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Yoav Kallus's user avatar
Yoav Kallus's user avatar
Yoav Kallus's user avatar
Yoav Kallus
  • Member for 12 years, 11 months
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Relative interior and dense subsets
Sure, why not. See below.
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Relative interior and dense subsets
I guess you're right that under your definition, the relative interior is non-empty. Sorry. But I think my second example still works.
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Relative interior and dense subsets
First of all, your assumption that $\operatorname{ri}(\operatorname{conv}(B))$ is non-empty is wrong, for example, when $B$ is a single point. Now, even if we assume that $\operatorname{ri}(\operatorname{conv}(B))$ is non-empty, consider the situation that $A$ is the interior of the unit square and that $B$ is an edge of the unit square. Neither (1) or (2) holds.
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Securing privacy of "who communicates with whom" under Orwell-like conditions
Shouldn't your Muscovites be called Cyril and Dimitry?
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Job sites for applied/interdisciplinary Mathematics?
This is not at all helpful for you, but I wanted to mention how pleased I was to see that in the mathjobs.org job listing page, next to all the alphabetical letters to narrow the list, there is also an "applied" option to narrow the list. I was equally disappointed to see that it just gives you a list of places you have sent an application to.
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Bi-spherical polyhedra
Unless I have miscalculated, for the pyramid the incenter and circumcenter do not coincide (except for a special height).
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Bi-spherical polyhedra
I.e. with a regular polygon as a base.
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Bi-spherical polyhedra
Your list of non-interesting bispherical polyhedra should also include dipyramids (of the right height). The more interesting case of course is when the incenter and circumcenter do not coincide. One example are pyramids of regular polygons.
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2-layer tilings with a center-of-gravity constraint
If there is a lower bound on the distance from the center of mass to the boundary in some "isotropic" position (seems like there should be), that would give you a lower bound on the double coverage under the assumption that each layer is a lattice and one layer is a 180-degree rotation of the other (like all the coverings in this question so far).
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Symmetric black-hole curves
Actually, scratch that. What I said doesn't make any sense at all. The phase space area can't overlap itself because of reversibility. Liouville's theorem just doesn't apply here for the reason you said, which is the volume is zero to start with.
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Symmetric black-hole curves
Yes, :). Still, in this case the rays keep coming back to the space of horizontal rays, so in the measure restricted to that space what I said still makes sense (I think).
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Symmetric black-hole curves
Very cool. This is very similar to the ellipsoid paradox (see dx.doi.org/10.1119/1.3596430 ), but with one focus taken to infinity. I see now how the physical objections are circumvented. Specifically, with Liouville's theorem, a "packet" of particles gets horizontally spread out as it gets vertically squished and eventually the front of the packet overlaps the back of it so the "phase space" volume taken up by the packet is no longer conserved.
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Symmetric black-hole curves
Geometric optics is a particular limit of a wave equation. Certainly time reversibility of that equation and thermodynamics should tell you that the steady state solution where a constant stream of parallel rays approaches from the left must result in an equal amount of power leaving (or else we can time-reverse the solution and have a free power source). Same goes for any finite fraction of rays being captured.
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