[Paper Review] A short proof of w_1^n(Hom(C_{2r+1}, K_{n+2}))=0 for all n and a graph colouring theorem by Babson and Kozlov

Carsten Schultz|ArXiv.org|Jul 17, 2005

Topological and Geometric Data Analysis11 references6 citations

TL;DR

This paper provides a concise topological proof that the $n$-th power of the first Stiefel-Whitney class vanishes for the ${\mathbb{Z}}_2$-space $\operatorname{Hom}(C_{2r+1}, K_{n+2})$, confirming a conjecture by Babson and Kozlov. Using an equivariant map into a product of spheres and Poincaré duality, the author shows $w_1^n(\operatorname{Hom}(C_{2r+1}, K_{n+2})) = 0$ for all $n \geq 0$, thereby establishing the strong form of their graph coloring theorem.

ABSTRACT

We show that the n-th power of the first Stiefel-Whitney class of the Z_2-operation on the graph complex Hom(C_{2r+1},K_{n+2})$ is zero, confirming a conjecture by Babson and Kozlov. This proves the strong form of their graph colouring theorem, which they had only proven for odd n. Our proof is also considerably simpler than their proof of the weak form of the theorem, which is also known as the Lovász conjecture.

Motivation & Objective

• To prove the conjecture by Babson and Kozlov that $w_1^n(\operatorname{Hom}(C_{2r+1}, K_{n+2})) = 0$ for all $n \geq 0$.
• To provide a unified, short proof valid for all $n$, overcoming the prior distinction between odd and even $n$ in earlier proofs.
• To establish the strong form of the Babson-Kozlov graph coloring theorem using characteristic classes.
• To demonstrate that the vanishing of $w_1^n$ implies non-(n+2)-colorability of graphs via the ${\mathbb{Z}}_2$-action.

Proposed method

• Construct an equivariant map $f: \operatorname{Hom}(C_{2r+1}, K_{n+2}) \to X_{n,r} \setminus A_{n,r}$, where $X_{n,r}$ is a product of $r$ copies of $S^n$ with a free ${\mathbb{Z}}_2$-action.
• Define $A_{n,r}$ as a ${\mathbb{Z}}_2$-invariant subspace of $X_{n,r}$, and show that $f$ misses $A_{n,r}$.
• Use Poincaré duality over $\mathbb{Z}_2$ to analyze the cohomology of the quotient spaces $X_{n,r}/{\mathbb{Z}}_2$ and $(X_{n,r} \setminus A_{n,r})/{\mathbb{Z}}_2$.
• Define classes $c_i \in H_{(2r-1)n}(X_{n,r}; \mathbb{Z}_2)$ as products of sphere fundamental classes, and show $c_i + c_{i+1} \in \operatorname{im} l_*$.
• Prove $\pi_*(c_r) = 0$ using the fact that the quotient map $Y \to Y/{\mathbb{Z}}_2$ has even degree, implying trivial image in homology.
• Conclude $w_1^n(X_{n,r} \setminus A_{n,r}) = 0$ via duality and the vanishing of the Poincaré dual of $\bar{p}_0^*(\omega)$.

Experimental results

Research questions

• RQ1 Does $w_1^n(\operatorname{Hom}(C_{2r+1}, K_{n+2})) = 0$ hold for all $n \geq 0$, including even $n$, as conjectured by Babson and Kozlov?
• RQ2 Can the strong form of the Babson-Kozlov graph coloring theorem be proven using characteristic classes without cohomology calculations?
• RQ3 Is there a unified topological proof for the vanishing of $w_1^n$ that works uniformly across all $n$, avoiding case distinctions?
• RQ4 Can the equivariant topology of $\operatorname{Hom}(C_{2r+1}, K_{n+2})$ be captured via a map into a product of spheres with a controlled invariant subspace?

Key findings

• The $n$-th power of the first Stiefel-Whitney class vanishes: $w_1^n(\operatorname{Hom}(C_{2r+1}, K_{n+2})) = 0$ for all $r \geq 1$ and $n \geq 0$, confirming the Babson-Kozlov conjecture.
• The proof avoids the complex cohomology calculations used in earlier works, offering a significantly simpler argument than the original proof of the weak form.
• The vanishing is established via an equivariant map into a product of $r$ $n$-spheres with a free ${\mathbb{Z}}_2$-action, missing a specific ${\mathbb{Z}}_2$-invariant subspace $A_{n,r}$.
• Using Poincaré duality over $\mathbb{Z}_2$, the proof shows that the Poincaré dual of the relevant cohomology class lies in the image of a boundary map, implying its class is zero.
• The key step is showing $\pi_*(c_r) = 0$ in homology, which follows from the even degree of the quotient map $Y \to Y/{\mathbb{Z}}_2$.
• The result implies the strong graph coloring theorem: if $w_1^n(\operatorname{Hom}(C_{2r+1}, G)) \neq 0$, then $G$ is not $(n+2)$-colorable.