Problems

Sperner’s lemma in dimension \(2\).
Subdivide a triangle \(ABC\) arbitrarily into a triangulation consisting of smaller triangles meeting edge to edge. Define Sperner coloring of the triangulation as an assignment of three colors to the vertices of the triangulation such that:

• Each of the three vertices \(A, B,\) and \(C\) of the initial triangle has a distinct color

• The vertices that lie along any edge of triangle \(ABC\) have only two colors, the two colors at the endpoints of the edge. For example, each vertex on \(AC\) must have the same color as \(A\) or \(C.\)

Here is an example of Sperner’s triangulation

Prove that every Sperner coloring of every triangulation has at least one "rainbow triangle", a smaller triangle in the triangulation that has its vertices colored with all three different colors. More precisely, there must be an odd number of rainbow triangles.

Prove Sperner’s lemma in dimension \(1\), namely on a line.
The simplex in this case is just a segment, the triangulation is subdivision of the segment into multiple small segments, and the conditions of a Sperner’s coloring are the following:

• There are only two colors;

• The opposite ends of the main segment are colored differently;

Then one needs to prove that there exists a small segment with two ends colored in different colors. In particular there is an odd number of such small segments.

Draw a Sperner’s coloring for the following \(3\)-dimensional simplex. The blue segments are visible, the grey ones are inside the tetrahedron. The point \(F\) is on the face \(ABC\), point \(E\) is on the face \(BCD\), point \(G\) is on the face \(ACD\) and the point \(H\) is on the face \(ABD\).

We want to prove Monsky’s theorem as a corollary of Sperner’s lemma: it is not possible to dissect a square into an odd number of triangles of equal area. After scaling we can consider the square with coordinates: \((0,0),(0,1),(1,0),(1,1)\), which we want to cut into \(n\) triangles with area \(\frac{1}{n}\) each for an odd \(n\). Consider the following coloring of all the points with rational coordinates \((\frac{p}{q},\frac{r}{s})\) inside the square:
We look at the powers of \(2\) in the fractions \((\frac{p}{q},\frac{r}{s})\), first of all the numbers \(p,q\) are coprime, and thus only one of them is divisible by \(2\), same with \(r,s\). Then the following possibilities might occur:

1. Neither \(q\) nor \(s\) is divisible by \(2\). In this case we color the point red.

2. \(\frac{r}{s}\) is divisible by a larger or equal power of \(2\) than \(\frac{p}{q}\) and \(p\) is not divisible by \(2\). In this case we color the point blue.

3. \(\frac{p}{q}\) is divisible by a strictly larger power of \(2\) than \(\frac{r}{s}\) and \(r\) is not divisible by \(2\). In this case we color the point green.

Under an assumption (which you do not have to prove) that the area of any rainbow triangle is at least \(\frac{1}{2}\) prove the Monsky’s theorem.

Prove the general version of Sperner’s lemma: Consider an \(n\)-dimensional simplex \(\mathcal{A} = A_1A_2...A_{n+1}\). Strictly speaking a simplex is a convex linear combination of \(n+1\) points in general position (when \(k\) points are never in one subspace of dimension \(k-1\)). One can view it as an \(n\)-dimensional tetrahedron or a body spanned over vertices \((0,0,...,0), (1,0,0,...,0), (0,1,0,0...,0), ... (0,0,...0,1)\). \[\mathcal{A} = \{\sum_{i=1}^{n+1}a_i(0,0,...,1,...,0), \,\,\, a_i \geq 0, \,\,\,\, \sum_{i=1}^{n+1}a_i = 1\}.\]

A simplicial subdivision of an \(n\)-dimensional simplex \(\mathcal{A}\) is a partition of \(\mathcal{A}\) into small simplices (cells) of the same dimension, such that any two cells are either disjoint, or they share a full face of a certain dimension.
Define a Sperner’s coloring of a simplicial subdivision as an assignment of \(n+1\) colors to the vertices of the subdivision, so that the vertices of \(\mathcal{A}\) receive all different colors, and points on each face of \(\mathcal{A}\) use only the colors of the vertices defining the respective face of \(\mathcal{A}\).
Prove that every Sperner’s coloring of any subdivision of an \(n\)-dimensional simplex contains an odd number of rainbow simplexes, namely whose vertices are colored using all \(n+1\) colors.

Draw Sperner’s coloring for the following triangulation. Try to avoid rainbow triangles at all costs.

Consider an \(n\)-dimensional simplex \(\mathcal{A} = A_1A_2...A_{n+1}\), namely a body spanned over vertices \((0,0,...,0), (1,0,0,...,0), (0,1,0,0...,0), ... (0,0,...0,1)\). \[\mathcal{A} = \{\sum_{i=0}^{n}a_i(0,0,...,1,...,0), \,\,\, a_i \geq 0, \,\,\,\, \sum_{i=1}^{n+1}a_i = 1\}.\] Where next to \(a_i\) there is a point with coordinate where \(1\) is in \(i\)-th place. The point \((0,0,...,0)\) belongs to the simplex as well.

A simplicial subdivision of an \(n\)-dimensional simplex \(\mathcal{A}\) is a partition of \(\mathcal{A}\) into small simplices (cells) of the same dimension, such that any two cells are either disjoint, or they share a full face of a certain dimension.
Define a Sperner’s coloring of a simplicial subdivision as an assignment of \(n+1\) colors to the vertices of the subdivision, so that the vertices of \(\mathcal{A}\) receive all different colors, and points on each face of \(\mathcal{A}\) use only the colors of the vertices defining the respective face of \(\mathcal{A}\).
Consider a simplicial subdivision given by pairwise connected middles of all the segments in the original simplex. Assign the numbers \(0,1,2...,n\) to the subdivision vertices in such a way as to conduct a Sperner’s coloring in such a way that you will have only one rainbow simplex.