### Colored Bárány's Theorem

Carnegie Mellon University
30th July 2015.

### Point selection problem

How to define "median" for point set on plane?

Is there always a point in many triangles?

### The planar case

Given $n$ points on the plane in general position.

Boros–Füredi Theorem

There exists a point in $\ge$$\frac{2}{9}$${n\choose 3}-O(n^2)$ triangles.

Bukh–Matoušek–Nivasch Theorem

The constant $\frac{2}{9}$ is the best possible by stretched grid!

### Higher dimensional case

Given $n$ points in $\mathbb{R}^d$ in general position.

A simplex is the convex hull of $d+1$ points.

Bárány's Theorem

There exists a point in $\ge c_d{n \choose d+1}$ simplices.

Bukh–Matoušek–Nivasch Theorem

We cannot hope for $c_d > \frac{(d+1)!}{(d+1)^d}$ by stretched grid.

Bárány: a point is in $\ge$ $c_d$$n\choose d+1 simplices.  1982 Bárány \frac{1}{(d+1)^d} 2003 Wagner \frac{(d^2+1)}{(d+1)^{d+1}} 2010 Gromov \frac{2d}{(d+1)(d+1)!} < \frac{(d+1)!}{(d+1)^d} ### Plan of talk  ▶ Introduce a colored variant Karasev's proof with a twist ### A colored variant Given n red points, n green points and n blue points. Colored Bárány's Theorem There is a point in at least pn^3 colorful triangles. ### Probabilistic equivalence Given n red points, n green points and n blue points. Probabilistic Equivalence A point in random colorful triangle with probability \ge p. Given 3 (continuous) probability measures m_0, m_1 and m_2. Continuous Equivalence A point in random colorful triangle with probability \ge p. ### Timeline (cont.)  2010 Gromov \frac{2}{9} \to \frac{2d}{(d+1)(d+1)!} m_i = m_j 2012 Karasev \frac{1}{6} \to \frac{1}{(d+1)!} \frac{2}{9} \to \frac{2d}{(d+1)(d+1)!} m_1 = m_2 2014 J. \frac{2}{9} \to \frac{2d}{(d+1)(d+1)!} ### Plan of talk  ▶ Introduce a colored variant ▶ Karasev's proof with a twist ### Karasev's proof Given 3 (continuous) probability measures m_0, m_1 and m_2. Proof by contradiction. Assume for every point v, \operatorname{Pr}\big(v \text{ is in} \big) < p := 2/9. Plane (plus a point at infinity) and sphere are the same. \infty \equiv \infty ### Preparation • Choose a sufficiently fine triangulation of sphere. • Add "buttress" inside the sphere. ### Auxiliary map Gradually build a map from "sphere with buttress" to sphere: \overset{f}{\longrightarrow} such that: 1. it is identity when restricted to the sphere; 2. it is economical: e.g., image of each "spike" is small. ### Build economical map gradually Step 1: Vertex O (sphere center) \mapsto \infty (the north pole). Step 2: Want \overline{PO}\mapsto \overline{P\infty}; need to find line segment \overline{P\infty}.  Claim \operatorname{Pr}\big(\overline{P\infty} \text{ intersects } \big) < p for some \overline{P\infty}.  Obs is in \Longleftrightarrow intersects . \operatorname{Pr}\big( intersects \big)=\operatorname{Pr}\big( is in \big) < p \implies \operatorname{Pr}\big( intersects \;\big\vert\;$$\big) < p$ for some

Step 3: Want to define $f$ on $OPQ$. Knew $f$ on $OP$, $OQ$, $PQ$.
$\overset{f}{\longrightarrow}$ $\overset{\mathrm{mod2}}{=}$

Set $a := m_1(A)$, $b := m_2(A)$, $x := \overline{m}(A) = (a + b)/2$.

$2x(1-x)\le$ $a(1-b)+b(1-a) \le$ $\operatorname{Pr}\big($ int. $\big)$ $< 2p$

$\implies x(1-x) < p = 2/9$. WLOG assume $\overline{m}(A) = x < 1/3$.

### Degree argument

We have found a (continous) map $f$ such that

$\overline{m}\big(f(OPQ)\mathrm{mod2}\big)<\frac{1}{3}$ $\implies \overline{m}\big(f(\text{spike})\mathrm{mod2}\big) < 1$.

 Def Degree of map $f$ is $\left|f^{-1}(\text{generic point})\right|$.

Note there are plenty of points not in $f(\text{spike})\mathrm{mod2}$.

 Obs $\sum$deg of $f$ on each spike is even. Cor $\text{deg on sphere} + 2\sum \text{deg on }OPQ$ is even. ! Degree of identity map is even. Impossible!

Carnegie Mellon University
zj@cmu.edu