Vanishing of all equivariant obstructions and the mapping degree

TL;DR

This paper proves the existence of symmetric, equivariant maps avoiding the diagonal in Euclidean space for certain group actions, using a novel approach that bypasses complex obstruction theory calculations.

Contribution

It introduces a new method to show the vanishing of all equivariant obstructions simultaneously, simplifying the proof of the existence of equivariant maps.

Findings

All obstructions vanish under specified conditions.

Classified degrees of equivariant maps from boundary of simplex.

Applied results to envy-free division problem.

Abstract

Suppose that $n\neq p^k$ and $n\neq 2p^k$ for all $k$ and all primes $p$. We prove that for any Hausdorff compactum $X$ with a free action of the symmetric group $\mathfrak S_n$ there exists an $\mathfrak S_n$-equivariant map $X \to {\mathbb R}^n$ whose image avoids the diagonal $\{(x,x\dots,x)\in {\mathbb R}^n|x\in {\mathbb R}\}$. Previously, the special cases of this statement for certain $X$ were usually proved using the equivartiant obstruction theory. Such calculations are difficult and may become infeasible past the first (primary) obstruction. We take a different approach which allows us to prove the vanishing of all obstructions simultaneously. The essential step in the proof is classifying the possible degrees of $\mathfrak S_n$-equivariant maps from the boundary $\partial\Delta^{n-1}$ of $(n-1)$-simplex to itself. Existence of equivariant maps between spaces is important for many questions arising from discrete mathematics and geometry, such as Kneser's conjecture, the Square Peg conjecture, the Splitting Necklace problem, and the Topological Tverberg conjecture, etc. We demonstrate the utility of our result applying it to one such question, a specific instance of envy-free division problem.