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.)

  • $\begingroup$ Is this somehow related to the concept of geometric quantization? To be more specific, does this have to do with the fact that to any compact Kähler manifold one can associate a finite dimensional quantum Hilbert space reflecting the size of the classical phase space ? $\endgroup$ – Konstantinos Kanakoglou Jan 17 at 2:29
  • $\begingroup$ Yes; I'd like this to be true for some work I'm doing on geometric quantization. $\endgroup$ – John Baez Jan 17 at 7:34
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    $\begingroup$ I think you can argue like this: $PH$ has a hyperplane bundle, which pulls back to a line bundle $L$ on $M$. The inclusion $M\subset PH$ can be written in homogeneous coordinates as $z\mapsto [s_0(z):s_1(z):...]$ where the $s_j$'s are global holomorphic sections of $L$. If the image of $M$ is not contained in any finite-dimensional linear subspace then the sections $s_j$'s would be linearly independent (and there are countably many of them). But the dimension of the space of global sections $H^0(M,L)$ is finite since $M$ is compact, contradiction. $\endgroup$ – YangMills Jan 18 at 0:08
  • $\begingroup$ That's what I've been thinking - especially after I 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
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    $\begingroup$ Note: That argument is in Remark 2.2.9 of Loi-Zedda (2018; arXiv). But I’m not (yet quite) convinced it can be turned into a watertight proof of the present conjecture — which they don’t claim. $\endgroup$ – Francois Ziegler Jan 19 at 3:44

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