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Orienting equal-order non-isomorphic targets
PDF p. 2: Section 3, ordinary classes, equal-order case
Statement
Let and be non-isomorphic finite simple graphs of the same order, each with at least one vertex. Their labels can be chosen so that contains no copy of .
The order of a graph is the number of its vertices. A copy of in is a subgraph of isomorphic to ; it need not be induced. A graph isomorphism is a bijection of vertex sets preserving and reflecting adjacency.
Facts & Assumptions
A copy of a finite simple graph in a finite simple graph exists exactly when an ordinary embedding exists. An ordinary embedding is an injective vertex map that preserves edges (Ordinary embeddings as injections preserving edges).
If two finite simple graphs admit ordinary embeddings in both directions, then they are isomorphic (Rigidity of mutual ordinary embeddings).
The order of a finite graph is the cardinality of its vertex set (Order, size, complement and independence number).
Proof
Given: Non-isomorphic finite simple graphs of equal order, each with at least one vertex.
Goal: Choose their labels so that has no copy of .
Ordinary embeddings cannot exist in both directions between the two graphs, since otherwise the rigidity fact would make them isomorphic, contrary to the hypothesis. Hence at least one direction has no ordinary embedding.
Label the source of a direction without an ordinary embedding and its target . By the copy-embedding equivalence, contains no copy of . Relabelling preserves the equal-order hypothesis.
Source notes
The cited article uses this orientation in the equal-order case. The argument above derives it from mutual-embedding rigidity; in fact that rigidity gives the orientation even without the equal-order hypothesis.
Depends on
Used by · 1
Dependency tree · 2 levels
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Sources
- Maria Axenovich and Patrick Schmitteckert, Non-isomorphic graphs have distinct vertex-Ramsey classes, arXiv:2610.03704v1 · PDF p. 2, Section 3, ordinary classes, equal-order case