2.1 Pairings and Standard Lattices

We use the notation of [Reference Pappas and Rapoport18]. Let $F_0$ be a complete discretely valued field with ring of integers $O_{F_0}$ , perfect residue field k of characteristic $\neq 2$ , and uniformizer $\pi _0$ . Let $F/F_0$ be a ramified quadratic extension and $\pi \in F$ a uniformizer with $\pi ^2 = \pi _0$ . Let V be a F-vector space of dimension $n =2m> 3$ and let

$$\begin{align*}\phi: V \times V \rightarrow F \end{align*}$$

be an $F/F_0$ -hermitian form. We assume that $\phi $ is split. This means that there exists a basis $e_1, \dots , e_n$ of V such that

$$\begin{align*}\phi(e_i,e_{n+1-j}) = \delta_{i,j} \quad \text{for all} \quad i,j = 1, \dots, n. \end{align*}$$

We attach to $\phi $ the respective alternating and symmetric $F_0$ -bilinear forms $V \times V \rightarrow F_0$

$$\begin{align*}\langle x, y \rangle = \frac{1}{2}\text{Tr}_{F /F_0} (\pi^{-1}\phi(x, y)) \quad \text{and} \quad ( x, y ) = \frac{1}{2}\text{Tr}_{F /F_0} (\phi(x, y)). \end{align*}$$

For any $O_F$ -lattice $\Lambda $ in V, we denote by

$$\begin{align*}\hat{\Lambda} = \{v \in V | \langle v, \Lambda \rangle \subset O_{F_0} \} \end{align*}$$

the dual lattice with respect to the alternating form and by

$$\begin{align*}\hat{\Lambda}^s = \{v \in V | ( v, \Lambda ) \subset O_{F_0} \} \end{align*}$$

the dual lattice with respect to the symmetric form. We have $ \hat {\Lambda }^s = \pi ^{-1}\hat {\Lambda }.$ Both $\hat {\Lambda }$ and $\hat {\Lambda }^s$ are $O_F$ -lattices in V, and the forms $\langle \, , \, \rangle $ and $ (\, , \,) $ induce perfect $O_{F_0}$ -bilinear pairings

(2.1.1) $$ \begin{align} \Lambda \times \hat{\Lambda} \xrightarrow{\langle \, , \, \rangle } O_{F_0}, \quad \Lambda \times \hat{\Lambda}^s \xrightarrow{ (\, , \,)} O_{F_0} \end{align} $$

for all $\Lambda $ . Also, the uniformizing element $\pi $ induces a $O_{F_0}$ - linear mapping on $\Lambda $ which we denote by t.

For $i= 0, \dots , n- 1$ , we define the standard lattices

$$\begin{align*}\Lambda_i = \text{span}_{O_F} \{\pi^{-1}e_1, \dots, \pi^{-1}e_i, e_{i+1}, \dots, e_n\}. \end{align*}$$

We consider nonempty subsets $I \subset \{0, \dots , m\}$ with the property

(2.1.2) $$ \begin{align} m-1 \in I \Longrightarrow m \in I \end{align} $$

(see [Reference Pappas and Rapoport18, §1.2.3(b)] for more details). We complete the $\Lambda _i$ with $i \in I$ to a self-dual periodic lattice chain by first including the duals $ \Lambda _{n-i}:=\hat {\Lambda }^s_i $ for $i \neq 0$ and then all the $\pi $ -multiples: For $j \in \mathbb {Z}$ of the form $j = k \cdot n \pm i$ with $i \in I$ , we put $\Lambda _{j} = \pi ^{-k}\Lambda _i$ . Then $ \{\Lambda _j\}_j$ form a periodic lattice chain $\Lambda _I$ (with $\pi \Lambda _j = \Lambda _{j-n}$ ) which satisfies $ \hat {\Lambda }_j = \Lambda _{-j}$ . We denote by $\mathcal {L}$ such a self-dual multichain. Observe that the lattice $\Lambda _0$ is self-dual for the alternating form $\langle \,,\,\rangle $ and $\Lambda _m$ is self-dual for the symmetric form $(\,,\,)$ .

2.2 Unitary Similitude Group and Parahoric Subgroups

We consider the unitary similitude group

$$\begin{align*}G:=\mathrm{GU}(V, \phi) = \{g \in GL_F (V ) \,\, | \,\, \phi(gx, gy) = c(g)\phi(x, y), \quad c(g) \in F^{\times}_0 \}, \end{align*}$$

and we choose a partition $n = r + s$ ; we refer to the pair $(r, s)$ as the signature. By replacing $\phi $ by $-\phi $ if needed, we can make sure that $s\leq r$ , and so we assume that $s \leq r$ (see [Reference Pappas and Rapoport18, §1.1] for more details). Identifying $G \otimes F \simeq GL_{n,F} \times \mathbb {G}_{m,F}$ , we define the cocharacter $ \mu _{r,s}$ as $ (1^{(s)}, 0^{(r)}, 1)$ of $D \times \mathbb {G}_m$ , where D is the standard maximal torus of diagonal matrices in $GL_n$ ; for more details, we refer the reader to [Reference Smithling26]. We denote by E the reflex field of $\{ \mu _{r,s}\}$ ; then $E = F_0$ if $r = s$ and $E = F$ otherwise (see [Reference Pappas and Rapoport18, §1.1]). We set $O := O_E$ .

We next recall the description of the parahoric subgroups of G from [Reference Pappas and Rapoport18, §1.2], which actually follows from the results on parahoric subgroups of $\mathrm {SU}(V,\phi )$ in [Reference Pappas and Rapoport19, §1.2]. Recall that $n=2m$ is even and I is a nonempty subset of $\{0,\dots ,m\}$ satisfying (2.1.2). Consider the subgroup

$$\begin{align*}P_I:=\{g\in G\mid g\Lambda_i=\Lambda_i , \quad \forall i \in I\}. \end{align*}$$

The subgroup $P_I$ is not a parahoric subgroup in general since it may contain elements with nontrivial Kottwitz invariant. Consider the kernel of the Kottwitz homomorphism:

$$\begin{align*}P_I^0:=\{g\in P_I\mid \kappa(g)=1\}, \end{align*}$$

where $\kappa _G : G(F_0) \twoheadrightarrow \mathbb {Z} \oplus (\mathbb {Z}/2\mathbb {Z})$ (see also [Reference Smithling26, §3] for more details). We have the following (see [Reference Pappas and Rapoport18, §1.2.3(b)]):

In this paper, we will focus on the special maximal parahoric subgroup when n is even. So, we fix $n=2m$ , $I=\{m\}$ , and we let $\mathcal {L}$ be the multichain defined in 2.1 for this choice of I. Denote by ${\mathcal G}$ the (smooth) group scheme of automorphisms of the polarized chain $\mathcal {L}$ over $O_{F_0}$ ; then ${\mathcal G}$ is the parahoric group model of G in the sense of Bruhat-Tits [Reference Bruhat and Tits5] with ${\mathcal G}(O_{F_0}) =P_{\{m\}}$ . It has connected fibers (see [Reference Pappas and Rapoport18, §1.5]).

4.1 An Affine Chart $_1{\mathcal U}_{r,s}$

Recall that $n=2m= r+s$ with $s\leq r$ . In this subsection, we are going to write down the equations that define $\mathcal {M}$ in a neighborhood $_1\mathcal {U}_{r,s}$ of $(\mathcal {F}_0, t\Lambda _m)$ , where

$$\begin{align*}\Lambda_m = \text{span}_{O_F} \{\pi^{-1}e_1, \dots, \pi^{-1}e_m, e_{m+1}, \dots, e_n\} = \end{align*}$$

$$\begin{align*}\text{span}_{O_{F_0}} \{\pi^{-1}e_1, \dots, \pi^{-1}e_m, e_{m+1}, \dots, e_n,e_1, \dots, e_m, \pi e_{m+1}, \dots, \pi e_n\}. \end{align*}$$

The matrix of $(\,,\, )$ in the latter basis is

$$\begin{align*}\left[\begin{array}{@{}c|c@{}} 0_n & J_n \\ \hline -J_n & 0_n \end{array} \right], \end{align*}$$

where

$$\begin{align*}J_n = \left[\begin{array}{@{}c|c@{}} 0_m & -H_m \\ \hline H_m & 0_m \end{array} \right], \end{align*}$$

and $H_m$ is the unit antidiagonal matrix (of size m). Observe that $ J^2_n = -I_n.$ Also, since $ s \leq r$ and $ n =2m =r+s$ , we get that $s\leq m$ .

In order to find the explicit equations that describe $_1{\mathcal U}_{r,s}$ , we use similar arguments as in the proof of [Reference Krämer14, Theorem 4.5] (see also [Reference Zachos28, §3]). In our case, we consider

$$\begin{align*}\mathcal{F}_1 = \left[\begin{array}{@{}c@{}} A\\ \hline I_n \end{array} \right], \quad \mathcal{F}_0 = X = \left[\begin{array}{@{}c@{}} X_1\\ \hline X_2 \end{array} \right], \end{align*}$$

where A is of size $ n \times n$ , X is of size $2n \times s$ and $X_1,X_2$ are of size $n\times s$ ; with the additional condition that $\mathcal {F}_0$ has rank s and so X has an invertible $s\times s$ -minor. We also ask that $(\mathcal {F}_0,\mathcal {F}_1)$ satisfy the following four conditions:

1. (1) $\mathcal {F}_1^{\bot } = \mathcal {F}_1$ ,

2. (2) $ \mathcal {F}_0 \subset \mathcal {F}_1$ ,

3. (3) $(t - \pi )\mathcal {F}_0 = (0)$ ,

4. (4) $(t + \pi ) \mathcal {F}_1 \subset \mathcal {F}_0$ .

Observe that

$$\begin{align*}M_t = \left[\begin{array}{@{}c|c@{}} 0_n & \pi_0I_n \\ \hline I_n & 0_n \end{array} \right] \end{align*}$$

of size $2n \times 2n $ is the matrix giving multiplication by t.

Condition (1) (i.e., $\mathcal {F}_1$ is isotropic) translates to $A^{t} = -J_nAJ_n$ . Condition (2) (i.e., $ \mathcal {F}_0 \subset \mathcal {F}_1$ ) implies that the generators of $\mathcal {F}_0$ can be expressed as a linear combination of the generators in $\mathcal {F}_1$ , which translates to

$$\begin{align*}\exists \, Y : X = \left[\begin{array}{@{}c@{}} A\\ \hline I_n \end{array} \right] \cdot Y, \end{align*}$$

where Y is of size $ n\times s$ . Thus, we have

$$\begin{align*}\left[\begin{array}{@{}c@{}} X_1\\ \hline X_2 \end{array} \right] = \left[\begin{array}{@{}c@{}} A\\ \hline I_n \end{array} \right] \cdot Y = \left[\begin{array}{@{}c@{}} A Y\\ \hline Y \end{array} \right], \end{align*}$$

and so $ X_1 = AY$ and $X_2 = Y.$ Condition (3) (i.e., $(t - \pi )\mathcal {F}_0 = (0)$ ) is equivalent to

$$\begin{align*}M_t \cdot X = \left[\begin{array}{@{}c@{}} \pi X_1\\ \hline \pi X_2 \end{array} \right], \end{align*}$$

which amounts to

$$\begin{align*}\left[\begin{array}{@{}c@{}} \pi_0 X_2\\ \hline X_1 \end{array} \right]= \left[\begin{array}{@{}c@{}} \pi X_1\\ \hline \pi X_2 \end{array} \right]. \end{align*}$$

Thus, $X_1 = \pi X_2$ , which translates to $A Y = \pi Y $ by condition $(2)$ . The last condition $(t + \pi ) \mathcal {F}_1 \subset \mathcal {F}_0$ translates to

$$\begin{align*}\exists \, Z : M_t \cdot \left[\begin{array}{@{}c@{}} A \\ \hline I_n \end{array} \right] + \left[\begin{array}{@{}c@{}} \pi A \\ \hline \pi I_n Y \end{array} \right] = X \cdot Z^t, \end{align*}$$

where Z is of size $n\times s$ . This amounts to

$$\begin{align*}\left[\begin{array}{@{}c@{}} \pi_0I_n + \pi A \\ \hline A + \pi I_n \end{array} \right] = \left[\begin{array}{@{}c@{}} X_1 Z^t \\ \hline X_2 Z^t \end{array} \right]. \end{align*}$$

From the above, we get $A + \pi I_n = X_2 Z^t$ , which by condition (2) translates to $ A = Y Z^t-\pi I_n. $ Thus, A can be expressed in terms of $Y,Z$ . In addition, by condition (2) and in particular by the relations $ X_1 = AY$ and $X_2 = Y$ , we deduce that the matrix X is given in terms of $Y,Z$ . Conversely, from $Y=X_2$ , we get that the matrix Y is given in terms of $A,X$ (Below we will also show that Z can be expressed in terms of $A,X$ ).

Observe from $X_1 = \pi X_2$ that all the entries of $X_1$ are in the maximal ideal and thus a minor involving entries of $X_1$ cannot be a unit. Recall that the matrix X has an invertible $s\times s$ -minor and $X_2=Y$ from condition (2). Thus, the matrix Y has a $s\times s$ -minor. Let

$$\begin{align*}Y = \left[\begin{array}{ccc} y_{1,1} &\cdots & y_{1,s} \\ \vdots & \ddots & \vdots \\ y_{n,1} & \cdots & y_{n,s} \end{array}\right] , \quad Z = \left[\begin{array}{ccc} z_{1,1} &\cdots & z_{1,s} \\ \vdots & \ddots & \vdots \\ z_{n,1} & \cdots & z_{n,s} \end{array}\right] = \left[\begin{array}{ccc} \mathbf{z}_1 \ \cdots \ \mathbf{z}_s \end{array}\right], \end{align*}$$

and

(4.1.1) $$ \begin{align} Y'=\left[\begin{array}{ccc} y_{i_1,1} &\cdots & y_{i_1,s} \\ \vdots & \ddots & \vdots \\ y_{i_s,1} & \cdots & y_{i_s,s} \end{array}\right] = I_s, \end{align} $$

where $1 \leq i_k \leq n$ for $k = 1, \dots ,s$ . Here, we want to mention how we can express Z in terms of $A,X$ . From the above, we have $Y Z^t=A+ \pi I_n$ and

$$\begin{align*}Y Z^t = \left[\begin{array}{c} \sum^s_{e=1} y_{1,e} \mathbf{z}^t_e \\ \vdots \\ \sum^s_{e=1} y_{n,e} \mathbf{z}^t_e \end{array}\right]. \end{align*}$$

By using (4.1.1), for any $1 \leq k \leq s$ , the $i_k$ -th row of the matrix $Y Z^t $ gives $\mathbf {z}^t_k $ . Since $Y Z^t=A+ \pi I_n$ , we can easily see that the matrix Z can be expressed in terms of $A,X$ .

We replace A by $Y Z^t-\pi I_n$ . Hence, conditions ( $1$ ) and ( $3$ ) are equivalent to

(4.1.2) $$ \begin{align} Z Y^t &= -J_nY Z^tJ_n, \end{align} $$

(4.1.3) $$ \begin{align} Y Z^t Y &= 2 \pi Y. \qquad\quad \end{align} $$

From the above, we deduce the following.

Lemma 4.1. An affine chart $_1{\mathcal U}_{r,s}$ of $\mathcal {M}$ around $(\mathcal {F}_0, t \Lambda _m)$ is given by $\mathrm {Spec \, } O_F[Y , Z ]/\mathscr {I} $ , where

$$\begin{align*}\mathscr{I} = (Y'-I_s,\ Z Y^t +J_nY Z^tJ_n,\ Y Z^t Y - 2 \pi Y) \end{align*}$$

and $Y'$ is a submatrix composed of s rows of Y (see the relation (4.1.1)).

Note that the affine chart $_1{\mathcal U}_{r,s}$ of $\mathcal {M}$ depends on the matrix $Y'$ we fixed (see Remark 4.4). Our next goal is to give a simplification of the above equations (see Theorem 4.2). We first study the relation $Z Y^t =-J_nY Z^tJ_n$ . Consider the matrix $P = (p_{i,j})$ of size $n=2m$ with $P^t=-J_nPJ_n$ . By a direct calculation, we get

$$\begin{align*}p_{i,j}=p_{n+1-j,n+1-i} \end{align*}$$

for $1\leq i, j\leq m$ or $m+1\leq i,j\leq n$ , and

$$\begin{align*}p_{i,j}=-p_{n+1-j,n+1-i} \end{align*}$$

for $1\leq i\leq m$ and $m+1\leq j\leq n$ , or $1\leq j\leq m$ and $m+1\leq i\leq n$ . By using this observation and by setting $P = Y Z^t$ , the relation $ Z Y^t=-J_nY Z^tJ_n$ gives

(4.1.4) $$ \begin{align} \sum^s_{k=1} z_{i,k}y_{j,k} = \pm \sum^s_{k=1} z_{n+1-j,k}y_{n+1-i,k} \end{align} $$

for $1\leq i,j \leq n$ , where the different signs $\pm $ depends on the position of $i,j$ as above. Consider the following 2 cases:

Case 1: Assume that $ 1\leq i_k \leq m$ for all $k \in \{1,\dots ,s\}.$ By setting $i = n+1-i_q $ and $j= i_p$ , the equation (4.1.4) gives $ z_{n+1-i_q,p} = -z_{n+1-i_p,q} .$ In particular, if $p=q$ , we get $z_{n+1-i_p,p} = 0 $ for all $p \in \{1,\dots ,s\}$ . By setting $ d_{p,q} = z_{n+1-i_p,q}$ , we get the matrix $D = (d_{p,q})_{p,q}$ of size $s \times s$ such that

(4.1.5) $$ \begin{align} D = \left[\begin{array}{ccc} d_{1,1} &\cdots & d_{s,1} \\ \vdots & \ddots & \vdots \\ d_{1,s} & \cdots & d_{s,s} \end{array}\right] = \left[\begin{array}{ccc} z_{n+1-i_1,1} &\cdots & z_{n+1-i_s,1} \\ \vdots & \ddots & \vdots \\ z_{n+1-i_1,s} & \cdots & z_{n+1-i_s,s} \end{array}\right]. \end{align} $$

The matrix D is skew-symmetric (i.e., $D = -D^t).$ Next, by setting $j = i_k$ in (4.1.4), we obtain

$$\begin{align*}z_{i,k} = \left\{ \begin{array}{ll} \sum^s_{e=1} d_{k,e}y_{n+1-i,e} & \mbox{if } 1 \leq i \leq m, \\ -\sum^s_{e=1} d_{k,e}y_{n+1-i,e} & \mbox{if } m+1\leq i \leq n. \end{array} \right. \end{align*}$$

Thus, we have $ Z = \widetilde {Y} \cdot D$ , where

$$\begin{align*}\widetilde{Y} = \left[\begin{array}{ccc} y_{n,1} & \cdots & y_{n,s} \\ \vdots & \ddots & \vdots \\ y_{m+1,1} & \cdots & y_{m+1,s} \\ \hline -y_{m,1} & \cdots & -y_{m,s} \\ \vdots & \ddots & \vdots \\ -y_{1,1} & \cdots & -y_{1,s} \\ \end{array}\right]. \end{align*}$$

Therefore, in this case, the equations (4.1.4) give $ Z = \widetilde {Y} \cdot D $ .

Case 2: In general, we let $ 1 \leq t \leq s$ and assume that

$$\begin{align*}1\leq i_1, \dots, i_t \leq m \quad \text{and} \quad m+1\leq i_{t+1}, \dots, i_s \leq n. \end{align*}$$

By setting $d_{p,q} = z_{n+1-i_p,q}$ and using the same arguments as above, we get the matrix $D = (d_{p,q})_{p,q}$ of size $s \times s$ such that

$$\begin{align*}d_{p,q}=-d_{q,p} \end{align*}$$

for $1\leq p, q \leq t$ or $t+1\leq p,q\leq n$ , and

$$\begin{align*}d_{p,q}=d_{q,p} \end{align*}$$

for $1\leq p\leq t$ and $t+1\leq q\leq n$ , or $1\leq q\leq t$ and $t+1\leq p\leq n$ . Next, by setting $j = i_k$ in (4.1.4) for $1\leq k \leq t$ , we obtain

$$\begin{align*}z_{i,k} = \left\{ \begin{array}{ll} \sum^s_{e=1} d_{k,e}y_{n+1-i,e} & \mbox{if } 1 \leq i \leq m, \\ -\sum^s_{e=1} d_{k,e}y_{n+1-i,e} & \mbox{if } m+1\leq i \leq n. \end{array} \right. \end{align*}$$

Also, by setting $j = i_k$ for $t+1\leq k \leq s$ , we get

$$\begin{align*}z_{i,k} = \left\{ \begin{array}{ll} -\sum^s_{e=1} d_{k,e}y_{n+1-i,e} & \mbox{if } 1 \leq i \leq m, \\ \sum^s_{e=1} d_{k,e}y_{n+1-i,e} & \mbox{if } m+1\leq i \leq n. \end{array} \right. \end{align*}$$

From the above, we have $ Z = \widetilde {Y} \cdot \widetilde {D}$ , where $\widetilde {D} = (\widetilde {d}_{i,j})_{i,j}$ is the $s \times s$ matrix defined as

$$\begin{align*}\widetilde{D} = \left[\begin{array}{ccc|ccc} d_{1,1} & \cdots & d_{t,1} & -d_{t+1,1} & \cdots & -d_{s,1}\\ \vdots & \ddots & \vdots & \vdots & \ddots & \vdots \\ d_{1,s} & \cdots & d_{t,s} & -d_{t+1,s} & \cdots & -d_{s,s}\\ \end{array}\right]. \end{align*}$$

It is easy to see that the matrix $ \widetilde {D}$ is skew symmetric (i.e., $\widetilde {d}_{i,j} = - \widetilde {d}_{j,i}).$ In particular, we have the following 4 cases: when $1 \leq i,j \leq t$ , we have $ \widetilde {d}_{i,j} = d_{i,j} = -d_{j,i} = - \widetilde {d}_{j,i}$ . For $1 \leq i \leq t$ and $t+1 \leq j \leq s $ , we have $ \widetilde {d}_{i,j} = -d_{i,j} = -d_{j,i} = - \widetilde {d}_{j,i}$ . When $1 \leq j \leq t$ and $t+1 \leq i \leq s $ , we get $ \widetilde {d}_{i,j} = d_{i,j} = d_{j,i} = - \widetilde {d}_{j,i}$ . Lastly, for $t+1 \leq i,j \leq s$ , we obtain $ \widetilde {d}_{i,j} = -d_{i,j} = d_{j,i} = - \widetilde {d}_{j,i}$ .

Combining the above cases, for any $1\leq i_1,\dots , i_s\leq n$ , the equations (4.1.4) show that there exists a skew symmetric matrix $\mathcal {D}$ of size $s \times s$ such that

$$\begin{align*}Z = \widetilde{Y} \cdot \mathcal{D}. \end{align*}$$

By abuse of notation, write $ D$ for $ \mathcal {D}$ . Next, define $ Q : =\widetilde {Y}^t \cdot Y $ of size $ s \times s.$ Note that $ Q^t = -Q.$ We are now ready to show the following:

Theorem 4.2. For any signature $(r,s)$ , there is an affine chart $_1{\mathcal U}_{r,s} \subset \mathcal {M}$ which is isomorphic to $\mathrm {Spec \, } O_F[ D, Y]/I $ , where

$$\begin{align*}I = \left(Y'-I_s,\ D+D^t,\ D\cdot Q + 2 \pi I_s\right) \end{align*}$$

for some choice of $Y'$ .

Proof. Recall from Lemma 4.1 that ${\mathcal U} = \mathrm {Spec \, } O_F[Y , Z ]/\mathscr {I} $ , where

$$\begin{align*}\mathscr{I} = (Y'-I_s,\ Z Y^t +J_nY Z^tJ_n,\ Y Z^t Y - 2 \pi Y). \end{align*}$$

From the above discussion, we have that the relation $Z Y^t =-J_nY Z^tJ_n $ gives $ Z = \widetilde {Y} \cdot D$ . Thus, it suffices to focus on the relation $ Y Z^t Y - 2 \pi Y=0$ . Observe that $ Y \cdot Z^t \cdot Y =- Y \cdot D \cdot Q.$ Thus,

$$\begin{align*}Y Z^t Y - 2 \pi Y = -Y\cdot( D \cdot Q + 2 \pi I_s)=0. \end{align*}$$

Now, by using the relation with the minors (4.1.1), we can easily see that the relation $Y Z^t Y - 2 \pi Y=0$ gives $D \cdot Q + 2 \pi I_s=0.$ Therefore, $_1{\mathcal U}_{r,s}$ is isomorphic to

$$\begin{align*}\mathrm{Spec \, } \frac{O_F[D,Y]}{\left(Y'-I_s,\ D+D^t, \ D\cdot Q + 2 \pi I_s\right)}.\\[-44pt] \end{align*}$$

In particular, when the signature $(r,s)=(n-2,2)$ , we have

$$\begin{align*}D=\left[\begin{array}{cc} 0&d\\ -d&0 \end{array}\right],\quad Q=\left[\begin{array}{cc} 0&q\\[2pt] -q&0 \end{array}\right], \end{align*}$$

where $d=d_{2,1}=-d_{12}$ and $q=\sum ^m_{j=1}y_{n+1-j,1}y_{j,2} - \sum ^n_{j=m+1}y_{n+1-j,1}y_{j,2}$ .

Corollary 4.3. When $(r,s)=(n-2,2)$ , there is an affine chart $_1\mathcal {U}_{r,s} \subset \mathcal {M}$ which is isomorphic to $V(I) = \mathrm {Spec \, } O_F[d,(y_{i,1})_{1 \leq i \leq n},(y_{i,2})_{1 \leq i \leq n} ]/ I$ , where

$$\begin{align*}I= ( y_{i_0,1}-1,\,\, y_{j_0,2} -1 ,\,\,y_{j_0,1},\,y_{i_0,2},\,\, d(\sum^m_{j=1}y_{n+1-j,1}y_{j,2} - \sum^n_{j=m+1}y_{n+1-j,1}y_{j,2}) -2 \pi) \end{align*}$$

for some $1\leq i_0,j_0 \leq n$ .

4.2 An Affine Chart $_2{\mathcal U}_{r,s}$

Keep the same notation as above. As in [Reference Pappas and Rapoport18], we set $\Lambda _m=\Lambda '\oplus \Lambda "$ , where

$$\begin{align*}\begin{array}{l} \Lambda'=\text{span}_{O_{F_0}}\{-\pi^{-1}e_1, -e_1,\dots,-e_m,\pi e_{m+1},\dots, \pi e_{n-1}\},\\ \Lambda"=\text{span}_{O_{F_0}}\{e_n, \pi e_n, -\pi^{-1}e_2, \dots, -\pi^{-1}e_m, e_{m+1}, \dots, e_{n-1}\}. \end{array} \end{align*}$$

In this subsection, we are going to write down the equations that define $\mathcal {M}$ in a neighborhood $_2\mathcal {U}_{r,s}$ of $(\mathcal {F}_0, \Lambda ')$ .

The matrix of the form $(\ ,\ )$ in this basis is

$$\begin{align*}\left[ \begin{array}{c|c} 0_n & S\\ \hline S^t & 0_n \end{array} \right],\end{align*}$$

where S is the skew-symmetric matrix of size n given by

$$\begin{align*}S=\left[ \begin{array}{c|c} J_2^t & 0\\ \hline 0 & J_{n-2} \end{array} \right]. \end{align*}$$

Observe that $S^2=-I_n$ since $J_n^2=-I_n$ . For any point $(\mathcal {F}_0,\mathcal {F}_1)\in \mathcal {M}$ , we ask that $ (\mathcal {F}_0,\mathcal {F}_1)$ satisfy the following conditions:

1. (1) $\mathcal {F}_1^{\bot } = \mathcal {F}_1$ ,

2. (2) $ \mathcal {F}_0 \subset \mathcal {F}_1$ ,

3. (3) $(t - \pi )\mathcal {F}_0 = (0)$ ,

4. (4) $(t + \pi ) \mathcal {F}_1 \subset \mathcal {F}_0$ .

Similar to §4.1, we consider

$$\begin{align*}\mathcal{F}_1=\left[\begin{array}{c} I_n\\ \hline A \end{array} \right],\quad \mathcal{F}_0=X=\left[\begin{array}{c} X_1\\ \hline X_2 \end{array} \right], \end{align*}$$

where $I_n, A$ are $n\times n$ -matrices, and $X_1, X_2$ are $n\times s$ -matrices. Recall that $\mathcal {F}_0$ has rank s, so X has a no-vanishing $s\times s$ minor. Write down the matrix A by

$$\begin{align*}A=\left[\begin{array}{c|c} T & B\\ \hline C & Y \end{array} \right], \end{align*}$$

where T is a $2\times 2$ -matrix, B (resp. C) is a $2\times (n-2)$ -matrix (resp. $(n-2)\times 2$ -matrix) and Y is of size $(n-2)\times (n-2)$ . Then condition (1) is equivalent to $A^t S=SA$ . This translates to

$$\begin{align*}T^t=-J_2TJ_2,\quad C=J_{n-2}B^tJ_2,\quad Y^t=-J_{n-2}YJ_{n-2}. \end{align*}$$

Note that the first equation immediately gives $T=\text {diag}(x,x)$ for some variable x, and the second equation shows that C is given in terms of B. Condition (2) is equivalent to

$$\begin{align*}\exists M: \left[\begin{array}{c} X_1\\ \hline X_2 \end{array} \right]= \left[\begin{array}{c} I_n\\ \hline A \end{array} \right]\cdot M, \end{align*}$$

where M is a $n\times s$ -matrix. This amounts to $X_1=M, X_2=AM$ . Let

$$\begin{align*}M=\left[\begin{array}{c} M_1\\ \hline M_2 \end{array} \right], \end{align*}$$

where $M_1$ is of size $2\times s$ and $M_2$ is of size $(n-2)\times s$ . We can see that $X_1, X_2$ can be expressed in terms of $A, M_1, M_2$ . More precisely, we get

$$\begin{align*}X_1=M=\left[\begin{array}{c} M_1\\ \hline M_2 \end{array} \right],\quad X_2=AM=\left[\begin{array}{c} xM_1+BM_2\\ \hline CM_1+YM_2 \end{array} \right]. \end{align*}$$

The map $t: \Lambda _m\rightarrow \Lambda _m$ is represented by the matrix

$$\begin{align*}M_t=\left[\begin{array}{cccc} t_0&&&\\ &&& I_{n-2}\\ && t_0 &\\ & \pi_0 I_{n-2} && \end{array} \right], \end{align*}$$

where $t_0$ is the $2\times 2$ -antidiag matirx $\left [\begin {array}{cc} & \pi _0\\ 1& \end {array} \right ]$ . Condition (3) translates to

$$\begin{align*}M_t\cdot\left[ \begin{array}{c} X_1\\ \hline X_2 \end{array}\right]=\left[ \begin{array}{c} \pi X_1\\ \hline \pi X_2 \end{array}\right]. \end{align*}$$

By setting $B_i$ as the i-th row of B and by using condition (2), the above relation amounts to

$$\begin{align*}M_1=\left[\begin{array}{c} \pi M_0\\ \hline M_0 \end{array} \right],\quad B_1M_2=\pi B_2M_2,\quad CM_1+YM_2=\pi M_2, \end{align*}$$

where $M_0$ is the second row of $M_1$ . Finally, condition (4) is equivalent to

$$\begin{align*}(M_t+\pi I_{2n})\cdot \left[\begin{array}{cc} I_2 &\\ & I_{n-2}\\ T & B\\ C & Y \end{array} \right]=\left[\begin{array}{c} M_1\\ M_2\\ xM_1+BM_2\\ CM_1+YM_2 \end{array} \right]\cdot \left[\begin{array}{cc} Z_1^t & Z_2^t \end{array} \right], \end{align*}$$

where $Z_1$ is of size $2\times s$ and $Z_2$ is of size $(n-2)\times s$ . This, in turn, gives

$$\begin{align*}\begin{array}{lll} M_0Z_1^t=[1\ \pi], & M_0Z_2^t=0, & M_2Z_1^t=C,\\ M_2Z_2^t=\pi I_{n-2} +Y, & Y^2= \pi_0 I_{n-2}, & B_1=B_2Y. \end{array} \end{align*}$$

From the above equations, we can see that A is given in terms of $x, C, Y$ by condition (1). Note that $C=M_2Z_1^t$ , and $Y=M_2Z_2^t-\pi I_{n-2}$ by condition (4). Thus, the matrix A is given in terms of $M, Z_1, Z_2$ and a variable x. Meanwhile, we have $X_1=M, X_2=AM$ by condition (2), so that $X_1, X_2$ are also given in terms of $M, Z_1, Z_2$ . However, we have $M=X_1$ , and we claim that $Z_1, Z_2$ can be expressed in terms of $A, X_1, X_2$ (we will show this later in this section).

Before we move on, let us simplify our equations first. By replacing $Y=M_2Z_2^t-\pi I_{n-2}$ , the equation $Y^t=-J_{n-2}YJ_{n-2}$ from condition (1) amounts to $(M_2Z_2^t)^t=-J_{n-2}(M_2Z_2^t)J_{n-2}$ , and the equations that involve Y from condition (3), (4) translate to

$$\begin{align*}CM_1+(M_2Z_2^t)M_2=2\pi M_2,\quad (M_2Z_2^t)^2=2\pi(M_2Z_2^t),\quad B_1+\pi B_2=B_2(M_2Z_2^t). \end{align*}$$

Finally, we replace C by $M_2Z_1^t$ , and deduce the following.

Lemma 4.5. The affine chart $_2{\mathcal U}_{r,s}$ of $\mathcal {M}$ around $(\mathcal {F}_0, \Lambda ')$ is given by

$$\begin{align*}\mathrm{Spec \, } O_F[x, B, M_0, M_2, Z_1, Z_2]/\mathscr{I}, \end{align*}$$

where $\mathscr {I}$ is the ideal generated by the entries of following equations:

$$\begin{align*}\begin{array}{lll} M_2Z_1^t=J_{n-2}B^tJ_2, & (M_2Z_2^t)^t=-J_{n-2}(M_2Z_2^t)J_{n-2}, & (M_2Z_2^t)^2=2\pi(M_2Z_2^t),\\ B_1M_2=\pi B_2M_2, &B_1+\pi B_2=B_2(M_2Z_2^t), & M_0Z_2^t=0,\\ M_0Z_1^t=[1\ \pi], &(M_2Z_1^t)M_1+(M_2Z_2^t)M_2=2\pi M_2. & \end{array} \end{align*}$$

Here, $B_i$ is the i-th row of B for $i=1,2$ .

Remark 4.6. Condition (3) also gives the equations:

$$\begin{align*}BC=0,\quad Ct_0+YC=0. \end{align*}$$

In [Reference Pappas and Rapoport18, §5.3], the authors show that these two equations are automatically true if $B,C,Y$ satisfy

$$\begin{align*}\begin{array}{ll} Y^2=\pi_0 I_{n-2}, & Y^t=-J_{n-2}YJ_{n-2},\\ B_1=B_2Y, & C=J_{n-2}B^tJ_2. \end{array} \end{align*}$$

4.2.1 The case $(r,s)=(n-1,1)$

In this and the next subsection, our goal is to give a simplification of the equations in Lemma 4.5. We first consider the affine chart $_2\mathcal {U}_{n-1,1} \subset \mathcal {M}$ for the signature $(n-1,1)$ . Note that $M_0$ is of size $1\times 1$ in this case. Since

$$\begin{align*}M_0 Z_1^t=[1\ \pi]\neq 0, \end{align*}$$

we get $M_0\neq 0$ . Without loss of generality, we assume $M_0=1$ as rank( $\mathcal {F}_0$ )=1.

Theorem 4.7. When the signature $(r,s)=(n-1,1)$ , the affine chart $_2\mathcal {U}_{n-1,1}$ is smooth and isomorphic to $\mathbb {A}_{O_F}^{n-1}$ .

Proof. From $M_0Z_1^t=[1\ \pi ]$ , $M_0Z_2^t=0$ , we obtain

$$\begin{align*}Z_1^t=\left[\begin{array}{cc} 1& \pi\\ \end{array} \right],\quad Z_2^t=0. \end{align*}$$

Let $M_2=[a_1\ \cdots \ a_{n-2}]^t$ be an $(n-2)\times 1$ vector. Then $C=M_2Z_1^t$ amounts to

$$\begin{align*}C=\left[\begin{array}{c|c} a_1 &\pi a_1\\ \vdots & \vdots \\ a_{m-1} &\pi a_{m-1}\\ \hline a_{m} &\pi a_{m}\\ \vdots & \vdots \\ a_{n-2}& \pi a_{n-2} \end{array} \right]. \end{align*}$$

Also, from $B=J_2C^tJ_{n-2}$ , we get

$$\begin{align*}B=\left[\begin{array}{ccc|ccc} -\pi a_{n-2} &\cdots &-\pi a_m& \pi a_{m-1}& \cdots &\pi a_1 \\ \hline a_{n-2} &\cdots & a_m& -a_{m-1}& \cdots & -a_1 \end{array}\right]. \end{align*}$$

It is easy to see that the relation $Y=M_2Z_2^t-\pi I_{n-2}$ gives $Y=-\pi I_{n-2}$ and $B_1+\pi B_2=B_2(M_2Z_2^t)=0$ . Finally, we check that

$$\begin{align*}\begin{array}{l} B_1M_2=\pi B_2M_2=\pi(\sum_{i=1}^{m-1}a_ia_{n-1-i}-\sum_{i=1}^{m-1}a_ia_{n-1-i})=0,\\ CM_1+(M_2Z_2^t)M_2=2\pi\cdot [a_1\ \cdots\ a_{n-2}]^t=2\pi M_2. \end{array} \end{align*}$$

From the above, the conditions on the $a_i$ are automatically satisfied. We deduce that the affine chart $_2\mathcal {U}_{n-1,1}$ is isomorphic to

$$\begin{align*}\mathrm{Spec \, } O_F[x, a_1, \dots, a_{n-2}]=\mathbb{A}_{O_F}^{n-1}.\\[-38pt] \end{align*}$$

4.2.2 The cases for $s\geq 2$

In this subsection, we consider the explicit equations of $_2\mathcal {U}_{r,s}$ for the signature $(r,s)$ with $s\geq 2$ . Observe that $\mathcal {F}_1=\{v+A(v)\mid v\in \Lambda '\otimes {\mathcal O}_S \}$ . In the special fiber (where ${\mathcal O}_S=k$ ), we have $t\mathcal {F}_1\subset \mathcal {F}_0$ , and $t(\Lambda '\otimes k)=\text {span}_{O_{F_0}}\{-e_1\}$ , which implies $-e_1+tA(v)\in \mathcal {F}_0$ . Since

$$\begin{align*}\mathcal{F}_0=X=\left[\begin{array}{c} M_1\\ M_2 \\ \hline xM_1+BM_2\\ CM_1+YM_2 \end{array} \right], \end{align*}$$

we can always assume that $-e_1+tA(v)$ is the first generating element of $\mathcal {F}_0$ . In the matrix language, this is equivalent to assume that the second row of X (i.e., $M_0$ ) is equal to $[1\ 0 \cdots 0]$ . Thus, we have

$$\begin{align*}M_1=\left[\begin{array}{cccc} \pi & 0&\cdots& 0\\ 1& 0&\cdots &0 \end{array} \right]. \end{align*}$$

By using $M_0Z_1^t=[1 \ \pi ]$ , $M_0Z_2^t=0$ , we set

$$\begin{align*}Z_1^t=\left[\begin{array}{cc} 1& \pi \\ z_{1,2} & z_{2,2}\\ \vdots &\vdots\\ z_{1,s} & z_{2,s} \end{array} \right], \quad Z_2^t=\left[\begin{array}{ccc} 0&\cdots & 0 \\ z_{1,2}'&\cdots & z_{n-2,2}'\\ \vdots & \ddots & \vdots \\ z_{1,s}'&\cdots & z_{n-2,s}' \end{array} \right], \end{align*}$$

and

$$\begin{align*}M_2=\left[\begin{array}{cccc} \mathbf{a_1} & \cdots &\mathbf{a_s} \end{array}\right]= \left[\begin{array}{ccc} a_{1,1}& \cdots & a_{1,s} \\ \vdots & \ddots &\vdots \\ a_{n-2,1} & \cdots & a_{n-2,s} \end{array} \right]. \end{align*}$$

We claim that $Z_1, Z_2$ can be expressed in terms of $A, X_1, X_2$ . Observe from $CM_1+YM_2=\pi M_2$ that all the entries of $CM_1+YM_2$ are in the maximal ideal, and thus a minor involving entries of $CM_1+YM_2$ cannot be a unit. Similarly, the entries of the first row of $xM_1+BM_2$ are also in the maximal ideal. By replacing the order of basis $\{\pi e_m,\dots , \pi e_n\}$ if necessary, we can assume that the matrix $M_2$ has an invertible $(s-1)\times (s-1)$ -minor. Let $M_2'$ be a submatrix of $M_2$ composing $(s-1)$ rows along with the corresponding s columns. We set

$$\begin{align*}M_2'=\left[\begin{array}{cccc} a_{i_2,1}& a_{i_2,2} &\cdots & a_{i_2,s} \\ \vdots & \vdots &\ddots &\vdots \\ a_{i_s,1} & a_{i_s,2}&\cdots & a_{i_s,s} \end{array} \right]= \left[\begin{array}{cccc} 0& 1 & \cdots & 0 \\ \vdots & \vdots & \ddots &\vdots \\ 0 & 0 & \cdots & 1 \end{array} \right] \end{align*}$$

for $1\leq i_2, \cdots , i_s\leq n-2$ . Consider the matrix $C=M_2Z_1^t$ . We have

$$\begin{align*}C=\left[\begin{array}{cc} \mathbf{a_1}+z_{1,2}\mathbf{a_2}+\cdots +z_{1,s}\mathbf{a_s} & \pi \mathbf{a_1}+z_{2,2}\mathbf{a_2}+\cdots +z_{2,s}\mathbf{a_s} \end{array}\right]. \end{align*}$$

By setting $i=i_2, \dots , i_s$ , it is easy to see that $Z_1$ can be expressed in terms of C. Similarly, $Z_2$ can be expressed in terms of Y as $M_2Z_2^t=\pi I_{n-2}+Y$ . Thus, $Z_1, Z_2$ are given in terms of $A, X_1, X_2$ .

Set

$$\begin{align*}D=\left[\begin{array}{ccc} d_{2,2}&\cdots & d_{2,s}\\ \vdots &\ddots &\vdots\\ d_{s,2}&\cdots & d_{s,s} \end{array}\right],\quad Q=\left[\begin{array}{ccc} q_{2,2}&\cdots & q_{2,s}\\ \vdots &\ddots &\vdots\\ q_{s,2}&\cdots & q_{s,s} \end{array}\right], \end{align*}$$

where $q_{i,j}=\sum _{k=1}^{m-1}(a_{k,i}a_{n-1-k,j}-a_{n-1-k,i}a_{k,j})$ . The matrix Q depends on $M_2$ . Note that $Q^t=-Q$ by definition.

Theorem 4.8. For any signature $(r,s)$ with $s\geq 2$ , there is an affine chart $_2\mathcal {U}_{r,s}\subset \mathcal {M}$ which is isomorphic to

$$\begin{align*}\mathbb{A}_{O_F}^{s-1}\times \mathrm{Spec \, } O_F[D, M_2]/I, \end{align*}$$

where

$$\begin{align*}I=(M_2'-[\mathbf{0}\mid I_{s-1}],\ D+D^t,\ D\cdot Q+2\pi I_{s-1}) \end{align*}$$

for some choice of $M^{\prime }_2$ . $[\mathbf {0}\mid I_{s-1}]$ is of size $(s-1)\times s$ , with the first column $0$ .

Proof. The proof is similar to the proof of Theorem 4.2. Consider $(M_2Z_2^t)^t=-J_{n-2}(M_2Z_2^t)J_{n-2}$ first. We claim that $Z_2^t$ depends on $M_2$ and D. Since $M_2Z_2^t=(\sum _{k=2}^{s}a_{i,k}z_{j,k}')_{ij}$ , the relation $(M_2Z_2^t)^t=-J_{n-2}(M_2Z_2^t)J_{n-2}$ amounts to

$$\begin{align*}a_{i,2}z_{j,2}'+\cdots+a_{i,s}z_{j,s}'=\pm(a_{n-1-j,2} z_{n-1-i,2}'+\cdots +a_{n-1-j,s}z_{n-1-i,s}'), \end{align*}$$

where the different sign $\pm $ depends on the position of $i_2,\dots , i_s$ . We obtain

$$\begin{align*}z_{n-1-i_2,2}'=z_{n-1-i_3,3}'=\dots =z_{n-1-i_s,s}'=0, \end{align*}$$

by letting $i=i_k$ , $j=n-1-i_k$ for $k=2,\dots ,s$ . To write down $Z_2^t$ explicitly, we need to consider the positions of $i_2,\dots ,i_s$ .

Case 1: Assume that $1\leq i_2,\dots ,i_s\leq m-1$ . In this case, we have $z_{n-1-i_p,q}'=-z_{n-1-i_q,p}'$ by letting $i=i_{p}$ , $j=n-1-i_{q}$ . Set $d_{p,q}=z_{n-1-i_p,q}'$ . We obtain

$$\begin{align*}\left[\begin{array}{ccc} z_{n-1-i_2,2}'& \cdots & z_{n-1-i_2,s}'\\ \vdots &\ddots & \vdots\\ z_{n-1-i_s,2}'& \cdots & z_{n-1-i_s,s}' \end{array}\right] = \left[\begin{array}{ccc} d_{2,2}&\cdots & d_{2,s}\\ \vdots &\ddots &\vdots\\ d_{s,2}&\cdots & d_{s,s} \end{array}\right]=D, \end{align*}$$

where $d_{p,q}=-d_{q,p}$ for $2\leq p,q\leq s$ . This implies that D is skew-symmetric.

For $i=i_2$ , we have

$$\begin{align*}z_{j,2}'=\pm(a_{n-1-j,2}d_{2,2}+\cdots +a_{n-1-j,s}d_{2,s}). \end{align*}$$

Since $1\leq i_2\leq m-1$ , the sign $\pm $ is positive when $1\leq j\leq m-1$ , and is negative when $m\leq j\leq n-2$ . Similar to $i=i_k$ for $k=3,\dots , s$ . We have

$$\begin{align*}Z_2^t= \left[\begin{array}{ccc} 0&\cdots &0\\ d_{2,2}&\cdots & d_{2,s}\\ \vdots &\ddots &\vdots\\ d_{s,2}&\cdots & d_{s,s} \end{array}\right]\cdot \left[\begin{array}{ccc|ccc} a_{n-2,2}&\cdots & a_{m,2} & -a_{m-1,2}&\cdots & -a_{1,2}\\ \vdots & \ddots & \vdots &\vdots &\ddots &\vdots\\ a_{n-2,s}&\cdots & a_{m,s} & -a_{m-1,s} & \cdots &-a_{1,s} \end{array} \right]. \end{align*}$$

Case 2: In general, we assume that

$$\begin{align*}1\leq i_2,\dots, i_t\leq m-1 \quad \text{and} \quad m\leq i_{t+1},\dots, i_s\leq n-2. \end{align*}$$

Set $d_{p,q}=z_{n-1-i_p,q}'$ . By letting $i=i_p, j=n-1-i_q$ , we obtain

$$\begin{align*}\begin{array}{l} d_{p,q}=-d_{q,p}, \quad \text{for}~ 2\leq p,q\leq t, ~\text{or}~ t+1\leq p,q\leq s,\\ d_{p,q}=d_{q,p}, \quad \text{for}~ 2\leq p\leq t, t+1\leq q\leq s, ~\text{or}~ t+1\leq p\leq s, 2\leq q\leq t. \end{array} \end{align*}$$

Thus, $z_{j,k}'=\pm (a_{n-1-j,k}d_{k,2}+\cdots +a_{n-1-j,s}d_{k,s})$ , where the sign $\pm $ is positive if $1\leq j\leq m$ , $2\leq k\leq t$ , or $m\leq j\leq n-2$ , $t+1\leq k\leq s$ , and negative otherwise. Therefore, $Z_2^t$ can be expressed as

$$\begin{align*}Z_2^t= \left[\begin{array}{ccc} 0&\cdots &0\\ d_{2,2}&\cdots & d_{2,s}\\ \vdots &\ddots &\vdots\\ d_{t,2}&\cdots & d_{t,s}\\ \hline -d_{t+1,2}&\cdots & -d_{t+1,s}\\ \vdots &\ddots &\vdots\\ -d_{s,2}&\cdots & -d_{s,s} \end{array}\right]\cdot \left[\begin{array}{ccc|ccc} a_{n-2,2}&\cdots & a_{m,2} & -a_{m-1,2}&\cdots & -a_{1,2}\\ \vdots &\ddots & \vdots &\vdots &\ddots &\vdots\\ a_{n-2,s}&\cdots & a_{m,s} & -a_{m-1,s} & \cdots &-a_{1,s} \end{array} \right]. \end{align*}$$

Set

$$\begin{align*}\widetilde{D}=[\tilde{d}_{p,q}]_{2\leq p,q\leq s}=\left[\begin{array}{ccc} d_{2,2}&\cdots & d_{2,s}\\ \vdots &\ddots &\vdots\\ d_{t,2}&\cdots & d_{t,s}\\ \hline -d_{t+1,2}&\cdots & -d_{t+1,s}\\ \vdots &\ddots &\vdots\\ -d_{s,2}&\cdots & -d_{s,s} \end{array}\right]. \end{align*}$$

For $2\leq p,q\leq t$ , we have $\tilde {d}_{p,q}=d_{p,q}=-d_{q,p}=-\tilde {d}_{q,p}$ . Similarly, it is easy to see that $\tilde {d}_{p,q}=-\tilde {d}_{q,p}$ for all the other positions of $p,q$ . Hence, the matrix $\widetilde {D}$ is skew-symmetric.

Combining the above cases, for any $1\leq i_2,\dots , i_s\leq n-2$ , there exists a skew-symmetric matrix $\mathcal {D}$ of size $(s-1)\times (s-1)$ such that

(4.2.1) $$ \begin{align} Z_2^t= \left[\begin{array}{c}0\cdots 0\\ \mathcal{D}\end{array}\right]\cdot \left[\begin{array}{ccc|ccc} a_{n-2,2}&\cdots & a_{m,2} & -a_{m-1,2}&\cdots & -a_{1,2}\\ \vdots &\ddots & \vdots &\vdots &\ddots &\vdots\\ a_{n-2,s}&\cdots & a_{m,s} & -a_{m-1,s} & \cdots &-a_{1,s} \end{array} \right]. \end{align} $$

By abuse of notation, write $ D$ for $ \mathcal {D}$ . Our next goal is to show that the second column of $Z_1^t$ depends on the first column of $Z_1^t$ and $M_2$ . Recall that

$$\begin{align*}C=\left[\begin{array}{cc} \mathbf{a_1}+z_{1,2}\mathbf{a_2}+\cdots +z_{1,s}\mathbf{a_s} & \pi \mathbf{a_1}+z_{2,2}\mathbf{a_2}+\cdots +z_{2,s}\mathbf{a_s} \end{array}\right] \end{align*}$$

and $B=J_2C^tJ_{n-2}$ . By using (4.2.1), the $(n-1-i_k)$ -th columns ( $k=2,\dots , s$ ) of the relation $B_1+\pi B_2=B_2(M_2Z_2^t)$ give

(4.2.2) $$ \begin{align} \left[\begin{array}{c} z_{2,2}\\ \vdots\\ z_{2,s} \end{array}\right]=\left[\begin{array}{ccc} d_{2,2}&\cdots & d_{2,s}\\ \vdots &\ddots & \vdots\\ d_{s,2}&\cdots & d_{s,s} \end{array}\right] \left[\begin{array}{ccc} q_{2,1}&\cdots & q_{2,s}\\ \vdots &\ddots & \vdots\\ q_{s,1}&\cdots & q_{s,s} \end{array}\right] \left[\begin{array}{c} 1\\ z_{1,2}\\ \vdots\\ z_{1,s} \end{array}\right]+\pi \left[\begin{array}{c} z_{1,2}\\ \vdots\\ z_{1,s} \end{array}\right], \end{align} $$

where $q_{i,j}=\sum _{k=1}^{m-1}(a_{k,i}a_{n-1-k.j}-a_{n-1-k,i}a_{k,j})$ . It is easy to see that the rest columns in $B_1+\pi B_2= B_2(M_2Z_2^t)$ are automatically satisfied when (4.2.2) holds.

Finally, consider $CM_1+(M_2Z_2^t)M_2=2\pi M_2$ . By using the relations (4.2.1) and (4.2.2), it is equivalent to

$$\begin{align*}D\cdot Q=-2\pi I_{s-1}, \end{align*}$$

where

$$\begin{align*}D=\left[\begin{array}{ccc} d_{2,2}&\cdots & d_{2,s}\\ \vdots &\ddots &\vdots\\ d_{s,2}&\cdots & d_{s,s} \end{array}\right],\quad Q=\left[\begin{array}{ccc} q_{2,2}&\cdots & q_{2,s}\\ \vdots &\ddots &\vdots\\ q_{s,2}&\cdots & q_{s,s} \end{array}\right]. \end{align*}$$

There are two relations we still need to consider: $B_1M_2=\pi B_2M_2$ , and $(M_2Z_2^t)^2=2\pi (M_2Z_2^t)$ . We leave it to the reader to verify that these conditions are already satisfied when $D\cdot Q=-2\pi I_{s-1}$ .

From all the above, we deduce that an affine neighborhood around $(\mathcal {F}_0,\Lambda ')$ is given by $ _2\mathcal {U}_{r,s}\cong \mathrm {Spec \, } O_F[x,z_{1,2},\dots ,z_{1,s}, D, M_2]/I$ , where

$$\begin{align*}I=(M_2'-[\mathbf{0}\mid I_{s-1}],\ D+D^t,\ D\cdot Q+2\pi I_{s-1}). \end{align*}$$

Set $\mathbb {A}_{O_F}^{s-1}=\mathrm {Spec \, } O_F[x,z_{1,2},\dots ,z_{1,s}]$ . For $s\geq 2$ , we have

$$\begin{align*}_2{\mathcal U}_{r,s}\cong \mathbb{A}_{O_F}^{s-1}\times \mathrm{Spec \, } O_F [D,M_2]/I.\\[-41pt] \end{align*}$$

In particular, for the signature $(r,s)=(n-3,3)$ , we have

$$\begin{align*}D=\left[\begin{array}{cc} 0 & d\\ -d& 0 \end{array}\right],\quad Q=\left[\begin{array}{cc} 0 & q\\ -q& 0 \end{array}\right], \end{align*}$$

where we set $d=d_{2,3}=-d_{3,2}, q=q_{2,3}=-q_{3,2}$ . Recall that $q=\sum _{k=1}^{m-1}(a_{k,2}a_{n-1-k,3}-a_{n-1-k,2}a_{k,3})$ . The equation $D\cdot Q=-2\pi I_{s-1}$ amounts to $d\cdot q-2\pi =0$ .

Corollary 4.9. When $(r,s)=(n-3,3)$ , there is an affine chart $_2{\mathcal U}_{n-3,3}$ which is isomorphic to

$$\begin{align*}V(I)=\mathrm{Spec \, } O_F[x,z_{1,2}, z_{1,3}, d, (a_{k,1})_{1\leq k\leq n-2}, (a_{k,2})_{1\leq k\leq n-2}, (a_{k,3})_{1\leq k\leq n-2}]/I, \end{align*}$$

where

$$\begin{align*}I=(a_{i_2,1}, a_{i_3,1}, a_{i_2,2}-1, a_{i_3,2}, a_{i_2,3}, a_{i_3,3}-1, d\cdot \sum_{k=1}^{m-1}(a_{k,2}a_{n-1-k,3}-a_{n-1-k,2}a_{k,3})-2\pi) \end{align*}$$

for some $ 1 \leq i_2,i_3 \leq n-2$ .

Remark 4.10. As in Remark 4.4, for different choices of $M^{\prime }_2$ , we get different equations from $D\cdot Q = -2 \pi I_s$ and so different affine charts $_2\mathcal {U}_{r,s}$ which in general are not isomorphic. For instance, in Corollary 4.9, there are two different affine charts up to isomorphism when we choose different positions for $i_2, i_3$ (see §5.3 for details). In Proposition 5.7, we show that for any such a choice, the affine scheme $V (I)$ has semi-stable reduction over $O_F$ .

5.1 Naive Splitting Models $\mathcal {M}$

By using the explicit description of the two affine charts $_1{\mathcal U}_{r,s}$ , $_2{\mathcal U}_{r,s}\subset \mathcal {M}$ , that we obtained in §4, we will show that $\mathcal {M}$ is not flat over $\mathrm {Spec \, } O_F$ .

Proof. It is enough to show a) When s is odd, the generic fiber $_1{\mathcal U}_{r,s}\otimes _{O_F}F$ is empty and so the point $(\mathcal {F}_0, t\Lambda _m)$ does not lift to characteristic zero, and b) When s is even, the generic fiber $_2{\mathcal U}_{r,s}\otimes _{O_F}F$ is empty and so the point $(\mathcal {F}_0, \Lambda ')$ does not lift to characteristic zero.

Observe that for a skew-symmetric matrix P of size $s\times s$ , the determinant of P is equal to $0$ if s is odd. Recall that we denote by $_1{\mathcal U}_{r,s}$ the affine chart around the point $(\mathcal {F}_0, t\Lambda _m)\in \mathcal {M}$ . By Theorem 4.2, the affine chart $_1{\mathcal U}_{r,s}$ is isomorphic to $\mathrm {Spec \, } O_F[D,Y]/I$ , where

$$\begin{align*}I=(Y'-I_s, \ D+D^t, \ D\cdot Q+2\pi I_s), \end{align*}$$

for any signature $(r,s)$ . Here, D is a skew-symmetric matrix of size $s\times s$ . Thus, when s is odd, we have $\det (D)=0$ . In this case, we can see that the generic fiber $_1{\mathcal U}_{r,s}\otimes _{O_F}F$ is empty since

$$\begin{align*}0=\det(D)\det(Q)=\det(D\cdot Q)=\det(-2\pi I_s)=(-2\pi)^s. \end{align*}$$

It follows that the point $(\mathcal {F}_0, t\Lambda _m)$ does not lift to characteristic zero when s is odd.

Recall that we denote by $_2{\mathcal U}_{r,s}$ the affine chart around the point $(\mathcal {F}_0, \Lambda ')\in \mathcal {M}$ . By Theorem 4.8, the affine chart $_2{\mathcal U}_{r,s}$ is isomorphic to $\mathbb {A}^{s-1}_{O_F}\times \mathrm {Spec \, } O_F[D,M_2]/I$ , where

$$\begin{align*}I=(M_2'-[0\mid I_s], \ D+D^t, \ D\cdot Q+2\pi I_{s-1}). \end{align*}$$

When s is even, we see that the generic fiber $_2{\mathcal U}_{r,s}\otimes _{O_F}F$ is empty since D is of size $(s-1)\times (s-1)$ and $0=\det (D\cdot Q)=(-2\pi )^{s-1}$ . Therefore, the point $(\mathcal {F}_0, \Lambda ')$ does not lift to characteristic zero when s is even.

Recall that there is a forgetful morphism $\tau : \mathcal {M}\rightarrow \mathrm {M}^\wedge \otimes _O O_F$ . From [Reference Pappas and Rapoport18, Remark 5.3], [Reference Pappas, Rapoport and Smithling20, Remark 2.6.10], the authors show that the wedge local model $\mathrm {M}^\wedge $ is not topological flat for any signature when n is even. Consider the local model $\mathrm {M}^{\mathrm {loc}}$ to be the scheme-theoretic closure in $\mathrm {M}^\wedge $ of its generic fiber. We have a closed morphism $\mathrm {M}^{\mathrm {loc}}\hookrightarrow \mathrm {M}^\wedge $ . Define the splitting model as the inverse image of $\mathrm {M}^{\mathrm {loc}}$ ; that is,

$$\begin{align*}\mathrm{M}^{\mathrm{spl}}:=\tau^{-1}(\mathrm{M}^{\mathrm{loc}})=\mathcal{M}\times_{\mathrm{M}^\wedge} \mathrm{M}^{\mathrm{loc}}. \end{align*}$$

The restriction of the forgetful morphism $\tau $ on $\mathrm {M}^{\mathrm {spl}}$ gives

$$\begin{align*}\tau: \mathrm{M}^{\mathrm{spl}}\rightarrow \mathrm{M}^{\mathrm{loc}}. \end{align*}$$

5.2 The Case $(r,s)=(n-2,2)$

By using the explicit description of $ _1\mathcal {U}_{n-2,2}$ in §4.1, we show the following.

Proof. From Corollary 4.3, it is enough to show that the scheme $V(I) $ has semi-stable reduction over $O_F$ . It suffices to prove that $V(I) $ is regular and its special fiber is reduced with smooth irreducible components that have smooth intersections with correct dimensions (see [Reference de Jong13, §2.16]).

First, we rename our variables as follows: $x_i := y_{i,1}$ and $y_i := y_{i,2}$ for $1 \leq i \leq n.$ Thus, $V(I) = \mathrm {Spec \, } O_F[d,\underline x, \underline y]/ I$ , where

$$\begin{align*}I= ( x_{i_0}-1,\, y_{j_0} -1 ,\,x_{j_0},\,y_{i_0},\, d(\sum^m_{j=1}y_{n+1-j}x_{j} - \sum^n_{j=m+1}y_{n+1-j}x_{j}) -2 \pi). \end{align*}$$

By symmetry, as above, there two cases to consider: Case 1 when $1 \leq i_0,\,j_0 \leq m$ and Case 2 when $ 1 \leq i_0 \leq m$ and $ m+1 \leq j_0 \leq n $ . We consider these two cases separately in this proof.

In Case 1, we can assume, for simplicity, that $i_0=1$ and $ j_0=2$ . Thus, $V(I)$ is isomorphic to $V(\mathcal {I}) = \mathrm {Spec \, } R/\mathcal {I}$ , where

$$\begin{align*}R = O_F[d,(x_i)_{3\leq i \leq n}, (y_i)_{3\leq i \leq n}], \end{align*}$$

and

$$\begin{align*}\mathcal{I} = (d(x_{n-1} +\sum^{n-3}_{i=m+1}x_iy_{n+1-i}- y_n - \sum^m_{i=3}x_iy_{n+1-i}) -2\pi ). \end{align*}$$

Over the special fiber ( $\pi =0$ ), we have $ V(\mathcal {I} _s) \cong \mathrm {Spec \, } \bar {R}/\mathcal {I}_s $ , where

$$\begin{align*}\bar{R} = k[d,(x_i)_{3\leq i \leq n}, (y_i)_{3\leq i \leq n}], \quad \mathcal{I}_s= (d(x_{n-1} +\sum^{n-3}_{i=m+1}x_iy_{n+1-i}- y_n - \sum^m_{i=3}x_iy_{n+1-i}) ). \end{align*}$$

It has two smooth irreducible components $V(\mathcal {I}_i) = \mathrm {Spec \, } \bar {R}/\mathcal {I}_i$ , where

$$\begin{align*}\mathcal{I}_1 = (d), \quad \mathcal{I}_2 = (x_{n-1} +\sum^{n-3}_{i=m+1}x_iy_{n+1-i}- y_n - \sum^m_{i=3}x_iy_{n+1-i}), \end{align*}$$

that are isomorphic to $\mathbb {A}_k^{2n-4}$ . Their intersection is also smooth and of dimension $2n-5$ . Next, we prove that $V(\mathcal {I} _s)$ is reduced by showing that $\mathcal {I}_s =\mathcal {I}_1 \cap \mathcal {I}_2.$ Clearly, $\mathcal {I} _s \subset \mathcal {I}_1 \cap \mathcal {I}_2$ . Let $g \in \mathcal {I}_1 \cap \mathcal {I}_2 .$ Thus, $g \in \mathcal {I}_1$ and $g = f_1 d$ for $f_1 \in \bar {R}$ . Also, $g \in \mathcal {I}_2$ and so $f_1 d \in \mathcal {I}_2$ . $\mathcal {I}_2$ is a prime ideal and $d \notin \mathcal {I}_2$ . Thus, $f_1 \in \mathcal {I}_2$ and

$$\begin{align*}g = df_1 \equiv d (x_{n-1} +\sum^{n-3}_{i=m+1}x_iy_{n+1-i}- y_n - \sum^m_{i=3}x_iy_{n+1-i}) f_2 \equiv 0 \, \text{mod} \, \mathcal{I}_s \end{align*}$$

for $f_2 \in \bar {R}$ . Thus, $g \in \mathcal {I}_s$ and so $\mathcal {I}_s = \mathcal {I}_1 \cap \mathcal {I}_2 .$ Lastly, we can easily see that the ideals $ \mathcal {I}_1,\mathcal {I}_2$ are principal over $V(\mathcal {I})$ . From the above, we deduce that $V(\mathcal {I}) $ is regular (see [Reference Hartl10, Remark 1.1.1]).

In Case 2, we set $\Delta =\{1,\dots ,n-2\}/ \{m, m+1\}$ and let $i_0=m, j_0=m+1$ . Then $V(I)$ is isomorphic to $V(\mathcal {J}) = \mathrm {Spec \, } R'/\mathcal {J}$ , where $R' = O_F[d,(x_i)_{i \in \Delta }, (y_i)_{i \in \Delta }]$ and

$$\begin{align*}\mathcal{J} = (d(1+ \sum^{m-1}_{i=1}x_iy_{n+1-i}- \sum^n_{i= m+2}x_iy_{n+1-i})-2\pi). \end{align*}$$

To see that $V(\mathcal {J}) $ has semi-stable reduction over $O_F$ , one proceeds exactly as in Case 1. Over the special fiber ( $\pi =0$ ), we have $ V(\mathcal {J} _s) \cong \mathrm {Spec \, } \bar {R}'/\mathcal {J}_s $ where $ \bar {R} = k[d,(x_i)_{i \in \Delta }, (y_i)_{i \in \Delta }]$ and

$$\begin{align*}\mathcal{J}_s = (d(1+ \sum^{m-1}_{i=1}x_iy_{n+1-i}- \sum^n_{i= m+2}x_iy_{n+1-i})). \end{align*}$$

It has two smooth irreducible components $V(\mathcal {J}_i) = \mathrm {Spec \, } R/\mathcal {J}_i$ , where

$$\begin{align*}\mathcal{J}_1 = (d), \quad \mathcal{J}_2 = (1+ \sum^{m-1}_{i=1}x_iy_{n+1-i}- \sum^n_{i= m+2}x_iy_{n+1-i}) \end{align*}$$

of dimension $2n-4$ . (Note that in this case, the second component $V(\mathcal {J}_2)$ is not isomorphic to the affine space $\mathbb {A}_k^{2n-4}$ as in Case 1.) Their intersection is also smooth and of dimension $2n-5$ . By using similar arguments as Case 1, we can show that $V(\mathcal {J}_s)$ is reduced and $V(\mathcal {J})$ is regular.

Since $_1{\mathcal U}_{n-2,2}$ has semi-stable reduction, and is therefore flat over $O_F$ , we have $_1{\mathcal U}_{n-2,2}\subset \mathrm {M}^{\mathrm {spl}}$ . Our next goal is to prove that $\mathrm { M}^{\mathrm {spl}}$ has semi-stable reduction over $O_F$ for the signature $(r,s)= (n-2,2)$ . From §4.1, it is enough to show that $\mathcal {G}$ -translates of $_1{\mathcal U}_{n-2,2}$ cover $\mathrm {M}^{\mathrm {spl}}$ (see Theorem 5.4).

Assume that s is even. Recall from §4.1 the ${\mathcal G}$ -equivariant morphism $\tau : \mathrm {M}^{\mathrm {spl}} \rightarrow \mathrm {M}^{\mathrm {loc}}\otimes _{O} O_F$ which is given by $(\mathcal {F}_0,\mathcal {F}_1) \mapsto \mathcal {F}_1$ . Let $\tau _s : \mathrm {M}^{\mathrm {spl}}\otimes k \rightarrow \mathrm {M}^{\mathrm {loc}}\otimes k$ be the morphism over the special fiber. Let $(\mathcal {F}_0,\mathcal {F}_1)$ be point of the special fiber of $\mathrm {M}^{\mathrm {spl}}$ . Over the special fiber the condition (4) amounts to $t\mathcal {F}_0 = (0)$ and so $ (0) \subset \mathcal {F}_0 \subset t \Lambda _m \otimes k$ (see §3 for the definition of the naive splitting models). Hence, we consider the morphism

$$\begin{align*}\pi : \mathrm{M}^{\mathrm{spl}}\otimes k \rightarrow \mathrm{Gr}(s,n)\otimes k \end{align*}$$

given by $ (\mathcal {F}_0,\mathcal {F}_1) \mapsto \mathcal {F}_0.$ Here, $\mathrm {Gr}(s,n)\otimes k$ is the finite-dimensional Grassmannian of s-dimensional subspaces of $t \Lambda _m \otimes k.$ This has a section

$$\begin{align*}\phi : \mathrm{Gr}(s,n)\otimes k \rightarrow \mathrm{M}^{\mathrm{spl}}\otimes k \end{align*}$$

given by $ \mathcal {F}_0 \mapsto (\mathcal {F}_0,\mathcal {F}_1)$ with $\mathcal {F}_1 = t\Lambda _m \otimes k.$ The image of the section $\phi $ is an irreducible component of $ \mathrm {M}^{\mathrm { spl}}\otimes _{O_F} k$ which is the fiber $\tau ^{-1}_s (t\Lambda _m\otimes k)$ over the ‘worst point’ of $\mathrm {M}^{\mathrm {loc}}\otimes k$ – that is, the unique closed ${\mathcal G}$ -orbit supported in the special fiber (see [Reference Pappas and Rapoport19, §5]). Hence, $\tau ^{-1}_s(t\Lambda _m)$ is isomorphic to the Grassmannian $ \mathrm {Gr}(s,n)\otimes k $ of dimension $s(n-s)$ .

We equip $t \Lambda _m \otimes k$ with the nondegenerate alternating form $ \langle \, , \, \rangle '$ which is defined as $\langle tv,tw \rangle ' := ( v,tw ) = v^tJ_n w$ . Note that this is well defined (see [Reference Richarz25, §5.2] for more details). Denote by $\mathcal {Q}(s,n)$ the closed subscheme of $\phi (\mathrm {Gr}(s,n)\otimes k) $ of dimension $ s(2n-3s+1)/2$ which parametrizes all the isotropic s-subspaces $\mathcal {F}_0$ in the n-space $t\Lambda _m \otimes k$ .

Next, we give a couple of useful observations concerning the special fiber of the affine chart $_1{\mathcal U}_{r,s}$ (see §4.1 for the notation), which will be used in the proof of the following theorem.

Remark 5.3. Recall from Theorem 4.2 that over the special fiber of $_1{\mathcal U}_{r,s}$ , we have $D=-D^t$ , $D\cdot Q=0$ and the minor relations from (4.1.1).

a) Observe that if $D =0$ , then $Z=0$ , which gives $A=0$ since $A = Y Z^t$ over the special fiber. If $A=0$ , then $\mathcal {F}_1 = t \Lambda _m$ . Hence, $_1{\mathcal U}_{r,s} \cap \tau ^{-1}(t\Lambda _m)$ is a smooth irreducible component of dimension $s(n-s)$ .

b) Recall that

$$\begin{align*}\mathcal{F}_0= \left[\begin{array}{c} \pi Y \\ Y \end{array}\right] \quad \text{and} \quad Y = \left[\begin{array}{ccc} \mathbf{y}_1 \ \cdots \ \mathbf{y}_s \end{array}\right], \end{align*}$$

where $ \mathbf {y}_i$ are the columns of the matrix $Y.$ By direct calculations, we get that the entries $q_{i,j}$ of the matrix Q are given by $q_{i,j} = \langle \mathbf {y}_i,\mathbf {y}_j \rangle '$ . Thus, over the special fiber, we have $\langle \mathcal {F}_0,\mathcal {F}_0 \rangle ' = Q $ , and so from the rank of $ Q$ , we read how isotropic $\mathcal {F}_0$ is with respect to $\langle \,,\,\rangle '$ . When the rank of the matrix Q is zero, which actually occurs, we have $\langle \mathcal {F}_0,\mathcal {F}_0 \rangle ' = 0 $ .

c) From (a) and (b), we can easily see that $_1{\mathcal U}_{r,s}$ contains points $(\mathcal {F}_0, \mathcal {F}_1)$ where $ \mathcal {F}_0 \in \mathcal {Q}(s, n)$ and $\mathcal {F}_1 = t\Lambda _m$ .

Proof. From §5.1, we have the forgetful ${\mathcal G}$ -equivariant morphism $\tau : \mathrm {M}^{\mathrm {spl}} \rightarrow \mathrm {M}^{\mathrm {loc}}\otimes _{O} O_F$ given by $(\mathcal {F}_0,\mathcal {F}_1) \mapsto \mathcal {F}_1$ . As in [Reference Pappas15, §4] and [Reference Pappas and Rapoport18, §2.4.2, 5.5], the worst point of $\mathrm {M}^{\mathrm {loc}}\otimes _{O} O_F$ is given by the k-valued point $t\Lambda _m\otimes k$ . From the construction of splitting models and local models, in order to show that $\mathcal {G}$ -translates of $_1{\mathcal U}_{n-2,2}$ cover $\mathrm {M}^{\mathrm {spl}}$ , it is enough to prove that $\mathcal {G}$ -translates of $_1{\mathcal U}_{n-2,2}$ cover $\tau ^{-1}(t\Lambda _m)$ .

Recall that $ \tau ^{-1}_s(t\Lambda _m\otimes k) \cong \mathrm {Gr}(2,n)\otimes k$ and $\mathcal {G}_k$ acts via its action by reduction to $t\Lambda _m\otimes k/t^2\Lambda _m\otimes k$ . This action factors through the symplectic group $Sp(n)_k$ of the symplectic form $\langle \,,\,\rangle '$ on $ t \Lambda _m\otimes k $ and gives the map $ \mathcal {G}_k \rightarrow Sp(n)_k$ . As in [Reference Pappas and Rapoport19, §4], $\mathcal {G}_k$ has $Sp(n)_k$ as its maximal reductive quotient. Therefore, the map $\mathcal {G}_k \rightarrow Sp(n)_k$ is surjective.

Next, the $Sp(n)_k$ -action on $\mathrm {Gr}(2, n)$ has two orbits. More precisely, the orbits are

$$\begin{align*}Q(0) = \{ \mathcal{F}_0 \in \mathrm{Gr}(2, n) \mid \mathcal{F} _0~\text{contains no isotropic vectors}\} \end{align*}$$

and

$$\begin{align*}Q(2) = \{ \mathcal{F}_0 \in \mathrm{Gr}(2, n) \mid \mathcal{F} _0~\text{is totally isotropic}\}. \end{align*}$$

Observe that $Q(2)$ is contained in the (Zariski) closure of $Q(0)$ , and so $Q(2) = \mathcal {Q}(2,n)$ is the unique closed orbit (see, for example, [Reference Barbasch and Evens3, §3.1] and [Reference Arbarello, Cornalba, Griffiths and Harris1]). Thus, $\mathcal {Q}(2,n)$ is contained in the closure of each orbit $Q(i)$ , $i=0,2$ .

Lastly, from Remark 5.3, we have that $_1{\mathcal U}_{n-2,2}\otimes k$ contains points $(\mathcal {F}_0,t\Lambda _m)$ with $\mathcal {F}_0 \in \mathcal {Q}(2,n) $ and so $_1{\mathcal U}_{n-2,2}$ contains points from all the orbits. Therefore, from all the above, we deduce that the $\mathcal {G}$ -translates of $_1{\mathcal U}_{n-2,2}$ cover $\tau ^{-1}(t\Lambda _m)$ .

5.3 The Case $(r,s)=(n-1,1)$ and $(r,s)=(n-3,3)$

By using the explicit description of $_2{\mathcal U}_{n-3,3}$ in 4.2, we show

Proof. Recall from §4.2.2 that the quadratic polynomial $\sum _{k=1}^{m-1}(a_{k,2}a_{n-1-k,3}-a_{n-1-k,2}a_{k,3})$ depends on the positions of $i_2, i_3$ . By symmetry, we only need to consider the case when $1\leq i_2, i_3\leq m-1$ and the case when $1\leq i_2\leq m-1, m\leq i_3\leq n-2$ .

We consider these two cases separately in this proof. In the first case, for simplicity, we can set $i_2=1, i_3=2$ . Thus, $_2\mathcal {U}_{n-3,3}$ is isomorphic to $V(\mathcal {I})=\mathrm {Spec \, } R/\mathcal {I}$ , where

$$\begin{align*}R=O_F[x,d,z_{1,2},z_{1,3},(a_{i,1})_{3\leq i\leq n-2}, (a_{i,2})_{3\leq i\leq n-2}, (a_{i,3})_{3\leq i\leq n-2}], \end{align*}$$

and

$$\begin{align*}\mathcal{I}=(d(a_{n-2,3}+\sum_{k=3}^{m-1}a_{n-1-k,3} a_{k,2}-\sum_{k=m}^{n-4}a_{n-1-k,3}a_{k,2}-a_{n-3,2})-2\pi). \end{align*}$$

It is easy to see that we have two irreducible components in the special fiber, where

$$\begin{align*}I_1=(d),\quad I_2=(a_{n-2,3}+\sum_{k=3}^{m-1}a_{n-1-k,3} a_{k,2}-\sum_{k=m}^{n-4}a_{n-1-k,3}a_{k,2}-a_{n-3,2}). \end{align*}$$

Both irreducible components are isomorphic to $\mathbb {A}_k^{3(n-3)}$ , so they are smooth with the correct dimension. Their intersection is isomorphic to $\mathbb {A}_k^{3(n-3)-1}$ , which is also smooth. Since $I_1, I_2$ are prime ideals, it is enough to prove that the special fiber $V(\mathcal {I}_s)$ is reduced by showing $\mathcal {I}_s=I_1\cap I_2$ . By using similar arguments as in Proposition 5.2, we get that $V(\mathcal {I}_s)$ is reduced and $V(\mathcal {I})$ is regular.

In the second case, we can set $\Delta =\{1,\dots ,n-2\}/ \{m-1, m\}$ , and let $i_2=m-1, i_3=m$ . Then $_2\mathcal {U}_{n-3,3}$ is isomorphic to $V(\mathcal {J})=\mathrm {Spec \, } R'/\mathcal {J}$ , where

$$\begin{align*}R'=O_F[x,d,z_{1,2},z_{1,3},(a_{i,1})_{i\in\Delta}, (a_{i,2})_{i\in\Delta}, (a_{i,3})_{i\in\Delta}], \end{align*}$$

and

$$\begin{align*}\mathcal{J}=(d(1+\sum_{k=1}^{m-2}a_{n-1-k,3}a_{k,2}-\sum_{k=m+1}^{n-2}a_{n-1-k,3}a_{k,2})-2\pi). \end{align*}$$

Again, we have two irreducible components over the special fiber:

$$\begin{align*}\mathcal{J}_1=(d),\quad \mathcal{J}_2=(1+\sum_{k=1}^{m-2}a_{n-1-k,3}a_{k,2}-\sum_{k=m+1}^{n-2}a_{n-1-k,3}a_{k,2}). \end{align*}$$

Both of them are smooth and of dimension $3(n-3)$ (Note that in this case, the second component $V(\mathcal {J}_2)$ is not isomorphic to $\mathbb {A}_k^{3(n-3)}$ ). Their intersection is also smooth and of dimension $3n-10$ . By using the same arguments as above, we can show that $V(\mathcal {J}_s)$ is reduced and $V (\mathcal {J} )$ is regular. Thus, $_2\mathcal {U}_{n-3,3}$ has semi-stable reduction over $O_F$ .

From Theorem 4.7 and Proposition 5.7, we have that $_2{\mathcal U}_{n-1,1}$ is smooth and $_2{\mathcal U}_{n-3,3}$ has semi-stable reduction. Thus, both $ _2{\mathcal U}_{n-1,1}$ and $_2{\mathcal U}_{n-3,3} $ are $O_F$ -flat. As in §5.2, we deduce that these affine charts are open subschemes of the corresponding splitting model $\mathrm {M}^{\mathrm { spl}}$ . We will prove that $\mathrm {M}^{\mathrm {spl}}$ is smooth when $(r,s)=(n-1,1)$ and $\mathrm {M}^{\mathrm {spl}}$ has semi-stable reduction when $(r,s)=(n-3,3)$ by showing that ${\mathcal G}$ -translates of $_2{\mathcal U}_{n-1,1}$ and $_2{\mathcal U}_{n-3,3}$ cover $\mathrm {M}^{\mathrm {spl}}$ .

Recall from §3 the ${\mathcal G}$ -equivariant morphism $\tau : \mathrm {M}^{\mathrm {spl}}\rightarrow \mathrm {M}^{\mathrm {loc}} \otimes _{O}O_F$ which is given by $(\mathcal {F}_0,\mathcal {F}_1)\mapsto \mathcal {F}_1$ and let $\tau _s: \mathrm {M}^{\mathrm {spl}}\otimes k\rightarrow \mathrm {M}^{\mathrm {loc}} \otimes k$ be the morphism over the special fiber. When s is odd, the unique closed ${\mathcal G}$ -orbit of $\mathrm {M}^{\mathrm {loc}} \otimes k$ is the orbit of $\Lambda '\otimes k$ (see §3). From the construction of the local model, it is enough to show that ${\mathcal G}_k$ -translates of $_2\mathcal {U}_{n-1,1}\otimes k$ and $_2\mathcal {U}_{n-3,3}\otimes k$ cover $\tau _s^{-1}(\Lambda '\otimes k)$ .

When $(r,s)=(n-1,1)$ , the inverse image $\tau _s^{-1}(\Lambda '\otimes k)$ contains points $(\mathcal {F}_0,\Lambda '\otimes k)$ satisfying $t\Lambda '\subset \mathcal {F}_0$ . Observe that $t\Lambda '=\text {span}_{k}\{-e_1\}$ and $\mathcal {F}_0$ has rank $1$ . Thus, $\tau _s^{-1}(\Lambda '\otimes k)=(\text {span}_{k}\{-e_1\}, \Lambda '\otimes k)$ is just a unique point (Note that it is different from the case when s is even, see §5.2, where we have $t\Lambda _m=0$ ). From §4.2, we see that $_2\mathcal {U}_{n-1,1}\otimes k$ contains that point and so ${\mathcal G}_k$ -translates of $_2\mathcal {U}_{n-1,1}\otimes k$ cover $\tau _s^{-1}(\Lambda '\otimes k)$ . Hence, ${\mathcal G}$ -translates of $_2\mathcal {U}_{n-1,1}$ cover $\mathrm {M}^{\mathrm {spl}}$ .

When s is odd and $\geq 3$ , we have

$$\begin{align*}\tau_s^{-1}(\Lambda'\otimes k)=\{(\mathcal{F}_0,\Lambda')\mid \text{span}_{k}\{-e_1\}\subset \mathcal{F}_0\subset \Lambda'\otimes k,\quad t\mathcal{F}_0=(0)\}. \end{align*}$$

Set $\mathcal {F}_0=\text {span}_k\{e_1, v_2, \dots , v_s\}$ . Then $t\mathcal {F}_0=0$ and $\mathcal {F}_0\subset \Lambda '\otimes k$ imply that $v_2, \dots , v_s$ are a linear combination of $\{-e_2,\dots , -e_m, \pi e_{m+1},\dots , \pi e_{n-1}\}$ . Thus, we consider the morphism

$$\begin{align*}\pi: \tau_s^{-1}(\Lambda'\otimes k)\rightarrow \mathrm{Gr}(s-1,n-2)\otimes k \end{align*}$$

given by $(\mathcal {F}_0,\Lambda ')\mapsto \text {span}_k\{v_2, \cdots , v_s\}$ . Here, $\mathrm {Gr}(s-1,n-2)\otimes k$ is the Grassmannian of $(s-1)$ -dimensional subspaces of

$$\begin{align*}\Lambda'/\{-\pi^{-1}e_1, -e_1\}=\text{span}_k\{-e_2,\dots, -e_m, \pi e_{m+1},\dots, \pi e_{n-1}\}. \end{align*}$$

We also have a section

$$\begin{align*}\phi: \mathrm{Gr}(s-1,n-2)\otimes k\rightarrow \mathrm{M}^{\mathrm{spl}}\otimes k, \end{align*}$$

given by $W\rightarrow (\text {span}_k(e_1, W),\Lambda ')$ . Similar to §5.2, it is easy to see that the image of the section $\phi $ is $\tau _s^{-1}(\Lambda '\otimes k)$ . Hence, $\tau _s^{-1}(\Lambda '\otimes k)$ is isomorphic to $\mathrm {Gr}(s-1,n-2)\otimes k$ . As in §5.2, we can equip $\tau _s^{-1}(\Lambda '\otimes k)$ with a nondegenerate alternating form; that is, $\langle tv,tw\rangle ':=(tv,w)=v^tJ_{n-2}w$ for $tv, tw\in \tau _s^{-1}(\Lambda '\otimes k)$ . Denote by $\mathcal {Q}(s-1,n-2)$ the closed subscheme of $\mathrm {Gr}(s-1,n-2)\otimes k$ which parametrizes all isotropic $(s-1)$ -subspaces with respect to $\langle \ , \ \rangle '$ .

The following observations concern the special fiber of $_2{\mathcal U}_{r,s}$ when s is odd. We keep the same notation as in §4.2.

Remark 5.8. (1) Set $\mathbf {a}_i$ the i-th column of the matrix $M_2$ ; $\mathbf {a}_i$ can be expressed as the element $tv\in \tau _s^{-1}(\Lambda '\otimes k)$ , where $tv=[ 0~0~a_{1,i}~\cdots ~a_{n-2,i}\mid ~0~\cdots ~ 0]^t$ . Thus,

$$ \begin{align*} \langle \mathbf{a}_i,\mathbf{a}_j\rangle' &=[a_{1,i}~\cdots ~a_{n-2,i}]J_{n-2}[a_{1,j}~\cdots ~a_{n-2,j}]^t \\ &=\sum_{k=1}^{m-1}(a_{k,i}a_{n-1-k.j}-a_{n-1-k.i}a_{k,j})\\ &= q_{i,j}. \end{align*} $$

(2) For the special fiber $\mathrm {M}^{\mathrm {spl}}\otimes k$ , consider $(_2\mathcal {U}_{r,s}\otimes k) \cap \tau _s^{-1}(\Lambda '\otimes k)$ . We get $C=0$ since $A=0$ . The first column of C is $\mathbf {a}_1+z_{1,2}\mathbf {a}_2+\cdots +z_{1,s}\mathbf {a}_s$ . By setting $i=i_2, \dots , i_s$ , this gives us $z_{1,2}=\cdots =z_{1,s}=0$ and so $\mathbf {a}_1=\mathbf {0}$ . Thus,

$$\begin{align*}\mathcal{F}_0=\left[\begin{array}{c} M_1\\ M_2\\ \hline 0\\ 0 \end{array}\right], \end{align*}$$

where

$$\begin{align*}M_1=\left[\begin{array}{cccc} 0& 0&\cdots & 0\\ 1& 0&\cdots &0 \end{array}\right],~ M_2=\left[\begin{array}{cccc} \mathbf{0} & \mathbf{a}_2 &\cdots &\mathbf{a}_s \end{array}\right]= \left[\begin{array}{cccc} 0 & a_{1,2}&\cdots & a_{1,s}\\ \vdots &\vdots &&\vdots\\ 0& a_{n-2,2}& \cdots & a_{n-2,s}\\ \end{array}\right]. \end{align*}$$

By definition, $Q=[q_{i,j}]=0$ is equivalent to $\langle \mathbf {a}_i,\mathbf {a}_j \rangle '=0$ for $2\leq i, j\leq s$ . Therefore, $\pi (\mathcal {F}_0)$ is totally isotropic under $\langle ~,~ \rangle '$ if and only if $Q=0$ (which actually occurs). Thus,

$$\begin{align*}\phi(\mathcal{Q}(s-1,n-2))\subset (_2\mathcal{U}_{r,s}\otimes k)\cap \tau_s^{-1}(\Lambda'\otimes k). \end{align*}$$

Proof. It is enough to prove that ${\mathcal G}$ -translates of $_2\mathcal {U}_{n-3,3}$ cover $\tau ^{-1}(\Lambda ')$ . We use similar arguments as in the proof of Theorem 5.4. Since $\mathcal {G}_k$ acts on $\tau _s^{-1}(\Lambda '\otimes k)\simeq \mathrm {Gr}(2,n-2)\otimes k$ , this action factors through the symplectic group $Sp(n-2)_k$ of the symplectic form $\langle \,,\,\rangle '$ on $ \Lambda '/ \{-\pi ^{-1}e_1, -e_1\} $ , and gives the surjective map $ \mathcal {G}_k \rightarrow Sp(n-2)_k$ , where

$$\begin{align*}Sp(n-2)_k=\{g\in SL(n-2)_k\mid g^tJ_{n-2}g=J_{n-2}\}. \end{align*}$$

By [Reference Barbasch and Evens3, §3], the $Sp(n-2)_k$ -action on $\mathrm {Gr}(2,n-2)\otimes k$ has two orbits: $Q(0)$ and $Q(2)$ , where

$$\begin{align*}Q(i)=\{\mathcal{F}_0\in \mathrm{Gr}(2,n-2)\mid \dim(rad(\mathcal{F}_0))=i\}, \end{align*}$$

for $i=0, 2$ . Similar to the proof of Theorem 5.4, we have $Q(2)\subset \overline {Q(0)}$ . Thus, $Q(2)$ is the unique closed orbit. Observe that $Q(2)$ contains all totally isotropic subspaces $\pi (\mathcal {F}_0)$ , and so $Q(2)= \mathcal {Q}(2,n-2)$ .

From Remark 5.8 (2), we have that $_2{\mathcal U}_{n-3,3}\otimes k$ contains points $(\mathcal {F}_0,\Lambda '\otimes k)$ with $\pi (\mathcal {F}_0) \in \mathcal {Q}(2,n-2) $ . By identifying $\tau _s^{-1}(\Lambda '\otimes k)$ with $\mathrm {Gr}(2,n-2)\otimes k$ , we get that $_2{\mathcal U}_{n-3,3}$ contains points from all the orbits. Therefore, from all the above, we deduce that the $\mathcal {G}$ -translates of $_2{\mathcal U}_{n-3,3}$ cover $\tau ^{-1}(\Lambda ')$ .

From Propositions 4.7 and 5.7 and Theorem 5.9, we have the following:

Remark 5.11. For s odd and $\geq 5$ , note that we have

$$\begin{align*}\pi: \tau_{s}^{-1}(\Lambda'\otimes k)\stackrel{\sim}{\rightarrow} \mathrm{Gr}(s-1,n-2)\otimes k, \end{align*}$$

and

$$\begin{align*}\mathcal{Q}(s-1,n-2)\subset (_2{\mathcal U}_{r,s}\otimes k)\cap \tau^{-1}_s(\Lambda\otimes k). \end{align*}$$

If we can show that the affine chart $_2{\mathcal U}_{r,s}$ is flat, then we have $_2{\mathcal U}_{r,s}\subset \mathrm {M}^{\mathrm {spl}}$ . By using a similar argument as the proof of Theorem 5.9, we could prove that ${\mathcal G}$ -translates of $_2{\mathcal U}_{r,s}$ cover $\mathrm {M}^{\mathrm {spl}}$ .