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Question: Is there a characterization of graphs that arise as intersections of a family of complete $k$-partite graphs on the same (finite) set of vertices?

It is clear that every such graph is $k$-colorable. Moreover, every $k$-colorable graph can be extended to a complete $k$-partite graph.

Maybe the answer to my question is "$k$-colorable graphs" but I somehow doubt it.

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up vote 4 down vote accepted

For each $k$-coloring of a graph $G$, there is a complete $k$-partite graph in which $G$ embeds as a spanning subgraph. Suppose $u$ and $v$ are vertices of $G$ such that in any $k$-colouring, they are assigned different colours. Then they will be adjacent in any complete $k$-partite extension. (We could say that these vertices are "morally adjacent", but note that this depends on $k$ implicitly.)

Any uniquely $k$-colourable graph that is not complete $k$-partite is not the intersection of its $k$-partite embeddings.

For another example, consider the graph constructed as follows. Let $H$, with vertex set $\{1,2,3,4\}$, be the graph obtained from $K_4$ by deleting the edge $34$. Then in any 3-colouring the vertices $3$ and $4$ get the same colour. Extend $H$ by joining a new vertex $5$ to $4$; call the new graph $H_5$. Then in any 3-colouring, the vertices 3 and 4 get the same colour, and vertices 4 and 5 get different colours. So 3 and 5 get different colours in any 3-colouring, although they are not adjacent. It follows that the intersection of the complete 3-partite embeddings of $G$ contains the edge $35$. (The graph $H_5$ is not uniquely 3-colourable.)

I do not see any way of characterizing the graphs that contain a morally adjacent pair of vertices, but the answer to the question is not "$k$-colourable graphs".

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Thank you. So this settles the subquestion (and proves that I am a weak counterexample-maker). Do you think that the characterization problem is tractable? – Gejza Jenča Apr 3 '13 at 16:39
@GejzaJenča: My feeling (for what it's worth) is that it would be hard to find a "good characterization", but there might nonetheless be an interesting one. – Chris Godsil Apr 3 '13 at 17:06

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