Graph charts and tangent directions for complex rank-r projectors

Since $P_0^2=P_0$, every $x\in {\mathbb{C}}^n$ decomposes as $P_{0}x+(I-P_0)x$ with the first term in $U$ and the second in $\ker P_0$ by [F4]. If $u=P_{0}y\in U$ and $z\in \ker P_0$, then $⟨u,z⟩=⟨P_{0}y,z⟩=⟨y,P_{0}z⟩=0$ by self-adjointness. Hence $\ker P_0=U^{\perp }$, giving the stated orthogonal block decomposition.

For any $X:U\to U^{\perp }$, $J_X^{\ast }{J}_X=I_U+X^{\ast }X$ is positive definite because $⟨(I_U+X^{\ast }X)u,u⟩=\mathrm{\parallel }u{\mathrm{\parallel }}^2+\mathrm{\parallel }Xu{\mathrm{\parallel }}^2>0$ for $u\mathrm{\ne }0$. The displayed $P(X)$ is Hermitian, and $P(X{)}^2=J_X(J_X^{\ast }{J}_X{)}^{-1}{J}_X^{\ast }{J}_X(J_X^{\ast }{J}_X{)}^{-1}{J}_X^{\ast }=P(X)$. Its range is $\mathrm{range}{J}_X=V_X$, of dimension $r$; [F2] gives $\mathrm{tr}P(X)=r$.

Let $T={\pi }_U{\mathrm{\mid }}_V:V\to U$, whose adjoint is $T^{\ast }=P_V{\mathrm{\mid }}_U:U\to V$. The block $A={\pi }_U{P}_V{\mathrm{\mid }}_U$ is $T{T}^{\ast }$, so it is invertible exactly when $T$ is invertible: $T{T}^{\ast }$ is positive definite exactly when $T^{\ast }$ is injective, and the domain and codomain of $T$ have the same finite dimension. Thus the chart condition is the open condition that $A$ be invertible; it is open because $P_V\mapsto A$ and determinant are continuous, and at $P_0$ we have $A=I_U$. In that domain, projection from $V$ to $U$ is bijective, so every $u\in U$ has a unique vector $u+Xu\in V$; this defines a unique linear $X$. In blocks $P_V=\left(\begin{array}{cc}A& B^{\ast }\\ B& C\end{array}\right)$, that map is $X=B{A}^{-1}$. The formula in step 1.2 is the unique orthogonal projector onto this graph. Matrix inversion is smooth on invertible matrices by [F5], so the graph map and $P(X)$ are inverse smooth coordinates.

Let $P(t)$ be a differentiable curve in ${\mathcal{P}}_r$ with $P(0)=P_0$ and derivative $H$. Differentiating $P(t{)}^{\ast }=P(t)$ and $P(t{)}^2=P(t)$ gives $H=H^{\ast }$ and $P_{0}H+H{P}_0=H$. In block form these equations force $H=\left(\begin{array}{cc}0& B^{\ast }\\ B& 0\end{array}\right)$. Conversely, for any such $B$, the graph curve $P(tB)$ from step 1.2 has derivative $\left(\begin{array}{cc}0& B^{\ast }\\ B& 0\end{array}\right)$ at $t=0$, since $(I+t^2{B}^{\ast }B{)}^{-1}=I+O(t^2)$. Thus these are exactly all tangent directions.

By [F3], each frame $Y$ with $Y^{\ast }Y=I$ and range $U$ exists. If $Y$ and $Z$ are two such frames, then $Q=Y^{\ast }Z$ satisfies $YQ=Y{Y}^{\ast }Z=P_{0}Z=Z$ and $Q^{\ast }Q=Z^{\ast }Y{Y}^{\ast }Z=Z^{\ast }Z=I$. Thus $Q$ is unitary and $Z=YQ$. Conversely, $YQ(YQ{)}^{\ast }=Y{Y}^{\ast }$ for every unitary $Q$. Hence projectors identify exactly the unitary changes of occupied frame.