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I need to compute a series of Tristram-Levine signatures for a family of torus knots. I was wondering if this has already been done or whether there is a good way to streamline the computation.

I am aware that Tristram's original paper on the signature invariants provides a formula for the $e^{\frac{2\pi i}{3}}$-signature of $(2, odd)$ knots, and I am looking for generalizations. The most relevant reference I found is the following paper by Maciej Borodzik and Krzysztof Oleszkiewicz: http://arxiv.org/abs/1002.4500 . It seems to me that a less complex (if less general) solution should exist.

To make the question a little more specific, the simplest example I'm working with is the $e^{\frac{2\pi i}{3}}$-signature of the $(5,6)$-torus knot, which has a $20\times20$ Seifert Matrix. In general, I'd like to be able to find the $e^{\frac{2\pi i}{3}}$- signature of various $(m, mn +1)$-torus knots where $m$ and $n$ are any integers.

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  • $\begingroup$ I suppose with a little work one could make the software Regina compute Tristram-Levine signatures. All the core code is implemented but I haven't set up a single procedure to do those computations in precisely that form. But the intersection forms and abelian covering spaces are all set-up. The webpage is here: regina.sourceforge.net $\endgroup$ May 7, 2014 at 1:43

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In general, closed formulas for such things tend to be rather slow, computationally. I would suggest that you look at the paper of Litherland: Signatures of iterated torus knots. Topology of low-dimensional manifolds, pp. 71–84, Lecture Notes in Math., 722, Springer, Berlin, 1979. Litherland starts with a formula, due to Brieskorn, for the signature function of a torus knot on the circle, that involves counting lattice points; this is easily implemented on a computer.

On the unit circle (where you are computing Levine-Tristram signatures) the function is piecewise constant with jumps at certain roots of unity. Litherland computes the jump function, which gives the signature function. I believe that this is computationally much faster. It's quite possible that for special classes of knots, such as the $(m,mn+1)$ torus knots you mention, that you can get a simple explicit answer.

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