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We are given a $d$-dimensional convex shape $S$ inscribed in the hypercube $[-1,1]^d$. We want find an approximation of its volume based only on a set of curves given by the intersection of the $S$ boundary and a finite number of $2$-planes.

We denote by $\gamma_{i,j}$ the curve given by the intersection of the $S$ boundary with the $2$-plane $\mathrm{span}\left(\mathbf{e}_i,\mathbf{e}_j\right)$ for all $1\le i<j\le d$, where $\mathbf{e}_1,\mathbf{e}_2,\ldots,\mathbf{e}_d$ are the standard basis vectors.

Question: Knowing only the curves $\gamma_{i,j}$ for all $1\le i<j\le d$, is it possible to provide an estimation $V'$ of the volume $V$ of $S$ such that there exist two positive constants $\alpha$ and $\beta$ for which $\alpha\le\frac{V'}{V}\le\beta$?

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    $\begingroup$ You cannot have "${d\choose 2}$ pairwise orthogonal randomly selected $2$-planes" if $d>3$. Did you instead mean the planes $\mathrm{span}\left(\mathbf{e}_i',\mathbf{e}_j'\right)$ for $i\ne j$? $\endgroup$
    – Iosif Pinelis
    Commented Oct 24, 2022 at 20:46
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    $\begingroup$ Perhaps some ideas can be gleaned from Shepp's work on computerized tomography; see e.g. faculty.wharton.upenn.edu/wp-content/uploads/2012/04/… $\endgroup$
    – Iosif Pinelis
    Commented Oct 24, 2022 at 20:54
  • $\begingroup$ Thank you @IosifPinelis . I changed that part. Yes, I mean the $2$-planes spanning those vectors. $\endgroup$
    – Penelope Benenati
    Commented Oct 24, 2022 at 20:59
  • $\begingroup$ Thank you for the reference @IosifPinelis ! Maybe I am wrong, but I have the feeling that this can be useful for $d=3$. However, I am very interested in the case $d\gg 1$. $\endgroup$
    – Penelope Benenati
    Commented Oct 24, 2022 at 21:05
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    $\begingroup$ Technically all you can say about your body is that it contains the convex hull of your curves and is contained in the intersection of the corresponding cylinders, which is a bit too little to get constants independent of $d$. Indeed, suppose that you have all squares. Then the best possible upper bound is $2^d$ (the volume of the cube) while the lower bound is not better than the volume of the intersection of the cube with the cross-polytope $|x_1|+\dots+|x_d|\le 2$, i.e., $\le 4^d/d!$. $\endgroup$
    – fedja
    Commented Oct 25, 2022 at 0:08

1 Answer 1

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Those constants don't exist for any $d\geq4$, here is an idea of why.

For each $\varepsilon>0$ let $A_\varepsilon=\{(x_1,\dots,x_d)\in[-1,1]^d;\lvert (d-1)x_d-\sum_{i=1}^{d-1} x_i\rvert\leq\varepsilon\text{ and }\lvert\frac{(d-1)(d-2)}{2}x_d-\sum_{i=1}^{d-1}ix_i\rvert\leq\varepsilon\}$. Note that there is a big constant $K$ independent of $\varepsilon$ such that for all $i,j$, the intersection of $A_\varepsilon$ with span$(e_i,e_j)$ is contained in the ball $B(0,K\varepsilon)$.

Also note that $v=(1,1,\dots,1)\in A_\varepsilon$, and let $B_\varepsilon=\{x\in A_\varepsilon;d(x,span(v))\leq K\varepsilon\}$.

Then both $A_\varepsilon$ and $B_\varepsilon$ are convex and inscribed in the cube, and their intersection with span$(e_i,e_j)$ is the same $\forall i,j$, but when $\varepsilon\to0$, the volume of $A_\varepsilon$ is, up to some constant, proportional to $\varepsilon^2$ and the volume of $B_\varepsilon$ is proportional up to some constant to $\varepsilon^{d-1}$.

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  • $\begingroup$ Thank you for your answer. Do you think it is at least possible to find $V'$ such that $\alpha\le\frac{V'}{V}\le\beta$ holds in expectation if we randomly rotate the standard basis, viz., if we use the basis formed by the pairwise orthogonal unit vectors $e_i'$ instead of $e_i$ for $i\in\{1,2,\ldots,d\}$, where $e_1'$ is selected u.a.r. from the unit $d$-sphere $S^d$ centered at the origin and, for all $j\in\{2,\ldots, d\}$, $e_j'$ is a selected u.a.r. from the unit $(d-j+1)$-sphere $S^{d-j+1}$ which is formed by the points of $S^d$ orthogonal to $e_1', e_2', \ldots, e_{i-1}'$? $\endgroup$
    – Penelope Benenati
    Commented Oct 25, 2022 at 13:26
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    $\begingroup$ By "holds in expectation" do you mean that for each basis you would find an estimate for the volume and then you would integrate over the bases, and the result should be the volume of $K$ up to a constant? $\endgroup$
    – Saúl RM
    Commented Oct 25, 2022 at 13:32
  • $\begingroup$ Yes, I am asking whether we can even disprove the existence of a function $f\left(\{\gamma_{i,j}: 1\le i<j\le d\}\right)$ to estimate $V$ given the whole set of curves, such that $\mathbb{E}\left[f\left(\{\gamma_{i,j}: 1\le i<j\le d\}\right)\right]:=\mathbb{E}\left[V'\right]\in \left[\alpha V,\beta V\right]$ holds for two positive constants $\alpha$ and $\beta$ with $\alpha < \beta$, where the expectation is taken over the random choice of the basis vectors, as you wrote now. $\endgroup$
    – Penelope Benenati
    Commented Oct 25, 2022 at 15:57
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    $\begingroup$ In that case it seems you don't even need curves and "poking the convex set with vectors" should work. If I am not mistaken, calculating the volume of a compact $K$ (where $0\in K$, which I think happens if $K$ is inscribed in the cube) in polar coordinates gives $Vol(K)=\frac{1}{d}\int_{\mathbb{S}^{d-1}}g(v)^ddv$, where $g(v):=\sup\{x\in\mathbb{R};x\in K\}$. So $f(\{\gamma_{i,j}\})=V(\mathbb{B}^{d})\cdot|\gamma_{1,1}(1,0)|^d$ should work. $\endgroup$
    – Saúl RM
    Commented Oct 25, 2022 at 17:44
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    $\begingroup$ Yes sorry, that's what I tried to mean. I was considering the curve to be parametrized in terms of $\mathbb{S}^1\subseteq\mathbb{R}^2\equiv span(e_i,e_j)$. Also, it should be $\gamma_{1,2}$, not $\gamma_{1,1}$ $\endgroup$
    – Saúl RM
    Commented Oct 25, 2022 at 19:06

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