3
$\begingroup$

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.)

$\endgroup$
  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.