= Solution
In weight one, the <Hodge numbers> are $h^{1,0}=h^{0,1}=r$, and the <Hodge filtration> is determined by the $r$-plane $H=F^1$. The bilinear orthogonality condition is $\psi(H,H)=0$. An isotropic $r$-plane in a symplectic $2r$-space is a <Lagrangian subspace>, so
$$
\boxed{\check D=\operatorname{SpGr}(r,2r)
=\operatorname{LGr}(\mathbb C^{2r},\psi).}
$$
The ordinary <Grassmannian> has dimension $r^2$. Restriction of the alternating form to an $r$-plane imposes $r(r-1)/2$ independent conditions. Equivalently, near a fixed Lagrangian plane, transverse Lagrangians are graphs of <symmetric matrices>, with $r(r+1)/2$ independent entries. Therefore
$$
\boxed{\dim_{\mathbb C}\check D=\frac{r(r+1)}2.}
$$
Write $\mathbb C^{2r}=\mathbb C^r\oplus\mathbb C^r$ in the block order of the given matrix. On $D$, positivity is $i\psi(v,\bar v)>0$ for every nonzero $v\in H$. The plane $H$ cannot contain $(x,0)\ne0$, because that vector has $\psi(v,\bar v)=0$. Projection onto the second coordinate factor is therefore <injective> on $H$ and, by equal dimensions, an <isomorphism>. There is a unique matrix $Z$ such that
$$
H=H_Z=\{(Zu,u):u\in\mathbb C^r\}.
$$
For $v=(Zu,u)$ and $v'=(Zw,w)$, direct evaluation gives
$$
\psi(v,v')=u^{\mathsf T}(Z-Z^{\mathsf T})w.
$$
Thus isotropy is exactly $Z=Z^{\mathsf T}$. For such $Z$,
$$
i\psi(v,\bar v)
=i\,u^{\mathsf T}(\bar Z-Z)\bar u
=2u^{\mathsf T}(\operatorname{Im}Z)\bar u.
$$
Hence positivity is precisely <positive definiteness> of $\operatorname{Im}Z$. Conversely, symmetry and this inequality define a positive Lagrangian plane; it is opposed to its conjugate because $(Z-\bar Z)u=0$ forces $u=0$. We have explicitly identified
$$
\boxed{D\cong\mathfrak H_r=
\{Z\in M_r(\mathbb C):Z=Z^{\mathsf T},\
\operatorname{Im}Z>0\},}
$$
the <Siegel upper half-space>. The choice of the graph $(Zu,u)$, rather than $(u,Zu)$, is what matches the sign of the given <symplectic form>.
Let an integral element of the <symplectic group> have block matrix
$$
g=\begin{pmatrix}A&B\\ C&D_0\end{pmatrix}.
$$
Acting on graph vectors gives
$$
g(Zu,u)=((AZ+B)u,(CZ+D_0)u).
$$
The image remains a positive Lagrangian plane, so its second projection is invertible. Thus $CZ+D_0$ is invertible and the induced action is
$$
\boxed{g\cdot Z=(AZ+B)(CZ+D_0)^{-1}.}
$$
The symplectic identities give symmetry of this matrix, and preservation of $i\psi(v,\bar v)$ gives
$$
\operatorname{Im}(g\cdot Z)
=(C\bar Z+D_0)^{-\mathsf T}
(\operatorname{Im}Z)(CZ+D_0)^{-1}>0.
$$
This is a Hermitian congruence by an invertible matrix, so it proves preservation of the positive domain as well as describing the action. These are coordinates on marked <Hodge structures>; passing to the integral-group quotient forgets the symplectic marking.
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