\section{Even orthogonal descent in characteristic two}
\label{fl:sec:char2-d}

Section~\ref{fl:sec:odd-symplectic} produced, in odd characteristic,
a nonzero generic descent from $\operatorname{GSpin}_{4n+1}$ to
split $\operatorname{GSpin}_{2n}$. The step that fails in
characteristic two is the Fourier--Jacobi argument: it moves a
Fourier coefficient from frequency zero to a nonzero frequency on
the unipotent subgroup of the $\operatorname{SL}_2$ in the Jacobi
group, a long root group, and it uses a quadratic exponent divided
by two. This
section gives a replacement. Its result is
Proposition~\ref{fl:prop:char2-direct-generic} together with
Corollary~\ref{fl:cor:char2-descent-split}; the realization
theorems for orthogonal groups, in all characteristics, are then
deduced in Section~\ref{rw:sec:orthogonal-realization}.

Throughout this section $p=2$. The argument has three parts.
\begin{enumerate}
\item In characteristic two the short root element contains a
square, and the commutator that should move the frequency has
character $\psi(sa^2)$ instead of $\psi(sa)$. The map
$a\mapsto a^2$ is not surjective on $F$, so this looks weaker. But
$a\mapsto\psi(sa^2)$ is an additive character of $a$, and the
Cartier operator computes it: it equals $\psi(C_\psi(s)a)$ for an
explicit additive map $C_\psi:F\to F$ that is surjective. This
allows us to prescribe the frequency
(Lemmas~\ref{fl:lem:cartier-character-formula}
and~\ref{fl:lem:char2-cartier}).
\item Iterating this along the rows of the unipotent radical moves
the Siegel--Whittaker coefficient of the residual representation
to a coefficient with a prescribed short-root frequency
(Lemma~\ref{fl:lem:char2-row-iteration}).
\item A second Fourier expansion then produces a descent whose
target is the special orthogonal group of a quadratic space
containing a plane $x^2+xy+c_0^2y^2$. Here the frequency $c_0$
cannot be prescribed, and the plane may a priori be anisotropic.
The local Jacquet module of
Proposition~\ref{fl:prop:uniform-b-jacquet} vanishes for an
anisotropic plane, which forces the Arf invariant of the plane to
be trivial (Proposition~\ref{fl:prop:char2-direct-generic} and
Corollary~\ref{fl:cor:char2-descent-split}).
\end{enumerate}
The matrix identities of this section and its statements about
sets of roots were also checked by computer, in small rank. These
checks are not part of the proofs; Appendix~\ref{rw:app:computer}
lists them, with the programs and the ranges covered.

\subsection{Conventions in characteristic two}

We use the quadratic space with basis
$e_1,\ldots,e_m,u,f_m,\ldots,f_1$ and form
\[
 Q\Bigl(zu+\sum_i(x_ie_i+y_if_i)\Bigr)=z^2+\sum_ix_iy_i .
\]
Its polar form $B$ has radical spanned by $u$. As before, $E_{a,b}$
denotes the matrix sending the basis vector $b$ to $a$ and the
other basis vectors to zero. The short root elements and their
coordinates are
\[
 x_{e_i}(a)=1+aE_{u,f_i}+a^2E_{e_i,f_i},\qquad
 x_{-e_i}(a)=1+aE_{u,e_i}+a^2E_{f_i,e_i},\qquad
 c_i(g)=g_{u,f_i}.
\]
The root $e_i$ has coroot $2e_i^*$ in the special orthogonal group
and $2e_i^*-e_0^*$ in $\operatorname{GSpin}$; the quadratic entry
$a^2$ does not affect this. The map $a\mapsto x_{e_i}(a)$ is a
homomorphism, and the short root groups of two different
hyperbolic pairs commute, because $B(u,u)=0$.

Fix a separating element $t\in F$, so that $dt$ is a basis of
$\Omega^1_{F/\mathbb F_q}$, and write the nontrivial additive
character $\psi$ of $F\backslash\mathbb A_F$ as
$\psi=\psi_{\eta\,dt}$ with $\eta\in F^\times$:
\[
 \psi(a)=\prod_v\psi_v(a_v),\qquad
 \psi_v(a)=(-1)^{\operatorname{Tr}_{\kappa(v)/\mathbb F_2}
                         \operatorname{Res}_v(a\eta\,dt)}.
\]
For a unipotent $F$-group $N$ write
$[N]=N(F)\backslash N(\mathbb A_F)$, with Haar measure of volume
one. For a character $\zeta$ of $N(\mathbb A_F)$ trivial on $N(F)$,
the Fourier coefficient of an automorphic function $\phi$ is
\[
 \mathcal F_{N,\zeta}\phi(g)=\int_{[N]}\phi(ng)\zeta(n)^{-1}\,dn.
\]
We say that the coefficient $(N,\zeta)$ is \emph{nonzero on} a
space of automorphic functions stable under right translation if
$\mathcal F_{N,\zeta}\phi(g)\ne0$ for some $\phi$ in the space and
some $g$.

\subsection{The Cartier operator and characters of squares}

\begin{lemma}[The character of a square]
\label{fl:lem:cartier-character-formula}
Every element of $F$ can be written uniquely as $A^2+tB^2$ with
$A,B\in F$. Put
\[
 C_t(A^2+tB^2)=B,\qquad C_\psi(s)=\eta^{-1}C_t(\eta s).
\]
Then for $s\in F$ and $a\in\mathbb A_F$,
\begin{equation}\label{fl:eq:cartier-character}
 \psi(sa^2)=\psi(C_\psi(s)a),
\end{equation}
and $C_\psi(t\eta\beta^2)=\beta$ for every $\beta\in F$. In
particular $C_\psi:F\to F$ is surjective.
\end{lemma}
\begin{proof}
Since $F$ is a function field in one variable over a perfect field
of characteristic two, $[F:F^2]=2$, and $t\notin F^2$ because
$dt\ne0$. So $1,t$ is a basis of $F$ over $F^2$. The same holds
for each completion: $[F_v:F_v^2]=2$ by the expansion of Laurent
series into even and odd parts, and $t$ is not a square in $F_v$,
since otherwise $F=F^2(t)$ would be contained in $F_v^2$, although
$F$ contains an element of valuation one at $v$.

Write $\eta s=A^2+tB^2$ with $A,B\in F$. Let $\mathcal C$ be the
Cartier operator on differentials of $F_v$; it is additive and
satisfies $\mathcal C(g^2\,dt)=0$ and
$\mathcal C(g^2t\,dt)=g\,dt$. For $a\in F_v$ this gives
\[
 \mathcal C(sa^2\eta\,dt)
 =\mathcal C\bigl((Aa)^2\,dt+(Ba)^2t\,dt\bigr)
 =aB\,dt=C_\psi(s)a\,\eta\,dt.
\]
In a local uniformizer $z$ the Cartier operator sends
$\sum_ja_jz^j\,dz$ to $\sum_ja_{2j+1}^{1/2}z^j\,dz$, so the
residue of $\mathcal C\omega$ is the square root of the residue of
$\omega$. The trace from $\kappa(v)$ to $\mathbb F_2$ is invariant
under the square root. Hence
$\psi_v(sa^2)=\psi_v(C_\psi(s)a)$ at every place $v$, including
those where $\eta\,dt$ has a zero or a pole, and taking the product
gives \eqref{fl:eq:cartier-character}. Finally
$\eta\cdot t\eta\beta^2=t(\eta\beta)^2$, so
$C_\psi(t\eta\beta^2)=\eta^{-1}\eta\beta=\beta$.
\end{proof}

\begin{lemma}[Prescribing a frequency]
\label{fl:lem:char2-cartier}
Let $N$ be a unipotent group with a character $\zeta$ trivial on
$N(F)$, and let $X=x(\mathbb G_a)$ and $Y=y(\mathbb G_a)$ be
one-parameter groups that normalize $N$ and preserve $\zeta$.
Assume that $X\cap N=Y\cap N=1$, that $[X,Y]\subset N$, and that
\[
 \zeta([x(s),y(a)])=\psi(sa^2).
\]
If the coefficient $(N,\zeta)$ is nonzero on a space $\mathcal V$
of automorphic functions stable under right translation, then for
every $\beta\in F$ the coefficient on $NY$ of the character
extending $\zeta$ by $y(a)\mapsto\psi(\beta a)$ is nonzero on
$\mathcal V$.
\end{lemma}
\begin{proof}
For $\phi\in\mathcal V$ put $h=\mathcal F_{N,\zeta}\phi$ and, for
$\beta\in F$,
\[
 h_\beta(g)=\int_{F\backslash\mathbb A_F}
                   h(y(a)g)\psi(\beta a)\,da.
\]
By the hypotheses on $Y$ this is the coefficient of $\phi$ in the
statement. For fixed $g$ the function $a\mapsto h(y(a)g)$ on
$F\backslash\mathbb A_F$ is locally constant, so it is a finite sum
of characters $\psi(\alpha a)$ with $\alpha\in F$. If $h\ne0$,
choose $\phi$, $g$ and $\alpha$ with $h_\alpha(g)\ne0$.

Let $s\in F$. Since $\phi$ is left invariant under $x(s)$ and
$x(s)$ normalizes $N$ and preserves $\zeta$, we have
$h(x(s)g')=h(g')$ for all $g'$. Moving $x(s)$ past $y(a)$ produces
the commutator, an element of $N$ on which $\zeta$ takes the value
$\psi(sa^2)$; signs play no role, since $\psi$ takes the values
$\pm1$. With Lemma~\ref{fl:lem:cartier-character-formula} we obtain
\begin{equation}\label{fl:eq:cartier-frequency-translation}
 h_\beta(x(s)g)
 =\int_{F\backslash\mathbb A_F}h(y(a)g)
     \psi\bigl((\beta+C_\psi(s))a\bigr)\,da
 =h_{\beta+C_\psi(s)}(g).
\end{equation}
Given $\beta$, take $s=t\eta(\alpha+\beta)^2$. Then
$C_\psi(s)=\alpha+\beta$ and
$h_\beta(x(s)g)=h_\alpha(g)\ne0$. The translation is by an element
of $G(F)$, so the nonzero coefficient is a coefficient of a
function in $\mathcal V$ over $F$.
\end{proof}

\subsection{Moving the frequency along the rows}

Let $G$ be $\operatorname{SO}_{2m+1}$ or $\operatorname{GSpin}_{2m+1}$,
let $U$ be its upper triangular maximal unipotent subgroup in the
basis above, and put $U_{\rm last}^0=\{g\in U:c_m(g)=0\}$, the
subgroup generated by the positive root groups other than
$U_{e_m}$. Matrix entries are indexed by the basis; we write
$g_{i,j}$ for the entry at $(e_i,e_j)$ and $g_{i,-j}$ for the entry
at $(e_i,f_j)$. For $1\le r<m-1$ define
\[
 \begin{split}
 U_{{\rm chain},r}&=\{g\in U_{\rm last}^0:g_{i,m}=0\ (r<i<m)\},\\
 U_{{\rm row},r}&=\{g\in U_{\rm last}^0:g_{r,j}=0\ (r<j<m)\},\\
 \lambda_r&=\sum_{i<r}g_{i,i+1}+g_{r,m}
              +\sum_{i=r+1}^{m-2}g_{i,i+1}+g_{m-1,-m}.
 \end{split}
\]
In terms of roots, $\lambda_r$ is the sum of the coordinates on the
chain $e_1-e_2,\ldots,e_{r-1}-e_r$, on $e_r-e_m$, on the chain
$e_{r+1}-e_{r+2},\ldots,e_{m-2}-e_{m-1}$, and on $e_{m-1}+e_m$. For
$r=m-1$ both groups are defined to be $U_{\rm last}^0$, and
$\lambda_{m-1}=\sum_{i<m}g_{i,i+1}$. We call
$(U_{\rm last}^0,\psi\lambda_{m-1})$ the \emph{endpoint
coefficient}. The Siegel--Whittaker coefficient $\mathcal{SW}$ of
Section~\ref{fl:sec:common-descent} is its integral over the
remaining root group $U_{e_m}$, so if $\mathcal{SW}$ is nonzero on
a space of functions, the endpoint coefficient is too.

\begin{lemma}[Row iteration]
\label{fl:lem:char2-row-iteration}
Let $\mathcal V$ be a space of automorphic functions on $G$, stable
under right translation, on which the Siegel--Whittaker coefficient
is nonzero. Then for every $r$ with $1\le r\le m-1$ the coefficient
$(U_{{\rm chain},r},\psi\lambda_r)$ is nonzero on $\mathcal V$.
Moreover, for every $r$ with $1\le r<m-1$ and every $\beta\in F$,
the coefficient
$(U_{{\rm chain},r},\psi(\lambda_r+\beta c_{m-1}))$ is nonzero on
$\mathcal V$.
\end{lemma}
\begin{proof}
The case $r=m-1$ is the endpoint coefficient. We show, for
$2\le r\le m-1$ and $\beta\in F$, that if
$(U_{{\rm chain},r},\psi\lambda_r)$ is nonzero on $\mathcal V$ then
so is
$(U_{{\rm chain},r-1},\psi(\lambda_{r-1}+\beta c_{m-1}))$. Taking
$\beta=0$ and descending induction on $r$ gives the first
assertion, and a general $\beta$ in the last step gives the second.

The step which is special to characteristic two is the second.
The commutator of the two root groups is quadratic in one of the
variables, and Lemma~\ref{fl:lem:char2-cartier} turns the
character $\psi(sa^2)$ which results into a character which is
linear in $a$. Steps~3 and~5 are verifications on roots.

\step{1}{from the chain group to the row group} By
Lemma~\ref{fl:lem:uniform-row-separation} with $b=0$, the
coefficients of $\psi\lambda_r$ on $U_{{\rm chain},r}$ and on
$U_{{\rm row},r}$ vanish identically together. Put
\[
 Z_r=U_{e_r},\qquad X_r=U_{e_r+e_m},\qquad Y=U_{-e_m},
\]
and let $N_r$ be generated by the positive root groups other than
those of
\[
 e_m,\quad e_r,\quad e_r+e_m,\quad e_r-e_j\ \ (r<j<m).
\]
Then $U_{{\rm row},r}=N_rZ_rX_r$. The character $\psi\lambda_r$ is
trivial on $Z_r$ and on $X_r$; for $r=m-1$ this uses that
$\lambda_{m-1}$ has no term on a root $e_i+e_m$. Integrating over
fewer variables, we see that the coefficient
$(N_rZ_r,\psi\lambda_r)$ is nonzero on $\mathcal V$.

\step{2}{the commutator} We apply
Lemma~\ref{fl:lem:char2-cartier} to $N=N_rZ_r$, $X=X_r$ and $Y$.
In the ordered basis $(e_r,e_m,u,f_m,f_r)$ of the relevant
five-dimensional space,
\[
 x_{e_r+e_m}(s)=1+s(E_{e_r,f_m}+E_{e_m,f_r}),\qquad
 x_{-e_m}(a)=1+aE_{u,e_m}+a^2E_{f_m,e_m},
\]
and a direct computation gives
\begin{equation}\label{fl:eq:char2-root-commutator}
 [x_{e_r+e_m}(s),x_{-e_m}(a)]
       =x_{e_r}(sa)\,x_{e_r-e_m}(sa^2):
\end{equation}
both sides fix $e_r,u,f_m$ and send
\[
 e_m\longmapsto e_m+sa^2e_r,\qquad
 f_r\longmapsto f_r+sa\,u+sa^2f_m+s^2a^2e_r.
\]
The first factor on the right lies in $Z_r$, where the character
is trivial, and the second has character $\psi(sa^2)$, because
$\lambda_r$ contains the coordinate on $e_r-e_m$ with coefficient
one. So $[X_r,Y]\subset N_rZ_r$ with the commutator character
required by Lemma~\ref{fl:lem:char2-cartier}.

\step{3}{normalization} We check that $X_r$ and $Y$ normalize
$N_rZ_r$ and preserve its character.

Conjugation by $Y=U_{-e_m}$ moves a root group $U_{e_i+e_m}$ into
$U_{e_i}U_{e_i-e_m}$. For $i\ne r$ both of these lie in $N_r$ and
carry the trivial character, and $U_{e_r+e_m}$ is not in $N_rZ_r$.
The remaining positive root groups in $N_rZ_r$ commute with $Y$:
the only positive root that could interact with $-e_m$ otherwise is
$e_m$, which is excluded, and the short root groups of the other
hyperbolic pairs commute with $Y$ in characteristic two.

Conjugation by $X_r=U_{e_r+e_m}$ can only produce root groups
$U_{e_i+e_m}$, from $e_i-e_r$, and $U_{e_i+e_r}$, from $e_i-e_m$.
These lie in $N_r$ and carry the trivial character. The one root
of this form on which $\lambda_r$ is not trivial, $e_{m-1}+e_m$,
would have to come from $e_{m-1}-e_r$, which is not a positive
root since $r\le m-1$.

\step{4}{prescribing the frequency on $Y$}
Lemma~\ref{fl:lem:char2-cartier} now shows that the coefficient on
$N_rZ_rY$ of the character $\psi\lambda_r$ extended by
$y(a)\mapsto\psi(\beta a)$ is nonzero on $\mathcal V$, for every
$\beta\in F$. Since $Z_r$ commutes with $Y$ and carries the trivial
character, the same holds for the group $N_rY$.

\step{5}{identification with the previous row} Let $w_r$ be the
element of the Weyl group, represented over $F$, that fixes
$e_i,f_i$ for $i<r$ and $u$, and satisfies
\[
 w_r(e_i)=e_{i+1}\ \ (r\le i\le m-2),\qquad
 w_r(e_{m-1})=f_m,\qquad w_r(e_m)=e_r,
\]
with the corresponding action on the dual vectors. We claim that
\[
 w_rU_{{\rm chain},r-1}w_r^{-1}=N_rY
\]
and that conjugation carries
$\psi(\lambda_{r-1}+\beta c_{m-1})$ to the character of step 4. For
$r\le m-2$ the roots on which $\lambda_{r-1}+\beta c_{m-1}$ is
nontrivial are transformed as follows.
\[
\begin{array}{c|c}
\text{root of }U_{{\rm chain},r-1}&\text{image under }w_r\\ \hline
 e_i-e_{i+1}\ (i<r-1)&e_i-e_{i+1}\\
 e_{r-1}-e_m&e_{r-1}-e_r\\
 e_i-e_{i+1}\ (r\le i\le m-3)&e_{i+1}-e_{i+2}\\
 e_{m-2}-e_{m-1}&e_{m-1}+e_m\\
 e_{m-1}+e_m&e_r-e_m\\
 e_{m-1}&-e_m
\end{array}
\]
The images in the first five rows are exactly the roots of
$\lambda_r$, and the last row carries the coordinate $c_{m-1}$ to
the coordinate of $Y$. For $r=m-1$ the last two roots of
$\lambda_{m-2}$ are carried to $e_{m-2}-e_{m-1}$ and
$e_{m-1}-e_m$, the last two roots of the endpoint character.

For the groups we compare the sets of roots. On the indices, $w_r$
is the map $i\mapsto i$ for $i<r$, $i\mapsto i+1$ for
$r\le i\le m-2$, $m-1\mapsto-m$ and $m\mapsto r$. It maps the
indices $1,\ldots,m-2$ onto the indices other than $r$ and $m$,
preserving their order; so the roots of $U_{{\rm chain},r-1}$
supported on the first $m-2$ indices give all the positive roots
which involve neither $r$ nor $m$. The roots $e_i\pm e_{m-1}$ with
$i\le m-2$ give the roots $e_j\mp e_m$ with $j\notin\{r,m\}$, and
$e_{m-1}$ gives $-e_m$. The sums $e_i+e_m$ with $i\le m-2$ give
the sums which involve $e_r$ and not $e_m$, the sum
$e_{m-1}+e_m$ gives $e_r-e_m$, and the differences $e_i-e_m$ with
$i<r$, the only ones in $U_{{\rm chain},r-1}$, give $e_i-e_r$.
These are the roots of $N_rY$. The matrix of $w_r$ permutes the
vectors of the basis, so it maps each root subgroup of the table
to its image with the same parameter, also in characteristic two.
The two characters are homomorphisms which agree on these root
subgroups, hence they agree.

Since the functions in $\mathcal V$ are left invariant under the
rational element $w_r$, the coefficient
$(U_{{\rm chain},r-1},\psi(\lambda_{r-1}+\beta c_{m-1}))$ of $\phi$
at $g$ equals the coefficient of step 4 at $w_rg$. It is therefore
nonzero on $\mathcal V$.
\end{proof}

\subsection{The descent and its terminal plane}

Now take $m=2n$ with $n\ge2$, so that $G$ is
$\operatorname{SO}_{4n+1}$ or $\operatorname{GSpin}_{4n+1}$. Let
$V_{\rm tail}$ be the span of the basis vectors other than
$e_1,\ldots,e_n,f_1,\ldots,f_n$. A linear form $\lambda$ on
$V_{\rm tail}$ with $\lambda(u)=1$ defines the descent character
\[
 \psi_\lambda(g)=\psi\Bigl(\sum_{i<n}g_{i,i+1}
       +\lambda\bigl(\operatorname{pr}_{V_{\rm tail}}(gf_n)\bigr)\Bigr)
\]
on the unipotent radical of the parabolic subgroup stabilizing the
flag spanned by $e_1,\ldots,e_n$. Its connected stabilizer in the
Levi subgroup is $\operatorname{GSpin}(\ker\lambda)$, and the
\emph{descent} with terminal covector $\lambda$ is the Fourier
coefficient with respect to $\psi_\lambda$, viewed as a function
on that group. The descent is \emph{generic} if it has a nonzero
Whittaker coefficient.

The construction below involves two linear forms on the last
three-dimensional space $V_{\rm term}=\langle e,u,f\rangle$, where
$e=e_m$ and $f=f_m$. One, called $\lambda_{c_0}$, defines the
group to which we descend; the other, called $\kappa_{\rm W}$,
defines a Whittaker character on that group.

\begin{proposition}[A generic descent in characteristic two]
\label{fl:prop:char2-direct-generic}
Let $\mathcal V$ be a space of automorphic functions on
$\operatorname{GSpin}_{4n+1}$, stable under right translation, on
which the Siegel--Whittaker coefficient is nonzero. Then there is
$c_0\in F$ such that $\mathcal V$ has a nonzero generic descent of
length $n$ whose terminal covector is
\[
 \lambda_{c_0}(zu+xe+yf)=z+x+c_0y
\]
on $V_{\rm term}$, extended by zero on the other hyperbolic pairs
of $V_{\rm tail}$. The restriction of $Q$ to
$\ker\lambda_{c_0}\cap V_{\rm term}$ is the binary form
$x^2+xy+c_0^2y^2$, whose Arf invariant is the class of $c_0$ in
$F/\{s^2+s:s\in F\}$. If the connected centre acts on
$\mathcal V$ through a character, it acts on the descent through
the same character.
\end{proposition}
\begin{proof}
The proof transforms the coefficient provided by row iteration
into a Whittaker coefficient of a descent, through a sequence of
Fourier exchanges
(Lemmas~\ref{fl:lem:odd-exchange} and~\ref{fl:lem:uniform-interleave})
and one Fourier expansion. Each exchange preserves the property of
being nonzero on $\mathcal V$. The expansion, in step~3, is the one
step which is not an equivalence: it produces a frequency $c_0$
which cannot be prescribed, and this is why
Corollary~\ref{fl:cor:char2-descent-split} is needed afterwards.

Table~\ref{rw:tab:char2-route} records the group and the character
after each step. They have the same shape throughout. The weights
of a basis of $\langle e_1,\ldots,e_{m-1}\rangle$, the standard
basis in steps~1 to~3 and a new basis from step~4 on, are divided
into two chains. Write $N$ for the
unipotent radical of the parabolic subgroup with Levi subgroup
$\operatorname{GL}_{m-1}\times\operatorname{SO}(V_{\rm term})$; it is
generated by the root groups of the sums of two weights of the
chains and by the terminal rows $r_i(V_{\rm term})$ of step~5. The
group is generated by $N$, by the root groups of the positive
differences within each chain, and by the root groups of certain
differences between the two chains. The character is $\psi$ of the
sum of the coordinates on the adjacent roots of each chain and of
one linear form on the terminal row of the last weight of each
chain, written in the coordinates $v=zu+xe+yf$ of $V_{\rm term}$. In
this description the roots $A_{n-1}-\epsilon_{\rm last}$,
$B_n+\epsilon_{\rm last}$ and $B_n$ of step~1 are the coordinates
$y$ on the row of $A_{n-1}$, and $x$ and $z$ on the row of $B_n$.

\begin{table}[bp]
\centering
\small
\renewcommand{\arraystretch}{1.3}
\begin{tabular}{@{}l
 >{\raggedright\arraybackslash}p{0.19\textwidth}
 >{\raggedright\arraybackslash}p{0.23\textwidth}
 >{\raggedright\arraybackslash}p{0.19\textwidth}
 >{\raggedright\arraybackslash}p{0.23\textwidth}@{}}
\toprule
step & the two chains & differences between the chains
 & forms on the last rows & obtained by\\
\midrule
1 & $A_1,\ldots,A_{n-1}$ and $B_1,\ldots,B_n$ &
 $A_i-B_j$ with $i<n-1$ &
 $y$ on $A_{n-1}$,\newline $z+x$ on $B_n$ &
 row iteration with $r=n-1$ and $\beta=1$; row separation\\
2 & the same &
 $A_i-B_j$ with $j>i+1$,\newline
 $B_j-A_i$ with $i\ge2$, $j\le i$ &
 the same &
 interleaving with $d=-1$\\
3 & the same &
 those of step 2 and $B_1-A_1$ &
 the same; in addition $c_0\delta$ on $B_1-A_1$ &
 Fourier expansion along $\delta$; this produces $c_0$\\
4 & $A'_1,\ldots,A'_n$ and $B'_1,\ldots,B'_{n-1}$ &
 $A'_i-B'_j$ with $j>i-1$,\newline
 $B'_j-A'_i$ with $j<i-1$ &
 $z+x+c_0y$ on $A'_n$,\newline $y$ on $B'_{n-1}$ &
 the same group in the basis changed by $g_{\rm pair}$\\
4 & the same &
 $A'_i-B'_j$ with $i<n$ &
 the same &
 interleaving with $d=1$, in the opposite direction\\
6 & the same, renamed $A_1,\ldots,A_n$ and $B_1,\ldots,B_{n-1}$ &
 all $A_i-B_j$ &
 the same &
 exchange of $U_{B_j}$ for $U_{A_n-B_j}$, $j<n$\\
\bottomrule
\end{tabular}
\caption{The group and the character after each step of the proof of
Proposition~\ref{fl:prop:char2-direct-generic}; step~4 has two
stages. In steps~1 to~3,
$A_i=e_i$ and $B_j=e_{n-1+j}$. In step~4 the new weights are those of
the vectors $b_1,c_0a_1+b_2,\ldots,c_0a_{n-1}+b_n$ for the chain
$A'$ and $a_1,\ldots,a_{n-1}$ for the chain $B'$; the term
$c_0\delta$ of step~3 has become part of the coordinate on the
adjacent root $A'_1-A'_2$. The forms are
$\lambda_{c_0}=z+x+c_0y$ and $\kappa_{\rm W}=y$. In the last line
the terminal rows of the weights $B_j$ are restricted to
$L_{c_0}=\ker\lambda_{c_0}$, so that the group is no longer
generated by $N$ and root groups; it is the group $R$ of step~6.}
\label{rw:tab:char2-route}
\end{table}

Comparing the first line with the last: the chain of length $n$
carries the form $z+x$ at the start and $z+x+c_0y$ at the end, the chain of
length $n-1$ carries the form $y$ throughout, and the differences
between the chains, which at the start are of the form (short
chain) minus (long chain), are at the end all the differences (long
chain) minus (short chain). The last configuration is the one of a
descent along the long chain followed by a Whittaker coefficient
along the short chain. Figure~\ref{rw:fig:char2-cross} shows the
differences between the chains at each step for $n=4$.

\input{sections/fig-char2}

\step{1}{the starting coefficient} By
Lemma~\ref{fl:lem:char2-row-iteration} with $r=n-1$ and
$\beta=1$, the coefficient
$(U_{{\rm chain},n-1},\psi(\lambda_{n-1}+c_{m-1}))$ is nonzero on
$\mathcal V$. Rename the coordinate weights
\[
 A_i=e_i\ (i<n),\qquad B_j=e_{n-1+j}\ (j\le n),\qquad
 \epsilon_{\rm last}=e_m,
\]
so that $B_n=e_{m-1}$. After the row separation of
Lemma~\ref{fl:lem:uniform-row-separation}, the character has
coefficient one on the adjacent roots of the two chains
$A_1,\ldots,A_{n-1}$ and $B_1,\ldots,B_n$, and on
$A_{n-1}-\epsilon_{\rm last}$, $B_n+\epsilon_{\rm last}$ and $B_n$.
The group is generated by the positive root groups other than
those of $\epsilon_{\rm last}$ and of $A_{n-1}-B_j$.

\step{2}{interleaving the two chains} The chains have lengths
$r=n-1$ and $b=n=r+1$, so
Lemma~\ref{fl:lem:uniform-interleave} applies with $d=-1$. The
exchanges it performs are
\[
 U_{A_i-B_j}\longleftrightarrow U_{B_j-A_{i+1}},
 \qquad i=n-2,\ldots,1,\quad j=1,\ldots,i+1.
\]
Its auxiliary group is here the unipotent radical of the parabolic
subgroup with Levi
$\operatorname{GL}_{m-1}\times\operatorname{SO}(V_{\rm term})$. We
verify the hypothesis on this group. Conjugation by a root group
$U_{A_i-B_j}$ in the list changes only the row of $A_i$, with
$i<n-1$, and conjugation by its partner $U_{B_j-A_{i+1}}$ changes
only the row of $B_j$, with $j\le i+1<n$. Neither is one of the two
rows, those of $A_{n-1}$ and $B_n$, on which the character of the
auxiliary group is nontrivial, so that character is preserved.
After the exchanges, the cross roots in the group are
\[
 A_i-B_j\quad(j>i+1),\qquad
 B_j-A_i\quad(i\ge2,\ j\le i).
\]
For $n=2$ there are no exchanges and this description holds from
the start.

\step{3}{a Fourier expansion and the frequency $c_0$} Introduce a
grading: $B_1$ has weight $2n$, both $A_i$ and $B_{i+1}$ have
weight $2(n-i)$, $\epsilon_{\rm last}$ has weight zero, and dual
vectors have the opposite weights. Let $U_{>0}$ be generated by
the root groups of positive weight. The positive cross weights are
those of $A_i-B_j$ with $j>i+1$ and of $B_j-A_i$ with $j\le i$.
Comparing with step 2, the group obtained there is the kernel in
$U_{>0}$ of the coordinate $\delta$ on the single root
$B_1-A_1$. Every positive weight is at least two, so $\delta$,
which has weight two, is an additive character of $U_{>0}$; the
squares of the coordinates on short roots have weight at least
four, so they do not affect it. Expanding along $\delta$, we find $c_0\in F$ such that
the coefficient on $U_{>0}$ of the character extended by
$\psi(c_0\delta)$ is nonzero on $\mathcal V$. The expansion gives
no control over $c_0$.

\step{4}{change of basis in the Levi subgroup} The Levi subgroup of
the grading has blocks $B_1$, the pairs $(A_i,B_{i+1})$, and
$V_{\rm term}$. On the part of weight two, the character is given
by the vector $(c_0,1)$ in the first row, by the trace on each
intermediate $2\times2$ block, and on the last block by the matrix
\[
 C_{\rm term}=\begin{pmatrix}1&0&0\\0&1&1\end{pmatrix},
\]
whose columns correspond to the roots with $-\epsilon_{\rm last}$,
with $+\epsilon_{\rm last}$, and to the short root. Apply the
matrix
$g_{\rm pair}=\left(\begin{smallmatrix}c_0&1\\1&0\end{smallmatrix}\right)$,
of determinant one, in each block $(A_i,B_{i+1})$. Then
\[
 g_{\rm pair}^{-1}\binom{c_0}{1}=\binom10,\qquad
 g_{\rm pair}^tC_{\rm term}
 =\begin{pmatrix}c_0&1&1\\1&0&0\end{pmatrix},
\]
and the trace characters on the intermediate blocks are unchanged.
In the new basis there is a long chain
$b_1,\,c_0a_1+b_2,\ldots,\,c_0a_{n-1}+b_n$ and a short chain
$a_1,\ldots,a_{n-1}$, with the dual basis transformed by the
inverse transpose. Call the vectors of the long chain $A'_i$,
$1\le i\le n$, and those of the short chain $B'_j$,
$1\le j\le n-1$. Their weights are $2(n-i+1)$ and $2(n-j)$. So the
root $A'_i-B'_j$ has weight $2(j-i+1)$: the positive roots between
the two chains are the $A'_i-B'_j$ with $j>i-1$ and the
$B'_j-A'_i$ with $j<i-1$. This is the situation at the end of
Lemma~\ref{fl:lem:uniform-interleave} with $d=1$. Each exchange of
that lemma involves two roots $A'_i-B'_j$ and $B'_j-A'_{i+1}$ with
$i<n$ and $j\le i-1$. Their root subgroups act on the auxiliary
radical only in the rows of $A'_i$ and of $B'_j$, which are not
the last rows $A'_n$ and $B'_{n-1}$ of the two chains; so they
preserve the character on these last rows. We apply the lemma in
the opposite direction, and get a nonzero coefficient in which the long
chain ends with the linear form $\lambda_{c_0}$ on $V_{\rm term}$
and the short chain ends with
\[
 \kappa_{\rm W}(zu+xe+yf)=y.
\]
The Levi subgroup acts linearly on the part of weight two,
including the short root coordinate, so these are identities
between characters and not only between their differentials.

\step{5}{vector groups in the terminal space} To compare this
coefficient with a descent followed by a Whittaker coefficient we
need the following subgroups. For a hyperbolic index $i$ and
$v\in V_{\rm term}$ define $r_i(v)$ by
\[
 r_i(v)f_i=f_i+v+Q(v)e_i,\qquad
 r_i(v)w=w+B(w,v)e_i\ \ (w\in V_{\rm term}),\qquad
 r_i(v)e_i=e_i,
\]
and by the identity on the other basis vectors. Since
$Q(v+w)=Q(v)+Q(w)+B(v,w)$, the map $v\mapsto r_i(v)$ is a
homomorphism, and one checks on the basis that
\begin{align}
 [r_i(v),r_j(w)]&=x_{e_i+e_j}(B(v,w)),
 \label{rw:eq:terminal-commutator}\\
 x_{e_i-e_j}(s)\,r_j(v)\,x_{e_i-e_j}(s)^{-1}
   &=r_j(v)\,r_i(sv)\,x_{e_i+e_j}(sQ(v)).
 \label{rw:eq:terminal-conjugation}
\end{align}
The plane $L_{c_0}=\ker\lambda_{c_0}\subset V_{\rm term}$ has
equation $z=x+c_0y$, so $Q$ restricts to $x^2+xy+c_0^2y^2$ on it.
The vector $e+u$ lies in $L_{c_0}$, has $Q(e+u)=1$, and satisfies
$B(e+u,v)=y$ for $v=zu+xe+yf$; that is, it represents
$\kappa_{\rm W}|_{L_{c_0}}$. Consequently the adjacent roots of the
short chain, together with $\kappa_{\rm W}$ on its last row, define
a generic character of the maximal unipotent subgroup of the
quasi-split group $\operatorname{GSpin}(\ker\lambda_{c_0})$,
whether or not $L_{c_0}$ is isotropic.

\step{6}{descent followed by a Whittaker coefficient} Rename the
long and short chains $A_1,\ldots,A_n$ and $B_1,\ldots,B_{n-1}$.
Taking the descent with terminal covector $\lambda_{c_0}$ along
the long chain and then the Whittaker coefficient along the short
chain is a Fourier coefficient on the group $R$ generated by all
root groups $U_{e_i+e_j}$ among the chain indices, the positive
differences within each chain, all $U_{A_i-B_j}$, and the terminal
rows $r_{A_i}(V_{\rm term})$ and $r_{B_j}(L_{c_0})$. Its character
is nontrivial on terminal rows only for $A_n$ and $B_{n-1}$, where
it is $\lambda_{c_0}$ and $\kappa_{\rm W}$. We pass from $R$ to the
group of step 4 by the exchanges, for $j=1,\ldots,n-1$,
\[
 X_j=U_{A_n-B_j},\qquad Y_j=r_{B_j}(Fu)=U_{B_j},
\]
and verify the hypotheses of Lemma~\ref{fl:lem:odd-exchange}.

When $X_j$ is removed, the remaining generators still generate a
group with these root subgroups. Among the differences, $A_n-B_j$
could reappear as a commutator only from $A_n-B_i$ and $B_i-B_j$
with $i<j$, and $A_n-B_i$ has been removed at an earlier step.
Commutators with the terminal rows give terminal rows and sums,
never a difference. The group $X_j$ normalizes what remains, and
preserves the character. Indeed, conjugating by $X_j$ a difference
$A_i-A_n$ or $B_j-B_k$ which is present produces the root
$A_i-B_j$ with $i<n$, or $A_n-B_k$ with $k>j$; both are present
and carry the trivial character. Conjugating a sum produces sums.
And by \eqref{rw:eq:terminal-conjugation} the action of $X_j$ on
the terminal rows adds to the row of $A_n$ a vector of the row of
$B_j$; this row is still $L_{c_0}$, since it is enlarged only
after the exchange, and $\lambda_{c_0}$ vanishes on $L_{c_0}$.

The group $Y_j$ commutes with all the terminal rows, by
\eqref{rw:eq:terminal-commutator} and because $B(u,\cdot)=0$. Its
commutators with differences come from $B_i-B_j$ with $i<j$ and
from $A_i-B_j$ with $i<n$. They lie in the rows of the $B_i$ with
$i<j$, which have already been enlarged, and in the rows of the
$A_i$ with $i<n$, on all of which the character is trivial; the
sums which appear carry the trivial character as well. Moreover
$Y_j$ meets the common subgroup trivially, because
$Fu\cap L_{c_0}=0$.
Finally, by \eqref{rw:eq:terminal-conjugation} with $v=au$,
\[
 [X_j(s),Y_j(a)]=r_{A_n}(sa\,u)\,x_{A_n+B_j}(sa^2),
\]
on which the character is $\psi(\lambda_{c_0}(sa\,u))=\psi(sa)$.
This is a perfect pairing, so the exchange applies. It replaces
the row $r_{B_j}(L_{c_0})$ by $r_{B_j}(V_{\rm term})$ and removes
$X_j$. Since $\kappa_{\rm W}(u)=0$, the character on the last row
of the short chain is unchanged.

After all $n-1$ exchanges we have the group and character of step
4. Hence the coefficient on $R$ is nonzero on $\mathcal V$. Since
all the groups involved have compact quotients, Fubini's theorem
identifies this coefficient with the Whittaker coefficient of the
descent of a function in $\mathcal V$, which is therefore nonzero.

\step{7}{central character and Arf invariant} Unipotent elements
lift uniquely to $\operatorname{GSpin}$. The rational elements of
Levi subgroups and the Weyl elements used above lift because
$H^1(F,\mathbb G_m)=0$, and two lifts differ by a rational point
of the centre, which acts trivially on automorphic functions. So
the construction
preserves the character through which the connected centre acts.
Finally the Arf invariant of
$x^2+xy+c_0^2y^2$ is the class of $c_0^2$, and
$c_0^2=c_0+(c_0^2+c_0)$ has the same class as $c_0$.
\end{proof}

\begin{corollary}[The descent is to the split group]
\label{fl:cor:char2-descent-split}
In Proposition~\ref{fl:prop:char2-direct-generic}, suppose that
$\mathcal V$ is an irreducible automorphic representation whose
local components at almost all places are subquotients of the
representation induced from $\operatorname{GL}_2^n$ considered in
Proposition~\ref{fl:prop:uniform-b-jacquet}. Then $c_0=s^2+s$ for
some $s\in F$, and the generic descent is a nonzero generic
automorphic function on split $\operatorname{GSpin}_{2n}$ over $F$.
\end{corollary}
\begin{proof}
Suppose that $c_0$ is not of the form $s^2+s$. Then $z^2+z=c_0$
defines a separable quadratic extension of $F$, and by Chebotarev's
theorem there is a place $v$, among the places where the local
component is as in the statement, that does not split in it. At
such a place the plane $L_{c_0}$ is anisotropic, and the twisted
Jacquet module of
Proposition~\ref{fl:prop:uniform-b-jacquet} is zero. But the
nonzero global descent coefficient gives a nonzero functional on
that Jacquet module. Indeed, translate a vector with nonzero
descent on the right so that the integral which defines the
descent is not zero at the identity, write it as a sum of pure
tensors, one of which has a nonzero integral, and fix the
components of this tensor at the places other than $v$. A change
of variable in the integral shows that the functional obtained on
the component at $v$ has the required equivariance. This
contradiction shows that $c_0=s^2+s$.

For such $c_0$, the vectors
\[
 e_{\rm spl}=su+s^2e+f,\qquad
 f_{\rm spl}=(s+1)u+(s^2+1)e+f
\]
lie in $L_{c_0}$, satisfy
$Q(e_{\rm spl})=Q(f_{\rm spl})=0$ and
$B(e_{\rm spl},f_{\rm spl})=1$, and $\kappa_{\rm W}$ takes the
value one on each. So $L_{c_0}$ is a hyperbolic plane over $F$.
The isometry that fixes $u$ and carries this pair to the standard
hyperbolic pair identifies the target of the descent with the
split group, and the generic character with a generic character of
its standard maximal unipotent subgroup.
\end{proof}

\begin{remark}[Two different uses of frequencies]
\label{rw:rem:two-frequencies}
The proof uses Fourier expansion in two ways. In row iteration the frequency on a short root group can be
prescribed, by Lemma~\ref{fl:lem:char2-cartier}. In step 3 of
Proposition~\ref{fl:prop:char2-direct-generic} the expansion only
provides some frequency $c_0$, and no rational conjugation is
available to change its class modulo $\{s^2+s\}$: such a
conjugation would amount to the surjectivity of $s\mapsto s^2+s$ on
$F$. That the class of $c_0$ is trivial is proved by the local
argument of Corollary~\ref{fl:cor:char2-descent-split}.
\end{remark}