Post #2153520
2025-10-10 18:58 UTC
Replies (3)
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@tao@mathstodon.xyz 2025-10-10 18:58
Previous comments on the problem had indicated that the convex case was too easy to be interesting, so I decided to look at the slightly larger class of star-shaped objects. Here I suspected that one could express the hypothesis and conclusion of the problem in terms of various integrals on the surface, and I was hoping to use some integral inequalities (e.g., Sobolev embedding) to then proceed. However, my differential geometry was rather rusty, so I asked the AI to perform these calculations for me. Somewhat to my surprise, the AI not only computed all the quantities I requested, but actually gave a complete proof of the problem in the star-shaped case. The proof manipulated the various integrals that arose using various inequalities and identities, some of which I recognized (Stokes' theorem and the Willmore inequality / Gauss-Bonnett), but there was one which was new to me (Minkowski's first integral formula). With all of these inequalities (and also the arithmetic mean-geometric mean inequality relating mean curvature to Gauss curvature), the proof of the star-shaped case was in fact a one-line argument. (2/8)
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@anupamasridhar@mathstodon.xyz 2025-10-12 16:41
@tao@mathstodon.xyz Not really my field at all but I tried to take a crack at this one: If we drop the convexity requirement but keep genus-0 and bounded curvatures ≤ 1, can we achieve volume smaller than the unit ball (4π/3) starting from a small volume genus 2 bowl, then cap the two handles with identical “corks” that are solids of revolution. If smoothing preserves |\kappa_i|\le 1, the result is a genus 0 surface with total volume below 4\pi/3. Cork seed in the plane. Take the equilateral triangle with vertices (0,0),(2,0),(1,\sqrt3). Remove three unit radius sectors of angle \theta=\pi/3. Call the remainder \mathcal H. Rotate \mathcal H about the x axis to obtain the cork \mathcal C. Volume by first moment. V(\mathcal C)=2\pi M_x(\mathcal H) with M_x=\iint y\,dA. Triangle: \operatorname{Area}=\sqrt3, centroid =\sqrt3/3, so M_x(T)=1. Two bottom sectors contribute 2\cdot(\pi/6)\cdot(1/\pi)=1/3. Top sector contributes (\pi/6)(\sqrt3-2/\pi)=\pi\sqrt3/6-1/3. Hence M_x(\mathcal H)=1-\pi\sqrt3/6 and V_{\text{cork}}=2\pi\Bigl(1-\frac{\pi\sqrt3}{6}\Bigr)=\frac{\pi}{3}\,(6-\sqrt3\,\pi)\approx 0.585. Patchwise curvature before smoothing. For a surface of revolution with meridian r(x), \kappa_{\mathrm{mer}}=\frac{|r’’|}{(1+r’^2)^{3/2}},\qquad \kappa_{\mathrm{par}}=\frac{1}{r\sqrt{1+r’^2}}. Spherical patches give \kappa_1=\kappa_2=1. Conical patches with slope \tan 60^\circ=\sqrt3 give \kappa_{\mathrm{mer}}=0 and \kappa_{\mathrm{par}}=1/(2r)\le 1/\sqrt3 since truncation enforces r\ge \sqrt3/2. Reparametrize the meridian by arclength s with angle \alpha(s): x_s=\cos\alpha, r_s=\sin\alpha, \kappa_{\mathrm{mer}}=|\alpha’|, \kappa_{\mathrm{par}}=\cos\alpha/r. Define F=r-\cos\alpha. Then F’=\sin\alpha\,(1+\alpha’).
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@KLAL1982@mathstodon.xyz 2025-10-22 02:10
@tao@mathstodon.xyz I'd be curious how chatGPT Pro would perform if you just asked it the question without giving it your suggestion.