The Drinfeld stratification for ${{\,\mathrm{GL}\,}}_n$

Abstract

We define a stratification of Deligne–Lusztig varieties and their parahoric analogues which we call the Drinfeld stratification. In the setting of inner forms of ${{\,\mathrm{GL}\,}}_n$, we study the cohomology of these strata and give a complete description of the unique closed stratum. We state precise conjectures on the representation-theoretic behavior of the stratification. We expect this stratification to play a central role in the investigation of geometric constructions of representations of p-adic groups.

Appendix A. The geometry of the fibers of projection maps

In this section, we study the fibers of the projection maps $X_h \rightarrow X_{h-1}$. This is a technical computation which we perform by first using the isomorphism $X_h \cong X_h(b,b_{\,\mathrm{cox}})$ for a particular choice of b which we call the special representative. This is the first time in this paper that we see the convenience of having the alternative presentations of $X_h$ discussed in Sects. 3.2 and 4.5.

1.1 A.1 The special representative

We first recall the content of Sect. 4.5 in the context of a particular representative of the $\sigma$-conjugacy class corresponding to the fixed integer $\kappa$.

Definition A.1.1

The special representative $b_{\,\mathrm{sp}}$ attached to $\kappa$ is the block-diagonal matrix of size $n \times n$ with $(n_0 \times n_0)$-blocks of the form $\left( \begin{matrix} 0 &{} \varpi \\ 1_{n_0-1} &{} 0 \end{matrix}\right) ^\kappa$.

By [9, Lemma 5.6], there exists a $g_0 \in G_{x,0}({\mathcal {O}}_{\breve{k}})$ such that $b_{\,\mathrm{sp}}= g_0 b_{\,\mathrm{cox}}\sigma (g_0)^{-1}$. Observe further that since $b_{\,\mathrm{sp}}, b_{\,\mathrm{cox}}$ are $\sigma$-fixed and $b_{\,\mathrm{sp}}^n = b_{\,\mathrm{cox}}^n = \varpi ^{kn}$,

$$\begin{aligned} \sigma ^n(g_0) = g_0. \end{aligned}$$

Therefore $b_{\,\mathrm{sp}}$ satisfies the conditions of Lemma 4.5.3. Recall from Sect. 4.5 that we have

$$\begin{aligned} X_h \cong X_h(b_{\,\mathrm{sp}},b_{\,\mathrm{cox}}) \cong \{v \in {\mathscr {L}}_h : \sigma (\det g_{b_{\,\mathrm{sp}}}(v)) = (-1)^{n-1} \det g_{b_{\,\mathrm{sp}}}(v) \in {\mathbb {W}}_h^\times \}, \end{aligned}$$

(A.1)

where

$$\begin{aligned} {\mathscr {L}}_h&= \big ({\mathbb {W}}_h \oplus (V {\mathbb {W}}_{h-1})^{\oplus n_0 - 1}\big )^{\oplus n'} \subset {\mathbb {W}}_h^{\oplus n} \\ g_{b_{\,\mathrm{sp}}}(v)&= \big ( v_1 \, \big | \, v_2 \, \big | \, v_3 \, \big | \, \cdots \, \big | \, v_n\big ) \\ \text {where } v_i&= \varpi ^{\lfloor (i-1)k_0/n_0 \rfloor } \cdot (b_{\,\mathrm{sp}}\sigma )^{i-1}(v) \text { for }1 \le i \le n-1. \end{aligned}$$

In this section, we will work with

$$\begin{aligned} X_h^+ :=\{v \in {\mathscr {L}}_h^+ : \sigma (\det g_{b_{\,\mathrm{sp}}(v)}) = \det g_{b_{{\,\mathrm{sp}}}}(v) \in {\mathbb {W}}_h^\times \} \end{aligned}$$

(A.2)

where ${\mathscr {L}}_h^+$ is now the subquotient of ${\mathbb {W}}_{h+1}^{\oplus n}$

$$\begin{aligned} {\mathscr {L}}_h^+ :=\big ({\mathbb {W}}_h \oplus (V {\mathbb {W}}_h)^{\oplus n_0 -1}\big )^{\oplus n'}, \end{aligned}$$

and $g_{b_{\,\mathrm{sp}}}(v)$ is defined as before. Note that (A.1) differs from (A.2) in that the former takes place in $G_{x,0}/G_{x,(h-1)+}$ and the latter takes place in $G_{x,0}/G_{x,h}$. A straightforward computation shows that the defining equation of $X_h^+$ does not depend on the quotient ${\mathscr {L}}_h^+/{\mathscr {L}}_h = {\mathbb {A}}^{n-n'}$.

Observe that $\det g_{b_{\,\mathrm{sp}}}(\zeta v) = {{\,\mathrm{Nm}\,}}(\zeta ) \cdot \det g_{b_{\,\mathrm{sp}}}(\zeta v)$ where ${{\,\mathrm{Nm}\,}}(\zeta ) = \zeta \cdot \sigma (\zeta ) \cdot \sigma ^2(\zeta ) \cdots \sigma ^{n-1}(\zeta )$. Picking any $\zeta$ such that $\sigma ({{\,\mathrm{Nm}\,}}(\zeta )) = (-1)^{n-1} {{\,\mathrm{Nm}\,}}(\zeta )$ allows us to undo the $(-1)^{n-1}$ factor in the defining equation in (A.1). In particular, this means

$$\begin{aligned} H_c^i(X_h^+, {\overline{{\mathbb {Q}}}}_\ell ) = H_c^{i + 2(n-n')}(X_h, \overline{{\mathbb {Q}}}_\ell ), \qquad \text {for all }i \ge 0. \end{aligned}$$

For each divisor $r \mid n'$, we define the rth Drinfeld stratum $X_{h,r}^+$ of $X_h^+$ to be the preimage of $X_{h,r}$ under the natural surjection $X_h^+ \rightarrow X_h$.

1.2 A.2 Fibers of $X_{h,r}^+ \rightarrow X_{h-1,r}^+$

For notational convenience, we write $b = b_{\,\mathrm{sp}}.$ We may identify ${\mathscr {L}}_h^+ = {\mathbb {A}}^{n(h-1)}$ with coordinates $x = \{x_{i,j}\}_{1 \le i \le n, \, 0 \le j \le h-1}$ which we typically write as $x = ({\widetilde{x}}, x_{1,h-1}, x_{2,h-1}, \ldots , x_{n,h-1}) \in {\mathscr {L}}_{h-1}^+ \times {\mathbb {A}}^n$; here, an element $v = (v_1, \ldots , v_n) \in {\mathscr {L}}_h^+$ is such that $v_i = [x_{i,0}, x_{i,1}, \ldots , x_{i,n}]$ if $i \equiv 1 \pmod {n_0}$ and $v_i = [0, x_{i,0}, x_{i,1}, \ldots , x_{i,n}]$ if $i \not \equiv 1 \pmod {n_0}$.

In this section, fix a divisor $r \mid n'$. From the definitions, $X_{h,r}^+$ can be viewed as the subvariety of $X_{h-1,r}^+ \times {\mathbb {A}}^n$ cut out by the equation

$$\begin{aligned} 0 = P_0(x)^q - P_0(x), \end{aligned}$$

where $P_0$ is the coefficient of $\varpi ^{h-1}$ in the expression $\det g_b^{{{\,\mathrm{red}\,}}}(v)$. Let c denote the polynomial consisting of the terms of $P_0(x)$ which only depend on ${\widetilde{x}}$. An explicit calculation shows that there exists a polynomial $P_1$ in x such that

$$\begin{aligned} P_0(x) = c({\widetilde{x}}) + \sum _{i=0}^{n_0-1} P_1(x)^{q^i}. \end{aligned}$$

(A.3)

Therefore $X_{h,r}^+$ is the subvariety of $X_{h-1,r}^+ \times {\mathbb {A}}^n$ cut out by

$$\begin{aligned} P_1(x)^{q^{n_0}} - P_1(x) = c({\widetilde{x}}) - c({\widetilde{x}})^q. \end{aligned}$$

One can calculate $P_1$ explicitly (see [9, Proposition 7.5]):

Lemma A.2.1

Explicitly, the polynomial $P_1$ is

$$\begin{aligned} P_1(x) = \sum _{1 \le i,j \le n^{\prime }} m_{ji} x_{1 + n_0(i-1),h-1}^{q^{(j-1)n_0}}, \end{aligned}$$

where $m :=(m_{ji})_{j,i}$ is the adjoint matrix of $\overline{g_b}({\bar{x}})$ and ${{\bar{x}}}$ denotes the image of x in $\overline{V} = {\mathscr {L}}_0/{\mathscr {L}}_0^{(1)}$. Explicitly, $m \cdot \overline{g_b}({\bar{x}}) = \det \overline{g_b}({\bar{x}})$ and the (j, i)th entry of m is $(-1)^{i+j}$ times the determinant of the $(n^{\prime }-1)\times (n^{\prime }-1)$ matrix obtained from $\overline{g_b}({\bar{x}})$ by deleting the ith row and jth column.

The main result of this section is:

Proposition A.2.2

There exists an $X_{h-1,r}^+$-morphism

$$\begin{aligned} M_r :X_{h-1,r}^+ \times {\mathbb {A}}^n \rightarrow X_{h-1,r}^+ \times {\mathbb {A}}^n \end{aligned}$$

(the left ${\mathbb {A}}^n$ in terms of the coordinates $\{x_{i,h-1}\}_{i=1}^n$ and the right ${\mathbb {A}}^n$ in terms of new coordinates $\{z_i\}_{i=1}^n$) satisfying the following properties:

• (i) $M_r$ is a composition of $X_{h-1,r}^+$-isomorphisms and purely inseparable $X_{h-1,r}^+$-morphisms.

• (ii) $M_r(X_{h,r}^+)$ is the closed subscheme defined by the equation

  $$\begin{aligned} z_1^{q^{n_0 r}} - z_1 = c({\widetilde{x}}) - c({\widetilde{x}})^q, \end{aligned}$$

  where c is as in (A.3).

• (iii) $M_r$ is ${\mathbb {W}}_h^{h-1}({\mathbb {F}}_{q^n})$-equivariant after equipping the left $X_{h-1,r}^+ \times {\mathbb {A}}^n$ with the ${\mathbb {W}}_h^{h-1}({\mathbb {F}}_{q^n})$-action

  $$\begin{aligned} 1 + \varpi ^{h-1} a :x_{i,h-1} \mapsto x_{i,h-1} + x_{i,0} a, \qquad \text {for all }1 \le i \le n, \end{aligned}$$

  and the right $X_{h-1,r}^+ \times {\mathbb {A}}^n$ with the ${\mathbb {W}}_h^{h-1}({\mathbb {F}}_{q^n})$-action

  $$\begin{aligned} 1 + \varpi ^{h-1} a :z_i \mapsto {\left\{ \begin{array}{ll} z_1 + {{\,\mathrm{Tr}\,}}_{{\mathbb {F}}_{q^n}/{\mathbb {F}}_{q^{n_0r}}}(a) &{}\quad \text {if }i = 1, \\ z_2 + a &{}\quad \text {if }r \ne n'\text { and }i = 2, \\ z_i &{}\quad \text {otherwise.} \end{array}\right. } \end{aligned}$$

In the rest of this section we prove Proposition A.2.2. To simplify the notation we will first establish the proposition in the case $\kappa = 0$ (i.e. $G = {{\,\mathrm{GL}\,}}_n$), and at the end generalize it to all $\kappa$. The first part of the proof of Proposition A.2.2 is given by the lemma below. Before stating it, we establish some notation. For an ordered basis ${\mathscr {B}}$ of V and $v \in V$, let $v_{{\mathscr {B}}}$ denote the coordinate vector of v in the basis ${\mathscr {B}}$. For two ordered bases ${\mathscr {B}}, {\mathscr {C}}= \{c_i\}_{i=1}^n$ of V, let $M_{{\mathscr {B}},{\mathscr {C}}}$ denote the base change matrix between them, that is, the ith column vector of $M_{{\mathscr {B}},{\mathscr {C}}}$ is $c_{i, {\mathscr {B}}}$. It is clear that

• $M_{{\mathscr {C}},{\mathscr {B}}} = M_{{\mathscr {B}},{\mathscr {C}}}^{-1}$,

• for any $v\in V$, $M_{{\mathscr {B}},{\mathscr {C}}} v_{{\mathscr {C}}} = v_{{\mathscr {B}}}$,

• for a third ordered basis ${\mathscr {D}}$ of V, one has $M_{{\mathscr {B}},{\mathscr {C}}} M_{{\mathscr {C}},{\mathscr {D}}} = M_{{\mathscr {B}},{\mathscr {D}}}$.

For a linear map $f :V \rightarrow V$, let $M_{{\mathscr {B}},{\mathscr {C}}}(f)$ denote the matrix representation of f; that is, $M_{{\mathscr {B}},{\mathscr {C}}}(f)\cdot v_{{\mathscr {C}}} = f(v)_{{\mathscr {B}}}$. In V we have the two ordered bases:

$$\begin{aligned} {\mathscr {E}}&:= \text { the standard basis of }V,\text { arising from the basis} \{e_i\}\text { of the lattice }{\mathscr {L}}_0, \\ {\mathscr {B}}_x&:= \{ \sigma _b^{i-1}(x) \}_{i=1}^n, \text {attached to the given }x \in X_0^+. \end{aligned}$$

We identify V with $\overline{{\mathbb {F}}}_q^n$ via the standard basis ${\mathscr {E}}$ and write $v = v_{{\mathscr {E}}}$ for all $v \in V$.

Lemma A.2.3

Assume $\kappa = 0$. There exists an $X_{h-1,r}^+$-isomorphism $X_{h-1,r}^+ \times {\mathbb {A}}^n {\mathop {\rightarrow }\limits ^{\sim }} X_{h-1, r}^+ \times {\mathbb {A}}^n$ given by a linear change of variables $x_{i,h-1} \rightsquigarrow x_{i,h-1}^{\prime }$, such that $P_1$ in the new coordinates $x_{i,h-1}^{\prime }$ takes the form

$$\begin{aligned} P_1 = x_{1,h-1}^{\prime } + x_{1,h-1}^{\prime ,q} + \dots + x_{1,h-1}^{\prime , q^{n-1}} + \sum _{j=0}^s \sum _{\lambda = i_j + 1}^{i_{j+1}} x_{s+2-j,h-1}^{\prime , q^{\lambda }}, \end{aligned}$$

and the action of $1 + \varpi ^{h-1} a \in W_h^{h-1}({\mathbb {F}}_{q^n})$ on the coordinates $x_{i,h-1}^{\prime }$ is given by

$$\begin{aligned} x_{i,h-1}^{\prime } \mapsto {\left\{ \begin{array}{ll} x_{1,h-1}^{\prime } + a &{}\quad \text {if }i=1, \\ x_{i,h-1} &{}\quad \text {if }i \ge 2. \end{array}\right. } \end{aligned}$$

(A.4)

Proof of Lemma A.2.3

We have to find a morphism $C := (c_{ij}) :X_{h-1,r}^+ \rightarrow {{\,\mathrm{GL}\,}}(V) = {{\,\mathrm{GL}\,}}_{n,{\mathbb {F}}_q}$ (this identification uses the standard basis ${\mathscr {E}}$ of V) such that the corresponding linear change of coordinates

$$\begin{aligned} x_{i,h-1} = c_{i,1} x_{1,h-1}^{\prime } + c_{i,2} x_{2,h-1}^{\prime } + \dots + c_{i,n} x_{n,h-1}^{\prime }, \text { for all }1 \le i \le n. \end{aligned}$$

(A.5)

brings $P_1$ to the requested form. Moreover, it suffices to do this fiber-wise by first determining $C({\tilde{x}})$ for any point $\tilde{x} \in X_{h-1,r}^+$ and then seeing that ${\tilde{x}} \mapsto C(\tilde{x})$ is in fact an algebraic morphism.

Fix $\tilde{x} \in X_{h-1,r}^+$ with image $x \in X_1^+$, and write C instead of $C(\tilde{x})$ to simplify notation. Let $C_i$ denote the ith column of C. Our coordinate change replaces $P_1$ by the polynomial (after dividing by the irrelevant non-zero constant $\det g_b(x) \in {\mathbb {F}}_q^{\times }$)

$$\begin{aligned} P_1&= x_{1,h-1}^{\prime } (m_1 \cdot C_1) + x_{1,h-1}^{\prime ,q} (m_2 \cdot \sigma _b(C_1)) + x_{1,h-1}^{\prime , q^2} (m_3 \cdot \sigma _b^2(C_1)) + \dots + x_{1,h-1}^{\prime , q^{n-1}} (m_n \cdot \sigma _b^{n-1}(C_1)) \nonumber \\&\quad + x_{2,h-1}^{\prime } (m_1 \cdot C_2) + x_{2,h-1}^{\prime ,q} (m_2 \cdot \sigma _b(C_2)) + x_{2,h-1}^{\prime ,q^2} (m_3 \cdot \sigma _b^2(C_2)) + \dots + x_{2,h-1}^{\prime ,q^{n-1}} (m_n \cdot \sigma _b^{n-1}(C_2)) \nonumber \\&\quad + \cdots + \nonumber \\&\quad + x_{n,h-1}^{\prime } (m_1 \cdot C_n) + x_{n,h-1}^{\prime ,q} (m_2 \cdot \sigma _b(C_n)) + x_{n,h-1}^{\prime ,q^2} (m_3 \cdot \sigma _b^2(C_n)) + \dots + x_{n,h-1}^{\prime ,q^{n-1}} (m_n \cdot \sigma _b^{n-1}(C_n)) \end{aligned}$$

(A.6)

in the indeterminates $\{x_{i,h-1}^{\prime }\}_{i=1}^n$. Here, we write $m_i$ to mean the ith row of the matrix m (adjoint to $g_b(x)$) from Lemma A.2.1. For $z \in V$, we put

$$\begin{aligned} m*z = \sum _{i=1}^{n} (m_i \cdot (b\sigma )^{i-1}(z)) e_i. \end{aligned}$$

(A.7)

The intermediate goal is to describe the map $m *:V \rightarrow V$ in terms of a coordinate matrix. Of course, $m*$ is not linear, but its composition with the projection on the ith component (corresponding to the ith standard basis vector) is $\sigma ^{i-1}$-linear. Thus we instead will describe the linear map $(m*)^{\prime } :V \rightarrow V$, which is the composition of $m*$ and the map $\sum _i v_i e_i \mapsto \sum _i \sigma ^{-(i-1)}(v_i) e_i$. This is done by the following lemma.

Lemma A.2.4

Assume $\kappa = 0$. We have

where the $y_i$’s are defined by the equation

$$\begin{aligned} (b\sigma )^n({\mathfrak {v}}) = v + \sum _{i=1}^{n-1} y_i (b\sigma )^i({\mathfrak {v}}). \end{aligned}$$

More precisely, if $\mu _{i,j}$ denotes the (i, j)th entry of $\det (g_b({\bar{x}}))^{-1} M_{{\mathscr {E}}, {\mathscr {B}}_x}((m*)^{\prime })$, then for $1 \le i,j \le n$ we have

$$\begin{aligned} \mu _{i,j} = {\left\{ \begin{array}{ll} 1 &{}\quad \text {if }j = 1 \\ 0 &{}\quad \text {if }i+j \le n+1\text { and }j > 1 \\ \sigma ^{-(i-1)}(y_{i-1}) &{} \quad \text {if }i + j = n+2 \\ \mu _{i-1,j} + \sigma ^{-(i-1)}(y_{i-1}) \sigma ^{n-(i-1)}(\mu _{n,j+i-(n+1)}) &{}\quad \text {if }i+j \ge n+3\text { and }i \ge 3. \end{array}\right. } \end{aligned}$$

In particular, if $i+j \ge n+3$ and $y_{i-1} = 0$, then $\mu _{i,j} = \mu _{i-1,j}$.

Proof of Lemma A.2.4

Let $z = \sum _{i=1}^n z_i (b\sigma )^{i-1}(x)$ be a generic element of V, written in ${\mathscr {B}}_x$-coordinates, that is $z_{{\mathscr {B}}_x}$ is the n-tuple $(z_i)_{i=1}^n$. The (i, j)th entry of $M_{{\mathscr {E}}, {\mathscr {B}}_x}((m*)^{\prime })$ is equal to $\sigma ^{-(i-1)}$ applied to the coefficient of $\sigma ^{i-1}(z_j)$ in the ith entry of $(b\sigma )^{i-1}(z)_{{\mathscr {B}}_x}$ ($=$ the ith entry of $m*z$).

The coordinate matrix of the $\sigma$-linear operator $b\sigma :V \rightarrow V$ in the basis ${\mathscr {B}}_x$,

$$\begin{aligned} M_{{\mathscr {B}}_x,{\mathscr {B}}_x}(b\sigma ) = \begin{pmatrix} 0 &{}\quad 0 &{}\quad \cdots &{}\quad 0 &{}\quad 1 \\ 1 &{}\quad 0 &{}\quad \cdots &{} 0 &{}\quad y_1 \\ 0 &{}\quad 1 &{}\quad \ddots &{}\quad \vdots &{}\quad y_2 \\ \vdots &{}\quad \ddots &{}\quad \ddots &{}\quad 0 &{}\quad \vdots \\ 0 &{}\quad \cdots &{}\quad 0 &{}\quad 1 &{}\quad y_{n-1} \end{pmatrix}. \end{aligned}$$

That is, for any $z \in V$,

$$\begin{aligned} b\sigma (z)_{{\mathscr {B}}_x} = M_{{\mathscr {B}}_x,{\mathscr {B}}_x}(b\sigma ) \cdot \sigma (z_{{\mathscr {B}}_x}), \end{aligned}$$

(A.8)

where the last $\sigma$ is applied entry-wise. Explicitly, the first entry of $b\sigma (z)_{{\mathscr {B}}_x}$ is $\sigma (z_n)$, and for $2 \le i \le n$ the ith entry of $b\sigma (z)_{{\mathscr {B}}_x}$ is $\sigma (z_{i-1}) + y_{i-1}\sigma (z_n)$. This allows to iteratively compute $b\sigma ^i(z)$ for all i, which we do to finish the proof.

First, we see that $z_1$ can occur in the nth (i.e. last) entry of $(b\sigma )^{\lambda - 1}(z)_{{\mathscr {B}}_x}$ only if $\lambda \ge n$; hence its contribution to the ith entry of $(b\sigma )^{i-1}(z)_{{\mathscr {B}}_x}$ for $i \le n$ is simply $\sigma ^{i-1}(z_1)$. This shows that the first column of $M_{{\mathscr {E}}, {\mathscr {B}}_x}((m*)^{\prime })$ consists of 1’s. Assume now $j \ge 2$. Then there is a smallest (if any) $i_0$ , such that $z_j$ occurs in the $i_0$th entry of $(b\sigma )^{i_0-1}(z)_{{\mathscr {B}}_x}$. Note that as $j\ge 2$, one has $i_0 \ge 2$. Then $z_j$ must have been occurred in the nth entry of $(b\sigma )^{i_0 - 2}(z)_{{\mathscr {B}}_x}$. As $z_j$ occurs in $z_{{\mathscr {B}}_x}$ in exactly the jth entry, and it needs $(n-j)$ times to apply $b\sigma$ to get it to the nth entry, we must have $i_0 - 2 \ge n - j$. This shows that the (i, j)th entry of $M_{{\mathscr {E}}, {\mathscr {B}}_x}((m*)^{\prime })$ is 0, unless $i \ge n+2-j$. The same consideration shows that if $i = n+2-j$, then $\sigma ^{i-1}(z_j)$ has the coefficient $y_{i-1}$ in $\sigma _b^{i - 1}(z)_{{\mathscr {B}}_x}$. This gives the entries of $M_{{\mathscr {E}}, {\mathscr {B}}_x}((m*)^{\prime })$ on the diagonal $i = n+2-j$. It remains to compute the entries below it, so assume $i > n + 2 - j$. Again, by the characterization of the entries of $M_{{\mathscr {E}}, {\mathscr {B}}_x}((m*)^{\prime })$ in the beginning of the proof and by the explicit description of how $\sigma _b$ acts (in the ${\mathscr {B}}_x$-coordinates), it is clear that the (i, j)th entry of $M_{{\mathscr {E}}, {\mathscr {B}}_x}((m*)^{\prime })$ is just the sum of the $(i-1,j)$th entry and $\sigma ^{-(i-1)}(y_{i-1})\sigma ^{n-(i-1)}((n,j-1)\text {th entry})$. This finishes the proof of Lemma A.2.4. $\square$

Now we continue the proof of Lemma A.2.3. Let ${\mathscr {C}}$ denote the ordered basis of V consisting of columns $C_1, C_2, \dots , C_n$ of C. We have $M_{{\mathscr {B}}_x,{\mathscr {C}}} = (\det g_b(x))^{-1} m \cdot C$. In particular, to give the invertible matrix C it is equivalent to give the invertible matrix $M_{{\mathscr {B}}_x,{\mathscr {C}}}$. But the ith column of $M_{{\mathscr {B}}_x,{\mathscr {C}}}$ is the coordinate vector of $C_i$ in the basis ${\mathscr {B}}_x$, i.e., what we denoted $C_{i,{\mathscr {B}}_x}$. We now show that one can find an invertible $M_{{\mathscr {B}}_x,{\mathscr {C}}}$, such that for its columns $C_{i,{\mathscr {B}}_x}$ we have

$$\begin{aligned} m *C_{1,{\mathscr {B}}_x}&= \textstyle \sum \limits _{\lambda =1}^n e_{\lambda }&\nonumber \\ m *C_{s+2 - j,{\mathscr {B}}_x}&= \textstyle \sum \limits _{\lambda = i_j + 1}^{i_j} e_{\lambda }&\quad \text {for }s \ge j \ge 0,\nonumber \\ m *C_{j^{\prime },{\mathscr {B}}_x}&= 0&\quad \text {if } j^{\prime }>s+2. \end{aligned}$$

(A.9)

Taking into account Eq. (A.6) and the definition of $m *$ in (A.7), this (plus the fact that $x \mapsto M_{{\mathscr {B}}_x,{\mathscr {C}}}$ will in fact an algebraic morphism) finishes the proof of Lemma A.2.3, except for the claim regarding the $W_h^{h-1}({\mathbb {F}}_{q^n})$-action.

To find $M_{{\mathscr {B}}_x,{\mathscr {C}}}$ satisfying (A.9), first observe that by Lemma A.2.4, there is some invertible matrix S depending on $\tilde{x} \in X_{h-1,r}^+$ (in fact, only on its image $x \in X_1^+$), such that $M_{{\mathscr {E}},{\mathscr {B}}_x}((m*)^{\prime }) \cdot S$ has the following form: its first column consists of 1’s; its ith column is 0, unless $i = n + 1 - i_j$ for some $s \ge j \ge 0$; for $s \ge j \ge 0$, the $\lambda$th entry of its $(n + 1 - i_j)$th column is 1 if $i_j +1 \le \lambda \le i_{j+1}$ (we put $i_{s+1} := n$ here) and zero otherwise. (To show this, use the general shape of $M_{{\mathscr {E}},{\mathscr {B}}_x}((m*)^{\prime })$ provided by Lemma A.2.4, and then consecutively apply row operations to it and use the last statement of Lemma A.2.4). Moreover, it is also clear from Lemma A.2.4 that S will be upper triangular with the upper left entry $= 1$.

Secondly, let T be a matrix such that: the first row has 1 in the first position and zeros otherwise; all except for the first entry of the first column are 0; for $s \ge j \ge 0$, the $(n + 1 - i_j)$th row has 1 in the $(s + 2 - j)$th position and 0’s otherwise; the remaining rows can be chosen arbitrarily. Obviously, T can be chosen to be a permutation matrix with entries only 0 or 1, and in particular invertible and independent of x. Finally, put $M_{{\mathscr {B}}_x,{\mathscr {C}}} := S \cdot T$. Explicitly the columns of the matrix

$$\begin{aligned} M_{{\mathscr {E}},{\mathscr {B}}_x}((m*)^{\prime }) \cdot M_{{\mathscr {B}}_x,{\mathscr {C}}} = (M_{{\mathscr {E}},{\mathscr {B}}_x}((m*)^{\prime }) \cdot S) \cdot T \end{aligned}$$

(A.10)

are as follows: the first column consist of 1’s; for $s \ge j \ge 0$, the the $\lambda$th entry of the $(s + 2 - j)$th column is 1 if $i_j +1 \le \lambda \le i_{j+1}$, and zero otherwise; all other columns consist of 0’s. On the other side, the jth column of of $M_{{\mathscr {E}},{\mathscr {B}}_x}((m*)^{\prime }) \cdot M_{{\mathscr {B}}_x,{\mathscr {C}}}$ is precisely $m *C_{j,{\mathscr {B}}_x}$ (up to the unessential $\sigma ^{-*}$-twist in each entry). This justifies (A.9).

The action of $1 + \varpi ^h a \in W_h^{h-1}({\mathbb {F}}_{q^n})$ on the coordinates $x_{i,h}$ is given by $(x_{i,h})_{i=1}^n \mapsto (x_{i,h} + ax_{i,0})_{i=1}^n$. We determine the action $1 + \varpi ^h a$ in the coordinates $x^{\prime }_{i,h}$. Indeed, let $C^{-1} = (d_{i,j})_{1\le i,j\le n}$. Then $1 +\varpi ^h a$ acts on $x_{i,h}^{\prime }$ by

$$\begin{aligned} x_{i,h}^{\prime } = \sum _{j=1}^n d_{i,j}x_{j,h} \mapsto \sum _{j=1}^n d_{i,j} (x_{j,h} + ax_{j,0}) = x_{i,h}^{\prime } + a\sum _{j=1}^n d_{i,j} x_{j,0}. \end{aligned}$$

Organizing the $x_{i,h}$ for $1 \le i \le n$ in one (column) vector, we can rewrite this as

$$\begin{aligned} 1 +\varpi ^h a :(x_{i,h}^{\prime })_{i=1}^n \mapsto (x_{i,h}^{\prime })_{i=1}^n + a C^{-1} \cdot x. \end{aligned}$$

We determine $C^{-1} \cdot x$. As $M_{{\mathscr {B}}_x,{\mathscr {C}}} = \det (g_b(x))^{-1} m C = g_b(x)^{-1}C$ (as $\det (g_b(x))^{-1} m = g_b(x)^{-1}$), we have $C^{-1} = M_{{\mathscr {B}}_x,{\mathscr {C}}}^{-1}g_b(x)^{-1}$. But x is the first column of $g_b(x)$, thus

$$\begin{aligned} C^{-1} \cdot x = M_{{\mathscr {B}}_x,{\mathscr {C}}}^{-1}g_b(x)^{-1}\cdot x = M_{{\mathscr {B}}_x,{\mathscr {C}}}^{-1} \cdot (1, 0, \ldots , 0)^\intercal , \end{aligned}$$

so $C^{-1} \cdot x$ is the first column of $M_{{\mathscr {B}}_x,{\mathscr {C}}}^{-1} = (ST)^{-1} = T^{-1} S^{-1}$. But S is upper triangular with upper left entry $=1$, so the first column of $M_{{\mathscr {B}}_x,{\mathscr {C}}}^{-1}$ is the first column of $T^{-1}$, which is $(1, 0, \ldots , 0)^\intercal$. This finishes the proof of the lemma. $\square$

The second part of the proof is given by the following lemma.

Lemma A.2.5

Assume $\kappa = 0$. There exists a $X_{h-1,r}^+$-morphism $X_{h-1,r}^+ \times {\mathbb {A}}^n \rightarrow X_{h-1,r}^+ \times {\mathbb {A}}^n$ such that if $\{z_i\}$ denotes the coordinates on ${\mathbb {A}}^n$ on the target ${\mathbb {A}}^n$, then the image of $X_{h,r}^+$ in $X_{h-1,r}^+ \times {\mathbb {A}}^n$ and the action of ${\mathbb {W}}_h^{h-1}({\mathbb {F}}_{q^n})$ on $z_i$ are given by Proposition A.2.2(ii),(iii). Moreover, such a morphism is given by the composition of the change-of-variables $x_{i,h}'$ and purely inseparable morphisms of the form $x_{i,h-1}' \mapsto x_{h-1}^{\prime ,q^{-j}}$ for appropriate i, j.

Proof

If $r = n$, this is literally Lemma A.2.3. Assume $r<n$. First, for $s \ge j \ge 0$, replace $x_{s+2-j}^{\prime }$ by $x_{s+2-j}^{\prime , q^{i_j + 1}}$. Then, by applying a series of iterated changes of variables of the form $x_c^{\prime } =: x_c^{\prime } + x_d^{\prime , q^{\lambda }}$ for appropriate $2 \le c,d \le s+2$ and $\lambda$ (essentially following the Euclidean algorithm to find the gcd of the integers $(i_{j+1} - i_j)$ (this gcd is equal to r)), we transform $P_1$ from Lemma A.2.3 to the form

$$\begin{aligned} P_1 = \sum _{i=0}^{n-1} x_{1,h}^{\prime , q^i} + \sum _{i=0}^{r-1} x_{2,h}^{\prime ,q^i}. \end{aligned}$$

As these operations does not involve $x_{1,h}^{\prime }$, the formulas (A.4) remain true. Now make the change of variables given by $z_1 := x_{2,h}^{\prime } + \sum _{j=0}^{\frac{n}{r} - 1} x_{1,h}^{\prime ,q^{rj}}$ and $z_2 := x_{1,h-1}^{\prime }$. In this coordinates, $P_1 = \sum _{i=0}^{r-1} z_1^{q^i}$ and the action is as claimed. $\square$

We are now ready to complete the proof of Proposition A.2.2.

Proof of Proposition A.2.2

Combining Lemmas A.2.3 and A.2.5 we obtain Proposition A.2.2 in the case $\kappa = 0$. Now let $\kappa$ be arbitrary. It is clear that the proof of Lemma A.2.3 can be applied to this more general situation. One then obtains the same statement, with the only difference being that now our change of variables does not affect the variables $x_{i,h-1}$ for $i \not \equiv 1 \mod n_0$ (these are exactly the variables which do not show up in $P_1$). That is, the right-hand side $X_{h-1,r}^+ \times {\mathbb {A}}^n$ will have the coordinates $\{x_{i,h-1}^{\prime } :i\equiv 1 \mod n_0, 1\le i\le n \} \cup \{x_{i,h-1} :i \not \equiv 1 \mod n_0, 1\le i\le n \}$ and the polynomial defining $X_{h,r}^+$ as a relative $X_{h-1,r}^+$ hypersurface in $X_{h-1,r}^+ \times {\mathbb {A}}^n$ is

$$\begin{aligned} P_1 = x_{1,h-1}^{\prime } + x_{1,h-1}^{\prime ,q^{n_0}} + \dots + x_{1,h-1}^{\prime , q^{n_0(n^{\prime }-1)}} + \sum _{j=0}^s \sum _{\lambda = i_j + 1}^{i_{j+1}} x_{s+2-j,h-1}^{\prime , q^{n_0\lambda }}, \end{aligned}$$

and the ${\mathbb {W}}_h^{h-1}({\mathbb {F}}_{q^n})$-action is given by

$$\begin{aligned} 1 + \varpi ^{h-1} a :x_{i,h-1}^{\prime } \mapsto {\left\{ \begin{array}{ll} x_{1,h-1}^{\prime } + a &{}\quad \text {if }i=1 \\ x_{i,h-1}^{\prime } &{}\quad \text {if }i \equiv 1 \mod n_0 \text {and }i > 1 \\ x_{i,h-1} + x_{i,0} a &{}\quad \text {if }i \not \equiv 1 \mod n_0. \end{array}\right. } \end{aligned}$$

We now apply the change of variables replacing $x_{i,h-1}$ by $x_{i,h-1}' :=x_{h-1} - x_{i,0} x_{1,h-1}'$ for all $i \not \equiv 1 \mod n_0$. This exactly gives us Lemma A.2.3 for arbitrary $\kappa$ (the only difference being the $q^{n_0}$-powers occurring in $P_1$). Now Lemma A.2.5 can be applied as in the case $\kappa = 0$, and this finishes the proof of Proposition A.2.2. $\square$