If we take a separable complex Hilbert space $H$, its projective space $PH$ is an infinite-dimensional Kähler manifold in a fairly obvious sense (see below). Suppose $M \subset PH$ is a finite-dimensional compact Kähler submanifold of $PH$. Must there be a finite-dimensional linear subspace $V \subset H$ such that $M \subset PV$?

(I don't know a good reference for infinite-dimensional Kähler manifolds but the relevant concept here seems clear. $PH$ is a smooth Hilbert manifold that can be covered with charts modeled on a *complex* Hilbert space, with transition functions that are holomorphic (given in a neighborhood of each point by an absolutely convergent power series). This makes each tangent space of $PH$ into a complex vector space, and this complex vector space structure extends to a complex Hilbert space structure, which varies smoothly---in fact real-analytically---from point to point. The imaginary part of the inner product gives $PH$ a symplectic structure, and the real part gives $PH$ a Riemannian structure, both of which are strongly nondegenerate: i.e. they each give an isomorphism $T_p PH \to T^*_p PH$ of the underlying real Hilbert spaces.)

afterI posted my question. This argument seems sort of magical to me because it leverages some finiteness of the intrinsic geometry of $M$ (the finite-dimensionality of $H^0(M,L)$ to get finiteness of its extrinsic geometry (the finite-dimensional of the smallest projective space in which $M$ sits.) $\endgroup$ – John Baez Jan 19 at 1:18