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Let $K$ be a closed convex cone in an n-dimensional Euclidean space. Suppose $K$ has non-empty interior. For $t > 0$ form the subcone $K_t$ consisting of all points in $K$ which lie a distance $t$ or greater from the boundary of $K$.

True or False? $K_t$ is a translate of $K$.

In other words, can we obtain $K_t$ by pushing $K$ into its interior?

Motivation: this problem arose while trying to establish the ‘optimal worst-case escape rate’ from the collision locus in the $N$-body problem. In the collinear equal mass $N$-body problem, the $N-1$ dimensional cone $K$ of interest is the one defined by the inequalities $ x_1 \le x_2 \le \dotsb \le x_N$ within the hyperplane $\sum x_i = 0$ of ${\mathbb R}^N$. (You might call this the “$A_N$ cone” since it is a Weyl chamber for the $A_N$ Lie algebra.)

What do I know?
Not much beyond two dimensions and a lot of special cases. In two dimensions the assertion is true. $K$ is a sector. $K_t$ is the translate of $K$ by a vector directed along the bisector of the angle formed by the sector's boundary. It follows that the assertion also holds for three-dimensional trihedral cones: i.e. those $K$'s in 3-space whose affine cross-section is a triangle. It seems a dissection + limit argument ought lead from here to the general 3-dimensional result but I did not get even this.

Convex geometry terminology? If the assertion is true, then the length of the translation vector taking $K$ to $K_1$ measures of the “sharpness” of the cone $K$. This length is the reciprocal of $m(K)= \max_{q \in \operatorname{int}(K)} \operatorname{dist}(q, \partial K)/\|q\|$ as $q$ varies over the interior of $K$. Does any one know the convex geometry term for this $m(K)$ or its reciprocal, the “sharpness” of $K$?

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  • $\begingroup$ TeX note: the usual TeX command for sums is \sum, not \Sigma (compare $\Sigma x_i = 0$ \Sigma x_i = 0 to $\sum x_i = 0$ \sum x_i = 0). MathJax note: backticks are parsed by the MathJax parser, not by TeX, so having too many of them in a paragraph can result in the intervening text being effectively wrapped in <pre>, as happened in this post. It's better to use " " or “ ” on MO. I edited accordingly for both. $\endgroup$
    – LSpice
    2 days ago

2 Answers 2

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$K_t$ need not be a translate of $K$. Let $A=[-4,4]\times[-1,1]\subseteq\mathbb{R}^2$ and consider the convex cone $K=\{t\cdot v;t\in[0,\infty),v\in A\times\{1\}\}\subseteq\mathbb{R}^3$. Note that $\partial K$ is contained in the union of the four planes $x_2=\pm x_3$ and $x_1=\pm4x_3$, so points in $K_1$ will be points of $K$ at distance $\geq1$ of those four planes.

Clearly $K$ has a unique point with minimal $x_3$ coordinate, the point $(0,0,0)$. However, the set $K_1$ only contains points with $x_3$ coordinate at least $\sqrt{2}$ (see (*) below), and all the points $(t,0,\sqrt{2})$ for $t\in[-1,1]$ are in $K_1$ (as they are at distance $\geq1$ of the four planes mentioned above). So $K_1$ is not a translate of $K$.

(*) To see why $K_1$ only contains points with $x_3$ coordinate at least $\sqrt{2}$, note that if $X:=\{(x,y)\in\mathbb{R}^2;y>|x|\}$, then all points in $X$ at distance $\geq1$ of $\partial X$ have $y$ coordinate $\geq\sqrt{2}$; similarly, all points of $K\subseteq\mathbb{R}\times X$ at distance $>1$ of the planes $x_2=\pm x_3$ have $x_3$ coordinate at least $\sqrt{2}$.

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  • $\begingroup$ Thanks! But the minimal value of $x_3$ on $K_1$ is not your $\sqrt{2}$, but rather is $x_* = \sqrt{5}/2 = 1.18..$ which is less than $\sqrt{2}$. To see this, form the sector obtained by intersecting the plane $x_2 = 0$ with your $K$. You get a sector whose angle $\theta$ satisfies $tan(\theta/2) = 2/1$ and so $sin(\theta/2) = 1/x_*$. The point $P_* = (0,0, x_*)$ is achieved by propagating the bounding rays of this sector in one unit. (Your $\sqrt{2}$ comes from the angle made by the orthogonal sector made by intersecting with the plane $x_1 = 0$.) I believe $K_1 = P_* + K$. $\endgroup$ 2 days ago
  • $\begingroup$ It seems the point $(0,0,x_*)$ is at distance $<1$ of the point $\left(0,\frac{x_*}{2},\frac{x_*}{2}\right)$, which is in the boundary of $K$. So $(0,0,x_*)$ cannot be in $K_1$, right? $\endgroup$
    – Saúl RM
    2 days ago
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    $\begingroup$ I also made an arithmetic error. Set $x_* = \sqrt{17}/4$. The triangle with half angle $\theta/2$ has, from your measurement specifications $\tan(\theta/2) = 4/1$ whereas I had set the tangent to $2/1$ by mistake. The correct triangle is thus a $1:4: \sqrt{17}$ right triangle and $x_*$ comes from this. $\endgroup$ 2 days ago
  • $\begingroup$ Verification: The eqns for $K_t$ can be obtained from the unit inward pointing normals of $K$. These yield the following eqns for $K_t$: $$a(y+z) \ge t$$ $$a(-y+ z) \ge t$$ $$b (x + 4z) \ge t$$ $$b (-x + 4z) \ge t$$ with $a = 1/\sqrt{2}$ and $b = 1/\sqrt{17}$. Setting $x = y = 0$ and $t =1$ and the $\ge$' to $=$' we see that $(0,0,\sqrt{17}/4)$ is on the boundary of $K_1$. $\endgroup$ 2 days ago
  • $\begingroup$ Yes, it seemed strange to me that something like $\arctan(1/4)$ didn't appear somewhere in the argument. I didn't comment on that because the same idea still applies. E.g. the point $(0,0,\sqrt{17}/4)$ is at distance $<1$ of the point $(0,\sqrt{17}/8,\sqrt{17}/8)$, which is in the boundary of $K$ (In particular, I don't think it is in the boundary of $K_1$) $\endgroup$
    – Saúl RM
    2 days ago
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Conversations with friends today led to a solution. A generic 4-sided convex polyhedral cone in 3-space led to a counterexample. To be specific, suppose the 3-dimensional convex polyhedral cone $K$ be defined by $x \ge 0, y \ge 0, z \ge 0$ and $ax + by - cz = 0$ where $(a,b,c)$ is a unit vector with $a, b , c > 0$. One verifies that the four faces of $K$ are all two-sided, each being bounded by two rays.

The polyhedron $K_t$, $t > 0$ is defined by the inequalities $x \ge t, y \ge t, z \ge t$ and $ax + by -cz \ge t$ since $x, y, z, ax + by -cz$ represent the distance from the four faces. In particular the face $z = 1$ of $K_{1}$ is coordinatized by $(x,y)$ and is characterized by the inequalities $x \ge 1, y \ge 1$ and $a x + by \ge 1 + c$. One verifies that for `most' choices of $a, b,c$ this particular face of $K_1$ is three-sided. For example, if $a = b = c = 1/\sqrt{3}$ we have $a + b - c = 1/\sqrt{3} < 1 + 1/ \sqrt{3}$ which means that the line $a x + by = 1+ c$ crosses into the interior $x > 1, y >1$ of this face of $K_1$,

We cannot translate a polyhedron all of whose faces are two-sided and end up with one having a three-sided face. So $K_1$ cannot be a translate of $K$.

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    $\begingroup$ Does this imply that you think the solution I posted yesterday is not right? If there is some misunderstanding/you do not agree with my last comment that the point $(0,0,\sqrt{17}/4)$ is at distance $<1$ of the boundary (and so is $(0,0,x)$ for all $x<\sqrt{2}$, which is at distance $<1$ of the point $(0,x/2,x/2)$ in the boundary), we can discuss it further $\endgroup$
    – Saúl RM
    yesterday
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    $\begingroup$ You are right Saul. I was wrong. The distance from the origin to K_1 is root 2. as you say. Your sharper angle (90 degrees) made by the plane x=0 yields the furthest point. Somehow I got the distance contributions from this angle and the other (y=0; shallower (arctan(1/4)) confused. $\endgroup$ 14 hours ago

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