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Jul 26, 2016 at 15:04 history edited Emilio Pisanty CC BY-SA 3.0
Changed image to https.
Mar 9, 2016 at 16:59 history edited Emilio Pisanty CC BY-SA 3.0
added 528 characters in body
Mar 9, 2016 at 10:13 history edited Emilio Pisanty CC BY-SA 3.0
Improved definition to deal unambiguously with degenerate eigenvalues.
S Feb 24, 2016 at 14:22 history suggested emiliocba
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Feb 24, 2016 at 14:00 review Suggested edits
S Feb 24, 2016 at 14:22
Feb 5, 2016 at 19:34 comment added Emilio Pisanty @Delio Oooooh, ok, that's an interesting sort of thing to ask. And yeah, it does sound like it should have connections.
Feb 5, 2016 at 19:10 comment added Delio Mugnolo @EmilioPisanty A good starting point, with a survey of earlier results and more numerical experiments than proofs is journals.cambridge.org/action/…
Feb 5, 2016 at 14:58 history edited Emilio Pisanty CC BY-SA 3.0
added 48 characters in body; edited title
Feb 5, 2016 at 12:06 comment added Emilio Pisanty @Delio That's the first time I hear of that theory - have you got a good introduction to the topic? (Say, physicist-grade.)
Feb 5, 2016 at 11:02 comment added Delio Mugnolo Very nice question and I wonder whether there is any connection to the theory of spectral maximal partitions.
Feb 4, 2016 at 16:53 answer added Emilio Pisanty timeline score: 37
Jan 6, 2016 at 0:11 history edited Emilio Pisanty CC BY-SA 3.0
added 190 characters in body
Jan 5, 2016 at 23:51 answer added Carlo Beenakker timeline score: 13
Jan 5, 2016 at 19:31 comment added Liviu Nicolaescu If you know enough of the eigenfunctions you can recover the whole domain.
Jan 5, 2016 at 19:19 comment added Emilio Pisanty @LiviuNicolaescu The goal here is a global isometry, or lack thereof. If knowing the eigenfunctions locally gets you local knowledge of the metric, then that's interesting but it's not quite what I'm asking.
Jan 5, 2016 at 19:16 history edited Emilio Pisanty CC BY-SA 3.0
Clarified the question.
Jan 5, 2016 at 19:15 history edited Nik Weaver CC BY-SA 3.0
you can hear the *sound* of a drum by hitting it anywhere
Jan 5, 2016 at 19:12 comment added Liviu Nicolaescu By shape I understand the Riemann metric defining the Laplacian. In the case at hand the metric is Euclidean so there is not much to say.
Jan 5, 2016 at 18:54 history edited Emilio Pisanty CC BY-SA 3.0
added 92 characters in body
Jan 5, 2016 at 18:49 history edited Emilio Pisanty CC BY-SA 3.0
Fixed reference. Added alternative description.
Jan 5, 2016 at 16:38 history edited Emilio Pisanty CC BY-SA 3.0
deleted 7 characters in body
Jan 5, 2016 at 16:38 comment added Jason Starr @LiviuNicolaescu: Should "the shape of that region $R$" be "the shape of $D$"?
Jan 5, 2016 at 16:36 comment added Liviu Nicolaescu If you know the spectrum of the Laplacian and the restrictions of the eigenfunctions to a tiny region $R\subset D$, then you can determine the shape of that region $R$. www3.nd.edu/~lnicolae/RandMorseSpecGeom.pdf
Jan 5, 2016 at 16:29 history asked Emilio Pisanty CC BY-SA 3.0