text:
\section{Splittings}
\label{sec:weil-hodge}

\subsection{Setup}
\label{ss:weil-hodge-setup}

Let $F$ be a number field and $B$ a quaternion division algebra over $F$.
Let $E$ be a quadratic extension of $F$ which embeds into $B$.
Let $*$ be the main involution on $B$ and $\rho$ the nontrivial Galois automorphism of $E$ over $F$.
We write $E = F + F \i$ and $B = E + E \j$ for some trace zero elements $\i \in E^\times$ and $\j \in B^\times$.
Let $\pr : B \rightarrow E$ be the associated projection.
Put $u = \i^2 \in F^\times$ and $J = \j^2 \in F^\times$.
Fix a nontrivial additive character $\psi$ of $\A/F$ and a character $\chi$ of $\A_E^\times / E^\times$ such that $\chi|_{\A^\times} = \xi_E$, where $\xi_E$ is the quadratic character of $\A^\times / F^\times$ associated to $E/F$ by class field theory.

We consider an $m$-dimensional right $B$-space $V$ equipped with a skew-hermitian form $\langle \cdot, \cdot \rangle : V \times V \rightarrow B$ given by
\begin{equation}
\label{eq:condition-skew-herm}
 \langle e_1 x_1 + \dots + e_m x_m, e_1 y_1 + \dots + e_m y_m \rangle
 = x_1^* \cdot \kappa_1 \i \cdot y_1 + \dots + x_m^* \cdot \kappa_m \i \cdot y_m
\end{equation}
for some basis $e_1,\dots,e_m$ of $V$ and some $\kappa_1, \dots, \kappa_m \in F^\times$.
We denote by $\GU(V)$ the unitary similitude group of $V$ and by $\nu : \GU(V) \rightarrow F^\times$ the similitude character:
\[
 \GU(V) = \{ g \in \GL(V) \, | \, \langle gv, gv' \rangle = \nu(g) \cdot \langle v, v' \rangle \text{ for all } v, v' \in V \},
\]
where $\GL(V)$ acts on $V$ on the left.
We have a natural embedding
\[
 E^\times \hookrightarrow \GU(V),
\]
where we may regard $\alpha \in E^\times$ as an element in $\GU(V)$ given by $e_i \mapsto e_i \alpha$ for all $i$.

Let $W = B$ be a left $B$-space equipped with a hermitian form $\langle \cdot, \cdot \rangle : W \times W \rightarrow B$ given by
\[
 \langle x, y \rangle = x \cdot y^*.
\]
We denote by $\GU(W)$ the unitary similitude group of $W$ and by $\nu : \GU(W) \rightarrow F^\times$ the similitude character:
\[
 \GU(W) = \{ h \in \GL(W) \, | \, \langle wh, w'h \rangle = \nu(h) \cdot \langle w, w' \rangle \text{ for all } w, w' \in W \},
\]
where $\GL(W)$ acts on $W$ on the right.
Then we have $\GU(W) \cong B^\times$.

Let $\V = V \otimes_B W$ be a $4m$-dimensional $F$-space equipped with a symplectic form
\[
 \llangle \cdot, \cdot \rrangle := \frac{1}{2} \tr_{B/F} \left( \langle \cdot, \cdot \rangle \otimes \langle \cdot, \cdot \rangle^* \right).
\]
Then we have a natural homomorphism
\begin{equation}
\label{eq:hom-GU(V)-GU(W)}
 \G(\U(V) \times \U(W)) \longrightarrow \Sp(\V),
\end{equation}
where
\[
 \G(\U(V) \times \U(W)) = \{ (g, h) \in \GU(V) \times \GU(W) \, | \, \nu(g) = \nu(h) \}
\]
and $\GL(V) \times \GL(W)$ acts on $\V$ on the right:
\[
 (v \otimes w) \cdot (g, h) := g^{-1} v \otimes w h.
\]
Let $\cG$ be a subgroup of $\G(\U(V) \times \U(W))$ defined by
\[
 \cG = \{ (g, h) \in \GU(V)^0 \times \GU(W) \, | \, \nu(g) = \nu(h) \in \N_{E/F}(E^\times) \}.
\]
We take a complete polarization $\V = \X \oplus \Y$ defined by
\begin{align*}
 \X & = F \cdot e_1 \otimes 1 + \dots + F \cdot e_m \otimes 1 + F \cdot e_1 \otimes \j + \dots + F \cdot e_m \otimes \j, \\
 \Y & = F \cdot e_1 \otimes \i + \dots + F \cdot e_m \otimes \i + F \cdot e_1 \otimes \i\j + \dots + F \cdot e_m \otimes \i\j.
\end{align*}

\subsection{Splitting over $\cG$}

For each place $v$ of $F$, let $\Mp(\V_v)$ be the metaplectic group over $F_v$:
\[
 1 \longrightarrow \C^1 \longrightarrow \Mp(\V_v) \longrightarrow \Sp(\V_v) \longrightarrow 1.
\]
Then $\Mp(\V_v)$ can be realized by a $2$-cocycle $z_{\Y_v}$ relative to $\Y_v$ and $\psi_v$ (see e.g.~\cite{rangarao}, \cite[\S 3.2.2]{periods1}).
For almost all $v$, there exists a map $s_{\Y_v}:K_v \rightarrow \C^1$, where $K_v$ is the standard maximal compact subgroup of $\Sp(\V_v)$, such that
\[
 z_{\Y_v}(k_1,k_2) = \frac{s_{\Y_v}(k_1 k_2)}{s_{\Y_v}(k_1) s_{\Y_v}(k_2)}
\]
for $k_1, k_2 \in K_v$ (see e.g.~\cite[\S 3.2.3]{periods1}).

\begin{prop}
\label{prop:spl-hodge}
For all $v$, there exists a map $s_v:\cG_v \rightarrow \C^1$ satisfying the following conditions:
\begin{enumerate}
\item For $\g_1, \g_2 \in \cG_v$, we have
\[
 z_{\Y_v}(\g_1,\g_2) = \frac{s_v(\g_1 \g_2)}{s_v(\g_1) s(\g_2)}.
\]
Here, by abuse of notation, we write $\g_i$ on the left-hand side for the image of $\g_i$ in $\Sp(\V_v)$ under \eqref{eq:hom-GU(V)-GU(W)}.
\item For $\z = (z, z)$ with $z \in F_v^\times$ and $\g \in \cG_v$, we have
\[
 s_v(\z \g) = \xi_{E_v}(z)^m \cdot s_v(\g).
\]
\item For almost all $v$, we have
\[
 s_v|_{\cG_v \cap K_v} = s_{\Y_v}|_{\cG_v \cap K_v}.
\]
\item For $\gamma \in \cG(F)$, we have
\[
 \prod_v s_v(\gamma) = 1.
\]
\end{enumerate}
\end{prop}

As in \cite[\S 3.3]{periods1}, Proposition \ref{prop:spl-hodge} enable us to define a Weil representation $\omega_\psi$ of $\cG(\A)$ on the Schwartz space $\SS(\X(\A))$.
Moreover, for any $\varphi \in \SS(\X(\A))$, the associated theta function
\[
 \Theta_{\varphi}(\g) := \sum_{x \in \X} \omega_{\psi}(\g) \varphi(x)
\]
on $\cG(\A)$ is left $\cG(F)$-invariant.

\begin{rem}
Suppose that $V$ is the $3$-dimensional skew-hermitian right $B$-space as in \S \ref{sec:constr-gl-ex-isom}.
Then $V$ satisfies the condition \eqref{eq:condition-skew-herm} and we may apply the above construction.
Note that $\nu(\GU(V_v)^0) = \N_{E_v/F_v}(E_v^\times)$ for all $v$, so that
\[
 \cG(\A) = \G(\U(V) \times \U(W))^0(\A).
\]
\end{rem}

The proof of Proposition \ref{prop:spl-hodge} will be given in \S \ref{ss:doubling-U(V)}--\S \ref{ss:proof-spl-hodge} below.
From now on, we fix a place $v$ of $F$ and suppress the subscript $v$ from the notation.

\subsection{The doubling method for $\U(V)$}
\label{ss:doubling-U(V)}

We consider the doubled space $V^\square = V \oplus V$ equipped with a skew-hermitian form
\[
 \langle (v_1,v_2), (v_1',v_2') \rangle := \langle v_1, v_1' \rangle - \langle v_2, v_2' \rangle.
\]
Then we have a natural embedding
\[
 \iota : \G(\U(V) \times \U(V)) \longrightarrow \GU(V^\square).
\]
If $\V^\square = \V \oplus \V$ is the doubled space equipped with a symplectic form defined similarly as above, then we have a natural embedding
\[
 \iota : \Sp(\V) \times \Sp(\V) \longrightarrow \Sp(\V^\square)
\]
and an identification
\[
 \V^\square = V^\square \otimes_B W.
\]
We take a complete polarization $\V^\square = \V^\bigtriangledown \oplus \V^\triangle$ defined by
\[
 \V^{\bigtriangledown} = \{ (x,-x) \, | \, x \in \V \}, \qquad
 \V^{\triangle} = \{ (x,x) \, | \, x \in \V \}.
\]
Under the above identification, we have
\[
 \V^{\bigtriangledown} = V^{\bigtriangledown} \otimes_B W, \qquad
 \V^{\triangle} = V^{\triangle} \otimes_B W,
\]
where $V^\square = V^\bigtriangledown \oplus V^\triangle$ is the complete polarization over $B$ defined similarly as above.

Now we recall Kudla's splitting over $\U(V^\square)$, where we regard $\U(V^\square)$ as a subgroup of $\Sp(\V^\square)$ via the natural embedding.
As in \cite[\S C.3]{periods1}, we regard $V^\square$ as a left $B$-space and let $\GL(V^\square)$ act on $V^\square$ on the right:
\begin{align*}
 x \cdot v & := v \cdot x^*, & x & \in B, \\
 v \cdot g & := g^{-1} \cdot v, & g & \in \GL(V^\square).
\end{align*}
Similarly, we regard $W$ as a right $B$-space and let $\GL(W)$ act on $W$ on the left.
Then we have an identification
\[
 \V^\square = W \otimes_B V^\square.
\]
Put
\[
 \v_i = \frac{1}{2 \kappa_i \i} \cdot (e_i, -e_i), \qquad
 \v_i^* = (e_i, e_i),
\]
so that $\langle \v_i, \v_j^* \rangle = \delta_{ij}$.
Using a basis $\v_1, \dots, \v_m, \v_1^*, \dots, \v_m^*$ of $V^\square$, we identify $\U(V^\square)$ with
\[
 \left\{ g \in \GL_{2m}(B) \, \left| \,
 g \begin{pmatrix} & \1_m \\ -\1_m & \end{pmatrix} {}^t g^*
 = \begin{pmatrix} & \1_m \\ -\1_m & \end{pmatrix}
 \right. \right\}.
\]
Let $P_{V^\triangle}$ be the maximal parabolic subgroup of $\U(V^\square)$ stabilizing $V^\triangle$:
\[
 P_{V^\triangle} = \left\{ \left.
 \begin{pmatrix} a & * \\ & ({}^t a^*)^{-1} \end{pmatrix}
 \, \right|  \, a \in \GL_m(B) \right\}.
\]
We define a map
\[
 \hat{s}_1 : \U(V^\square) \longrightarrow \C^1
\]
as follows:
\begin{itemize}
\item If $B$ is split, then we set
\[
 \hat{s}_1(g) = 1
\]
for $g \in \U(V^\square)$.
\item If $B$ is ramified, then we set
\[
 \hat{s}_1(g) = (-1)^j
\]
for $g \in P_{V^\triangle} \tau_j P_{V^\triangle}$ with
\[
 \tau_j =
 \begin{pmatrix}
  \1_{m-j} & & & \\
  & & & -\1_j \\
  & & \1_{m-j} & \\
  & \1_j & &
 \end{pmatrix}.
\]
\end{itemize}
By \cite[Theorem 3.1, cases $1_-$ and $2_+$]{kudla-splitting}, we have
\begin{equation}
\label{eq:spl-kudla-1}
 z_{\V^\triangle}(g_1,g_2) = \frac{\hat{s}_1(g_1 g_2)}{\hat{s}_1(g_1) \hat{s}_1(g_2)}
\end{equation}
for $g_1, g_2 \in \U(V^\square)$.

\begin{lem}
\label{lem:s-hat-1-E-conj}
For $\alpha \in E^{\times}$ and $g \in \U(V^\square)$, we have
\[
 \hat{s}_1(\alpha g \alpha^{-1}) = \hat{s}_1(g).
\]
\end{lem}

\begin{proof}
Since $\alpha p \alpha^{-1} \in P_{V^\triangle}$ for $p \in P_{V^\triangle}$ and $\alpha \tau_j \alpha^{-1} = \tau_j$, the assertion follows.
\end{proof}

\begin{lem}
\label{lem:s-hat-1-E^1}
Let $\alpha \in E^1$.
Then we have
\[
 \hat{s}_1(\iota(\alpha, 1)) = 1
\]
if $B$ is split, and
\[
 \hat{s}_1(\iota(\alpha, 1)) =
 \begin{cases}
  1 & \text{if $\alpha = 1$,} \\
  (-1)^m & \text{if $\alpha \ne 1$}
 \end{cases}
\]
if $B$ is ramified.
\end{lem}

\begin{proof}
We may assume that $B$ is ramified and $\alpha \ne 1$.
Then we have
\[
 \begin{bmatrix}
  \v_i \cdot \iota(\alpha,1) \\
  \v_i^* \cdot \iota(\alpha,1)
 \end{bmatrix}
 = A \cdot
 \begin{bmatrix}
  \v_i \\
  \v_i^*
 \end{bmatrix},
\]
where
\begin{align*}
  A & =
 \begin{pmatrix}
  \frac{1}{2}(\alpha+1) & \frac{1}{4 \kappa_i \i}(\alpha-1) \\
  \kappa_i \i (\alpha-1) & \frac{1}{2}(\alpha+1)
 \end{pmatrix} \\
 & =
 \begin{pmatrix}
  -\frac{1}{\kappa_i \i (\alpha^\rho - 1)} & \frac{1}{2} (\alpha+1) \\
  & \kappa_i \i (\alpha-1) \\
 \end{pmatrix} \cdot
 \begin{pmatrix}
  & -1 \\
  1 &
 \end{pmatrix} \cdot
 \begin{pmatrix}
  1 & \frac{\alpha+1}{2 \kappa_i \i (\alpha-1)} \\
  & 1
 \end{pmatrix}.
\end{align*}
This implies the assertion.
\end{proof}

\subsection{The doubling method for $\U(\WW)$}
\label{ss:doubling-U(WW)}

We consider a $2$-dimensional left $E$-space $\WW = B$ equipped with a skew-hermitian form
\[
 (x, y) = -\i \cdot \pr (x \cdot y^*).
\]
Then we have a natural embedding
\[
 \GU(W) \hookrightarrow \GU(\WW)
\]
and an isomorphism $\GU(\WW) \cong (B^{\times} \times E^{\times}) / F^{\times}$, where $B^{\times} \times E^{\times}$ acts on $\WW$ by
\[
 x \cdot (h, \alpha) = \alpha^{-1} \cdot x \cdot h.
\]
We write $[h,\alpha]$ for the image of $(h,\alpha)$ in $\GU(\WW)$.
Also, we consider the doubled space $\WW^\square = \WW \oplus \WW$ equipped with a skew-hermitian form
\[
 ((w_1,w_2), (w_1',w_2')) := (w_1, w_1') - (w_2, w_2').
\]
Then we have a natural embedding
\[
 \iota : \G(\U(\WW) \times \U(\WW)) \longrightarrow \GU(\WW^\square).
\]

Let $\VV = e_1 E + \dots + e_m E$ be an $m$-dimensional right $E$-space equipped with a hermitian form
\[
 (e_1 x_1 + \dots + e_m x_m, e_1 y_1 + \dots + e_m y_m)
 = x_1^\rho \cdot \kappa_1 \cdot y_1 + \dots + x_m^\rho \cdot \kappa_m \cdot y_m.
\]
Let $f: \VV \otimes_E \WW \rightarrow V \otimes_B W$ be the natural isomorphism.
Then we have
\begin{equation}
\label{eq:f-equivariance-U(WW)}
 f(v \otimes (w \cdot [h,\alpha])) = f(v \otimes w) \cdot (\alpha, h)
\end{equation}
for $h \in B^\times$ and $\alpha \in E^\times$, and
\[
 \llangle \cdot, \cdot \rrangle \circ (f \times f)
 = \frac{1}{2} \tr_{E/F} ( (\cdot, \cdot) \otimes (\cdot, \cdot)^{\rho} ).
\]
Hence we may identify $\VV \otimes_E \WW$ with $\V$ and omit $f$ from the notation.
Similarly, we identify $\VV \otimes_E \WW^\square$ with $\V^\square$.

Now we recall Kudla's splitting over $\U(\WW^\square)$, where we regard $\U(\WW^\square)$ as a subgroup of $\Sp(\V^\square)$ via the natural embedding.
We take a complete polarization $\WW^\square = \WW^\bigtriangledown \oplus \WW^\triangle$ over $E$ defined by
\[
 \WW^{\bigtriangledown} = \{ (w,-w) \, | \, w \in \WW \}, \qquad
 \WW^{\triangle} = \{ (w,w) \, | \, w \in \WW \}.
\]
Put
\[
 \w_1 = - \frac{1}{2\i} \cdot (1, -1), \qquad
 \w_2 = \frac{1}{2 J \i} \cdot (\j, -\j), \qquad
 \w_1^* = (1, 1), \qquad
 \w_2^* = (\j, \j),
\]
so that $(\w_i, \w_j^*) = \delta_{ij}$.
Using a basis $\w_1, \w_2, \w_1^*, \w_2^*$ of $\WW^\square$, we identify $\U(\WW^\square)$ with
\[
 \left\{ h \in \GL_4(E) \, \left| \,
 h \begin{pmatrix} & \1_2 \\ -\1_2 & \end{pmatrix} {}^t h^\rho
 = \begin{pmatrix} & \1_2 \\ -\1_2 & \end{pmatrix}
 \right. \right\}.
\]
Let $P_{\WW^\triangle}$ be the maximal parabolic subgroup of $\U(\WW^\square)$ stabilizing $\WW^\triangle$:
\[
 P_{\WW^\triangle} = \left\{ \left.
 \begin{pmatrix} a & * \\ & ({}^t a^\rho)^{-1} \end{pmatrix}
 \, \right|  \, a \in \GL_2(E) \right\}.
\]
We define a map
\[
 \hat{s}_2 : \U(\WW^\square) \longrightarrow \C^1
\]
by setting
\[
 \hat{s}_2(h) = \chi(x(h))^m \cdot \gamma^{-j}
\]
for $h = p_1 \tau_j p_2$ with
\[
 p_i =
 \begin{pmatrix}
  a_i & * \\
  & ({}^t a_i^\rho)^{-1}
 \end{pmatrix}
 \in P_{\WW^\triangle}, \qquad
 \tau_j =
 \begin{pmatrix}
  \1_{2-j} & & & \\
  & & & -\1_j \\
  & & \1_{2-j} & \\
  & \1_j & &
 \end{pmatrix},
\]
where
\[
 x(h) = \det(a_1 a_2) \mod \N_{E/F}(E^\times)
\]
and
\[
 \gamma = (u, \det \VV)_F \cdot \gamma_F(-u, \tfrac{1}{2} \psi)^m \cdot \gamma_F(-1, \tfrac{1}{2} \psi)^{-m}.
\]
Here, $(\cdot, \cdot)_F$ is the quadratic Hilbert symbol of $F$ and $\gamma_F(\cdot,\frac{1}{2}\psi)$ is the Weil index as in \cite[Appendix]{rangarao}, \cite[\S 3.1.1]{periods1}.
By \cite[Theorem 3.1, cases $3_+$]{kudla-splitting}, we have
\begin{equation}
\label{eq:spl-kudla-2}
 z_{\V^\triangle}(h_1,h_2) = \frac{\hat{s}_2(h_1 h_2)}{\hat{s}_2(h_1) \hat{s}_2(h_2)}
\end{equation}
for $h_1, h_2 \in \U(\WW^\square)$.

\begin{lem}
\label{lem:s-hat-2-E-conj}
For $\alpha \in E^{\times}$ and $h \in \U(\WW^\square)$, we have
\[
 \hat{s}_2(\iota([\alpha,1], [\alpha,1]) \cdot h \cdot \iota([\alpha,1], [\alpha,1])^{-1}) = \hat{s}_2(h).
\]
\end{lem}

\begin{proof}
Put $h_\alpha = \iota([\alpha,1], [\alpha,1])$.
Since
\[
 \begin{bmatrix}
  \w_1 \cdot h_\alpha \\
  \w_2 \cdot h_\alpha \\
  \w_1^* \cdot h_\alpha \\
  \w_2^* \cdot h_\alpha
 \end{bmatrix}
 =
 \begin{pmatrix}
  \alpha & & & \\
  & \alpha^\rho & & \\
  & & \alpha & \\
  & & & \alpha^\rho
 \end{pmatrix}
 \cdot
 \begin{bmatrix}
  \w_1 \\
  \w_2 \\
  \w_1^* \\
  \w_2^*
 \end{bmatrix},
\]
we have $x(h_\alpha p h_\alpha^{-1}) = x(p)$ for $p \in P_{\WW^\triangle}$ and $h_\alpha \tau_j h_\alpha^{-1} = \tau_j$.
Hence the assertion follows.
\end{proof}

\begin{lem}
\label{lem:s-hat-2-E-diag}
For $\alpha \in E^\times$, we have
\[
 \hat{s}_2(\iota([\alpha,\alpha], [\alpha,\alpha])) = \chi(\alpha)^{-2m}.
\]
\end{lem}

\begin{proof}
Put $h_\alpha = \iota([\alpha,\alpha], [\alpha,\alpha])$.
Since
\[
 \begin{bmatrix}
  \w_1 \cdot h_\alpha \\
  \w_2 \cdot h_\alpha \\
  \w_1^* \cdot h_\alpha \\
  \w_2^* \cdot h_\alpha
 \end{bmatrix}
 =
 \begin{pmatrix}
  1 & & & \\
  & \alpha^{-1} \alpha^\rho & & \\
  & & 1 & \\
  & & & \alpha^{-1} \alpha^\rho
 \end{pmatrix}
 \cdot
 \begin{bmatrix}
  \w_1 \\
  \w_2 \\
  \w_1^* \\
  \w_2^*
 \end{bmatrix}
\]
and $\chi(\alpha^\rho) = \chi(\alpha)^{-1}$, the assertion follows.
\end{proof}

\begin{lem}
\label{lem:s-hat-2-E^1}
Let $\alpha \in E^1$.
Then we have
\[
 \hat{s}_2(\iota([1,\alpha], 1)) = \chi(\alpha)^{-m}
\]
if $B$ is split, and
\[
 \hat{s}_2(\iota([1,\alpha], 1)) = \chi(\alpha)^{-m} \times
 \begin{cases}
  1 & \text{if $\alpha = 1$,} \\
  (-1)^m & \text{if $\alpha \ne 1$}
 \end{cases}
\]
if $B$ is ramified.
\end{lem}

\begin{proof}
We may assume that $\alpha \ne 1$.
Then we have $\tr_{E/F}(\alpha) \ne 2$ and hence $\alpha -1 \in E^\times$.
As in the proof of Lemma \ref{lem:s-hat-1-E^1}, we have
\[
 \begin{bmatrix}
  \w_1 \cdot \iota([1,\alpha],1) \\
  \w_2 \cdot \iota([1,\alpha],1) \\
  \w_1^* \cdot \iota([1,\alpha],1) \\
  \w_2^* \cdot \iota([1,\alpha],1)
 \end{bmatrix}
 = A \cdot
 \begin{bmatrix}
  \w_1 \\
  \w_2 \\
  \w_1^* \\
  \w_2^*
 \end{bmatrix},
\]
where
\begin{align*}
 A & =
\begin{pmatrix}
 \frac{1}{2}(\alpha^{-1}+1) & & - \frac{1}{4 \i}(\alpha^{-1}-1) & \\
 & \frac{1}{2}(\alpha^{-1}+1) & & \frac{1}{4 J \i}(\alpha^{-1}-1) \\
 - \i (\alpha^{-1}-1) & & \frac{1}{2}(\alpha^{-1}+1) & \\
 & J \i (\alpha^{-1}-1) & & \frac{1}{2}(\alpha^{-1}+1)
\end{pmatrix} \\
 & =
 \begin{pmatrix}
  \frac{1}{\i (\alpha - 1)} & & * & \\
  & -\frac{1}{J \i (\alpha - 1)} & & * \\
  & & - \i (\alpha^\rho-1) & \\
  & & & J \i (\alpha^\rho-1)
 \end{pmatrix}
 \cdot \tau_2 \cdot
 \begin{pmatrix}
  1 & & * & \\
  & 1 & & * \\
  & & 1 & \\
  & & & 1
 \end{pmatrix}.
\end{align*}
Hence we have
\[
 x(\iota([1,\alpha],1)) = - \frac{1}{u J (\alpha-1)^2} \equiv
\frac{J}{(\alpha-1)^2} \equiv - \frac{J}{\alpha} \mod \N_{E/F}(E^\times),
\]
so that
\begin{align*}
 \chi(x(\iota([1,\alpha],1))) & = \chi(\alpha)^{-1} \cdot \xi_E(-J) \\
 & = \chi(\alpha)^{-1} \cdot \xi_E(-1) \times
 \begin{cases}
  1 & \text{if $B$ is split,} \\
  -1 & \text{if $B$ is ramified.}
 \end{cases}
\end{align*}
Also, we have
\[
 \gamma^2 = (-1, -u)_F^m \cdot (-1, -1)_F^m = (-1,u)_F^m = \xi_E(-1)^m.
\]
This implies the assertion.
\end{proof}

\subsection{Splitting over $\cG^\sharp$}

Let $\cG^\sharp$ be a subgroup of $\GU(V) \times \GU(W) \times \G\U(V) \times \GU(W)$ defined by
\[
 \cG^\sharp = \{ (g,h,\alpha,\alpha) \in \GU(V)^0 \times \GU(W) \times E^\times \times E^\times
 \, | \, \nu(g) = \nu(h) = \N_{E/F}(\alpha) \}.
\]
Then we have a natural homomorphism
\begin{equation}
\label{eq:hom-G^sharp}
 \cG^\sharp \subset \G(\U(V) \times \U(W)) \times \G(\U(V) \times \U(W))
 \longrightarrow \Sp(\V) \times \Sp(\V) \subset \Sp(\V^\square).
\end{equation}
We define a map
\[
 \hat{s}^\sharp: \cG^\sharp \longrightarrow \C^1
\]
by setting
\[
 \hat{s}^\sharp(g,h,\alpha,\alpha)
 = \chi(\alpha)^{-m} \cdot \hat{s}_1(\iota(g \alpha^{-1},1)) \cdot \hat{s}_2(\iota(h \alpha^{-1}, 1))
 \cdot z_{\V^\triangle}(\iota(g \alpha^{-1},1), \iota(h \alpha^{-1},1)).
\]

\begin{lem}
\label{lem:spl-s-hat-sharp}
For $\g_1, \g_2 \in \cG^\sharp$, we have
\[
 z_{\V^\triangle}(\g_1,\g_2) = \frac{\hat{s}^\sharp(\g_1 \g_2)}{\hat{s}^\sharp(\g_1) \hat{s}^\sharp(\g_2)}.
\]
Here, by abuse of notation, we write $\g_i$ on the left-hand side for the image of $\g_i$ in $\Sp(\V^\square)$ under \eqref{eq:hom-G^sharp}.
\end{lem}

\begin{proof}
Write $\g_i = (g_i,h_i,\alpha_i,\alpha_i)$.
If $\alpha_1 = \alpha_2 = 1$, then the assertion follows from \eqref{eq:spl-kudla-1}, \eqref{eq:spl-kudla-2}, and \cite[Chapitre 2, II.5]{mvw}.
If $\alpha_1$ and $\alpha_2$ are arbitrary, put $\h_i = (g_i \alpha_i^{-1}, h_i \alpha_i^{-1}, 1,1)$ and $\ba_i = (\alpha_i,\alpha_i,\alpha_i,\alpha_i)$.
Let $P_{\V^\triangle}$ be the maximal parabolic subgroup of $\Sp(\V^\square)$ stabilizing $\V^\triangle$.
Then it follows from \cite[Theorem 4.1]{rangarao} that
\begin{equation}
\label{eq:rangarao-2-cocycle}
 z_{\V^\triangle}(p_1 \sigma p, p^{-1} \sigma' p_2) = z_{\V^\triangle}(\sigma, \sigma')
\end{equation}
for $p_1, p_2, p \in P_{\V^\triangle}$ and $\sigma, \sigma' \in \Sp(\V^\square)$ (see also \cite[\S 3.1.1]{periods1}).
Since $\g_1\g_2 = \h_1 \cdot \ba_1 \h_2 \ba_1^{-1} \cdot \ba_1 \ba_2$ and the image of $\ba_i$ in $\Sp(\V^\square)$ belongs to $P_{\V^\triangle}$, we have
\[
 z_{\V^\triangle}(\g_1, \g_2) = z_{\V^\triangle}(\h_1, \ba_1 \h_2 \ba_1^{-1})
 = \frac{\hat{s}^\sharp(\h_1 \ba_1 \h_2 \ba_1^{-1})}{\hat{s}^\sharp(\h_1) \hat{s}^\sharp(\ba_1\h_2\ba_1^{-1})}.
\]
On the other hand, by definition, we have
\[
 \frac{\hat{s}^\sharp(\g_1 \g_2)}{\hat{s}^\sharp(\g_1) \hat{s}^\sharp(\g_2)}
 = \frac{\hat{s}^\sharp(\g_1 \g_2 (\ba_1 \ba_2)^{-1})}{\hat{s}^\sharp(\g_1 \ba_1^{-1}) \hat{s}^\sharp(\g_2 \ba_2^{-1})}
 = \frac{\hat{s}^\sharp(\h_1 \ba_1 \h_2 \ba_1^{-1})}{\hat{s}^\sharp(\h_1) \hat{s}^\sharp(\h_2)}.
\]
It follows from Lemmas \ref{lem:s-hat-1-E-conj} and \ref{lem:s-hat-2-E-conj}, combined with \eqref{eq:rangarao-2-cocycle}, that
\[
 \hat{s}^\sharp(\ba_1 \h_2 \ba_1^{-1}) = \hat{s}^\sharp(\h_2).
\]
This completes the proof.
\end{proof}

\begin{lem}
\label{lem:compare-s-sharp-s-2}
For $\alpha \in E^1$, we have
\[
 \hat{s}^\sharp(1,1,\alpha,\alpha) = \hat{s}_2(\iota(1, [\alpha,\alpha])).
\]
\end{lem}

\begin{proof}
Put
\begin{align*}
 g_\alpha & = \iota(\alpha, 1) \in \U(V^\square), \\
 h_\alpha & = \iota([\alpha, 1], 1) \in \U(\WW^\square), \\
 k_\alpha & = \iota([1, \alpha], 1) \in \U(\WW^\square), \\
 m_\alpha & = \iota([\alpha, \alpha], [\alpha, \alpha]) \in \U(\WW^\square).
\end{align*}
By definition, we have
\[
 \hat{s}^\sharp(1,1,\alpha,\alpha)
 = \chi(\alpha)^{-m} \cdot \hat{s}_1(g_\alpha^{-1}) \cdot \hat{s}_2(h_\alpha^{-1})
 \cdot z_{\V^\triangle}(g_\alpha^{-1}, h_\alpha^{-1}).
\]
Since $m_\alpha = h_\alpha \cdot k_\alpha \cdot \iota(1, [\alpha,\alpha])$ and the image of $m_\alpha$ in $\Sp(\V^\square)$ belongs to $P_{\V^\triangle}$, it follows from \eqref{eq:spl-kudla-2}, \eqref{eq:rangarao-2-cocycle}, and Lemma \ref{lem:s-hat-2-E-diag} that
\begin{align*}
 \hat{s}_2(\iota(1, [\alpha,\alpha]))
 & = \hat{s}_2(k_\alpha^{-1} h_\alpha^{-1}) \cdot \hat{s}_2(m_\alpha)
 \cdot z_{\V^\triangle}(k_\alpha^{-1} h_\alpha^{-1}, m_\alpha) \\
 & = \hat{s}_2(k_\alpha^{-1} h_\alpha^{-1}) \cdot \chi(\alpha)^{-2m} \\
 & = \chi(\alpha)^{-2m} \cdot \hat{s}_2(k_\alpha^{-1}) \cdot \hat{s}_2(h_\alpha^{-1})
 \cdot z_{\V^\triangle}(k_\alpha^{-1},  h_\alpha^{-1}).
\end{align*}
By Lemmas \ref{lem:s-hat-1-E^1} and \ref{lem:s-hat-2-E^1}, we have
\[
 \hat{s}_2(k_\alpha^{-1}) = \chi(\alpha)^m \cdot \hat{s}_1(g_\alpha^{-1}).
\]
By \eqref{eq:f-equivariance-U(WW)}, the image of $k_\alpha$ in $\Sp(\V^\square)$ agrees with that of $g_\alpha$, so that
\[
 z_{\V^\triangle}(k_\alpha^{-1}, h_\alpha^{-1})
 = z_{\V^\triangle}(g_\alpha^{-1}, h_\alpha^{-1}).
\]
This completes the proof.
\end{proof}

\subsection{Proof of Proposition \ref{prop:spl-hodge}}
\label{ss:proof-spl-hodge}

Now we take a complete polarization $\V^\square = \X^\square \oplus \Y^\square$ defined by
\[
 \X^\square = \X \oplus \X, \qquad
 \Y^\square = \Y \oplus \Y.
\]
As in \cite[\S D.3]{periods1}, we have
\[
 z_{\Y^\square}(\iota(\sigma_1,\sigma_2), \iota(\sigma_1',\sigma_2'))
 = z_\Y(\sigma_1, \sigma_1') \cdot z_\Y(\sigma_2, \sigma_2')^{-1}
\]
for $\sigma_i, \sigma_i' \in \Sp(\V)$.
Fix $\sigma_0 \in \Sp(\V^\square)$ such that $\V^\bigtriangledown = \X^\square \cdot \sigma_0$ and $\V^\triangle = \Y^\square \cdot \sigma_0$.
Put
\[
 \mu(\sigma)
 = z_{\Y^\square}(\sigma_0, \sigma)^{-1} \cdot z_{\Y^\square}(\sigma_0 \sigma \sigma_0^{-1}, \sigma_0)
\]
for $\sigma \in \Sp(\V^\square)$.
Note that $\mu$ does not depend on the choice of $\sigma_0$.
Then, by \cite[Lemma 4.2]{kudla-splitting}, we have
\begin{equation}
\label{eq:compare-zY-zV}
 z_{\Y^\square}(\sigma, \sigma') = z_{\V^\triangle}(\sigma, \sigma')
 \cdot \frac{\mu(\sigma \sigma')}{\mu(\sigma) \mu(\sigma')}
\end{equation}
for $\sigma, \sigma' \in \Sp(\V^\square)$.

Put $s^\sharp = \hat{s}^\sharp \cdot \mu$ and $s_2 = \hat{s}_2 \cdot \mu$.
By Lemma \ref{lem:spl-s-hat-sharp} and \eqref{eq:spl-kudla-2}, we have
\[
 z_{\Y^\square}(\g_1, \g_2) = \frac{s^\sharp(\g_1 \g_2)}{s^\sharp(\g_1) s^\sharp(\g_2)}
\]
for $\g_1,\g_2 \in \cG^\sharp$ and
\[
 z_{\Y^\square}(h_1, h_2) = \frac{s_2(h_1 h_2)}{s_2(h_1) s_2(h_2)}
\]
for $h_1, h_2 \in \U(\WW^\square)$.
We define a map
\[
 s: \cG \longrightarrow \C^1
\]
by setting
\[
 s(g,h) = \frac{s^\sharp(g,h,\alpha,\alpha)}{s_2(\iota(1, [\alpha,\alpha]))},
\]
where we choose $\alpha \in E^\times$ such that $\nu(g) = \nu(h) = \N_{E/F}(\alpha)$.

\begin{lem}
\label{lem:s-well-def}
The map $s$ is well-defined, i.e., for $(g,h,\alpha,\alpha) \in \cG^\sharp$ and $\beta \in E^1$, we have
\[
 \frac{s^\sharp(g,h,\alpha \beta, \alpha \beta)}{s_2(\iota(1, [\alpha \beta, \alpha \beta]))}
 = \frac{s^\sharp(g,h,\alpha, \alpha)}{s_2(\iota(1, [\alpha, \alpha]))}.
\]
\end{lem}

\begin{proof}
First note that, by \eqref{eq:f-equivariance-U(WW)}, the image of $(\alpha, \alpha) \in \G(\U(V) \times \U(W))$ in $\Sp(\V)$ agrees with that of $[\alpha, \alpha] \in \U(\WW)$.
We have
\begin{align*}
 s^\sharp(g,h,\alpha \beta, \alpha \beta)
 & = s^\sharp(g,h,\alpha,\alpha) \cdot s^\sharp(1,1,\beta,\beta)
 \cdot z_{\Y^\square}((g,h,\alpha,\alpha), (1,1,\beta,\beta)) \\
 & = s^\sharp(g,h,\alpha,\alpha) \cdot s^\sharp(1,1,\beta,\beta)
 \cdot z_\Y((\alpha,\alpha), (\beta,\beta))^{-1}
\end{align*}
and
\begin{align*}
 s_2(\iota(1, [\alpha \beta, \alpha \beta]))
 & = s_2(\iota(1, [\alpha, \alpha])) \cdot s_2(\iota(1, [\beta, \beta]))
 \cdot z_{\Y^\square}(\iota(1, [\alpha, \alpha]), \iota(1, [\beta, \beta])) \\
 & = s_2(\iota(1, [\alpha, \alpha])) \cdot s_2(\iota(1, [\beta, \beta]))
 \cdot z_\Y([\alpha, \alpha], [\beta, \beta])^{-1}.
\end{align*}
By Lemma \ref{lem:compare-s-sharp-s-2}, we have
\[
 s^\sharp(1,1,\beta,\beta) = s_2(\iota(1, [\beta, \beta])).
\]
This implies the assertion.
\end{proof}

\begin{lem}
\label{lem:s-proof(i)}
The map $s$ satisfies the condition (i) of Proposition \ref{prop:spl-hodge}, i.e., for $\g_1, \g_2 \in \cG$, we have
\[
 z_\Y(\g_1,\g_2) = \frac{s(\g_1\g_2)}{s(\g_1) s(\g_2)}.
\]
\end{lem}

\begin{proof}
Write $\g_i = (g_i, h_i)$ and choose $\alpha_i \in E^\times$ such that $\nu(g_i) = \nu(h_i) = \N_{E/F}(\alpha_i)$.
Then we have
\begin{align*}
 \frac{s(\g_1\g_2)}{s(\g_1) s(\g_2)}
 & = \frac{s^\sharp(g_1 g_2, h_1 h_2, \alpha_1 \alpha_2, \alpha_1 \alpha_2)}{s^\sharp(g_1, h_1, \alpha_1, \alpha_1) s^\sharp(g_2, h_2, \alpha_2, \alpha_2)}
 \cdot \frac{s_2(\iota(1, [\alpha_1, \alpha_1])) s_2(\iota(1, [\alpha_2, \alpha_2]))}{s_2(\iota(1, [\alpha_1 \alpha_2, \alpha_1 \alpha_2]))} \\
 & = z_{\Y^\square}((g_1, h_1, \alpha_1, \alpha_1), (g_2, h_2, \alpha_2, \alpha_2))
 \cdot z_{\Y^\square}(\iota(1, [\alpha_1, \alpha_1]), \iota(1, [\alpha_2, \alpha_2]))^{-1} \\
 & = z_\Y((g_1, h_1), (g_2, h_2)) \cdot z_\Y((\alpha_1, \alpha_1), (\alpha_2, \alpha_2))^{-1}
 \cdot z_{\Y}([\alpha_1, \alpha_1], [\alpha_2, \alpha_2]) \\
 & = z_\Y((g_1, h_1), (g_2, h_2)).
\end{align*}
This completes the proof.
\end{proof}

By definition, the map $s$ also satisfies the other conditions of Proposition \ref{prop:spl-hodge}.
This completes the proof of Proposition \ref{prop:spl-hodge}.

\subsection{Independence of the choice of $\chi$}

To define the map $s$, we have used the fixed character $\chi$ of $E^\times$ such that $\chi|_{F^\times} = \xi_E$.
However, we have:

\begin{lem}
The map $s$ defined as above does not depend on the choice of $\chi$.
\end{lem}

\begin{proof}
Let $\chi_1$ and $\chi_2$ be two characters of $E^\times$ such that $\chi_1|_{F^\times} = \chi_2|_{F^\times} = \xi_E$.
We will write $s = s_{\chi_i}$, etc., to indicate the dependence on $\chi_i$.
For $(g,h) \in \Gc$, choose $\alpha \in E^\times$ such that $\nu(g) = \nu(h) = \N_{E/F}(\alpha)$.
Then, by definition, we have
\begin{align*}
 \frac{s_{\chi_1}(g,h)}{s_{\chi_2}(g,h)}
 & = \frac{s^\sharp_{\chi_1}(g,h,\alpha,\alpha)}{s^\sharp_{\chi_2}(g,h,\alpha,\alpha)}
 \cdot \frac{s_{2,\chi_2}(\iota(1, [\alpha,\alpha]))}{s_{2,\chi_1}(\iota(1, [\alpha,\alpha]))} \\
 & = \frac{\hat{s}^\sharp_{\chi_1}(g,h,\alpha,\alpha)}{\hat{s}^\sharp_{\chi_2}(g,h,\alpha,\alpha)}
 \cdot \frac{\hat{s}_{2,\chi_2}(\iota(1, [\alpha,\alpha]))}{\hat{s}_{2,\chi_1}(\iota(1, [\alpha,\alpha]))} \\
 & = \eta(\alpha)^{-m}
 \cdot \frac{\hat{s}_{2,\chi_1}(\iota(h \alpha^{-1}, 1))}{\hat{s}_{2,\chi_2}(\iota(h \alpha^{-1}, 1))}
 \cdot \frac{\hat{s}_{2,\chi_2}(\iota(1, [\alpha,\alpha]))}{\hat{s}_{2,\chi_1}(\iota(1, [\alpha,\alpha]))},
\end{align*}
where $\eta = \chi_1/\chi_2$.
On the other hand, for $k \in \U(\WW^\square)$, we have
\[
 \frac{\hat{s}_{2,\chi_1}(k)}{\hat{s}_{2,\chi_2}(k)}
 = \eta(x(k))^m
 = \tilde{\eta}(\det k)^m,
\]
where $\tilde{\eta}$ is the character of $E^1$ such that $\tilde{\eta}(x/x^\rho) = \eta(x)$ for $x \in E^\times$.
Since
\begin{align*}
 \det \iota(h \alpha^{-1}, 1) & = \nu(h) \N_{E/F}(\alpha)^{-1} = 1, \\
 \det \iota(1, [\alpha,\alpha]) & = \N_{E/F}(\alpha) \alpha^{-2} = \alpha^{-1} \alpha^\rho,
\end{align*}
we have
\[
 \frac{s_{\chi_1}(g,h)}{s_{\chi_2}(g,h)}
 = \eta(\alpha)^{-m} \cdot \tilde{\eta}(\alpha^{-1} \alpha^\rho)^{-m}
 = 1.
\]
This completes the proof.
\end{proof}

\subsection{Compatibility with seesaws}

We write $V = V' \oplus V''$ as an orthogonal direct sum of skew-hermitian right $B$-spaces
\[
 V' = e_1 B \oplus \dots \oplus e_{m'} B, \qquad
 V'' = e_{m'+1} B \oplus \dots \oplus e_m B.
\]
Let $\V' = V' \otimes_B W$ and $\V'' = V'' \otimes_B W$ be the symplectic $F$-spaces as in \S \ref{ss:weil-hodge-setup}.
Then we have $\V = \V' \oplus \V''$, which gives rise to a seesaw diagram
\[
 \xymatrix{
  \GU(V) \ar@{-}[dr] \ar@{-}[d] & \G(\U(W) \times \U(W)) \ar@{-}[dl] \ar@{-}[d] \\
  \G(\U(V') \times \U(V'')) & \GU(W)}.
\]
Let $\cG'$ and $\cG''$ be the subgroups of $\G(\U(V') \times \U(W))$ and $\G(\U(V'') \times \U(W))$, respectively, as in \S \ref{ss:weil-hodge-setup}.
Put
\[
 \cG''' = \{ (g', g'', h) \in \GU(V')^0 \times \GU(V'')^0 \times \GU(W)
 \, | \, \nu(g') = \nu(g'') = \nu(h) \in \N_{E/F}(E^\times) \}.
\]
We regard $\cG'''$ as subgroups of $\cG$ and $\cG' \times \cG''$ via the above seesaw diagram.
We take the complete polarizations $\V' = \X' \oplus \Y'$ and $\V'' = \X'' \oplus \Y''$ as in \S \ref{ss:weil-hodge-setup}, so that
\[
 \X = \X' \oplus \X'', \qquad
 \Y = \Y' \oplus \Y''.
\]
Let $s': \cG' \rightarrow \C^1$ and $s'':\cG'' \rightarrow \C^1$ be the maps trivializing $z_{\Y'}$ and $z_{\Y''}$, respectively, defined similarly as above.
Then, by construction, we have
\[
 s = s' \otimes s''
\]
on $\cG'''$.

\subsection{Compatibility with \cite{periods2}}
\label{ss:compatibility-periods2}

In this section, we compare the splitting $s$ with the standard one for unitary dual pairs when $\dim V = 1$.
In this case, using the notation of \S \ref{ss:doubling-U(WW)}, we have a seesaw diagram
\[
 \xymatrix{
  E^\times \hspace{-1.5cm} & \cong \hspace{-1cm} & \GU(V)^0 \ar@{-}[dr] \ar@{=}[d] & \GU(\WW) \ar@{-}[dl] \ar@{-}[d] & \hspace{-1cm} \cong & \hspace{-1cm} (B^\times \times E^\times)/F^\times \\
  E^\times \hspace{-1.5cm} & \cong \hspace{-1cm} & \GU(\VV) & \GU(W) & \hspace{-1cm} \cong & \hspace{-1cm} B^\times}.
\]
We define a map
\[
 s^\natural : \G(\U(V) \times \U(W))^0 \longrightarrow \C^1
\]
by setting
\[
 s^\natural(\alpha, h) = s_2(\iota([h, \alpha], 1))
\]
for $(\alpha, h) \in \G(\U(V) \times \U(W))^0$, where $[h, \alpha] \in \U(\WW)$ and $s_2 = \hat{s}_2 \cdot \mu$.
Then $s^\natural$ trivializes $z_\Y$ by \eqref{eq:f-equivariance-U(WW)}.
This splitting will be used in \cite{periods2}.

\begin{lem}
We have
\[
 s^\natural(\alpha, h) = s(\alpha, h) \cdot \chi(\alpha)^{-1}.
\]
\end{lem}

\begin{proof}
Recall that
\[
 s(\alpha, h) = \frac{s^\sharp(\alpha, h, \alpha, \alpha)}{s_2(\iota(1, [\alpha, \alpha]))}.
\]
We have
\begin{align*}
 s^\sharp(\alpha, h, \alpha, \alpha)
 & = s^\sharp(1, h \alpha^{-1}, 1, 1)
 \cdot s^\sharp(\alpha, \alpha, \alpha, \alpha)
 \cdot z_{\Y^\square}((1, h \alpha^{-1}, 1, 1), (\alpha, \alpha, \alpha, \alpha)) \\
 & = s^\sharp(1, h \alpha^{-1}, 1, 1)
 \cdot s^\sharp(\alpha, \alpha, \alpha, \alpha)
 \cdot z_{\Y}((1, h \alpha^{-1}), (\alpha, \alpha)).
\end{align*}
By definition and Lemma \ref{lem:s-hat-2-E-diag}, we have
\[
 s^\sharp(1, h \alpha^{-1}, 1, 1) = s_2(\iota([h \alpha^{-1}, 1], 1))
\]
and
\begin{align*}
 s^\sharp(\alpha, \alpha, \alpha, \alpha)
 & = \chi(\alpha)
 \cdot s_2(\iota([\alpha, \alpha], [\alpha, \alpha])) \\
 & = \chi(\alpha)
 \cdot s_2(\iota([\alpha, \alpha], 1))
 \cdot s_2(\iota(1, [\alpha, \alpha]))
 \cdot z_{\Y^\square}(\iota([\alpha, \alpha], 1), \iota(1, [\alpha, \alpha])) \\
 & = \chi(\alpha)
 \cdot s_2(\iota([\alpha, \alpha], 1))
 \cdot s_2(\iota(1, [\alpha, \alpha])).
\end{align*}
Hence we have
\begin{align*}
 s(\alpha, h)
 & = \chi(\alpha)
 \cdot s_2(\iota([h \alpha^{-1}, 1], 1))
 \cdot s_2(\iota([\alpha, \alpha], 1))
 \cdot z_{\Y}([h \alpha^{-1}, 1], [\alpha, \alpha]) \\
 & = \chi(\alpha)
 \cdot s_2(\iota([h \alpha^{-1}, 1], 1))
 \cdot s_2(\iota([\alpha, \alpha], 1))
 \cdot z_{\Y^\square}(\iota([h \alpha^{-1}, 1], 1), \iota([\alpha, \alpha], 1)) \\
 & = \chi(\alpha) \cdot s_2(\iota([h, \alpha], 1)).
\end{align*}
\end{proof}

\subsection{Compatibility with \cite{periods1}}
\label{ss:compatibility-periods1}

In this section, we compare the splitting $s$ with the one defined in \cite[Appendix C]{periods1}.
Suppose again that $F$ is a number field.
Let $V = B_1 \otimes_E B_2$ be the $2$-dimensional skew-hermitian right $B$-space as in \cite[\S 2.2]{periods1},
where $B_1$ and $B_2$ are quaternion algebras over $F$ such that $E$ embeds into $B_1$ and $B_2$, and such that $B_1 \cdot B_2 = B$ in the Brauer group.
We write $B_i = E + E \j_i$ for some trace zero element $\j_i \in B_i^\times$ and put $J_i = \j_i^2 \in F^\times$.
We may assume that
\[
 J_1 \cdot J_2 = J.
\]
Then the skew-hermitian form on $V$ is given by \eqref{eq:condition-skew-herm} with
\begin{align*}
 e_1 & = 1 \otimes 1, & \kappa_1 & = 1, \\
 e_2 & = \j_1 \otimes 1, & \kappa_2 & = -J_1.
\end{align*}
Recall the exact sequence
\[
 1 \longrightarrow F^\times \longrightarrow B_1^\times \times B_2^\times
 \longrightarrow \GU(V)^0 \longrightarrow 1,
\]
where $F^\times$ embeds into $B_1^\times \times B_2^\times$ by $z \mapsto (z, z^{-1})$ and $B_1^\times \times B_2^\times$ acts on $V$ on the left by
\[
 (g_1, g_2) \cdot (x_1 \otimes x_2) = g_1 x_1 \otimes g_2 x_2.
\]
We write $[g_1,g_2]$ for the image of $(g_1,g_2)$ in $\GU(V)^0$.
If we put
\[
 \tilde{\cG} = \{ (g_1, g_2, h) \in B_1^\times \times B_2^\times \times B^\times \, | \, \nu(g_1) \nu(g_2) = \nu(h) \in \N_{E/F}(E^\times) \},
\]
then we have a natural surjective map $\tilde{\cG} \twoheadrightarrow \cG$.
We take the complete polarization $\V = \X \oplus \Y$ as in \S \ref{ss:weil-hodge-setup}, which agrees with the one given in \cite[\S 2.2]{periods1}.
For each place $v$ of $F$, let
\[
 \tilde{s}_v: \GU(V_v)^0 \times \GU(W_v) \longrightarrow \C^1
\]
be the map trivializing $z_{\Y_v}$ defined in \cite[Appendix C]{periods1}.
Since both $\tilde{s}_v$ and $s_v$ trivialize $z_{\Y_v}$,
there exists a \emph{continuous} character $\Chi$ of $\cG(\A)$ such that
\[
 \tilde{s}_v|_{\cG_v} = s_v \cdot \Chi_v
\]
for all $v$.
Since both $\tilde{s}_v$ and $s_v$ satisfy the product formula, $\Chi$ is trivial on $\cG(F)$.
We regard $\Chi$ as a character of $\tilde{\cG}(\A)$.

\begin{prop}
Assume that
\begin{itemize}
 \item $F$ is totally real;
 \item $E$ is totally imaginary;
 \item $B_{1,v}$ and $B_{2,v}$ are split for some real place $v$ of $F$.
\end{itemize}
Then, for $(g_1, g_2, h) \in \tilde{\cG}(\A)$, we have
\[
 \Chi(g_1, g_2, h) = 1.
\]
Namely, we have
\[
 \tilde{s}_v|_{\cG_v} = s_v
\]
for all $v$.
\end{prop}

\begin{proof}
Since $\Chi$ is automorphic, it is trivial on
\[
 B_1^{(1)}(\A) \times B_2^{(1)}(\A) \times B^{(1)}(\A).
\]
Moreover, in \S \ref{ss:weil-hodge-computation} below, we will prove the following:
\begin{itemize}
\item For $\alpha \in \A_E^\times$, we have
\begin{equation}
\label{eq:weil-hodge-computation1}
 \Chi(1,\alpha,\alpha) = 1.
\end{equation}
\item For $\alpha \in \A_E^\times$, we have
\begin{equation}
\label{eq:weil-hodge-computation2}
 \Chi(\alpha, \alpha^{-1}, 1) = 1.
\end{equation}
\item Let $v$ be a real place of $F$ such that $B_{1,v}$ and $B_{2,v}$ are split.
Choose $t_{i,v} \in F_v^\times$ such that $J_i = t_{i,v}^2$.
Then we have
\begin{equation}
\label{eq:weil-hodge-computation3}
 \Chi_v(t_{1,v}^{-1} \cdot \j_1, t_{2,v}^{-1} \cdot \j_2, 1) = 1.
\end{equation}
Note that $\nu(t_{i,v}^{-1} \cdot \j_i) = -1 \notin \N_{E_v/F_v}(E_v^\times)$.
\end{itemize}
This implies the assertion.
\end{proof}

\subsection{Computation of splittings}
\label{ss:weil-hodge-computation}

We retain the notation of \S \ref{ss:compatibility-periods1}.
We fix a place $v$ of $F$ and suppress the subscript $v$ from the notation.
Recall that $\Chi$ is a \emph{continuous} character of $\tilde{\cG}$ such that
\[
 \Chi(g_1,g_2,h) = \frac{\tilde{s}(g_1,g_2,h)}{s(g_1,g_2,h)},
\]
where we regard $\tilde{s}$ and $s$ as maps on $\tilde{\cG}$.
To compute $\Chi$ explicitly, we need to introduce more notation.

\subsubsection{Notation}
\label{sss:notation}

We denote by $\GSp_{2n}(F)$ the symplectic similitude group and by $\nu : \GSp_{2n}(F) \rightarrow F^\times$ the similitude character:
\[
 \GSp_{2n}(F) = \left\{ \sigma \in \GL_{2n}(F) \, \left| \,
 \sigma \begin{pmatrix} & \1_n \\ -\1_n & \end{pmatrix} {}^t \sigma
 = \nu(\sigma) \cdot \begin{pmatrix} & \1_n \\ -\1_n & \end{pmatrix}
 \right. \right\}.
\]
Let $\Sp_{2n}(F) = \ker \nu$ be the symplectic group and $P$ the standard maximal parabolic subgroup of $\Sp_{2n}(F)$:
\[
 P = \{ \m(\a) \n(\b) \, | \, \a \in \GL_n(F), \b \in \Sym_n(F) \},
\]
where
\[
 \m(\a) = \begin{pmatrix} \a & \\ & {}^t \a^{-1} \end{pmatrix}, \qquad
 \n(\b) = \begin{pmatrix} \1_n & \b \\ & \1_n \end{pmatrix}.
\]
Put
\[
 d(\nu) = \begin{pmatrix} \1_n & \\ & \nu \cdot \1_n \end{pmatrix}, \qquad
 \tau_j =
 \begin{pmatrix}
  \1_{n-j} & & & \\
  & & & -\1_j \\
  & & \1_{n-j} & \\
  & \1_j & &
 \end{pmatrix}.
\]
If $\sigma = p_1 \tau_j p_2 \in \Sp_{2n}(F)$ with $p_i = \m(\a_i) \n(\b_i) \in P$, put
\[
 x(\sigma) = \det(\a_1 \a_2) \mod (F^{\times})^2, \qquad j(\sigma) = j.
\]
Note that
\begin{equation}
\label{eq:xj-conj}
 x(d(\nu) \cdot \sigma \cdot d(\nu)^{-1}) = \nu^{j(\sigma)} \cdot x(\sigma),
 \qquad
 j(d(\nu) \cdot \sigma \cdot d(\nu)^{-1}) = j(\sigma).
\end{equation}
We define a map
\[
 v: \Sp_{2n}(F) \times F^\times \longrightarrow \C^1
\]
by setting
\[
 v(\sigma, \nu) = (x(\sigma), \nu)_F \cdot \gamma_F(\nu, \tfrac{1}{2} \psi)^{-j(\sigma)},
\]
where $(\cdot, \cdot)_F$ is the quadratic Hilbert symbol of $F$ and $\gamma_F(\cdot,\frac{1}{2}\psi)$ is the Weil index as in \cite[Appendix]{rangarao}, \cite[\S 3.1.1]{periods1}.
Let $z$ be the $2$-cocycle on $\Sp_{2n}(F)$ realizing the metaplectic group (see e.g.~\cite{rangarao}, \cite[\S 3.2.2]{periods1}).
By \cite[Theorem 4.1 and Corollary 4.2]{rangarao}, we have:
\begin{itemize}
 \item $z(\sigma, \sigma^{-1}) = 1$ for $\sigma \in \Sp_{2n}(F)$;
 \item $z(p_1 \sigma p, p^{-1} \sigma' p_2) = z(\sigma, \sigma')$ for $p_1, p_2, p \in P$ and $\sigma, \sigma' \in \Sp_{2n}(F)$;
 \item $z(\tau_i, \tau_j) = 1$;
 \item $z(\tau_n, \n(\b) \tau_n) = \gamma_F(\tfrac{1}{2} \psi)^n \cdot \gamma_F(\det \b, \tfrac{1}{2} \psi) \cdot h_F(\b)$ for $\b \in \Sym_n(F) \cap \GL_n(F)$, where $h_F(\b)$ is the Hasse invariant of the nondegenerate symmetric bilinear form associated to $\b$.
\end{itemize}
We may extend $z$ to a $2$-cocycle on $\GSp_{2n}(F)$ (see e.g.~\cite[Appendix B]{periods1}).
Then, for $\sigma, \sigma' \in \GSp_{2n}(F)$ with $\nu(\sigma) = \nu$ and $\nu(\sigma') = \nu^{-1}$, we have
\[
 z(\sigma, \sigma') = z(\sigma \cdot d(\nu)^{-1}, d(\nu) \cdot \sigma') \cdot v(\sigma' \cdot d(\nu), \nu).
\]

Recall that $\V = V \otimes_B W$ is an $8$-dimensional symplectic $F$-space.
Let $\e_1, \dots, \e_4, \e_1^*, \dots, \e_4^*$ be the basis of $\V$ given in \cite[\S 2.2]{periods1}.
Then we have
\[
 \X = F \e_1 + \dots +  F \e_4, \qquad
 \Y = F \e_1^* + \dots +  F \e_4^*, \qquad
 \llangle \e_i, \e_j^* \rrangle = \delta_{ij}.
\]
Using this basis, we identify $\GSp(\V)$ with $\GSp_8(F)$.
Under this identification, we write $P_\Y$ and $z_\Y$ for $P$ and $z$, respectively.
We refer to \cite[\S C.1]{periods1} for an explicit description of the image of $B_1^\times \times B_2^\times \times B^\times$ in $\GSp(\V)$.
Also, using a basis
\[
 (\e_1,0), \dots, (\e_4, 0), (0,\e_1), \dots, (0,\e_4),
 (\e_1^*,0), \dots, (\e_4^*, 0), (0,-\e_1^*), \dots, (0,-\e_4^*)
\]
of $\V^{\square}$, we identify $\GSp(\V^{\square})$ with $\GSp_{16}(F)$.
For $1 \le i \le 4$, put
\begin{align*}
 \X_i & = F \e_i, &
 \X_i^{\square} & = \X_i \oplus \X_i, \\
 \Y_i & = F \e_i^*, &
 \Y_i^{\square} & = \Y_i \oplus \Y_i, \\
 \V_i & = \X_i \oplus \Y_i, &
 \V_i^{\square} & = \V_i \oplus \V_i.
\end{align*}
Then we have a natural embedding
\[
 \iota_i: \Sp(\V_i^\square) \longrightarrow \Sp(\V^\square).
\]
Using a basis $(\e_i,0), (0,\e_i), (\e_i^*,0), (0,-\e_i^*)$ of $\V^{\square}_i$, we identify $\GSp(\V^{\square}_i)$ with $\GSp_4(F)$.
Put
\[
 \sigma_0 =
  \begin{pmatrix}
  \frac{1}{2} \1_4 & - \frac{1}{2} \1_4 & & \\
  & & \frac{1}{2} \1_4 & \frac{1}{2} \1_4 \\
  & & \1_4 & -\1_4 \\
  -\1_4 & -\1_4 & &
 \end{pmatrix}
 \in \Sp(\V^{\square}).
\]
Then we have
\[
 \begin{bmatrix}
  \frac{1}{2}(\vec{\e}, -\vec{\e}) \\
  \frac{1}{2}(\vec{\e}^*, -\vec{\e}^*) \\
  (\vec{\e}^*, \vec{\e}^*) \\
  (-\vec{\e}, -\vec{\e})
 \end{bmatrix}
 = \sigma_0 \cdot
 \begin{bmatrix}
  (\vec{\e}, 0) \\
  (0, \vec{\e}) \\
  (\vec{\e}^*, 0) \\
  (0, -\vec{\e}^*)
 \end{bmatrix}, \qquad
 \vec{\e} =
 \begin{bmatrix}
  \e_1 \\
  \e_2 \\
  \e_3 \\
  \e_4
 \end{bmatrix}, \qquad
 \vec{\e}^* =
 \begin{bmatrix}
  \e_1^* \\
  \e_2^* \\
  \e_3^* \\
  \e_4^*
 \end{bmatrix},
\]
so that $\V^{\bigtriangledown} = \X^{\square} \cdot \sigma_0$ and $\V^{\triangle} = \Y^{\square} \cdot \sigma_0$.

\subsubsection{Proof of \eqref{eq:weil-hodge-computation1}}

In this section, we will show that
\[
 \Chi(1, \alpha, \alpha) = 1
\]
for $\alpha \in E^\times$.
We write $\alpha = a + b \i$ with $a, b \in F$ and put $\nu = a^2 - b^2 u$.
Since $\Chi$ is continuous, we may assume that
\[
 a \ne 0, \qquad b \ne 0.
\]

\begin{lem}
\label{lem:weil-hodge-computation1-1}
We have
\[
 \tilde{s}(1, \alpha, \alpha) = \gamma_F(J_1, \tfrac{1}{2} \psi) \cdot (-2abJ_2, J_1)_F.
\]
\end{lem}

\begin{proof}
Put $g = [1, \alpha] \in \GU(V)^0$ and $h = \alpha \in \GU(W)$.
Then we have $\tilde{s}(1, \alpha, \alpha) = \tilde{s}(g) \cdot \tilde{s}(h) \cdot z_\Y(g, h)$.
By \cite[Proposition C.18]{periods1}, we have
\[
 \tilde{s}(g) = (- \nu J_2, J_1)_F, \qquad
 \tilde{s}(h) = (J_2, J_1)_F.
\]
It remains to compute $z_{\Y}(g, h)$.

Recall that
\[
 z_\Y(g, h) = z_\Y(g \cdot d(\nu), d(\nu)^{-1} \cdot h) \cdot v(h \cdot d(\nu)^{-1}, \nu^{-1}).
\]
We have
\[
 g = \nu^{-1} \cdot
 \begin{pmatrix}
  a \cdot \1_4 & -bu \cdot \J_2 \\
  -b \cdot \J_2^{-1} & a \cdot \1_4
 \end{pmatrix}, \qquad
 h =
 \begin{pmatrix}
  a \cdot \1_4 & bu \cdot \J \\
  b \cdot \J^{-1} & a \cdot \1_4
 \end{pmatrix}
\]
in $\GSp(\V)$, where
\[
 \J_2 =
 \begin{pmatrix}
  1 & & & \\
  & -J_1 & & \\
  & & J_2 & \\
  & & & -J
 \end{pmatrix}, \qquad
 \J =
 \begin{pmatrix}
  1 & & & \\
  & -J_1 & & \\
  & & -J_2 & \\
  & & & J
 \end{pmatrix}.
\]
Since
\begin{align*}
 g & =
 \begin{pmatrix}
 - b^{-1} \cdot \J_2 & \nu^{-1} a \cdot \1_4 \\
 & - \nu^{-1} b \cdot \J_2^{-1}
 \end{pmatrix}
 \cdot \tau_4 \cdot \n(- a b^{-1} \cdot \J_2), \\
 h & = \n(a b^{-1} \cdot \J) \cdot \tau_4 \cdot
 \begin{pmatrix}
  b \cdot \J^{-1} & a \cdot \1_4 \\
  & \nu b^{-1} \cdot \J
 \end{pmatrix},
\end{align*}
we have
\[
 z_\Y(g \cdot d(\nu), d(\nu)^{-1} \cdot h)
 = z_\Y(\tau_4 \cdot \n(- \nu a b^{-1} \cdot \J_2), \n(\nu a b^{-1} \cdot \J) \cdot \tau_4)
 = z_\Y(\tau_4, \n(\b) \cdot \tau_4),
\]
where
\[
 \b = - \nu a b^{-1} \cdot \J_2 +  \nu a b^{-1} \cdot \J
 = 2 \nu a b^{-1} \cdot
 \begin{pmatrix}
  0 & & & \\
  & 0 & & \\
  & & -J_2 & \\
  & & & J
 \end{pmatrix}.
\]
If we put
\[
 \b' =
 2 \nu a b^{-1} \cdot
 \begin{pmatrix}
  - J_2 & \\
  & J
 \end{pmatrix},
\]
then we have
\[
 z_\Y(\tau_4, \n(\b) \cdot \tau_4) =
 \gamma_F(\tfrac{1}{2} \psi)^2 \cdot \gamma_F(\det \b', \tfrac{1}{2} \psi) \cdot h_F(\b').
\]
Hence, since $\det \b' \equiv -J_1 \bmod (F^{\times})^2$ and
\[
 h_F(\b') = (-2 \nu a b^{-1} J_2, 2 \nu a b^{-1} J)_F = (-2 \nu a b J_2, J_1)_F,
\]
we have
\begin{align*}
 z_\Y(g \cdot d(\nu), d(\nu)^{-1} \cdot h)
 & = \gamma_F(-1, \tfrac{1}{2} \psi)^{-1} \cdot \gamma_F(-J_1, \tfrac{1}{2} \psi)
 \cdot (-2 \nu a b J_2, J_1)_F \\
 & = \gamma_F(J_1, \tfrac{1}{2} \psi) \cdot (2 \nu a b J_2, J_1)_F.
\end{align*}
On the other hand, since $x(h \cdot d(\nu)^{-1}) \equiv 1 \bmod (F^{\times})^2$ and $j(h \cdot d(\nu)^{-1}) = 4$, we have
\[
 v(h \cdot d(\nu)^{-1}, \nu^{-1}) = 1.
\]
Thus we obtain
\[
 z_\Y(g, h) = \gamma_F(J_1, \tfrac{1}{2} \psi) \cdot (2 \nu a b J_2, J_1)_F.
\]
This completes the proof.
\end{proof}

Now we compute $s(1, \alpha, \alpha)$.
Note that $[1, \alpha] \in \GU(V)^0$ is the image of $\alpha$ under the embedding $E^\times \hookrightarrow \GU(V)$ as in \S \ref{ss:weil-hodge-setup}.
Hence, by definition, we have
\[
 s(1,\alpha,\alpha) = \frac{s^\sharp(\alpha, \alpha, \alpha, \alpha)}{s_2(\iota(1, [\alpha, \alpha]))},
\]
where $[\alpha, \alpha] \in \U(\WW)$.
Since
\begin{align*}
 s_2(\iota([\alpha, \alpha], [\alpha, \alpha]))
 & = s_2(\iota([\alpha, \alpha], 1)) \cdot s_2(\iota(1, [\alpha, \alpha]))
 \cdot z_{\Y^\square}(\iota([\alpha, \alpha], 1), \iota(1, [\alpha, \alpha])) \\
 & = s_2(\iota([\alpha, \alpha], 1)) \cdot s_2(\iota(1, [\alpha, \alpha])),
\end{align*}
we have
\begin{align*}
 s(1,\alpha,\alpha)
 & = \frac{s^\sharp(\alpha, \alpha, \alpha, \alpha)}{s_2(\iota([\alpha, \alpha], [\alpha, \alpha]))}
 \cdot s_2(\iota([\alpha, \alpha], 1)) \\
 & = \frac{\hat{s}^\sharp(\alpha, \alpha, \alpha, \alpha)}{\hat{s}_2(\iota([\alpha, \alpha], [\alpha, \alpha]))}
 \cdot \hat{s}_2(\iota([\alpha, \alpha], 1)) \cdot \mu(\iota([\alpha, \alpha], 1)).
\end{align*}
Hence, by definition and Lemma \ref{lem:s-hat-2-E-diag}, we have
\[
 s(1, \alpha, \alpha) = \chi(\alpha)^2
 \cdot \hat{s}_2(\iota([\alpha, \alpha], 1)) \cdot \mu(\iota([\alpha, \alpha], 1)).
\]

\begin{lem}
\label{lem:weil-hodge-computation1-2}
We have
\[
 \hat{s}_2(\iota([\alpha, \alpha], 1)) = \chi(\alpha)^{-2} \cdot (u, J_1)_F.
\]
\end{lem}

\begin{proof}
Put $h = \iota([\alpha, \alpha], 1) \in \U(\WW^\square)$ and $\beta = \alpha^{-1} \alpha^\rho$, so that $\beta - 1 \in E^\times$.
As in the proof of Lemma \ref{lem:s-hat-1-E^1}, we have
\[
 \begin{bmatrix}
  \w_1 \cdot h \\
  \w_2 \cdot h \\
  \w_1^* \cdot h \\
  \w_2^* \cdot h
 \end{bmatrix}
 = A \cdot
 \begin{bmatrix}
  \w_1 \\
  \w_2 \\
  \w_1^* \\
  \w_2^*
 \end{bmatrix},
\]
where
\begin{align*}
 A & =
 \begin{pmatrix}
  1 & & & \\
  & \frac{1}{2} (\beta+1) & & \frac{1}{4 J \i} (\beta-1) \\
  & & 1 & \\
  & J \i (\beta-1) & & \frac{1}{2} (\beta+1)
 \end{pmatrix} \\
 & =
 \begin{pmatrix}
  1 & & & \\
  & -\frac{1}{J \i (\beta^\rho - 1)} & & * \\
  & & 1 & \\
  & & & J \i (\beta-1)
 \end{pmatrix}
 \cdot \tau_1 \cdot
 \begin{pmatrix}
  1 & & & \\
  & 1 & & * \\
  & & 1 & \\
  & & & 1
 \end{pmatrix}.
\end{align*}
Hence we have
\[
 \hat{s}_2(\iota([\alpha, \alpha], 1)) = \chi(J \i (\beta -1))^2 \cdot \gamma^{-1},
\]
where
\begin{align*}
 \gamma & = (u, \det \VV)_F \cdot \gamma_F(-u, \tfrac{1}{2} \psi)^2 \cdot \gamma_F(-1, \tfrac{1}{2} \psi)^{-2} \\
 & = (u, -J_1)_F \cdot (-1, -u)_F \cdot (-1, -1)_F \\
 & = (u, J_1)_F.
\end{align*}
Since $\beta-1 = \alpha^{-1}(\alpha^\rho - \alpha) = - 2b \i \alpha^{-1}$,
we have $\chi(J \i (\beta -1))^2 = \chi(- 2b uJ \alpha^{-1})^2 = \chi(\alpha)^{-2}$.
This completes the proof.
\end{proof}

\begin{lem}
\label{lem:weil-hodge-computation1-3}
We have
\[
 \mu(\iota([\alpha, \alpha], 1)) = \gamma_F(J_1, \tfrac{1}{2} \psi) \cdot (- 2abuJ_2, J_1)_F.
\]
\end{lem}

\begin{proof}
We write $\alpha^{-1} \alpha^\rho = c + d \i$ with $c, d \in F$, so that
\[
 c = \frac{a^2 + b^2 u}{a^2 - b^2 u} \ne \pm 1, \qquad
 d = - \frac{2ab}{a^2 - b^2 u} \ne 0.
\]
Recall that
\[
 \mu(\iota([\alpha, \alpha], 1))
 = z_{\Y^\square}(\sigma_0, \sigma)^{-1} \cdot z_{\Y^\square}(\sigma_0 \sigma \sigma_0^{-1}, \sigma_0),
\]
where $\sigma$ is the image of $\iota([\alpha, \alpha], 1)$ in $\Sp(\V^\square)$.
We have
\[
 \sigma_0 = \prod_{i=1}^4 \iota_i(\tau_1 \cdot \m(\a_1)), \qquad
 \sigma = \prod_{i=3}^4 \iota_i(\sigma_i),
\]
where
\[
 \a_1 =
 \begin{pmatrix}
  \frac{1}{2} & - \frac{1}{2} \\
  - 1 & - 1
 \end{pmatrix}, \qquad
 \sigma_i =
 \begin{pmatrix}
  c & & d k_i u & \\
  & 1 & & \\
  \frac{d}{k_i} & & c & \\
  & & & 1
 \end{pmatrix}, \qquad
 k_i =
 \begin{cases}
  J_2 & \text{if $i=3$,} \\
  -J & \text{if $i=4$.}
 \end{cases}
\]
Since $\tau_1 \cdot \m(\a_1) \in P_{\Y_i^\square} \cdot \tau_1 \cdot \m(\a_2)$ and $\sigma_i \in \n(\b_{1,i}) \cdot \tau \cdot P_{\Y_i^\square}$, where
\[
 \a_2 =
 \begin{pmatrix}
  1 & \\
  1 & 1
 \end{pmatrix}, \qquad
 \b_{1,i} = \frac{c k_i}{d} \cdot
 \begin{pmatrix}
  1 & \\
  & 0
 \end{pmatrix}, \qquad
 \tau =
 \begin{pmatrix}
  & & -1 & \\
  & 1 & & \\
  1 & & & \\
  & & & 1
 \end{pmatrix},
\]
we have
\begin{align*}
 z_{\Y^\square}(\sigma_0, \sigma)
 & = \prod_{i=3}^4 z_{\Y_i^\square}(\tau_1 \cdot \m(\a_1), \sigma_i) \\
 & = \prod_{i=3}^4 z_{\Y_i^\square}(\tau_1, \m(\a_2) \cdot \n(\b_{1,i}) \cdot \tau).
\end{align*}
If we put
\[
 \b_{2,i} = \frac{c k_i}{d} \cdot
 \begin{pmatrix}
  1 & 1 \\
  1 &
 \end{pmatrix}, \qquad
 \b_{3,i} = \frac{c k_i}{d} \cdot
 \begin{pmatrix}
  0 & \\
  & 1
 \end{pmatrix},
\]
then we have $\m(\a_2) \cdot \n(\b_{1,i}) = \n(\b_{2,i}) \cdot \n(\b_{3,i}) \cdot \m(\a_2)$ and hence
\[
 z_{\Y^\square}(\sigma_0, \sigma)
 = \prod_{i=3}^4 z_{\Y_i^\square}(\tau_1 \cdot \n(\b_{2,i}), \n(\b_{3,i}) \cdot \m(\a_2) \cdot \tau).
\]
Since $\tau_1 \cdot \n(\b_{2,i}) \in P_{\Y_i^\square} \cdot \tau_1$ and $\n(\b_{3,i}) \cdot \m(\a_2) \cdot \tau \in \tau \cdot P_{\Y_i^\square}$, we have
\[
 z_{\Y^\square}(\sigma_0, \sigma) = \prod_{i=3}^4 z_{\Y_i^\square}(\tau_1, \tau) = 1.
\]

On the other hand, we have
\[
 \sigma_0 \sigma \sigma_0^{-1} = \prod_{i=3}^4 \iota_i(\sigma_i'),
\]
where
\[
 \sigma_i' =
 \begin{pmatrix}
  \frac{1}{2} (c+1) & \frac{d k_i u}{2} &
  \frac{d k_i u}{4} & -\frac{1}{4} (c-1) \\
  \frac{d}{2 k_i} & \frac{1}{2} (c+1) &
  \frac{1}{4} (c-1) & - \frac{d}{4 k_i} \\
  \frac{d}{k_i} & c-1 & \frac{1}{2} (c+1) & -\frac{d}{2 k_i} \\
  -c+1 & - d k_i u & - \frac{d k_i u}{2} & \frac{1}{2} (c+1)
 \end{pmatrix}.
\]
Since $\sigma_i' \in P_{\Y_i^\square} \cdot \tau_2 \cdot \n(\b_{4,i}) \cdot \n(\b_{5,i})$, where
\[
 \b_{4,i} = - \frac{d}{2(c-1) k_i} \cdot
 \begin{pmatrix}
  0 & \\
  & 1
 \end{pmatrix}, \qquad
 \b_{5,i} = \frac{d k_i u}{2(c-1)} \cdot
 \begin{pmatrix}
  1 & \\
  & 0
 \end{pmatrix},
\]
we have
\begin{align*}
 z_{\Y^\square}(\sigma_0 \sigma \sigma_0^{-1}, \sigma_0)
 & = \prod_{i=3}^4 z_{\Y_i^\square}(\sigma_i', \tau_1 \cdot \m(\a_1)) \\
 & = \prod_{i=3}^4 z_{\Y_i^\square}(\tau_2, \n(\b_{4,i}) \cdot \n(\b_{5,i}) \cdot \tau_1).
\end{align*}
Hence, since $\n(\b_{5,i}) \cdot \tau_1 \in \tau_1 \cdot P_{\Y_i^{\square}}$ and
\[
 \frac{d}{c-1} = - \frac{a}{bu},
\]
we have
\begin{align*}
 z_{\Y^\square}(\sigma_0 \sigma \sigma_0^{-1}, \sigma_0)
 & = \prod_{i=3}^4 z_{\Y_i^\square}(\tau_2, \n(\b_{4,i}) \cdot \tau_1) \\
 & = \prod_{i=3}^4 \left[ \gamma_F(\tfrac{1}{2} \psi) \cdot \gamma_F(2ab k_i u, \tfrac{1}{2} \psi) \right] \\
 & = \gamma_F(-1, \tfrac{1}{2} \psi)^{-1}
 \cdot \gamma_F(k_3 k_4, \tfrac{1}{2} \psi) \cdot (2ab k_3 u, 2ab k_4 u)_F \\
 & = \gamma_F(-k_3 k_4, \tfrac{1}{2} \psi) \cdot (-2ab k_3 u, -k_3 k_4)_F.
\end{align*}
This completes the proof.
\end{proof}

By Lemmas \ref{lem:weil-hodge-computation1-2} and \ref{lem:weil-hodge-computation1-3}, we have
\[
 s(1, \alpha, \alpha) = \gamma_F(J_1, \tfrac{1}{2} \psi) \cdot (- 2abJ_2, J_1)_F.
\]
Now \eqref{eq:weil-hodge-computation1} follows from this and Lemma \ref{lem:weil-hodge-computation1-1}.

\subsubsection{Proof of \eqref{eq:weil-hodge-computation2}}

In this section, we will show that
\[
 \Chi(\alpha, \alpha^{-1}, 1) = 1
\]
for $\alpha \in E^\times$.
We write $\alpha = a + b \i$ with $a, b \in F$ and put $\nu = a^2 - b^2 u$.
Since $\Chi$ is continuous, we may assume that
\[
 a \ne 0, \qquad b \ne 0.
\]

\begin{lem}
\label{lem:weil-hodge-computation2-1}
We have
\[
 \tilde{s}(\alpha, \alpha^{-1}, 1) = \gamma_F(J, \tfrac{1}{2} \psi) \cdot (-2abJ_1, J)_F.
\]
\end{lem}

\begin{proof}
Put $g_1 = [\alpha, 1] \in \GU(V)^0$ and $g_2 = [1, \alpha^{-1}] \in \GU(V)^0$.
Then we have $\tilde{s}(\alpha, \alpha^{-1}, 1) = \tilde{s}(g_1) \cdot \tilde{s}(g_2) \cdot z_\Y(g_1, g_2)$.
By \cite[Proposition C.18]{periods1}, we have
\[
 \tilde{s}(g_1) = (- \nu J_1, J_2)_F, \qquad
 \tilde{s}(g_2) = (- \nu J_2, J_1)_F.
\]
We have
\[
 g_1 = \nu^{-1} \cdot
 \begin{pmatrix}
  a \cdot \1_4 & -bu \cdot \J_1 \\
  -b \cdot \J_1^{-1} & a \cdot \1_4
 \end{pmatrix}, \qquad
 g_2 =
 \begin{pmatrix}
  a \cdot \1_4 & bu \cdot \J_2 \\
  b \cdot \J_2^{-1} & a \cdot \1_4
 \end{pmatrix}
\]
in $\GSp(\V)$, where
\[
 \J_1 =
 \begin{pmatrix}
  1 & & & \\
  & J_1 & & \\
  & & -J_2 & \\
  & & & -J
 \end{pmatrix}, \qquad
 \J_2 =
 \begin{pmatrix}
  1 & & & \\
  & -J_1 & & \\
  & & J_2 & \\
  & & & -J
 \end{pmatrix}.
\]
Hence, as in the proof of Lemma \ref{lem:weil-hodge-computation1-1}, we have
\[
 z_{\Y}(g_1, g_2) = \gamma_F(J, \tfrac{1}{2} \psi) \cdot (2 \nu abJ_1, J)_F.
\]
This completes the proof.
\end{proof}

Now we compute $s(\alpha, \alpha^{-1}, 1)$.
By definition, we have
\begin{align*}
 s(\alpha,\alpha^{-1}, 1)
 & = s^\sharp([\alpha, \alpha^{-1}], 1, 1, 1) \\
 & = \hat{s}^\sharp([\alpha, \alpha^{-1}], 1, 1, 1) \cdot \mu(\iota([\alpha, \alpha^{-1}], 1)) \\
 & = \hat{s}_1(\iota([\alpha, \alpha^{-1}], 1)) \cdot \mu(\iota([\alpha, \alpha^{-1}], 1)),
\end{align*}
where $[\alpha, \alpha^{-1}] \in \U(V)^0$.

\begin{lem}
\label{lem:weil-hodge-computation2-2}
We have
\[
 \hat{s}_1(\iota([\alpha, \alpha^{-1}], 1)) = (u, J)_F.
\]
\end{lem}

\begin{proof}
Put $g = \iota([\alpha, \alpha^{-1}], 1) \in \U(V^\square)$ and $\beta = \alpha^{-1} \alpha^\rho$, so that $\beta - 1 \in E^\times$.
As in the proof of Lemma \ref{lem:s-hat-1-E^1}, we have
\[
 \begin{bmatrix}
  \v_1 \cdot g \\
  \v_2 \cdot g \\
  \v_1^* \cdot g \\
  \v_2^* \cdot g
 \end{bmatrix}
 = A \cdot
 \begin{bmatrix}
  \v_1 \\
  \v_2 \\
  \v_1^* \\
  \v_2^*
 \end{bmatrix},
\]
where
\begin{align*}
 A & =
 \begin{pmatrix}
  1 & & & \\
  & \frac{1}{2} (\beta+1) & & - \frac{1}{4 J_1 \i} (\beta-1) \\
  & & 1 & \\
  & -J_1 \i (\beta-1) & & \frac{1}{2} (\beta+1)
 \end{pmatrix} \\
 & =
 \begin{pmatrix}
  1 & & & \\
  & \frac{1}{J_1 \i (\beta^\rho - 1)} & & * \\
  & & 1 & \\
  & & & -J_1 \i (\beta-1)
 \end{pmatrix}
 \cdot \tau_1 \cdot
 \begin{pmatrix}
  1 & & & \\
  & 1 & & * \\
  & & 1 & \\
  & & & 1
 \end{pmatrix}.
\end{align*}
Hence we have
\[
 \hat{s}_1(g) =
 \begin{cases}
  1 & \text{if $B$ is split,} \\
  -1 & \text{if $B$ is ramified.}
 \end{cases}
\]
This completes the proof.
\end{proof}

\begin{lem}
\label{lem:weil-hodge-computation2-3}
We have
\[
 \mu(\iota([\alpha, \alpha^{-1}], 1))
 = \gamma_F(J, \tfrac{1}{2} \psi) \cdot (-2abu J_1, J)_F.
\]
\end{lem}

\begin{proof}
Recall that
\[
 \mu(\iota([\alpha, \alpha^{-1}], 1))
 = z_{\Y^\square}(\sigma_0, \sigma)^{-1} \cdot z_{\Y^\square}(\sigma_0 \sigma \sigma_0^{-1}, \sigma_0),
\]
where $\sigma$ is the image of $\iota([\alpha, \alpha^{-1}], 1)$ in $\Sp(\V^\square)$.
If we write $\alpha^{-1} \alpha^\rho = c + d \i$ with $c, d \in F$, then we have
\[
 \sigma = \prod_{i=2}^3 \iota_i(\sigma_i),
\]
where
\[
 \sigma_i =
 \begin{pmatrix}
  c & &  d k_i u & \\
  & 1 & & \\
  \frac{d}{k_i} & & c & \\
  & & & 1
 \end{pmatrix}, \qquad
 k_i =
 \begin{cases}
  J_1 & \text{if $i=2$,} \\
  -J_2 & \text{if $i=3$.}
 \end{cases}
\]
Hence, as in the proof of Lemma \ref{lem:weil-hodge-computation1-3}, we have $z_{\Y^\square}(\sigma_0, \sigma)  = 1$ and
\[
 z_{\Y^\square}(\sigma_0 \sigma \sigma_0^{-1}, \sigma_0)
 = \gamma_F(-k_2 k_3, \tfrac{1}{2} \psi) \cdot (-2ab k_2 u, -k_2 k_3)_F.
\]
This completes the proof.
\end{proof}

By Lemmas \ref{lem:weil-hodge-computation2-2} and \ref{lem:weil-hodge-computation2-3}, we have
\[
 s(\alpha, \alpha^{-1}, 1) = \gamma_F(J, \tfrac{1}{2} \psi) \cdot (-2ab J_1, J)_F.
\]
Now \eqref{eq:weil-hodge-computation2} follows from this and Lemma \ref{lem:weil-hodge-computation2-1}.

\subsection{Proof of \eqref{eq:weil-hodge-computation3}}

Assume that $J_1, J_2 \in (F^\times)^2$.
Choose $t_i \in F^\times$ such that $J_i = t_i^2$ and put $\j_i^\natural = t_i^{-1} \cdot \j_i$.
In this section, we will show that
\[
 \Chi(\j_1^\natural, \j_2^\natural, 1) = 1.
\]

\begin{lem}
\label{lem:weil-hodge-computation3-1}
We have
\[
 \tilde{s}(\j_1^\natural, \j_2^\natural, 1) = 1.
\]
\end{lem}

\begin{proof}
Put $g_1 = [\j_1^\natural, 1] \in \GU(V)^0$ and $g_2 = [1, \j_2^\natural] \in \GU(V)^0$.
Then we have $\tilde{s}(\j_1^\natural, \j_2^\natural, 1) = \tilde{s}(g_1) \cdot \tilde{s}(g_2) \cdot z_\Y(g_1, g_2)$.
By \cite[Proposition C.18]{periods1}, we have
\[
 \tilde{s}(g_1) = \tilde{s}(g_2) = 1.
\]
It remains to compute $z_{\Y}(g_1, g_2)$.

Noting that $\nu(g_1) = \nu(g_2) = -1$, we have
\[
 z_\Y(g_1, g_2) = z_\Y(g_1 \cdot d(-1), d(-1) \cdot g_2) \cdot v(g_2 \cdot d(-1), -1).
\]
We have
\[
 g_1 = \m(\a_1) \cdot d(-1), \qquad
 g_2 = \m(\a_2) \cdot d(-1)
\]
in $\GSp(\V)$, where
\[
 \a_1 =
 \begin{pmatrix}
  & \frac{1}{t_1} & & \\
  t_1 & & & \\
  & & & \frac{1}{t_1} \\
  & & t_1 &
 \end{pmatrix}, \qquad
 \a_2 =
 \begin{pmatrix}
  & & \frac{1}{t_2} & \\
  & & & \frac{1}{t_2} \\
  t_2 & & & \\
  & t_2 & &
 \end{pmatrix}.
\]
Hence we have
\[
 z_\Y(g_1 \cdot d(-1), d(-1) \cdot g_2) = 1.
\]
On the other hand, since $x(g_2 \cdot d(-1)) = 1$ and $j(g_2 \cdot d(-1)) = 0$, we have
\[
 v(g_2 \cdot d(-1), -1) = 1.
\]
Thus we obtain $z_\Y(g_1, g_2) = 1$.
This completes the proof.
\end{proof}

Now we compute $s(\j_1^\natural, \j_2^\natural, 1)$.
By definition, we have
\begin{align*}
 s(\j_1^\natural, \j_2^\natural, 1)
 & = s^\sharp([\j_1^\natural, \j_2^\natural], 1, 1, 1) \\
 & = \hat{s}^\sharp([\j_1^\natural, \j_2^\natural], 1, 1, 1) \cdot \mu(\iota([\j_1^\natural, \j_2^\natural], 1)) \\
 & = \hat{s}_1(\iota([\j_1^\natural, \j_2^\natural], 1)) \cdot \mu(\iota([\j_1^\natural, \j_2^\natural], 1)),
\end{align*}
where $[\j_1^\natural, \j_2^\natural] \in \U(V)^0$.

\begin{lem}
\label{lem:weil-hodge-computation3-2}
We have
\[
 \hat{s}_1(\iota([\j_1^\natural, \j_2^\natural], 1)) = 1.
\]
\end{lem}

\begin{proof}
Since $B$ is split, the assertion follows.
\end{proof}

\begin{lem}
\label{lem:weil-hodge-computation3-3}
We have
\[
 \mu(\iota([\j_1^\natural, \j_2^\natural], 1)) = 1.
\]
\end{lem}

\begin{proof}
Recall that
\[
 \mu(\iota([\j_1^\natural, \j_2^\natural], 1))
 = z_{\Y^\square}(\sigma_0, \sigma)^{-1} \cdot z_{\Y^\square}(\sigma_0 \sigma \sigma_0^{-1}, \sigma_0),
\]
where $\sigma$ is the image of $\iota([\j_1^\natural, \j_2^\natural], 1)$ in $\Sp(\V^\square)$.
Since
\[
 \sigma = \m(\a_1),
\]
where
\[
 \a_1 =
 \begin{pmatrix}
  \a & \\
  & \1_4
 \end{pmatrix}, \qquad
 \a =
 \begin{pmatrix}
  & & & \frac{1}{t_1 t_2} \\
  & & \frac{t_1}{t_2} & \\
  & \frac{t_2}{t_1} & & \\
  t_1 t_2 & & &
 \end{pmatrix},
\]
we have $z_{\Y^\square}(\sigma_0, \sigma) = 1$.
On the other hand, we have
\[
 \sigma_0 \sigma \sigma_0^{-1} =
 \begin{pmatrix}
  \tfrac{1}{2} (\1_4 + \a) & & & \tfrac{1}{4} (\1_4 - \a) \\
  & \tfrac{1}{2} (\1_4 + {}^t \a) & - \tfrac{1}{4} (\1_4 - {}^t \a) & \\
  & - \1_4 + {}^t \a & \tfrac{1}{2} (\1_4 + {}^t \a) & \\
  \1_4 - \a & & & \tfrac{1}{2} (\1_4 + \a)
 \end{pmatrix}.
\]
Since $\sigma_0 \sigma \sigma_0^{-1} \in P_{\Y^\square} \cdot \tau \cdot \m(\a_2)$ and $\sigma_0 \in \tau_4 \cdot P_{\Y^\square}$, where
\[
 \a_2 =
 \begin{pmatrix}
  \a' & \\
  & \a''
 \end{pmatrix}, \qquad
 \a' =
 \begin{pmatrix}
  1 & & & \\
  & 1 & & \\
  & -\frac{t_2}{t_1} & 1 & \\
  -t_1 t_2 & & & 1
 \end{pmatrix}, \qquad
 \a'' =
 \begin{pmatrix}
  1 & & & \\
  & 1 & & \\
  & -\frac{t_1}{t_2} & 1 & \\
  -\frac{1}{t_1 t_2} & & & 1
 \end{pmatrix},
\]
and
\[
 \tau =
 \begin{pmatrix}
  \1_2 & & & & & & & \\
  & \0_2 & & & & -\1_2 & & \\
  & & \1_2 & & & & & \\
  & & & \0_2 & & & & -\1_2 \\
  & & & & \1_2 & & & \\
  & \1_2 & & & & \0_2 & & \\
  & & & & & & \1_2 & \\
  & & & \1_2 & & & & \0_2
 \end{pmatrix},
\]
we have
\[
 z_{\Y^\square}(\sigma_0 \sigma \sigma_0^{-1}, \sigma_0)
 = z_{\Y^\square}(\tau, \m(\a_2) \cdot \tau_4).
\]
Hence, since $\m(\a_2) \cdot \tau_4 \in \tau_4 \cdot P_{\Y^\square}$, we have
\[
 z_{\Y^\square}(\sigma_0 \sigma \sigma_0^{-1}, \sigma_0)
 = z_{\Y^\square}(\tau, \tau_4) = 1.
\]
This completes the proof.
\end{proof}

By Lemmas \ref{lem:weil-hodge-computation3-2} and \ref{lem:weil-hodge-computation3-3}, we have
\[
 s(\j_1^\natural, \j_2^\natural, 1) = 1.
\]
Now \eqref{eq:weil-hodge-computation3} follows from this and Lemma \ref{lem:weil-hodge-computation3-1}.

\subsection{Compatibility with \cite{hk-duke}}
\label{ss:compatibility-hk-duke}

Suppose that $F$ is local.
In this section, we compare the splitting $s$ with the standard one for symplectic-orthogonal dual pairs when $B$ is split.
In this case, we have $J \in \N_{E/F}(E^\times)$, so that we may write $J = k^2 - l^2 u$ for some $k, l \in F$.
We define an isomorphism $\ii:B \rightarrow \M_2(F)$ of quaternion $F$-algebras by
\[
 \ii(a + b \i + c \j + d \i \j) =
 \begin{pmatrix}
  a & b \\
  bu  & a
 \end{pmatrix}
 + \begin{pmatrix}
  c & d \\
  du  & c
 \end{pmatrix}
 \begin{pmatrix}
  k & -l \\
  lu  & -k
 \end{pmatrix}.
\]
Put
\begin{align*}
 e & = \frac{1}{2} + \frac{k}{2J} \j - \frac{l}{2J} \i \j, &
 e' & = \frac{1}{2} \i + \frac{lu}{2J} \j - \frac{k}{2J} \i \j, \\
 e'' & =  \frac{1}{2u} \i - \frac{l}{2J} \j + \frac{k}{2uJ} \i \j, &
 e^* & = \frac{1}{2} - \frac{k}{2J} \j + \frac{l}{2J} \i \j,
\end{align*}
so that
\[
 \ii(e) = \mat{1}{0}{0}{0}, \qquad
 \ii(e') = \mat{0}{1}{0}{0}, \qquad
 \ii(e'') = \mat{0}{0}{1}{0}, \qquad
 \ii(e^*) = \mat{0}{0}{0}{1}.
\]
In particular, we have
\[
 \begin{bmatrix}
  e \cdot x \\
  e' \cdot x
 \end{bmatrix}
 = \ii(x) \cdot
 \begin{bmatrix}
  e \\
  e'
 \end{bmatrix}
\]
for $x \in B$.

Let $V$ be an $m$-dimensional skew-hermitian right $B$-space as in \eqref{eq:condition-skew-herm}.
We consider the $2m$-dimensional quadratic $F$-space $V^\dagger := V e$ given by Morita theory (see \cite[Appendix C]{periods1} for details).
With respect to a basis $e_1 e, e_1 e'', \dots, e_m e, e_m e''$ of $V^\dagger$, the symmetric bilinear form on $V^\dagger$ is associated to
\[
 \frac{1}{2} \cdot
 \begin{pmatrix}
  \kappa_1 u & & & & \\
  & - \kappa_1& & & \\
  & & \ddots & & \\
  & & & \kappa_m u & \\
  & & & & - \kappa_m
 \end{pmatrix}.
\]
Similarly, we consider the $2$-dimensional symplectic $F$-space $W^\dagger := e W$.
Then, by \cite[Lemma C.2]{periods1}, we have an identification
\[
 \V = V^\dagger \otimes_F W^\dagger.
\]
We take a complete polarization $W^\dagger = X \oplus Y$ defined by
\[
 X = F e, \qquad Y = F e'.
\]
This induces a complete polarization $\V = \X' \oplus \Y'$, where
\[
 \X' = V^\dagger \otimes_F X, \qquad
 \Y' = V^\dagger \otimes_F Y.
\]
More explicitly, we have
\begin{align*}
 \X' & = F \cdot e_1 \otimes e + \dots + F \cdot e_m \otimes e + F \cdot e_1 \otimes e'' + \dots + F \cdot e_m \otimes e'', \\
 \Y' & = F \cdot e_1 \otimes e' + \dots + F \cdot e_m \otimes e' + F \cdot e_1 \otimes e^* + \dots + F \cdot e_m \otimes e^*.
\end{align*}

Now we recall the splitting defined in \cite[\S 5.1]{hk-duke}.
Using a basis $e, e'$ of $W^\dagger$, we identify $\GSp(W^\dagger)$ with $\GSp_2(F) = \GL_2(F)$.
We define a map
\[
 s^\dagger: \Sp(W^\dagger) \longrightarrow \C^1
\]
by setting
\[
 s^\dagger(h) = \xi_E(x(h))^m \cdot \gamma'^{-j(h)},
\]
where $x(h)$ and $j(h)$ are as in \S \ref{sss:notation}, and
\[
 \gamma' = \gamma_F(\tfrac{1}{2} \psi)^{2m} \cdot \gamma_F(\det V^\dagger, \tfrac{1}{2} \psi) \cdot h_F(V^\dagger).
\]
We extend $s^\dagger$ to a map
\[
 s^\dagger: \G(\O(V^\dagger) \times \Sp(W^\dagger)) \longrightarrow \C^1
\]
so that
\[
 s^\dagger(\g) = s^\dagger(h \cdot d(\nu(h))^{-1})
\]
for $\g = (g, h) \in \G(\O(V^\dagger) \times \Sp(W^\dagger))$.

\begin{lem}
\label{lem:spl-s-dagger}
For $\g_1, \g_2 \in \G(\O(V^\dagger) \times \Sp(W^\dagger))$, we have
\[
 z_{\Y'}(\g_1, \g_2) = \frac{s^\dagger(\g_1 \g_2)}{s^\dagger(\g_1) s^\dagger(\g_2)}.
\]
\end{lem}

\begin{proof}
If $\g_1, \g_2 \in \Sp(W^\dagger)$, then the assertion follows from \cite[Theorem 3.1, cases $1_+$]{kudla-splitting}.
By \cite[\S 5.1]{hk-duke}, this implies the general case.
Nevertheless, we include a direct argument for the convenience of the reader.

Let $\g_i = (g_i, h_i) \in \G(\O(V^\dagger) \times \Sp(W^\dagger))$.
Recall that
\[
 z_{\Y'}(\g_1, \g_2) = \gamma_F(\tfrac{1}{2} \psi \circ q(\Y' \cdot \g_1^{-1}, \Y' \cdot \g_2^{-1} \g_1^{-1}, \Y')),
\]
where $q$ denotes the Leray invariant (see e.g.~\cite{rangarao}, \cite[\S 3.1.2]{periods1}).
Put
\[
 \nu_i = \nu(h_i), \qquad
 h_i' = h_i \cdot d(\nu_i)^{-1}, \qquad
 h_2'' = d(\nu_1) \cdot h_2' \cdot d(\nu_1)^{-1},
\]
so that
\[
 h_1 h_2 = h_1' \cdot d(\nu_1) \cdot h_2' \cdot d(\nu_2) = h_1' h_2'' \cdot d(\nu_1 \nu_2).
\]
Since $\Y' \cdot (g, d(\nu)) = \Y'$ for $g \in \GO(V^\dagger)$ and $\nu \in F^{\times}$, we have $\Y' \cdot \g_1^{-1} = \Y' \cdot h_1'^{-1}$ and $\Y' \cdot \g_2^{-1} \g_1^{-1} = \Y' \cdot h_2''^{-1} h_1'^{-1}$, so that
\[
 q(\Y' \cdot \g_1^{-1}, \Y' \cdot \g_2^{-1} \g_1^{-1}, \Y')
 = q(\Y' \cdot h_1'^{-1}, \Y' \cdot h_2''^{-1} h_1'^{-1}, \Y').
\]
Hence we have
\[
 z_{\Y'}(\g_1, \g_2) = z_{\Y'}(h_1', h_2'') = \frac{s^\dagger(h'_1 h''_2)}{s^\dagger(h'_1) s^\dagger(h''_2)}.
\]
On the other hand, by definition, we have $s^\dagger(\g_i) = s^\dagger(h'_i)$ and $s^\dagger(\g_1 \g_2) = s^\dagger(h_1' h_2'')$.
By \eqref{eq:xj-conj}, and noting that $\nu(\GO(V^\dagger)) = \N_{E/F}(E^\times)$ if $m$ is odd, we have
\[
 s^\dagger(h_2'') = \xi_E(\nu_1)^{j(h_2') m} \cdot s^\dagger(h_2') = s^\dagger(h_2').
\]
This completes the proof.
\end{proof}

Fix $\varsigma_0 \in \Sp(\V)$ such that $\X = \X' \cdot \varsigma_0$ and $\Y = \Y' \cdot \varsigma_0$.
Put
\[
 \mu_0(\sigma) = z_{\Y'}(\varsigma_0, \sigma) \cdot z_{\Y'}(\varsigma_0 \sigma \varsigma_0^{-1}, \varsigma_0)^{-1}
\]
for $\sigma \in \Sp(\V)$.
Note that $\mu_0$ does not depend on the choice of $\varsigma_0$.
Then, by \cite[Lemma 4.2]{kudla-splitting}, we have
\[
 z_\Y(\sigma, \sigma') = z_{\Y'}(\sigma, \sigma')
 \cdot \frac{\mu_0(\sigma \sigma')}{\mu_0(\sigma) \mu_0(\sigma')}
\]
for $\sigma, \sigma' \in \Sp(\V)$.

Put $s_0 = s^\dagger \cdot \mu_0$.
Via the canonical isomorphisms $\GU(V) \cong \GO(V^\dagger)$ and $\GU(W) \cong \GSp(W^\dagger)$, we regard $s_0$ as a map
\[
 s_0 : \G(\U(V) \times \U(W)) \longrightarrow \C^1.
\]
By Lemma \ref{lem:spl-s-dagger}, we have
\[
 z_\Y(\g_1, \g_2) = \frac{s_0(\g_1 \g_2)}{s_0(\g_1) s_0(\g_2)}
\]
for $\g_1, \g_2 \in \G(\U(V) \times \U(W))$.

\begin{prop}
\label{prop:compare-s-0-s}
We have
\[
 s_0|_\cG = s.
\]
\end{prop}

The rest of this section is devoted to the proof of Proposition \ref{prop:compare-s-0-s}.

As in \S \ref{ss:doubling-U(V)}, we define the doubled space $W^{\dagger \square} = W^\dagger \oplus W^\dagger$ and take the complete polarization $W^{\dagger \square} = W^{\dagger \bigtriangledown} \oplus W^{\dagger \triangle}$.
Then we have identifications
\[
 \V^\square = V^\dagger \otimes_F W^{\dagger \square}, \qquad
 \V^\bigtriangledown = V^\dagger \otimes_F W^{\dagger \bigtriangledown}, \qquad
 \V^\triangle = V^\dagger \otimes_F W^{\dagger \triangle}.
\]
We also take complete polarizations $W^{\dagger \square} = X^\square \oplus Y^\square$ and $\V^\square = \X'^\square \oplus \Y'^\square$, where
\begin{align*}
 X^\square & = X \oplus X, &
 \X'^\square & = \X' \oplus \X' = V^\dagger \otimes_F X^\square, \\
 Y^\square & = Y \oplus Y, &
 \Y'^\square & = \Y' \oplus \Y' = V^\dagger \otimes_F Y^\square.
\end{align*}
As in \cite[\S D.3]{periods1}, we have
\[
 z_{\Y'^\square}(\iota(\sigma_1,\sigma_2), \iota(\sigma_1',\sigma_2'))
 = z_{\Y'}(\sigma_1, \sigma_1') \cdot z_{\Y'}(\sigma_2, \sigma_2')^{-1}
\]
for $\sigma_i, \sigma_i' \in \Sp(\V)$.
Using a basis $(e,0), (0,e), (e',0), (0,-e')$ of $W^{\dagger \square}$, we identify $\GSp(W^{\dagger \square})$ with $\GSp_4(F)$.
Put
\[
 h_0 =
 \begin{pmatrix}
  \frac{1}{2} & -\frac{1}{2} & & \\
  & & \frac{1}{2} & \frac{1}{2} \\
  & & 1 & -1 \\
  -1 & -1 & &
 \end{pmatrix}
 \in \Sp(W^{\dagger \square}).
\]
Then we have
\[
 \begin{bmatrix}
  \frac{1}{2}(e, -e) \\
  \frac{1}{2}(e', -e') \\
  (e', e') \\
  (-e, -e)
 \end{bmatrix}
 = h_0 \cdot
 \begin{bmatrix}
  (e, 0) \\
  (0, e) \\
  (e', 0) \\
  (0, -e')
 \end{bmatrix},
\]
so that $W^{\dagger \bigtriangledown} = X^\square \cdot h_0$ and $W^{\dagger \triangle} = Y^\square \cdot h_0$.
Put $\h_0 = \id \otimes h_0 \in \Sp(\V^\square)$ and
\[
 \mu'(\sigma) = z_{\Y'^\square}(\h_0, \sigma)^{-1} \cdot z_{\Y'^\square}(\h_0 \sigma \h_0^{-1}, \h_0)
\]
for $\sigma \in \Sp(\V^\square)$.
Since $\V^\bigtriangledown = \X'^\square \cdot \h_0$ and $\V^\triangle = \Y'^\square \cdot \h_0$, we have
\begin{equation}
\label{eq:compare-zY'-zV}
 z_{\Y'^\square}(\sigma, \sigma') = z_{\V^\triangle}(\sigma, \sigma')
 \cdot \frac{\mu'(\sigma \sigma')}{\mu'(\sigma) \mu'(\sigma')}
\end{equation}
for $\sigma, \sigma' \in \Sp(\V^\square)$.

As in \S \ref{ss:proof-spl-hodge}, we put $s^{\sharp \prime} = \hat{s}^\sharp \cdot \mu'$ and $s_2' = \hat{s}_2 \cdot \mu'$, and define a map
\[
 s': \cG \longrightarrow \C^1
\]
by setting
\[
 s'(g,h) = \frac{s^{\sharp \prime}(g,h,\alpha,\alpha)}{s'_2(\iota(1, [\alpha,\alpha]))},
\]
where we choose $\alpha \in E^\times$ such that $\nu(g) = \nu(h) = \N_{E/F}(\alpha)$.
As in Lemma \ref{lem:s-well-def}, $s'$ is well-defined.
Moreover, as in Lemma \ref{lem:s-proof(i)}, we have
\[
 z_{\Y'}(\g_1,\g_2) = \frac{s'(\g_1\g_2)}{s'(\g_1) s'(\g_2)}
\]
for $\g_1, \g_2 \in \cG$.

\begin{lem}
We have
\[
 s = s' \cdot \mu_0.
\]
\end{lem}

\begin{proof}
Put $\varsigma_{00} = \iota(\varsigma_0, \varsigma_0) \in \Sp(\V^\square)$ and
\[
 \mu_{00}(\sigma) = z_{\Y'^\square}(\varsigma_{00}, \sigma) \cdot z_{\Y'^\square}(\varsigma_{00} \sigma \varsigma_{00}^{-1}, \varsigma_{00})^{-1}
\]
for $\sigma \in \Sp(\V^\square)$.
Since $\X^\square = \X'^\square \cdot \varsigma_{00}$ and $\Y^\square = \Y'^\square \cdot \varsigma_{00}$, we have
\begin{equation}
\label{eq:compare-zY-zY'}
 z_{\Y^\square}(\sigma, \sigma') = z_{\Y'^\square}(\sigma, \sigma')
 \cdot \frac{\mu_{00}(\sigma \sigma')}{\mu_{00}(\sigma) \mu_{00}(\sigma')}
\end{equation}
for $\sigma, \sigma' \in \Sp(\V^\square)$.
Then it follows from \eqref{eq:compare-zY-zV}, \eqref{eq:compare-zY'-zV}, and \eqref{eq:compare-zY-zY'} that
\[
 \frac{\mu_{00}(\sigma \sigma')}{\mu_{00}(\sigma) \mu_{00}(\sigma')}
 = \frac{\mu(\sigma \sigma')}{\mu(\sigma) \mu(\sigma')}
 \cdot \frac{\mu'(\sigma) \mu'(\sigma')}{\mu'(\sigma \sigma')}
\]
for $\sigma, \sigma' \in \Sp(\V^\square)$.
Namely, $\mu_{00} \cdot \mu' / \mu$ is a character of $\Sp(\V^\square)$.
Since $[\Sp(\V^\square), \Sp(\V^\square)] = \Sp(\V^\square)$, this character must be trivial and hence
\[
 \mu_{00} = \mu/\mu'.
\]
Since
\begin{align*}
 \mu_{00}(\iota(\sigma,1))
 & = z_{\Y'^\square}(\iota(\varsigma_0, \varsigma_0), \iota(\sigma, 1)) \cdot z_{\Y'^\square}(\iota(\varsigma_0 \sigma \varsigma_0^{-1}, 1), \iota(\varsigma_0, \varsigma_0))^{-1} \\
 & = z_{\Y'}(\varsigma_0, \sigma) \cdot z_{\Y'}(\varsigma_0 \sigma \varsigma_0^{-1}, \varsigma_0)^{-1} \\
 & = \mu_0(\sigma)
\end{align*}
for $\sigma \in \Sp(\V)$, it suffices to show that
\[
 \frac{s(\g)}{\mu(\iota(\g, 1))} = \frac{s'(\g)}{\mu'(\iota(\g, 1))}
\]
for $\g \in \cG$.
Here, by abuse of notation, we write $\g$ in the denominator for the image of $\g$ in $\Sp(\V)$ under \eqref{eq:hom-GU(V)-GU(W)}, so that $\iota(\g, 1) \in \Sp(\V^\square)$.

For $\g = (g, h) \in \cG$, choose $\alpha \in E^\times$ such that $\nu(g) = \nu(h) = \N_{E/F}(\alpha)$.
Put $\g^\sharp = (g,h,\alpha,\alpha) \in \cG^\sharp$ and $\ba = [\alpha, \alpha] \in \U(\WW)$.
Note that the image of $\g^\sharp$ in $\Sp(\V^\square)$ agrees with $\iota(\g, 1) \cdot \iota(1,\ba)$.
By definition, we have
\[
 s(\g) = \frac{\hat{s}^\sharp(\g^\sharp)}{\hat{s}_2(\iota(1, \ba))} \cdot \frac{\mu(\g^\sharp)}{\mu(\iota(1, \ba))}, \qquad
 s'(\g) = \frac{\hat{s}^\sharp(\g^\sharp)}{\hat{s}_2(\iota(1, \ba))} \cdot \frac{\mu'(\g^\sharp)}{\mu'(\iota(1, \ba))}.
\]
Thus it remains to show that
\[
 \frac{\mu(\g^\sharp)}{\mu(\iota(\g,1)) \cdot \mu(\iota(1, \ba))}
 = \frac{\mu'(\g^\sharp)}{\mu'(\iota(\g,1)) \cdot \mu'(\iota(1, \ba))}.
\]
But the left-hand side is equal to
\[
 \frac{z_{\Y^\square}(\iota(\g, 1), \iota(1, \ba))}{z_{\V^\triangle}(\iota(\g, 1), \iota(1, \ba))} = \frac{1}{z_{\V^\triangle}(\iota(\g, 1), \iota(1, \ba))}
\]
by \eqref{eq:compare-zY-zV}, whereas the right-hand side is equal to
\[
 \frac{z_{\Y'^\square}(\iota(\g, 1), \iota(1, \ba))}{z_{\V^\triangle}(\iota(\g, 1), \iota(1, \ba))} = \frac{1}{z_{\V^\triangle}(\iota(\g, 1), \iota(1, \ba))}
\]
by \eqref{eq:compare-zY'-zV}.
This completes the proof.
\end{proof}

Thus, to finish the proof of Proposition \ref{prop:compare-s-0-s}, it remains to prove the following.

\begin{lem}
We have
\[
 s^\dagger|_\cG = s'.
\]
\end{lem}

\begin{proof}
Since both $s^\dagger$ and $s'$ trivialize $z_{\Y'}$, there exists a \emph{continuous} character $\Chi'$ of $\cG$ such that
\[
 s^\dagger|_\cG = s' \cdot \Chi'.
\]
We will show that $\Chi'$ is trivial.
Since $[\Sp(W^\dagger), \Sp(W^\dagger)] = \Sp(W^\dagger)$, $\Chi'$ is trivial on $\U(W) \cong \Sp(W^\dagger)$.
Let $\g = (g,1) \in \cG$ with $g \in \U(V)^0 \cong \SO(V^\dagger)$.
By definition, we have $s^\dagger(\g) = 1$ and
\[
 s'(\g) = s^{\sharp \prime}(g,1,1,1) = \hat{s}^\sharp(g,1,1,1) \cdot \mu'(\iota(\g, 1)) = \mu'(\iota(\g, 1)).
\]
Since $\iota(\g, 1)$ belongs to $P_{\Y'^\square}$ and commutes with $\h_0$, we have
\[
 \mu'(\iota(\g, 1)) = z_{\Y'^\square}(\h_0, \iota(\g,1))^{-1} \cdot z_{\Y'^\square}(\iota(\g, 1), \h_0) = 1.
\]
Hence we have $s'(\g) = 1$, so that $\Chi'(\g) = 1$.

Thus it remains to show that
\[
 \Chi'(\g) = 1
\]
for $\g = (\alpha,\alpha) \in \cG$ with $\alpha \in E^\times$.
We write $\alpha = a + b \i$ with $a, b \in F$ and put $\nu = a^2 - b^2 u$.
Since $\Chi'$ is continuous, we may assume that
\[
 a \ne 0, \qquad b \ne 0.
\]
By definition, we have $s^\dagger(\g) = s^\dagger(h)$, where
\[
 h =
 \begin{pmatrix}
  a & b \\
  bu & a
 \end{pmatrix}
 \cdot
 \begin{pmatrix}
  1 & \\
  & \nu^{-1}
 \end{pmatrix}.
\]
Since
\[
 h =
 \begin{pmatrix}
  \frac{1}{bu} & a \\
  & bu
 \end{pmatrix}
 \cdot
 \begin{pmatrix}
  & -1 \\
 1 &
 \end{pmatrix}
 \cdot
 \begin{pmatrix}
  1 & \frac{a}{\nu bu} \\
  & 1
 \end{pmatrix},
\]
we have
\[
 s^\dagger(h) = \xi_E(bu)^m \cdot \gamma'^{-1} = (-b, u)_F^m \cdot \gamma'^{-1}.
\]
Recall that
\begin{align*}
 \gamma'^{-1} & = \gamma_F(\tfrac{1}{2} \psi)^{-2m} \cdot \gamma_F(\det V^\dagger, \tfrac{1}{2} \psi)^{-1} \cdot h_F(V^\dagger) \\
 & = \gamma_F(-1, \tfrac{1}{2} \psi)^m \cdot \gamma_F((-u)^m, \tfrac{1}{2} \psi)^{-1} \cdot h_F(V^\dagger) \\
 & = \gamma_F(-1, \tfrac{1}{2} \psi)^m \cdot \gamma_F(-u, \tfrac{1}{2} \psi)^{-m} \cdot (-u,-u)_F^{(m-1)m/2} \cdot h_F(V^\dagger) \\
 & = \gamma_F(u, \tfrac{1}{2} \psi)^m \cdot (-u,-u)_F^{(m-1)m/2} \cdot h_F(V^\dagger).
\end{align*}
If we put $V_i^\dagger = F e_i e + F e_i e''$, then we have
\begin{align*}
 h_F(V^\dagger) & = \prod_{i=1}^m h_F(V_i^\dagger) \cdot \prod_{1 \le i < j \le m} (\det V_i^\dagger, \det V_j^\dagger)_F \\
 & = \prod_{i=1}^m (- 2\kappa_i, u)_F \cdot (-u, -u)_F^{(m-1)m/2}.
\end{align*}
Hence we have
\[
 s^\dagger(\g) = \gamma_F(u, \tfrac{1}{2} \psi)^m \cdot \prod_{i=1}^m (2 b \kappa_i, u)_F.
\]
On the other hand, by definition, we have
\begin{align*}
 s'(\g) & = \frac{s^{\sharp \prime}(\alpha,\alpha,\alpha,\alpha)}{s_2'(\iota(1, \ba))} \\
 & = \frac{s^{\sharp \prime}(\alpha,\alpha,\alpha,\alpha)}{s_2'(\iota(\ba, \ba))} \cdot \frac{s_2'(\iota(\ba, \ba))}{s_2'(\iota(1, \ba))} \\
 & = \frac{s^{\sharp \prime}(\alpha,\alpha,\alpha,\alpha)}{s_2'(\iota(\ba, \ba))} \cdot s_2'(\iota(\ba, 1)) \cdot z_{\Y'^\square}(\iota(\ba, 1), \iota(1, \ba)) \\
 & = \frac{s^{\sharp \prime}(\alpha,\alpha,\alpha,\alpha)}{s_2'(\iota(\ba, \ba))} \cdot s_2'(\iota(\ba, 1)) \\
 & = \frac{\hat{s}^\sharp(\alpha,\alpha,\alpha,\alpha)}{\hat{s}_2(\iota(\ba, \ba))} \cdot \hat{s}_2(\iota(\ba, 1)) \cdot \mu'(\iota(\ba, 1)),
\end{align*}
where $\ba = [\alpha, \alpha] \in \U(\WW)$.
By definition and Lemma \ref{lem:s-hat-2-E-diag}, we have
\[
 \frac{\hat{s}^\sharp(\alpha,\alpha,\alpha,\alpha)}{\hat{s}_2(\iota(\ba, \ba))} = \chi(\alpha)^m.
\]
Also, as in Lemma \ref{lem:weil-hodge-computation1-2}, we have
\[
 \hat{s}_2(\iota(\ba, 1)) = \chi(J \i (\beta - 1))^m \cdot \gamma^{-1} = \chi(2 b \alpha^{-1})^m \cdot \gamma^{-1},
\]
where $\beta = \alpha^{-1} \alpha^\rho$ and
\begin{align*}
 \gamma^{-1} & = (\det \VV, u)_F \cdot \gamma_F(-u, \tfrac{1}{2} \psi)^{-m} \cdot \gamma_F(-1, \tfrac{1}{2} \psi)^m \\
 & = \gamma(u, \tfrac{1}{2} \psi)^m \cdot \prod_{i=1}^m (\kappa_i, u)_F.
\end{align*}
Hence, noting that the image of $\g$ in $\Sp(\V)$ agrees with that of $\ba$, we have
\[
 s'(\g) = \gamma(u, \tfrac{1}{2} \psi)^m \cdot \prod_{i=1}^m (2b \kappa_i, u)_F \cdot \mu'(\iota(\g, 1)).
\]

Thus we are reduced to showing that
\[
 \mu'(\iota(\g, 1)) = 1
\]
for $\g = (\alpha, \alpha) \in \cG$ with $\alpha = a + b \i \in E^\times$ such that $a \ne 0$ and $b \ne 0$.
This is further reduced to the case $\dim V = 1$.
Then we may identify $V^\dagger$ with the $F$-space $Fe + Fe''$ equipped with a symmetric bilinear form
\[
 \langle x_1 e + x_2 e'', y_1 e + y_2 e'' \rangle^\dagger
 = \kappa u \cdot x_1 y_1 - \kappa \cdot x_2 y_2,
\]
where $\kappa = \kappa_1/2$.
We take a basis
\begin{align*}
 \x_1 & = e \otimes (e, 0), &
 \y_1 & = \frac{1}{\kappa u} \cdot e \otimes (e', 0), \\
 \x_2 & = e'' \otimes (e, 0), &
 \y_2 & = - \frac{1}{\kappa} \cdot e'' \otimes (e', 0), \\
 \x_3 & = e \otimes (0, e), &
 \y_3 & = \frac{1}{\kappa u} \cdot e \otimes (0, -e'), \\
 \x_4 & = e'' \otimes (0, e), &
 \y_4 & = - \frac{1}{\kappa} \cdot e'' \otimes (0, -e')
\end{align*}
of $\V^\square = V^\dagger \otimes_F W^{\dagger \square}$, so that
\[
 \X'^\square = F \x_1 + \dots + F \x_4, \qquad
 \Y'^\square = F \y_1 + \dots + F \y_4, \qquad
 \llangle \x_i, \y_j \rrangle = \delta_{ij}.
\]
Using this basis, we identify $\Sp(\V^\square)$ with $\Sp_8(F)$.
Then we have
\begin{align*}
 \h_0 & =
 \begin{pmatrix}
  \frac{1}{2} & & -\frac{1}{2} & & & & & \\
  & \frac{1}{2} & & -\frac{1}{2} & & & & \\
  & & & & \frac{\kappa u}{2} & & \frac{\kappa u}{2} & \\
  & & & & & -\frac{\kappa}{2} & & -\frac{\kappa}{2} \\
  & & & & 1 & & -1 & \\
  & & & & & 1 & & -1 \\
  -\frac{1}{\kappa u} & & -\frac{1}{\kappa u} & & & & & \\
  & \frac{1}{\kappa} & & \frac{1}{\kappa} & & & &
 \end{pmatrix}, \\
 \iota(\g, 1) & =
 \begin{pmatrix}
  \frac{1}{2}(c+1) & \frac{du}{2} & & & -\frac{d\kappa u}{2} & \frac{\kappa}{2}(c-1) & & \\
  \frac{d}{2} & \frac{1}{2}(c+1) & & & - \frac{\kappa}{2}(c-1) & \frac{d\kappa}{2} & & \\
  & & 1 & & & & & \\
  & & & 1 & & & & \\
  -\frac{d}{2\kappa} & -\frac{1}{2\kappa}(c-1) & & & \frac{1}{2}(c+1) & -\frac{d}{2} & & \\
  \frac{1}{2\kappa}(c-1) & \frac{du}{2\kappa} & & & -\frac{du}{2} & \frac{1}{2}(c+1) & & \\
  & & & & & & 1 & \\
  & & & & & & & 1
 \end{pmatrix},
\end{align*}
where
\[
 c = \frac{a^2 + b^2 u}{a^2 - b^2 u} \ne \pm 1, \qquad
 d = - \frac{2ab}{a^2 - b^2 u} \ne 0,
\]
so that $c^2 - d^2 u = 1$.
Recall that
\[
 \mu'(\iota(\g, 1)) = z_{\Y'^\square}(\h_0, \iota(\g, 1))^{-1} \cdot z_{\Y'^\square}(\h_0 \cdot \iota(\g, 1) \cdot \h_0^{-1}, \h_0).
\]
Since $\h_0 \in P_{\Y'^\square} \cdot \tau_2 \cdot \m(\a_1)$ and $\iota(\g, 1) = \n(\b_1) \cdot \tau \cdot P_{\Y'^\square}$, where
\[
 \a_1 =
 \begin{pmatrix}
  1 & & & \\
  & 1 & & \\
  1 & & 1 & \\
  & 1 & & 1
 \end{pmatrix}, \qquad
 \b_1 = \frac{d \kappa}{c-1} \cdot
 \begin{pmatrix}
  -u & & & \\
  & 1 & & \\
  & & 0 & \\
  & & & 0
 \end{pmatrix}, \qquad
 \tau =
 \begin{pmatrix}
  & & -\1_2 & \\
  & \1_2 & & \\
  \1_2 & & & \\
  & & & \1_2
 \end{pmatrix},
\]
we have
\begin{align*}
 z_{\Y'^\square}(\h_0, \iota(\g, 1))
 & = z_{\Y'^\square}(\tau_2 \cdot \m(\a_1), \n(\b_1) \cdot \tau) \\
 & = z_{\Y'^\square}(\tau_2, \m(\a_1) \cdot \n(\b_1) \cdot \tau).
\end{align*}
If we put
\[
 \b_2 = \frac{d \kappa}{c-1} \cdot
 \begin{pmatrix}
  -u & & -u & \\
  & 1 & & 1 \\
  -u & & & \\
  & 1 & &
 \end{pmatrix}, \qquad
 \b_3 = \frac{d \kappa}{c-1} \cdot
 \begin{pmatrix}
  0 & & & \\
  & 0 & & \\
  & & -u & \\
  & & & 1
 \end{pmatrix},
\]
then we have $\m(\a_1) \cdot \n(\b_1) = \n(\b_2) \cdot \n(\b_3) \cdot \m(\a_1)$ and hence
\[
 z_{\Y'^\square}(\h_0, \iota(\g, 1)) = z_{\Y'^\square}(\tau_2 \cdot \n(\b_2), \n(\b_3) \cdot \m(\a_1) \cdot \tau).
\]
Since $\tau_2 \cdot \n(\b_2) \in P_{\Y'^\square} \cdot \tau_2$ and $\n(\b_3) \cdot \m(\a_1) \cdot \tau \in \tau \cdot P_{\Y'^\square}$, we have
\[
 z_{\Y'^\square}(\h_0, \iota(\g, 1)) = z_{\Y'^\square}(\tau_2, \tau) = 1.
\]
On the other hand, we have $\h_0 \in \tau_2 \cdot P_{\Y'^\square}$ and
\begin{align*}
 & \h_0 \cdot \iota(\g, 1) \cdot \h_0^{-1} \\
 & =
\begin{pmatrix}
 \frac{1}{4}(c+3) & \frac{du}{4} & -\frac{d}{4} & -\frac{1}{4}(c-1) & -\frac{d \kappa u}{8} & \frac{\kappa}{8}(c-1) & -\frac{\kappa u}{8}(c-1) & \frac{d \kappa u}{8} \\
 \frac{d}{4} & \frac{1}{4}(c+3) & -\frac{1}{4u}(c-1) & -\frac{d}{4} & -\frac{\kappa}{8}(c-1) & \frac{d \kappa}{8} & -\frac{d \kappa u}{8} & \frac{\kappa}{8}(c-1) \\
 -\frac{du}{4} & -\frac{u}{4}(c-1) & \frac{1}{4}(c+3) & \frac{du}{4} & \frac{\kappa u}{8}(c-1) & -\frac{d \kappa u}{8} & \frac{d \kappa u^2}{8} & -\frac{\kappa u}{8}(c-1) \\
 -\frac{1}{4}(c-1) & -\frac{du}{4} & \frac{d}{4} & \frac{1}{4}(c+3) & \frac{d \kappa u}{8} & -\frac{\kappa}{8}(c-1) & \frac{\kappa u}{8}(c-1) & -\frac{d \kappa u}{8} \\
 -\frac{d}{2 \kappa} & -\frac{1}{2 \kappa}(c-1) & \frac{1}{2 \kappa u}(c-1) & \frac{d}{2 \kappa} & \frac{1}{4}(c+3) & -\frac{d}{4} & \frac{du}{4} & -\frac{1}{4}(c-1) \\
 \frac{1}{2 \kappa}(c-1) & \frac{du}{2 \kappa} & -\frac{d}{2 \kappa} & -\frac{1}{2 \kappa}(c-1) & -\frac{du}{4} & \frac{1}{4}(c+3) & -\frac{u}{4}(c-1) & \frac{du}{4} \\
 -\frac{1}{2 \kappa u}(c-1) & -\frac{d}{2 \kappa} & \frac{d}{2 \kappa u} & \frac{1}{2 \kappa u}(c-1) & \frac{d}{4} & -\frac{1}{4 u}(c-1) & \frac{1}{4}(c+3) & -\frac{d}{4} \\
 \frac{d}{2 \kappa} & \frac{1}{2 \kappa}(c-1) & -\frac{1}{2 \kappa u}(c-1) & -\frac{d}{2 \kappa} & -\frac{1}{4}(c-1) & \frac{d}{4} & -\frac{du}{4} & \frac{1}{4}(c+3)
\end{pmatrix} \\
 & \in P_{\Y'^\square} \cdot \tau \cdot \m(\a_2) \cdot \n(\b_4),
\end{align*}
where
\[
 \a_2 =
 \begin{pmatrix}
  1 & & & -1 \\
  & 1 & -\frac{1}{u} & \\
  & & 1 & \\
  & & & 1
 \end{pmatrix}, \qquad
 \b_4 = \frac{(c+1)\kappa}{d} \cdot
 \begin{pmatrix}
  0 & & & \\
  & 0 & & \\
  & & u & \\
  & & & -1
 \end{pmatrix}.
\]
Since $\tau \cdot \n(\b_4) \in P_{\Y'^\square} \cdot \tau$ and $\n(\b_4)^{-1} \cdot \m(\a_2) \cdot \n(\b_4) \cdot \tau_2 \in \tau_2 \cdot P_{\Y'^\square}$, we have
\begin{align*}
 z_{\Y'^\square}(\h_0 \cdot \iota(\g, 1) \cdot \h_0^{-1}, \h_0)
 & = z_{\Y'^\square}(\tau \cdot \m(\a_2) \cdot \n(\b_4), \tau_2) \\
 & = z_{\Y'^\square}(\tau \cdot \n(\b_4), \n(\b_4)^{-1} \cdot \m(\a_2) \cdot \n(\b_4) \cdot \tau_2) \\
 & = z_{\Y'^\square}(\tau, \tau_2) \\
 & = 1.
\end{align*}
This completes the proof.
\end{proof}

config: arxiv


TITLE: Inequality : $\sum_{cyc}\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}\geq \frac{3}{5}$
QUESTION [2 upvotes]: Let $a$, $b$ and $c$ be positive real numbers. Prove that:
  $$\frac{\sqrt{a^3c}}{2\sqrt{b^3a}+3bc}+\frac{\sqrt{b^3a}}{2\sqrt{c^3b}+3ca}+\frac{\sqrt{c^3b}}{2\sqrt{a^3c}+3ab}\geq \frac{3}{5}$$

My attempt :
Substitute $a=x^2, b=y^2, c=z^2$, we have to prove that
$\displaystyle\sum_{cyc}\frac{x^3z}{2y^3x+3y^2z^2} \geq \frac{3}{5}$
By C-S,
$\left(\displaystyle\sum_{cyc}\frac{x^3z}{2y^3x+3y^2z^2}\right)\left(\displaystyle\sum_{cyc}(2y^2x+3y^2z^2)\frac{x}{z}\right) \geq \left(\displaystyle\sum_{cyc}x^2\right)^2$
$\left(\displaystyle\sum_{cyc}\frac{x^3z}{2y^3x+3y^2z^2}\right)\left(\displaystyle\sum_{cyc}(x^2y^2+3y^2xz\right) \geq \left(\displaystyle\sum_{cyc}x^2\right)^2$

REPLY [2 votes]: Let $a = x^2, b = y^2, c=z^2$, where $x, y, z > 0$. By C-S, we have
$$ \sum_{cyc} \frac{x^3 z}{2 y^3 x + 3y^2z^2} \sum_{cyc} \frac{2yx + 3z^2}{zx} \geq \left( \sum_{cyc} \frac{x}{y} \right)^2.$$
But
$$ \sum_{cyc} \frac{2yx + 3z^2}{zx} = 5 \sum_{cyc} \frac{x}{y}.$$
Combining them we get
$$ \sum_{cyc} \frac{x^3 z}{2 y^3 x + 3y^2z^2} \geq \frac{1}{5} \sum_{cyc} \frac{x}{y}.$$
It remains to apply AM-GM.


TITLE: If the extended Riemann integral of f doesn't exist, then its possible to make a rearrengement
QUESTION [0 upvotes]: Set f as a continous function in $\Omega$, where f has positive and negative values. Proof that if the extended integral of f doesn't exist, then for each $\alpha\in\mathbb{R}$, it is possible to make a sequence of close bounded sets such as $U_n^\alpha\subset\Omega$ and $\bigcup_{n\in\mathbb{N}}U_n^\alpha$ generates $\Omega$, so that:
$$\lim_{n\to\infty}\int_{U_n^\alpha}f=\alpha$$
I have tried that, as f is not integrable in an extended way, then $\exists\{U_n\}_{n\in\mathbb{N}}$ a sequence of open sets so that $\bigcup_{n\in\mathbb{N}}U_n^\alpha=\Omega$ and $U_n\subset U_{n+1}$ that makes $\{\int_{U_n}|f|\}_{n\in\mathbb{N}}$ diverge. Then, as f gets negative and positive values, we can make $f_+\;,\;f_-$. As $|f|=f_++f_-$, then one of this functions has to diverge. Lets define $C_k:=U_n-(\bigcup_{i=1}^{n-1}U_i)$.
Look that:
$$\int_{U_n}f_+=\sum_{i=1}^n\int_{C_i}f_+$$
That's all I got, I don't know how to prove that this series is conditionally convergent, so that we can make the rearrengement.

REPLY [1 votes]: Here is a sketch of your ideas in a simple setting.  Please explain if this is what you intended (more or less).
Suppose $f:[0,\infty)\to\mathbb R$ is continuous.
Define the open sets
$G_1=\{x\in [0,\infty): f(x)>0\}$,  $G_2=\{x\in [0,\infty): f(x)<0\}$ and the closed set
$E =  \{x\in [0,\infty): f(x)=0\}$.
Since  $G_1$ is open,  we can write it as a union of its component intervals $\{(a_i,b_i)\}$.  Similarly $G_2$ is open and we can write it as a union of its component intervals $\{(c_j,d_j)\}$.  The set  $E$ is closed and $\int_E f(x)\,dx=0.$
We will have to assume${}^1$ that all component intervals of $G_1$ and $G_2$ are bounded.
We will have to assume${}^2$  that
$$\int_{G_1} f(x)\,dx = \infty \text{ and  } \int_{G_2} f(x)\,dx =- \infty .$$
That will guarantee that there are infinitely many component intervals for each of the sets and that
$$0 < \alpha_i =\int_{(a_i,b_i)} f(x)\,dx < \infty$$
and
$$0 > \beta_j =\int_{(c_j,d_j)} f(x)\,dx > - \infty$$
but that  $$\sum_i \alpha_i =\infty \text{ and  }\sum_j \beta_j =-\infty.$$
This is still not enough for your purposes so assume${}^3$  also that $\alpha_i\to 0$ and $\beta_j\to 0$.
Finally with all of these assumptions${}^{1,2,3}$  you appear to be in the situation that (I think) you desire.  Namely take any real number $r$ and rearrange the sums of the numbers $\alpha_i$ and the $\beta_j$ into a series that sums to $r$.
This appears to show that
$$\int_0^\infty f(x)\,dx= \int_E f(x)\,dx + \int_{G_1 \cup G_2}f(x)\,dx    = r$$
if we are allowed to rearrange the order of taking the components in the two open sets.
Final question?  Why?


text:
\section{Introduction}
\label{sect:introduction}

A challenging aspect of tactical aircraft trajectory planning\footnote{By \textit{tactical} we refer to 10 minutes to 1 hour prior to the potential hazard encounter.} is the avoidance of hazardous convective weather regions, more commonly referred to as storms or thunderstorms\footnote{Throughout this paper we will use the more general term \textit{thunderstorm} to describe convective weather regions.}. The inherently uncertain nature of thunderstorms causes major safety risks. Strong conflicting up- and downdrafts lead to heavy turbulence. Hail, severe icing and lightning can also inflict significant damage to aircraft equipment and windshields. In addition to increased safety risks, thunderstorms are also a leading cause of reduced time (\cite{NationalResearchCouncil}) and cost (\cite{Michalek2009a}) efficiencies. From 2008 to 2013, inclement weather caused 69\% of system-impacting delays (delays greater than  15 minutes), as recorded in the \textit{OPSNET Standard ``Delay by Cause" Reports}\footnote{{NextGen Weather - Weather Delays} \url{https://www.faa.gov/nextgen/programs/weather/faq/}}. Within those weather delays, thunderstorms emerging from atmospheric instabilities were responsible for around 25\%, turning them into the leading cause of flight delays in the US airspace.

Today, commonly used approaches to plan trajectories in the presence of thunderstorms are suboptimal and prone to human error. Thunderstorms are typically detected by satellite images or weather radar signals, which are processed into graphical images. The identified \textit{thunderstorm cells} are then classified according to severity. Regions of airspace that are considered to pose a significant safety hazard are recommended as unsafe-to-fly regions to air traffic controllers (ATCO) and pilots (\cite{Krozel2007}).

In many cases these regions are avoided by manual trajectory modifications. This involves increased workload and potential loss of situational awareness for both ATCOs and pilots, jeopardizing safety and air traffic capacity.

One of the key concepts of future Air Traffic Management (ATM) systems is the so-called \textit{trajectory-based operations} (TBO) described in the SESAR (\cite{sesar}) or NextGen (\cite{nextgen}) ATM master plans. The TBO concept will give aircraft the autonomy to update their trajectories to satisfy business needs, such as fuel or flight time savings, or to avoid potential conflicts or hazardous weather regions. Robust planning that accounts for the uncertainty associated to aircraft motion and thunderstorm development is paramount for a successful transition to TBO.

  In \cite{mcnally2015dynamic}, the authors reported two years of operational testing of a tool coined {\it Dynamic Weather Routes} at American Airlines. The tool is a ground-based real-time trajectory automation system that continuously analyzes en-route aircraft. It computes simple corrections to flight plans to avoid thunderstorms while ensuring conflict avoidance, respecting airspace constraints and minimizing delay. However, the tool neither accounts for weather uncertainty, nor does it use advanced control and optimization algorithms, thus leaving room for improvement in the system's efficiency, safety, and capacity.

The lack of probabilistic forecast data is one of the main bottlenecks in the development and full-scale deployment of automated trajectory planning tools that account for thunderstorm uncertainty. Today's state-of-the-art meteorological forecast products include current observations of the thunderstorm cells as well as predictions for their locations at discrete forecast horizons. A commonly used approach relies on deterministic short term forecasts, so-called nowcasts. They provide information on thunderstorm cell positions for up to a few hours ahead (\cite{wilson1998nowcasting}). Examples for such nowcast systems include the satellite data-based cloud tracker Cb-TRAM (Cumulonimbus Tracking and Monitoring) \cite{zinner2008cb}; and the radar-based Rad-TRAM (Radar Tracking and Monitoring), developed by \cite{kober2009tracking}.
Nevertheless, these forecast products do not provide any uncertainty quantification.

The integration of stochastic information in meteorological forecast products has indeed proven difficult, mainly due to the short lifespan of convective phenomena (around 30 minutes) and an imperfect knowledge of the atmosphere at macroscopic scale. Recent efforts in this direction include, e.g., \cite{mecikalski2015probabilistic, osinski2017short}. The former presented two statistical learning approaches that incorporate numerical weather prediction (NWP) input within satellite images to produce probabilistic forecasts. \cite{osinski2017short} presented an experimental product developed by Meteo-France. It combines the numerical weather prediction model AROME-NWC with simulated radar images and post-processing. An ensemble of time-lagged forecasts with thunderstorm development information is then produced. Besides these efforts in combining data-driven methods and NWP, pure data-driven methods can be also used to model uncertainties in forecast products through comparison of actual instances of thunderstorms and their predictions.

The authors in \cite{Radhakrishna2012} investigated statistical properties of thunderstorm cells' growth and decay, as well as their scale dependence and predictability. They concluded that forecast accuracy of deterministic, radar-based nowcasts can be improved by the extrapolation of growth and decay for a few hours.
In \cite{Sauer2014}, the authors analyzed the error of radar-based nowcasts and their increasing uncertainty with time by determining spatial deviations of the nowcast predictions compared to the observed thunderstorm realizations. The authors argued that thunderstorm cells could be enlarged by an uncertainty margin that is dependent on the desired probability of avoiding the respective thunderstorm cell.
In \cite{zhang2013probabilistic,zhang2017probabilistic}, the authors introduced a probabilistic analysis of weather forecasting. The analysis is based on ensemble weather forecasting data provided by the Center for Analysis and Prediction of Storm (CAPS) \cite{jung2008assimilation}.
Additionally, \cite{zhang2017probabilistic} presented a path planning algorithm to avoid the stochastic hazard, including a mission risk analysis. Nevertheless, additional efforts are needed in stochastic thunderstorm modeling and aircraft trajectory planning, e.g., considering the additional uncertainty introduced by the aircraft dynamic system.

Several methods to address aircraft path planning subject to uncertainties in system dynamics and obstacles (such as unsafe-to-fly regions) have been proposed. The problem of routing an aircraft to a target while avoiding uncertain hazardous weather zones in the presence of wind disturbance can be cast as the optimal control of a stochastic dynamic system through probabilistic obstacles, such as unsafe-to-fly regions. The following challenges arise: First, both the aircraft dynamics and the thunderstorm cells are nonlinear and non-convex. Second, the use of stochastic models is required, resulting in a theoretically and computationally challenging \textit{stochastic optimal control} problem.

A possible approach is to bound the uncertain obstacles with deterministic sets.
The authors in \cite{Kuwata2005} presented a receding horizon control strategy for aircraft trajectory planning in environments with unknown but bounded disturbances. The authors in \cite{Kamgarpour2010a} proposed a receding horizon framework for designing safe aircraft trajectories. They took into account a dynamic weather forecast product to determine deterministic bounds for the thunderstorm regions to be avoided. While this set-bounded approach offers a high degree of safety, the solutions often degrade overall performance due to high conservatism arising from the protection against low probability uncertainty outliers. More importantly, the optimization problem can quickly become infeasible, for example in a scenario featuring a high number of  obstacles. Such scenarios call for a solution that minimizes the risk of failure, instead of trying to eliminate it entirely.
In \cite{Liu2014}, a stochastic optimal control framework for mid-air conflict resolution was presented that could incorporate uncertainties in both aircraft and wind dynamics. However, in the studied aircraft-weather conflict, the weather hazard was considered deterministic. In \cite{Nilim2002}, the authors addressed the problem of generating aircraft trajectories using a Markov decision problem where the evolution of the thunderstorms was modeled as a Markov chain. The transition probabilities were extracted from historical data using maximum-likelihood estimators. However, these probabilities were assumed stationary. Moreover, delay was used as the objective to be minimized. Given the uncertainties in thunderstorm development, an important objective is to minimize the probability of intersecting a hazardous weather region.

A suitable approach to generate trajectories that reach a target set while minimizing the probability of intersecting probabilistic unsafe-to-fly regions is to formulate a \textit{stochastic reach-avoid problem}. The authors in \cite{Summers2011} considered a stochastic reach-avoid problem with a time-varying stochastic obstacle set. The  problem was reformulated as a finite horizon stochastic optimal control problem and solved with Bellman's Dynamic Programming principle. We proposed this method as a feasible approach to aircraft trajectory planning in the presence of a single hypothetical thunderstorm, modeled by a stochastic ellipse \cite{Hentzen2017}.  However, in this preliminary study, we assumed a priori knowledge of the stochastic unsafe-to-fly regions. No forecast data was incorporated to define stochastic unsafe-to-fly regions. Furthermore, the approach did not address trajectory planning in the presence of multiple thunderstorm regions.

The contribution of the paper is twofold: First, we propose a method for modeling thunderstorm in a stochastic manner. The method accounts for uncertainty in both the movement and, crucially, in the growth or decay of the thunderstorm cells. It is based on statistical analysis of  historical nowcast data and its comparison with actual thunderstorm observations. Second, we combine the developed stochastic thunderstorm model with an optimal trajectory planning algorithm based on the stochastic reach-avoid methodology by \cite{Summers2011}. The resulting aircraft trajectories  maximize the probability of reaching a given waypoint while avoiding multiple thunderstorms, taking into account uncertainties in both the system dynamics (due to wind disturbance) and the unsafe-to-fly regions (due to uncertainties in nowcast data on thunderstorms). The resulting trajectories are validated against actual realizations of thunderstorms.

The paper is structured as follows: The proposed stochastic thunderstorm model derived from nowcast data is presented in Section~\ref{sect:stoch_mod}. The stochastic reach-avoid framework, in which the aircraft trajectory generation is cast as a stochastic optimal control problem, is presented in Section~\ref{sect:stoch_reach}. The trajectory planner, resulting from the integration of the stochastic thunderstorm model and the optimal control algorithm into one framework, is applied to two flight scenarios using historical forecast data in Section~\ref{sect:case_studies}. Conclusions and future research directions are discussed in Section~\ref{sect:conclusion}.


TITLE: In a field of characteristic 0, for any integer $m$ and an element $x$, does there exist another element $y$ that $ym=x$?
QUESTION [0 upvotes]: As the title.
Or rather, for any integer $m$ which is not the characteristic, does such an 'integer division' exist?

REPLY [1 votes]: The crucial fact is the following: if $k$ is a field of characteristic zero, there is a natural embedding $e$ of $\mathbb{Q}$ into $k$.
To prove this, first note that subring of $k$ generated by $1_k$ is isomorphic to $\mathbb{Z}$ (since the characteristic is zero), yielding an embedding $d$ of $\mathbb{Z}$ into $k$. Now we can extend this embedding of $\mathbb{Z}$ to an embedding $e$ of $\mathbb{Q}$ as desired: $e({p\over q})=d(p)\times d(q)^{-1}$.
Division by an integer $n$ in $k$ is then conducted by multiplying by $e({1\over n}$).

Exercise: by a similar argument, embed $\mathbb{Z}/p\mathbb{Z}$ into any field of characteristic $p$.


TITLE: Is this an even or odd function?
QUESTION [0 upvotes]: The corrected exercise from my tutor states that the following is an odd function. I am pretty sure this is not true.
$$f: [-1,3] \rightarrow \mathbb{R}: x \mapsto \begin{equation}
  =\left\{
  \begin{array}{@{}ll@{}}
    x, & \text{if}\ x \in [-1,1) \\
    0, & \text{if}\ x \in [1,3)
  \end{array}\right.
\end{equation} $$
Clearly for $ -1=x \in D$ we have $f(-x)=f(-(-1))=f(1)=0\neq -f(x)=-f(-1)=1$
Thus f is not odd.
Also $f(-x)=f(-(-1))=f(1)=0\neq f(x)=f(-1)=-1$
This f is not even.
Is this assumption correct? The professor stated that f is uneven.

REPLY [0 votes]: For the reasons given in the question, the function is neither even nor odd.