The toughness conjecture for Kneser graphs

Let $n$ and $k$ be natural numbers with $n\geq 2k+1$. The vertex set of the Kneser graph $K(n,k)$ is $\binom{[n]}{k}$, with two vertices adjacent exactly when they are disjoint. For a connected graph $G$, define its toughness by

$$t(G)=\min_S\frac{|S|}{c(G\setminus S)},$$

where the minimum is over vertex cuts $S$ and $c(G\setminus S)$ is the number of remaining components. A maximum independent set is an independent set of largest possible size.

The toughness conjecture for Kneser graphs. If $k\geq 5$ and $n\geq 2k+1$, then

$$t(K(n,k))=\frac{n}{k}-1.$$

Moreover, if a vertex subset $S$ satisfies

$$t(K(n,k))=\frac{|S|}{c(K(n,k)\setminus S)},$$

then $S$ is the complement of a maximum independent set in $K(n,k)$.

The result is known for $k\in\{3,4\}$ for all $n\geq 2k+1$, and for $k\geq 5$ when $n$ is sufficiently large as a function of $k$. The conjecture asks for the remaining range $k\geq 5$ and $n\geq 2k+1$.