Picard group of symplectic group modulo orthogonal group Let $Sp(2n)$ be the group of complex symplectic $2n\times 2n$ matrices, and $O(2n)$ the group of complex orthogonal $2n\times 2n$ matrices.
Consider $Sp(2n)\cap O(2n)\subset Sp(2n)$ and the quotient $X=Sp(2n)/(Sp(2n)\cap O(2n))$. How could one compute the Picard group of $X$?
EDIT. Consider the action of $Sp(2n)$ on the projective space $\mathbb{P}^N$ of $2n\times 2n$ matrices modulo scalar given by $Sp(2n)\times\mathbb{P}^N\rightarrow\mathbb{P}^N$, $(A,Z)\mapsto AZA^t$. The stabilizer $H$ of the identity is then given by those matrices in $Sp(2n)$ such that $AA^t = \lambda I$ for some $\lambda\in\mathbb{C}^{*}$.  
Let $X = Sp(2n)/H$ be the orbit of the identity in $\mathbb{P}^N$.
How could one compute the Picard group of $X$?
Consider for instance the case $n = 1$. Since any $2\times 2$ symmetric matrix with non-zero determinant has a multiple that is symplectic the orbit $X$ is $\mathbb{P}^2\setminus C$ where $C\subset\mathbb{P}^2$ is the conic parametrizing matrices with zero determinant. So, in this case, $Pic(X) \cong \mathbb{Z}/2\mathbb{Z}$.
 A: With the suggested choice of the symplectic and orthogonal form, there is a direct sum decomposition of $\mathbb{C}^{2n}$ into the sum of two Lagrangian (with respect to the both forms) subspaces:
$$
L_1 = \langle e_k + ie_{n+k} \rangle_{k=1}^n,
\qquad 
L_2 = \langle e_k - ie_{n+k} \rangle_{k=1}^n.
$$ 
Moreover, the the pairings between $L_1$ and $L_2$ induced by the both forms are proportional. Therefore
$$
\mathrm{Sp}(2n) \cap \mathrm{O}(2n) \cong \mathrm{GL}_n
$$
which acts on $L_1 \oplus L_2$ by $A \mapsto (A,A^{-1})$.
Using this, it is easy to see that
$$
X = \mathrm{LGr}(2n) \times \mathrm{LGr}(2n) \setminus D,
$$
where $\mathrm{LGr}(2n)$ is the Lagrangian Grassmannian for the symplectic form, and $D \subset \mathrm{LGr}(2n) \times \mathrm{LGr}(2n)$ parameterizes pairs of intersecting Lagrangian subspaces. It is well known that $\mathrm{Pic}(\mathrm{LGr}(2n)) = \mathbb{Z}$ and it is easy to see that $D$ is a divisor of bidegree $(1,1)$. Therefore, $\mathrm{Pic}(X) = \mathbb{Z}$. 
A: 
Answer: ${\rm Pic\,} X={\Bbb Z}/2{\Bbb Z}$; see Corollary 4 below.
Theorem 1. Let $G$ be a simply connected semisimple group over a field $k$ of characteristic 0.
  Let $H\subset G$ be an algebraic subgroup defined over $k$, not necessarily connected. Set $X=G/H$.
  Then there is a canonical isomorphism ${\rm Pic\,} X={\widehat H}(k)$, where ${\widehat H}(k) ={\rm Hom}_k(H,{\Bbb G}_{m})$
  is the character group of $H$.

Proof. First assume that $H$ is connected. We deduce the theorem from results of the paper
J.-J. Sansuc, Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres.
J. Reine Angew. Math. 327 (1981), 12–80.
By Proposition 6.10 of this paper, there is a natural exact sequence of abelian groups
$${\widehat G}(k)\to {\widehat H}(k)\to{\rm Pic\,} X\to {\rm Pic\,} G.$$
Clearly we have ${\widehat G}(k)=0$. By Sansuc's Lemma 6.9(iv), we have ${\rm Pic\,} G=0$ (here Sansuc refers to a paper by Fossum and Iversen).
We obtain an isomorphism ${\widehat H}(k)= {\rm Pic\,} X$, as required.
Now we do not assume that $H$ is connected. We deduce Theorem 1 from a general result of
M. Borovoi and J. van Hamel, Extended equivariant Picard complexes and homogeneous spaces. Transform. Groups 17 (2012), 51-86.
Since  ${\rm Pic\,} G_{\bar k}=0$ and  $X$ has $k$-points, by Theorem 2 (Theorem 7.1) of this paper there is a canonical isomorphism
$$ {\rm Pic\,} X=H^1(k,[{\widehat G}({\bar k})\to {\widehat H}({\bar k})\rangle).$$
Here ${\bar k}$ is an algebraic closure of $k$, ${\widehat H}({\bar k})={\rm Hom}_{\bar k}(H,{\Bbb G}_{m})$, and similarly for ${\widehat G}({\bar k})$.
Further, $[{\widehat G}({\bar k})\to {\widehat H}({\bar k})\rangle$ denotes the complex of ${\rm Gal}({\bar k}/k)$-modules
$$\dots \to 0\to {\widehat G}({\bar k})\to {\widehat H}({\bar k})\to 0\to \dots$$
with ${\widehat H}({\bar k})$ in degree 1, and $H^1(k,[{\widehat G}({\bar k})\to {\widehat H}({\bar k})\rangle)$ denotes the first Galois hypercohomology of this complex.
In our case ${\widehat G}({\bar k})=0$, and therefore,
$$  {\rm Pic\,} X=H^1(k,[0\to {\widehat H}({\bar k})\rangle)=H^0(k,{\widehat H}({\bar k}))={\widehat H}(k),$$
as required.
This looks like  killing a fly with a bazooka, and there should be an elementary proof of Theorem 1.
Construction 2.
The class in ${\rm Pic\,} X$  corresponding to a character
$$\chi\colon H\to{\Bbb G}_m$$
is described as follows. We consider the direct product $G\times {\Bbb G}_m$ and the injective homomorphism
$$\iota_\chi\colon H\to G\times {\Bbb G}_m,\quad  h\mapsto (h,\chi(h)).$$
Further, we consider the quotient $Y_\chi:=(G\times {\Bbb G}_m)/\iota_\chi(H)$ and the projection map
\begin{gather*}\pi\colon\, Y_\chi=(G\times {\Bbb G}_m)/\iota_\chi(H)\,\longrightarrow\, G/H=X,\quad \\
[g,c]\,\mapsto\, [g]\quad \text{for }g\in G,\  c\in{\Bbb C}^\times.\end{gather*}
The group ${\Bbb G}_m$ acts on the fibers of $\pi$  by $c'\cdot [g,c]=[g,c'c]$  for $c'\in{\Bbb C}^\times$.
We see that $\pi\colon Y_\chi\to X$ is a principal ${\Bbb G}_m$-bundle over $X$.
To $\chi$ we associate the class of  $Y_\chi$ in ${\rm Pic\,} X$.
We compute  the character group $\widehat H$ of the stabilizer $H={\rm Sp}(2n)\cap{\rm GO}(2n)$, where
$$ {\rm GO}(2n)=\{A\in{\rm GL}(2n,{\Bbb C})\mid A^t A=\lambda_A I,\ \lambda_A\in{\Bbb C}^\times\}.$$

Proposition 3. For  $H={\rm Sp}(2n)\cap{\rm GO}(2n)$ 
  we have ${\widehat H}={\Bbb Z}/2{\Bbb Z}$.

Proof. 
We compute the group $H$.
We write the equations for $A\in H$:
$$ 
A^t A =\lambda_A I,\qquad A^t J A=J, \qquad\text{where } J=
\begin{pmatrix} 0 & I_n\\ -I_n &0 \end{pmatrix}.
$$
We obtain
$$\lambda_A A^{-1} J A=J, \quad\text{whence } \lambda_A J A=AJ.$$
Let $x$ be an eigenvector of $J$ with eigenvalue $\mu$.
Then 
$$ Jx=\mu x,$$
whence
$$AJx=\mu Ax,\qquad \lambda_A JAx=\mu Ax,\qquad Jy=\lambda_A^{-1} \mu y, \text{ where }y=Ax.$$ 
We see that $y$ is an eigenvector of the matrix $J$ with eigenvalue $\lambda_A^{-1}$.
Thus  $\lambda_A^{-1}\mu$ is an eigenvalue of $J$ as well.
Since our matrix $J$ has only two eigenvalues $i$ and $-i$, we conclude that $\lambda_A$ can take values only $1$ and $-1$. Thus we obtain a homomorphism
$$\lambda\colon H\to \mu_2,\quad A\mapsto \lambda_A.$$
Consider the matrix 
$$ S=i\begin{pmatrix} 0 & I_n \\ I_n & 0\end{pmatrix}. $$
An easy calsulation shows that
$$ S^t S=S^2=-I,\qquad S^t J S=SJS=J.$$
Thus $S\in H$, $\lambda_S=-1$.
We obtain a short exact sequence
$$ 1\to H_1\to H\to \mu_2\to 1,$$
where $H_1={\rm Sp}(2n)\cap{\rm SO}(2,n)$ and where 
the homomorphism $\lambda\colon H\to\mu_2$ is surjective because $\lambda_S=-1$.
We have $H=H_1\cup S\cdot H_1$.
The group $H_1$ was computed by Sasha in his answer: 
it is isomorphic to ${\rm GL}(n,{\Bbb C})$ acting on $V=L_1\oplus L_2$ by
$B\mapsto (B,B^{-1})$. The linear operator $S$ permutes the subspaces $L_1$ and $L_2$, and it acts on the normal subgroup $H_1$ of $H$ as follows:
$$ S\cdot (B,B^{-1}) \cdot S^{-1}=(B^{-1},B).$$
Hence
$$ S\cdot (B,B^{-1}) \cdot S^{-1}\cdot (B,B^{-1})^{-1}=(B^{-2},B^2).$$
It follows that the commutator subgroup $(H,H)$ of $H$ is $H_1$.
Thus 
$${\widehat H}=\widehat{H/H_1}=\widehat{\mu_2}={\Bbb Z}/2{\Bbb Z},$$
as required. The nontrivial element of the character group ${\widehat H}$ is the character 
$$\lambda\colon H\to \mu_2\hookrightarrow{\Bbb G}_m,\quad 
A\mapsto \lambda_A\in {\Bbb C}^\times.$$

Corollary 4. For $X={\rm Sp}(2n)/({\rm Sp}(2n)\cap {\rm GO}(2n))$ we have ${\rm Pic\,} X={\Bbb Z}/2{\Bbb Z}$.

