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Jan 20, 2019 at 22:44 comment added John Baez What are the possible problems?
Jan 19, 2019 at 3:44 comment added Francois Ziegler 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.
Jan 19, 2019 at 1:18 comment added John Baez 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.)
Jan 18, 2019 at 0:08 comment added YangMills 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.
Jan 17, 2019 at 7:34 comment added John Baez Yes; I'd like this to be true for some work I'm doing on geometric quantization.
Jan 17, 2019 at 2:29 comment added Konstantinos Kanakoglou 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 ?
Jan 17, 2019 at 0:47 history asked John Baez CC BY-SA 4.0