Compactness of the finite complex Grassmannian

If $P^{\ast }=P$ and $P^2=P$, then ${\mathbb{C}}^N=\mathrm{ran}P\oplus \ker P$: every $x$ equals $Px+(x-Px)$, and the intersection is zero. The two summands are orthogonal because $⟨Pu,y⟩=⟨u,P^{\ast }y⟩=0$ for $y\in \ker P$. An orthonormal basis adapted to this decomposition makes $P$ diagonal with one $1$ for each basis vector in $\mathrm{ran}P$ and zero otherwise. Trace is invariant under change of basis, so $\mathrm{tr}P={\dim }_{\mathbb{C}}\mathrm{ran}P$, while $\mathrm{\parallel }P{\mathrm{\parallel }}_F^2=\mathrm{tr}(P^{\ast }P)=\mathrm{tr}(P^2)=k$. Thus the stated equations are exactly the orthogonal projections onto $k$-planes, and every such plane has one unique projection.

For fixed $k$, the equations $P^{\ast }=P$, $P^2=P$, and $\mathrm{tr}P=k$ are closed conditions on the real and imaginary matrix entries, and step 1.1 bounds the Frobenius norm by $\sqrt{k}$. Since $2N^2\ge 1$, [F1] makes ${\mathrm{Gr}}_{\mathbb{C}}(k,N)$ compact. There are finitely many indices $k$; restricting any open cover of their union to each compact member gives a finite subcover for that member, and the union of these finitely many finite subfamilies is finite. Hence ${\mathcal{G}}_N$ is compact.