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Is it the case that, for any given knot $K$, there exists some graph $G$ whose every embedding into $\mathbb{R}^3$ (or into $\mathbb{S}^3$) contains a cycle that realizes $K$?

I know the famous Conway-Gordon result that the complete graph on seven vertices $K_7$ is intrinsically knotted in that every embedding contains a knotted cycle. And Joel Foisy proved that $K_{3,3,1,1}$ is also intrinsically knotted (J. Graph Theory, 2002, ACM link).

Instead, I am asking a Ramsey-like question: Whether it's known that, for every particular knot, there exists some graph for which that particular knot is unavoidable.

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    $\begingroup$ Trivial observation: since every graph is a subgraph of a complete graph, it suffices to consider complete graphs for the purposes of this question. $\endgroup$
    – John Pardon
    Commented Jul 15, 2014 at 13:20

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Yes. See this paper of Negami. The main result is that for any fixed knot (or link) of type $k$, there is a constant $R(k)$ such that every straight line embedding of $K_{R(k)}$ in $\mathbb{R}^3$ contains a knot (or link) equivalent to $k$.

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  • $\begingroup$ Very nice! But usually the notion of intrinsic knottedness relies on tame embeddings, rather than specifically straight-line embeddings. $\endgroup$
    – Joseph O'Rourke
    Commented Jul 15, 2014 at 13:16
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    $\begingroup$ @JosephO'Rourke a generalisation of Fary's theorem should be able to show these equivalent. $\endgroup$
    – John Dvorak
    Commented Jul 15, 2014 at 13:20
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    $\begingroup$ Certainly you cannot allow arbitrary embeddings into $\mathbb{R}^3$, because then you could embed each edge as a local knot and the theorem would be false. $\endgroup$
    – Tony Huynh
    Commented Jul 15, 2014 at 13:29
  • $\begingroup$ Does it make sense to ask whether, allowing arbitrary embeddings, there must be a cycle realising a knot which is the connected sum of $K$ and some other knot? $\endgroup$
    – Sean Eberhard
    Commented Jul 15, 2014 at 14:05
  • $\begingroup$ Negami obtains an upper bound on $R(k)$, but, as he says in his final sentence, this bound "is not however so worthy" because it depends upon as-yet unknown values of Ramsey numbers. $\endgroup$
    – Joseph O'Rourke
    Commented Jul 15, 2014 at 18:02
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The answer is no for the following silly reason. Take any graph $G$ and some embedding into $\mathbb R^3$. To any edge, we may perform the connected sum with a fixed knot $K'$ (that is, take a small straight part of the edge and connect sum to it $K'$ there). Do this operation separately to all of the edges. Then any cycle in $G$ has knot type of the form $K'\#(\text{something})$. We just need to pick a knot $K'$ so that the given $K$ is not of this form.

We could modify the question and ask whether a fixed knot $K$ exists in some weaker sense in any given embedding of $G$. One such (very) weak sense could be that there exists a knotted torus $T$ in knot type $K$ and a cycle in $G$ which is contained in $T$ and cannot be made disjoint from a compressing disk for $T$. I suspect this may also have a negative answer, however. It seems one can probably always find embeddings of $G$ which are arbitrarily complicated, and in particular which are not related in any way to the given knot $K$.

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    $\begingroup$ Tony Huynh said it first. $\endgroup$
    – John Pardon
    Commented Jul 15, 2014 at 13:34

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