$ .

Proof If $z_{\gamma }^2/{\pi }$ is not integral over $O_F$ , then $\operatorname {Orb}({\gamma }, f^{\prime }_{1/2}, s) = 0$ by Corollary 3.4 and there is nothing to prove. So assume that $z^2_{\gamma }/{\pi }$ is integral and consider the orbital integral factorization from Corollary 3.6,

$$ \begin{align*}\operatorname{Orb}({\gamma}, f^{\prime}_{1/2}, s) = \operatorname{Orb}({\gamma}_0, f^{\prime}_{1/2}, s)\operatorname{Orb}({\gamma}_1, f^{\prime}_{1/2}, s).\end{align*} $$

Since $\mathrm {Inv}({\gamma }) = \mathrm {Inv}({\gamma }_0)\mathrm {Inv}({\gamma }_1)$ , we have

$$ \begin{align*}\mathrm{ord}_{1/2}(\mathrm{Inv}({\gamma})) = \mathrm{ord}_{1/2}(\mathrm{Inv}({\gamma}_0)) + \mathrm{ord}_{1/2}(\mathrm{Inv}({\gamma}_1)).\end{align*} $$

It hence suffices to show (3.26) for ${\gamma }_0$ and ${\gamma }_1$ separately, meaning we may either assume $z_{\gamma }^2/{\pi }$ to be topologically nilpotent or integrally invertible.

If $z_{\gamma }^2/{\pi }$ is topologically nilpotent, then we can apply Theorem 2.11 (1): For any element $\widetilde {{\gamma }} \in G^{\prime }_{\mathrm {rs}}$ with $\mathrm {Inv}(\widetilde {{\gamma }}; T) = {\pi }^{-n}\mathrm {Inv}({\gamma }; {\pi }T)$ , we have $\operatorname {Orb}({\gamma }, f^{\prime }_{1/2}, s) = \operatorname {Orb}(\widetilde {{\gamma }}, 1_{GL_{2n}(O_F)}, s)$ . Using (3.25), the desired vanishing statement (3.26) follows directly from (3.24).

If $z_{\gamma }^2/{\pi }$ is integral over $O_F$ and $\det (z_{\gamma }^2/{\pi }) \in O_F^\times $ , then

$$ \begin{align*}\mathrm{ord}_{1/2}(\mathrm{Inv}({\gamma}; T)) = \mathrm{ord}_0({\pi}^{-n}\mathrm{Inv}({\gamma}; {\pi}T)) = 0,\end{align*} $$

because $z^2_{\gamma }/{\pi }$ then lies in $O_{L_{\gamma }}^\times $ and is hence a norm from $E\otimes _F L_{\gamma }$ . (This uses that $E/F$ is unramified.) The inequality $\mathrm {ord}_{s = 0} \operatorname {Orb}({\gamma }, f^{\prime }_{1/2}, s) \geq 0$ holds trivially and there is nothing to prove in this case. This completes the argument. (We remark that if $z_{\gamma }^2/{\pi }\in O_{L_{\gamma }}^\times $ , then Corollary 3.7 states that $\operatorname {Orb}({\gamma }, f^{\prime }_{1/2}, s)$ is a constant independent of s.)

Proof of Theorem 2.7

Let $\gamma \in G^\prime _{\mathrm {rs}}$ . We need to show the identity

(3.27) $$ \begin{align} \operatorname{Orb}({\gamma}, f^{\prime}_{1/2}) = \begin{cases} \operatorname{Orb}(g, f_{1/2}) & \text{if there exists a matching } g\in G_{1/2}\\ 0 & \text{otherwise.} \end{cases} \end{align} $$

We first note that if ${\pi }^{-n}\mathrm {Inv}({\gamma };{\pi }T) \notin O_F[T]$ , then both sides of the equation vanish. This follows from Corollary 3.4 and from Lemma 3.10, respectively. Hence, we assume from now on that ${\pi }^{-n}\mathrm {Inv}({\gamma };{\pi }T) \in O_F[T]$ or, equivalently, that $z^2_\gamma /\pi $ is integral. Consider $L_{\delta }$ and $B_{\delta }$ as in Equation (3.23) for $\delta = \mathrm {Inv}({\gamma };T)$ . Let $L_{\delta }=\prod _i L_i$ and $B_{\delta }=\prod _i L_i$ be their factorizations according to idempotents. By Corollary 2.8 of [Reference Li and Mihatsch11] there exists an element $g \in G_{1/2, \mathrm {rs}}$ that matches $\gamma $ if and only if $\mathrm {ord}_{1/2}({\delta }) = 0$ , see Definition 3.13. In particular, if there is no matching element $g \in G_{1/2, \mathrm {rs}}$ , then $\operatorname {Orb}({\gamma }, f^{\prime }_{1/2}) = 0$ by Corollary 3.14. This proves the vanishing part of (3.27).

We henceforth consider the case that there is an element $g \in G_{1/2, \mathrm {rs}}$ that matches $\gamma $ . Assuming that ${\gamma } = 1 + z_{\gamma }$ and that $g = 1 + z_g$ , let ${\gamma } = ({\gamma }_0, {\gamma }_1)$ and $g = (g_0, g_1)$ be the components of ${\gamma }$ and g such that $z^2_{{\gamma }_0}/{\pi }$ and $z^2_{g_0}/{\pi }$ are topologically nilpotent, and $z_{{\gamma }_1}^2/{\pi }$ and $z_{g_1}^2/{\pi }$ integrally invertible (Corollaries 3.7 and 3.12). Then ${\gamma }_0$ matches $g_0$ and ${\gamma }_1$ matches $g_1$ as can be seen by using the isomorphism $O_F[z^2_\gamma ]\cong O_F[z^2_g]$ that sends $z^2_\gamma $ to $z^2_g$ . By Lemmas 3.6 and 3.11, the two sides of (3.27) factor and the desired equality becomes

$$ \begin{align*}\operatorname{Orb}({\gamma}_0, f^{\prime}_{1/2, n_0})\operatorname{Orb}({\gamma}_1, f^{\prime}_{1/2, n_1}) \overset{?}{=} \operatorname{Orb}(g_0, 1_{GL_{n_0}(O_B)}) \operatorname{Orb}(g_1, 1_{GL_{n_1}(O_B)}).\end{align*} $$

Here, $B = B_{1/2}$ . We can prove this identity factor-by-factor, meaning we may assume that $z_{\gamma }^2/{\pi }$ is topologically nilpotent or that $z_{\gamma }^2/{\pi } \in O_{L_{\gamma }}^\times $ .

Assume first that $z_{\gamma }^2/{\pi }$ is topologically nilpotent. Then we may apply Theorem 2.11 (1) and (2): Let $\widetilde {{\gamma }} \in G^{\prime }_{\mathrm {rs}}$ and $\widetilde g \in G_{0, \mathrm {rs}}$ be such that $\mathrm {Inv}(\widetilde {{\gamma }}; T) = {\pi }^{-n}{\delta }({\pi }T) = \mathrm {Inv}(\widetilde g; T)$ . We obtain

$$ \begin{align*}\operatorname{Orb}({\gamma}, f^{\prime}_{1/2}) = \operatorname{Orb}(\widetilde{{\gamma}}, 1_{GL_{2n}(O_F)}) = \operatorname{Orb}(\widetilde g, 1_{GL_{2n}(O_F)}) = \operatorname{Orb}(g, 1_{GL_n(O_B)}),\end{align*} $$

where the middle equality is the Guo–Jacquet FL, i.e. the case ${\lambda } = 0$ of Theorem 2.7.

Assume now that $z_{\gamma }^2/{\pi } \in O_{L_{\gamma }}^\times $ . Without loss of generality, we may assume that ${\gamma } = {\gamma }(x)$ for some $x\in GL_n(F)$ and $g = 1 + x'\varpi $ for some $x'\in GL_n(E)$ . The fact that ${\gamma }$ and g match translates to the identity

$$ \begin{align*}\mathrm{char}_F(x/{\pi}; T) = \mathrm{char}_E(x'{\sigma}(x'); T).\end{align*} $$

In other words, $x/{\pi }$ and $x'$ match in the sense of the quadratic base change FL, see [Reference Kottwitz5]. We obtain from Lemmas 3.7 and 3.12, as well as the base change FL that

$$ \begin{align*}\operatorname{Orb}({\gamma}, f^{\prime}_{1/2}) = \operatorname{Orb}(x/{\pi}, 1_{GL_n(O_F)}) = \operatorname{Orb}^{\sigma}(x', 1_{GL_n(O_E)}) = \operatorname{Orb}(g, 1_{GL_n(O_B)}).\end{align*} $$

The proof of Theorem 2.7 is now complete.

Proof The first identity is a special case of Proposition 4.23 (2) from [Reference Li and Mihatsch11] which we may apply because $z_g$ is topologically nilpotent by assumption. The second identity follows directly from the definitions of $\mathcal {Z}(z_g)$ and $\mathcal {Z}(w)$ .

Proof Let $g\in G_{1/2, \mathbb {Y}, \mathrm {rs}}$ be an element whose invariant has the property that ${\pi }^{-n}\mathrm {Inv}(g; {\pi }T)$ lies in $T^n + {\pi }O_F[T]$ . We need to construct an element $\widetilde g \in G_{0, \mathbb {Y}, \mathrm {rs}}$ such that both $\mathrm {Inv}(\widetilde g, T) = {\pi }^{-n}\mathrm {Inv}(g, {\pi }T)$ and $\mathrm {Int}(g) = \mathrm {Int}(\widetilde g)$ .

Given g, we define $\widetilde g \in D_{0, \mathbb {Y}}$ by the following relation:

$$ \begin{align*}g = \begin{pmatrix} x & {\pi}y \\ y & x \end{pmatrix}\quad \text{and}\quad \ \widetilde g := \begin{pmatrix} x & y \\ y & x \end{pmatrix}.\end{align*} $$

Since g is regular semi-simple, $g_+ = \left (\begin {smallmatrix} x & \\ & x\end {smallmatrix} \right )$ and $g_- = \left (\begin {smallmatrix} & {\pi }y \\ y & \end {smallmatrix} \right )$ are both invertible. Then $\widetilde g_+ = g_+$ and $\widetilde g_- = \left (\begin {smallmatrix} & y \\ y & \end {smallmatrix} \right )$ are invertible as well. The assumption on $\mathrm {Inv}(g;T)$ implies that $\mathrm {Inv}(g;{\pi }) \neq 0$ , so $\widetilde g$ lies in $G_{0,\mathbb {Y}}$ and has invariant polynomial

$$ \begin{align*}\mathrm{Inv}(\widetilde g; T) = {\pi}^{-n}\mathrm{Inv}(g; {\pi}T).\end{align*} $$

This polynomial is separable by assumption on g, so $\widetilde {g}$ is regular semi-simple. Moreover, $z_{\widetilde {g}}^2$ is topologically nilpotent since $\mathrm {Inv}(\widetilde {g}; T) \equiv T^n$ modulo ${\pi }O_F[T]$ . We may now apply Lemma 4.2 twice to see that

(4.3) $$ \begin{align} \begin{aligned} \mathcal{I}(\widetilde g) & = \mathcal{N} \cap_{\mathcal{M}_0} \mathcal{Z}\left(\left(\begin{smallmatrix} & y \\ y & \end{smallmatrix}\right)\right)\\ & = \mathcal{Z}(y)\\ & = \mathcal{N} \cap_{\mathcal{M}_{1/2}} \mathcal{Z}\left(\left(\begin{smallmatrix} & {\pi}y \\ y & \end{smallmatrix}\right)\right)\\ & = \mathcal{I}(g). \end{aligned} \end{align} $$

Note that $z_g^2 = {\pi }z_{\widetilde g}^2$ , so the two F-algebras $L_g = F[z_g^2]$ and $L_{\widetilde g} = F[z_{\widetilde g}^2]$ agree as subalgebras of $C_{\mathbb {Y}}$ . The isomorphism in (4.3) is then equivariant with respect to the action of $L_g^\times = L_{\widetilde g}^\times $ . Choosing the same subgroup ${\Gamma }\subset L_g^\times = L_{\widetilde g}^\times $ in the definition of the intersection number (Definition 2.5), we obtain ${\Gamma }\backslash \mathcal {I}(g) = {\Gamma }\backslash \mathcal {I}(\widetilde g)$ and hence the identity

$$ \begin{align*}\mathrm{Int}(g) = \mathrm{Int}(\widetilde g),\end{align*} $$

as was to be shown.

Proof Every strict $O_F$ -module over $\mathbb {F}$ is, in a unique way, the product of its identity connected component and its maximal étale quotient. This decomposition is functorial in all respects, giving (4.4) from definitions.

Proof By the product structure on $\mathcal {I}(g)(\mathbb {F})$ from Lemma 4.4, $\mathcal {I}(g)$ being nonempty implies that both $\mathcal {I}(g^\circ )$ and $\mathcal {I}(g^{\mathrm {et}})$ are nonempty. Since $\mathbb {Y}^\circ $ has no étale factor by definition, Proposition 4.23 (1) of [Reference Li and Mihatsch11] applies and states that $z_g^\circ $ is topologically nilpotent. Then $\mathcal {I}(g^\circ ) = \mathcal {N}^\circ \cap \mathcal {Z}(z_g^\circ )$ by Lemma 4.2. Writing $g = \left (\begin {smallmatrix} x & {\pi }y \\ y & x\end {smallmatrix} \right )$ as in (4.1) (here we used that we consider the case ${\lambda } = 1/2$ ), we obtain from

(4.5) $$ \begin{align} z_g^\circ = \begin{pmatrix} & {\pi}w^\circ \\ w^\circ & \end{pmatrix},\quad\ \ w = x^{-1}y, \end{align} $$

that $\mathcal {I}(g^\circ ) = \mathcal {Z}(w^\circ )$ where the right hand side is a closed formal subscheme of $\mathcal {N}$ . Thus $\mathcal {I}(g^\circ ) \neq \emptyset $ implies that $w^\circ $ is integral. Since $(z_g^\circ )^2/{\pi } = (w^\circ )^2$ , this shows that $(z_g^\circ )^2/{\pi }$ is integral as claimed.

It is left to prove that $(z_g^{\mathrm {et}})^2/{\pi }$ is integral. Passing from $\mathbb {Y}^{\mathrm {et}}$ and $\mathbb {X}^{\mathrm {et}}$ to Tate modules, we may identify $\mathcal {I}(g^{\mathrm {et}})(\mathbb {F})$ with the set $\mathcal {L}(\varpi ^{-1} z_g^{\mathrm {et}})$ from Definition 3.8. By Lemma 3.10, this set being nonempty implies $(z_g^{\mathrm {et}})^2/{\pi }$ integral. The proof is now complete.

Proof We consider the two cases $n^\circ $ even or odd separately. Assume first that $n^\circ $ is even. Then $D_{1/2, \mathbb {Y}}^\circ $ is a CDA over F of degree $2n^\circ $ and Hasse invariant $(n^\circ + 1)/2n^\circ $ . Thus $L_g^\circ $ is a field extension of degree $n^\circ $ of F. As $E\otimes _F L_g^\circ $ embeds into $D_{1/2, \mathbb {Y}}^\circ $ , this tensor product has to be a field, so the inertia degree of $L_g$ over F is odd. This implies that its ramification index is even. Moreover, $B_g^\circ $ embeds into $D_{1/2, \mathbb {Y}}^\circ $ and is hence a division algebra. It follows that $(z_g^\circ )^2\in (L_g^\circ )^\times $ is not a norm from $E\otimes _F L_g^\circ $ and hence has odd valuation. Since the ramification index is even, $(z_g^\circ )^2/{\pi }$ has odd valuation as well. So $(z_g^\circ )^2/{\pi }\in O_{L_g^\circ }^\times $ is impossible as was to be shown.

Now we consider the case that $n^\circ $ is odd. Then $D_{1/2, \mathbb {Y}} \cong M_2(Q)$ where Q is a CDA over F of degree $n^\circ $ and Hasse invariant $(n^\circ + 1)/2n^\circ $ . The étale F-algebra $L_g^\circ $ , which has degree $n^\circ $ , again has to be a field because there cannot exist an embedding $L_g\to M_2(Q)$ otherwise. (This argument used that $n^\circ $ is odd.) Then one obtains that $B_g^\circ \cong M_2(L_g^\circ )$ because it equals the centralizer of $L_g^\circ $ in $M_2(Q)$ . This means that $(z_g^\circ )^2\in L_g^\circ $ is a norm from $E\otimes _FL_g^\circ $ , which is equivalent to $(z_g^\circ )^2$ having even valuation. The degree $[L_g^\circ :F]$ is odd, so it follows that $(z_g^\circ )^2/{\pi }$ has odd valuation and thus cannot lie in $O_{L_g^\circ }^\times $ as was to be shown.

Construction 4.7 Assume that $g = (g^\circ , g^{\mathrm {et}}) \in G_{1/2, \mathbb {Y}, \mathrm {rs}}$ satisfies that $z_g^2/{\pi }$ is integral. By Lemma 4.6, $(z_g^\circ )^2/{\pi }$ is even topologically nilpotent. Let $R = O_F[z_g^2/{\pi }] \subset C_{\mathbb {Y}}$ be the $O_F$ -algebra generated by $z_g^2/{\pi }$ . Using Hensel’s Lemma as in Lemma 3.5, there is a unique factorization $R = R_0 \times R_1$ such that, when writing $z_g^2/{\pi } = ({\zeta }_0, {\zeta }_1)$ , the component ${\zeta }_0$ is topologically nilpotent and ${\zeta }_1\in R_1^\times $ . Since $(z^\circ _g)^2/{\pi }$ is topologically nilpotent, the projection $R\to R_1$ factors through the projection map

$$ \begin{align*}F[z_g^2/{\pi}] = F[(z_g^\circ)^2/{\pi}]\times F[(z_g^{\mathrm{et}})^2/{\pi}] \longrightarrow F[(z_g^{\mathrm{et}})^2/{\pi}].\end{align*} $$

Let $\mathbb {Y} = \mathbb {Y}_0\times \mathbb {Y}_1$ be the decomposition of $\mathbb {Y}$ up to isogeny with respect to the idempotents defining $R = R_0\times R_1$ . By what was just said, $\mathbb {Y}_1$ is an étale ${\pi }$ -divisible $O_E$ -module. The centralizer in $G_{1/2, \mathbb {Y}, \mathrm {rs}}$ of the idempotent $(1,0)\in R$ is thus of the form $J_0 \times J_1$ with

$$ \begin{align*}J_0 = G_{1/2, \mathbb{Y}_0}\quad \text{and}\quad J_1 = G_{1/2, \mathbb{Y}_1} \cong GL_{n_1}(B).\end{align*} $$

Here, $n_1$ is the $O_E$ -height of $\mathbb {Y}_1$ . Let $(g_0, g_1)\in J_{0, \mathrm {rs}} \times J_{1, \mathrm {rs}}$ be a pair of regular semi-simple elements such that $H_{\mathbb {Y}} (g_0, g_1)H_{\mathbb {Y}} = H_{\mathbb {Y}} gH_{\mathbb {Y}}$ . Such a pair exists: For example, after an $H_{\mathbb {Y}}$ -translation, we may assume that $g = 1 + z_g$ in which case g commutes with $z_g$ . Then g itself lies in $J_0 \times J_1$ .

Proof All three quantities in (4.6) only depend on the invariants of the three elements g, $g_0$ and $g_1$ . So we may assume that, g has the form $\left (\begin {smallmatrix} 1 & {\pi }y \\ y & 1\end {smallmatrix}\right )$ . By Lemma 4.2,

$$ \begin{align*}\mathcal{I}(g) = \mathcal{N}\cap \mathcal{Z}\left(\left(\begin{smallmatrix} & {\pi}y \\ y & \end{smallmatrix}\right)\right) = \mathcal{Z}(y),\end{align*} $$

where $\mathcal {Z}(y)\subset \mathcal {N}$ is the subspace of all $(Y, {\iota }, {\rho })$ such that ${\rho }y{\rho }^{-1}\in \operatorname {End}(Y)$ . Moreover, since $y^2 = z_g^2/{\pi }$ , the ring R in Construction 4.7 agrees with $O_F[y^2]$ . Using its idempotents, every $(Y, {\iota }, {\rho })\in \mathcal {Z}(y)$ factors as $(Y_0, {\iota }_0, {\rho }_0) \times (Y_1, {\iota }_1, {\rho }_1)$ where the two triples lie in $\mathcal {I}(g_0)$ and $\mathcal {I}(g_1)$ . Furthermore, we may choose ${\Gamma }\subset L^\times _g = F[{\pi }y^2]^\times $ as ${\Gamma } = {\Gamma }_0\times {\Gamma }_1$ with ${\Gamma }_i \subset F[z_{g_i}^2]^\times $ . In this way, we obtain the factorization

(4.7) $$ \begin{align} {\Gamma}\backslash \mathcal{I}(g) \cong ({\Gamma}_0 \backslash \mathcal{I}(g_0)) \times_{\operatorname{Spf} O_{\breve F}} ({\Gamma}_1 \backslash \mathcal{I}(g_1)). \end{align} $$

Since $\mathbb {Y}_1$ is an étale ${\pi }$ -divisible $O_E$ -module, $\mathcal {I}(g_1)$ is étale over $\operatorname {Spf} O_{\breve F}$ . Moreover, passing to Tate modules, $\mathcal {I}(g_1)(\mathbb {F})$ may be identified with the set $\mathcal {L}(\varpi ^{-1} z_{g_1})$ from Definition 3.8. The formal scheme ${\Gamma }_1\backslash \mathcal {I}(g_1)$ is then a disjoint union of $\operatorname {Orb}(g_1, 1_{GL_{n_1}(O_B)})$ many copies of $\operatorname {Spf} O_{\breve F}$ . Thus (4.6) follows from (4.7).

Proof of Theorem 2.9 (Vanishing Part)

Let ${\gamma }\in G^{\prime }_{\mathrm {rs}}$ be regular semi-simple with invariant ${\delta } = \mathrm {Inv}({\gamma })$ . We claim that there exists a strict $O_E$ -module $\mathbb {Y}$ over $\mathbb {F}$ and a matching element $g\in G_{1/2, \mathbb {Y}, \mathrm {rs}}$ if and only if $\mathrm {ord}_{1/2}({\delta }) = 1$ . Indeed, this condition is by definition equivalent to

(4.8) $$ \begin{align} L_{\delta} \cong L_0 \times L_1,\quad B_{\delta} \cong (B_{1/2}\otimes_FL_0) \times B_1, \end{align} $$

where $L_0$ is an étale F-algebra, $L_1$ a field extension of F, and $B_1/L_1$ the quaternion algebra that is not isomorphic to $B_{1/2}\otimes _FL_1$ . Assuming this condition is met, let $\mathbb {Y}$ be the unique strict $O_E$ -module such that $n^\circ = [L_1:F]$ and $n^{\mathrm {et}} = [L_0:F]$ . Then

(4.9) $$ \begin{align} D_{1/2, \mathbb{Y}} \cong M_{[L_0:F]}(B) \times D^\circ_{1/2, \mathbb{Y}}. \end{align} $$

Here, $D^\circ := D_{1/2, \mathbb {Y}}^\circ $ is a CSA of degree $2n^\circ $ with Hasse invariant $(n^\circ + 1)/2n^\circ $ . There are two cases: If $n^\circ $ is even, then $B_1$ and $D^\circ $ are both division algebras. If $n^\circ $ is odd, then $B_1 = M_2(L_1)$ and $D^\circ $ is the ring of $(2\times 2)$ -matrices over a CDA. In both cases, there exists an F-algebra embedding $B_1\to D^\circ $ and hence an F-algebra embedding $B_{\delta }\to D_{1/2, \mathbb {Y}}$ . By Corollary 2.8 (2) of [Reference Li and Mihatsch11], resp. its extension to semi-simple F-algebras, this is equivalent to the existence of an element $g\in G_{1/2, \mathbb {Y}, \mathrm {rs}}$ with $\mathrm {Inv}(g;T) = {\delta }(T)$ .

Conversely, assume that there exists an embedding ${\beta }:B_{\delta } \to D_{1/2, \mathbb {Y}}$ where $D_{1/2, \mathbb {Y}}$ is as in (4.9). Then ${\beta }(B_{\delta })$ agrees with the centralizer of ${\beta }(L_{\delta })$ for dimension reasons, see Proposition 2.6 (3) of [Reference Li and Mihatsch11], which implies that $B_{\delta }$ takes the form in (4.8), and hence that $\mathrm {ord}_{1/2}({\delta }) = 1$ . This finishes the prove of our claim.

The vanishing statement is now obtained as follows. Assume there is no $\mathbb {Y}$ with matching $g\in G_{1/2, \mathbb {Y}, \mathrm {rs}}$ . By the claim, this means $\mathrm {ord}_{1/2}({\delta }) = 0$ or $\mathrm {ord}_{1/2}({\delta }) \geq 2$ . In the first case, we apply the functional equation

$$ \begin{align*}\operatorname{Orb}({\gamma}, f^{\prime}_{1/2}, -s) = {\varepsilon}_{1/2}({\gamma}) \operatorname{Orb}({\gamma}, f^{\prime}_{1/2}, s),\end{align*} $$

from (2.17). The sign ${\varepsilon }_{1/2}({\gamma })$ can be seen to equal $(-1)^{\mathrm {ord}_{1/2}({\delta })}$ . Thus $\operatorname {\partial Orb} ({\gamma }, f^{\prime }_{1/2}) = 0$ if $\mathrm {ord}_{1/2}({\delta }) = 0$ . In the second case, the vanishing of $\operatorname {\partial Orb}({\gamma }, f^{\prime }_{1/2})$ is implied by Corollary 3.14.

Proof of Theorem 2.9 (Reduction to the linear AFL)

Fix $\mathbb {Y}$ and a regular semi-simple element $g\in G_{1/2, \mathbb {Y}, \mathrm {rs}}$ . Let ${\gamma }\in G^{\prime }_{\mathrm {rs}}$ be a matching element. Assuming the linear AFL for all strict $O_E$ -modules with the same connected height $n^\circ $ , we need to see that

(4.10) $$ \begin{align} \operatorname{\partial Orb} ({\gamma}, f^{\prime}_{1/2}) = 2\,\mathrm{Int}(g)\log(q). \end{align} $$

The condition that ${\gamma }$ and g match is by definition equivalent to assuming that $z_{\gamma }^2$ and $z_g^2$ have the same characteristic polynomial. Thus, $z_{\gamma }^2/{\pi }$ is integral if and only if $z_g^2/{\pi }$ is integral. By Corollary 3.4 and by Lemma 4.5, both sides of (4.10) vanish if these elements are not integral. So from now on we assume that $z_g^2/{\pi }$ and $z_{\gamma }^2/{\pi }$ are integral.

We may assume that ${\gamma }$ and g take the form ${\gamma } = 1 + z_{\gamma }$ and $g = 1 + z_g$ . Let ${\gamma } = ({\gamma }_0, {\gamma }_1)$ and $g = (g_0, g_1)$ be the components of ${\gamma }$ and g such that $z^2_{{\gamma }_0}/{\pi }$ and $z^2_{g_0}/{\pi }$ are topologically nilpotent, and $z_{{\gamma }_1}^2/{\pi }$ and $z_{g_1}^2/{\pi }$ integrally invertible. Then ${\gamma }_0$ matches $g_0$ and ${\gamma }_1$ matches $g_1$ .

We write $f^{\prime }_{1/2, n_0}$ and $f^{\prime }_{1/2, n_1}$ for the Parahoric test functions on $GL_{2n_0}(F)$ and $GL_{2n_1}(F)$ . By Lemma 3.7, $\operatorname {\partial Orb} ({\gamma }_1, f^{\prime }_{1/2,n_1}) = 0$ . By Corollary 3.6, we hence obtain that

(4.11) $$ \begin{align} \operatorname{\partial Orb} ({\gamma}, f^{\prime}_{1/2}) = \operatorname{\partial Orb}({\gamma}_0, f^{\prime}_{1/2, n_0}) \operatorname{Orb}({\gamma}_1, f^{\prime}_{1/2, n_1}). \end{align} $$

By Proposition 4.8, we also have the factorization

(4.12) $$ \begin{align} \mathrm{Int}(g) = \mathrm{Int}(g_0)\operatorname{Orb}(g_1, 1_{GL_{n_1}(O_B)}). \end{align} $$

By the fundamental lemma (Theorem 2.7),

$$ \begin{align*}\operatorname{Orb}({\gamma}_1, f^{\prime}_{1/2, n_1}) = \operatorname{Orb}(g_1, 1_{GL_{n_1}(O_B)}).\end{align*} $$

Thus, Identity (4.10) follows if we can prove

$$ \begin{align*}\operatorname{\partial Orb}({\gamma}_0, f^{\prime}_{1/2, n_0}) = 2\,\mathrm{Int}(g_0)\log(q).\end{align*} $$

Since $z_{{\gamma }_0}^2/{\pi }$ and $z_{g_0}^2/{\pi }$ are topologically nilpotent, by Theorem 2.11, there are two elements $\widetilde {{\gamma }}_0 \in GL_{2n_0}(F)_{\mathrm {rs}}$ and $\widetilde g_0 \in G_{0, \mathbb {Y}_0, \mathrm {rs}}$ such that

$$ \begin{align*}\operatorname{\partial Orb}(\widetilde {{\gamma}}_0, 1_{GL_{2n_0}(O_F)}) = \operatorname{\partial Orb}({\gamma}_0, f^{\prime}_{1/2, n_0}) \quad \text{and}\quad \mathrm{Int}(\widetilde g_0) = \mathrm{Int}(g_0).\end{align*} $$

The linear AFL for $\mathbb {Y}_0$ , whose connected part has height $n^\circ $ , precisely states that

$$ \begin{align*}\operatorname{\partial Orb}(\widetilde {{\gamma}}_0, 1_{GL_{2n_0}(O_F)}) = 2\,\mathrm{Int}(\widetilde g_0)\log(q),\end{align*} $$

and the proof is complete.