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Let $\mathfrak a,\mathfrak b$ be cardinals (a cardinal being identified with the smallest ordinal of that cardinality), $\mathfrak b\subseteq\mathfrak a$. The Kneser graph $KG(\mathfrak a,\mathfrak b)$ is the simple graph with vertex set $V(KG(\mathfrak a,\mathfrak b))=\binom{\mathfrak a}{\mathfrak b}:=\{X\subseteq \mathfrak{a}:|X|=\mathfrak b\}$ and $XY\in E(KG(\mathfrak a,\mathfrak b))$ iff $X\cap Y=\emptyset$.

Do there exist cardinal pairs $(\mathfrak a_1,\mathfrak b_1)\neq(\mathfrak a_2,\mathfrak b_2)$ with $\mathfrak a_1,\mathfrak a_2$ infinite and $\mathfrak b_1,\mathfrak b_2\neq 0$ such that $KG(\mathfrak a_1,\mathfrak b_1)$ and $KG(\mathfrak a_2,\mathfrak b_2)$ are isomorphic, in short $KG(\mathfrak a_1,\mathfrak b_1)\cong KG(\mathfrak a_2,\mathfrak b_2)$?

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No. If $G=KG(\mathfrak a,\mathfrak b)$, where $\mathfrak a$ is infinite and $0\lt\mathfrak b\le\mathfrak a$, then $\mathfrak a$ and $\mathfrak b$ are determined by the structure of the graph $G$.

$\mathfrak a$ is the maximum cardinality of a clique (set of pairwise adjacent vertices) in $G$.

$\mathfrak b=1$ if and only if $G$ is a complete graph. If $\mathfrak b\gt1$, then $\mathfrak b$ is the maximum cardinality of a clique in $G$ for which there is a vertex of $G$ which is adjacent to no element of the clique.

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  • $\begingroup$ How do I even know that the maximum clique cardinality exists when it doesn't in $\sum K_n$? $\endgroup$ – SK19 Jun 11 at 9:48
  • $\begingroup$ @SK19 In the graph $KG(\mathfrak a,\mathfrak b)$ the maximum clique cardinality exists and is equal to $\mathfrak a$ because (1) tvhere is a clique of cardinality $\mathfrak a$, and (2) every clique has cardinality $\le\mathfrak a$. Do you need a proof of !1) or (2) or both? $\endgroup$ – bof Jun 11 at 14:45
  • $\begingroup$ @SK19 Anent (1), the fact that $KG(\mathfrak a,\mathfrak b)$ has a clique of cardinality $\mathfrak a$, i.e., that a set of cardinality $\mathfrak a$ contains a collection of $\mathfrak a$ pairwise disjoint subsets of cardinality $\mathfrak b$, follows from the fact that $\mathfrak a\mathfrak b=\mathfrak a$. $\endgroup$ – bof Jun 11 at 14:53
  • $\begingroup$ @SK19 On the other hand, let $\mathcal X$ be any clique in $KG(\mathfrak a,\mathfrak b)$. Let $f$ be a choice function on $\mathcal X$, i.e., $f(X)\in X$ for each $X\in\mathcal X$. Then $f:\mathcal X\to\mathfrak a$ is an injection, whence $|\mathcal X|\le\mathfrak a$. $\endgroup$ – bof Jun 11 at 15:00

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