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## Trivial pretzel links

Is there a simple rule to check whether a pretzel link P(n_1,...,n_k) is a trivial link? I am interested in the 2-component case but every information would be helpful.

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en.wikipedia.org/wiki/Pretzel_link for those who, like me, didn't know what a pretzel link is. In general one should try to define jargon, or provide link to background material, when asking a question on MO. – Willie Wong Oct 13 2010 at 19:26
Why specify to 2-components? Is there an easy way to tell whether a pretzel link has exactly two components? – Willie Wong Oct 13 2010 at 19:30

## 2 Answers

A pretzel link has as many components as the pretzel link with coefficients reduced $(\mod 2)$. So it will have two components if and only if there are precisely two even coefficients.

A conceptual classification is given by considering the 2-fold branched cover, or the $\pi$-orbifold, obtained by taking the orbifold quotient of the 2-fold branched cover. For pretzel links, the 2-fold branched cover is a connect sum of Seifert fibered spaces (this is more generally true for Montesinos links). Then a 2-component link is trivial if and only if the two-fold branched cover is $S^2\times S^1$. A theorem of Bonahon and Siebenmann classifies Seifert fibered orbifolds, so in principal you can classify pretzel links from their classification.

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 Ah, after you pointed it out in your first paragraph, it is actually completely obvious that it should be so. Thanks. – Willie Wong Oct 13 2010 at 22:57 Thanks for your answer. Actually I was looking for something like an algorithm but that's interesting. – Paolo Aceto Oct 14 2010 at 1:21

In principle, normal surface theory answers your question - it can detect if a link is split and can decide if a knot is unknotted. In practice, just draw your link in SnapPea. But neither if these could be called a "simple rule." You might look at Purcell's thesis work/early papers (eg "Volumes of highly twisted knots and links"), using hyperbolic geometry, to give a simple rule implying that a link is non-trivial.

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