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Sep 30, 2015 at 14:38 vote accept B K
Sep 29, 2015 at 14:15 answer added Liviu Nicolaescu timeline score: 1
Sep 21, 2015 at 21:46 comment added Paul Siegel $L^2(M)$ is a separable Hilbert space for any closed connected Riemannian manifold $M$, and any two separable Hilbert spaces are isomorphic. So by itself $L^2(M)$ knows nothing about the geometry of $M$. The general philosophy of spectral theory is that a bare Hilbert space is boring, but a Hilbert space equipped with an operator (or better yet, an algebra of operators) is very interesting. In particular, I know of no source of examples of subspaces $V_j$ that are not eigenspaces of some operator (generally elliptic pseudodifferential).
Sep 21, 2015 at 21:19 history asked B K CC BY-SA 3.0