Finding an isometry between two subspaces of a Hilbert space So, I'm given a Hilbert space which is the direct sum $H=H_1\oplus H_2$ of two separable Hilbert spaces $H_j$. There is a closed subspace $D\subseteq H$ which satisfies that it is not a subspace of either $H_j$ (i.e., it is only a subspace of their direct sum). Define orthogonal projections $P_j$ which map onto $H_j$, respectively. My aim is to construct a surjective, linear isometry $J:P_1D\rightarrow P_2D$ (here, $J$ is isometric if $\langle u,v\rangle=\langle Ju,Jv\rangle$ for any $u,v\in D(J)$), so that $D=\{u+Ju~|~u\in D(J):=P_1D\}$. I am almost totally stumped by this.
It should also be noted that due to the larger context of the question the following can be taken as given: for $u,v\in D$, we have $\langle P_1u,P_1v\rangle=\langle P_2u,P_2v\rangle$.
My first approach was to try to define an inverse of $P_1$ on $D(J)$, but this won't work unless $P_1$ is injective into $D(J)$ (which is certainly not guaranteed), so I gave up on that.
My second attempt involved separating $D$ using separability of $H$; but by the same general lack of injectivity of $P_j$, one cannot guarantee that, for a complete orthonormal sequence $(v_k)_{k=1}^N$ for $D$ (with $N\in\mathbb{N}\cup\{\infty\}$), $(P_jv_k)_{k=1}^N$ is also a complete orthonormal sequence for $P_jD$; furthermore throwing away any vectors $P_jv_k$ which destroy orthonormality of the sequence removes the possibility of defining the map $J$ by using this method.
Can anybody offer any pointers/hints on how to achieve my goal? Many thanks in advance!
 A: The existence of $J$ would imply that $P_1D$ and $P_2D$ are isometric. This is not the case in general: let $H_1=\mathbb C^2$, $H_2=\mathbb C^3$, and 
$$
D=\{(0,a)\oplus(b,c,0):\ a,b,c\in\mathbb C\}.
$$
Then $P_1D$ is one-dimensional, while $P_2D$ is two-dimensional. 
A: If such a $J$ exists, we must necessarily have $D \cap H_2 = \{0\}$. For $P_1(u+Ju) = P_1(u) = u$, so $d = P_1(d) + J(P_1(d))$ for all $d \in D$. $D\cap H_2 = \{0\}$ means the projection $P_1$ is injective on $D$, and $J = P_2 \circ (P_1\lvert_D)^{-1}$.
Conversely, if $D\cap H_2 = \{0\}$, $(P_1\lvert_D)^{-1}$ exists and $J = P_2 \circ (P_1\lvert_D)^{-1}$ is a linear isometry $J \colon P_1(D) \to P_2(D)$ - due to the premise $\langle P_1u, P_1v\rangle = \langle P_2 u, P_2 v\rangle$ for all $u,v\in D$. Since $(P_1\lvert_D)^{-1} \colon P_1(D) \to D$ is surjective, also $J \colon P_1(D) \to P_2(D)$ is surjective.
Summarising: A $J\colon P_1(D) \to P_2(D)$ with the desired properties exists if and only if $P_1\lvert_D$ is injective (which means, if and only if $D\cap H_2 = \{0\}$). The premise
$$\langle P_1u, P_1v\rangle = \langle P_2 u, P_2 v\rangle\tag{1}$$ for all $u,v\in D$ guarantees that $D\cap H_2 = \{0\}$, for if $v \in D \cap H_2$, then we have
$$0 =\lVert P_1 v\rVert^2 = \lVert P_2 v\rVert^2 = \lVert v\rVert^2$$
by $(1)$.
