Timeline for Is every knot unavoidable in the embeddings of some graph?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jul 18, 2014 at 1:20 | review | Suggested edits | |||
| Jul 18, 2014 at 1:35 | |||||
| Jul 15, 2014 at 23:30 | vote | accept | Joseph O'Rourke | ||
| Jul 15, 2014 at 18:02 | comment | added | Joseph O'Rourke | 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. | |
| Jul 15, 2014 at 14:05 | comment | added | Sean Eberhard | 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? | |
| Jul 15, 2014 at 13:29 | comment | added | Tony Huynh | 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. | |
| Jul 15, 2014 at 13:20 | comment | added | John Dvorak | @JosephO'Rourke a generalisation of Fary's theorem should be able to show these equivalent. | |
| Jul 15, 2014 at 13:16 | history | edited | Tony Huynh | CC BY-SA 3.0 |
deleted 3 characters in body
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| Jul 15, 2014 at 13:16 | comment | added | Joseph O'Rourke | Very nice! But usually the notion of intrinsic knottedness relies on tame embeddings, rather than specifically straight-line embeddings. | |
| Jul 15, 2014 at 13:07 | history | answered | Tony Huynh | CC BY-SA 3.0 |