Timeline for level sets of multivariate polynomials
Current License: CC BY-SA 2.5
9 events
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| Nov 9, 2009 at 22:04 | comment | added | fedja | Yes, because you are considering all lines and I'm considering only coordinate lines (i.e., lines in the direction of the coordinate axes). But $n$ or $\sqrt[]n$ doesn't really make any difference to you, does it? I mean, it is still a bound that grows fast with $n$ and you want no dependence on $n$ at all. | |
| Nov 9, 2009 at 15:46 | comment | added | ioannis.parissis | In our case this number is at most $2d$,and we can make the integral bigger by considering all lines that intersect the unit cube. We end up with $|S|\lesssim d n$, $n$ coming from the measure of lines that intersect $Q$. We have a $\sqrt{n}$ discrepancy here! | |
| Nov 9, 2009 at 15:46 | comment | added | ioannis.parissis | Let me rephrase your argument the way i understand it better. By crofton's formula for reasonable surfaced $S\subset \mathbb R ^n$,$ |S|=\int_{\mathcal L_S}n _{S}(l) d\nu(l)$ where $\mathcal {L_S}$ is the space of lines that intersect $S$, $\nu$ is the motion invariant measure on the set of lines and $n_S(l)$ is the number of intersections of $l$ with $S$. | |
| Nov 9, 2009 at 4:01 | comment | added | fedja | A small update: apparently, even the question about dimension-free estimates for the Gaussian perimeters of algebraic surfaces of fixed degree is open (your question is not easier because we can easily simulate independent Gaussians by long sums of coordinates). Even for $d=2$, the best that is known is that the Gaussian perimeters of all balls are uniformly bounded and the Gaussian perimeters of all origin-symmetric quadratic surfaces are uniformly bounded. Fascinating! It might make a nice polymath project... | |
| Nov 9, 2009 at 1:30 | comment | added | fedja | This other way is just noting that coordinate lines (they are enough to average over) are restricted to the cube (in other words, the area of the projection of your set to each coordinate hyperplane is bounded by $1$). As to the bound independent of $n$ in the cube, it holds when $d=1$ (sections of the cube by hyperplanes have area at most $\sqrt[]2$) but it'll take me some time (possibly, infinite) to figure out if it holds for other $d$ as well. That the cube is unit is crucial here, of course. Seems like a really nice question (unless there is some trivial counterexample)! | |
| Nov 9, 2009 at 1:28 | vote | accept | ioannis.parissis | ||
| Feb 18, 2016 at 15:05 | |||||
| Nov 9, 2009 at 0:45 | comment | added | ioannis.parissis | Thank you for the answer. I am a bit confused. So you consider all the lines that meet the set $E_ \alpha$. If you average over all the lines that meet the set you will get the surface measure of the boundary (which is what we want). So we need to compute this in another way as well. I'm missing this other way. To answer your question now, I need an estimate of the measure of the boundary that stays bounded in $n$ when $$d<<n$$. So we can suppose that $d$ is fixed and $n\rightarrow \infty$. I'm not sure if this is even possible of course. You should think that $\alpha \sim d$ | |
| Nov 9, 2009 at 0:31 | vote | accept | ioannis.parissis | ||
| Nov 9, 2009 at 1:28 | |||||
| Nov 8, 2009 at 23:50 | history | answered | fedja | CC BY-SA 2.5 |