I recently came upon a recursive formula for the (ordinary) signatures of torus knots. The formula, which I found in Murasugi's book "Knot Theory and Applications" (Springer, 2007), originally appeared in a paper by Gordon, Litherland and Murasugi, "Signatures of Covering Links" (http://www.maths.ed.ac.uk/~aar/papers/glm.pdf). It is apparent from the formula that torus knots have even signatures. I wondered if this is a property specific to torus knots, so I went and poked around the Knot Atlas for a while. I did not see any knot of odd signature. (I did not check every knot, but checked enough to get discouraged.)

My question is: does there exist a knot of odd signature?

Secondly, why do all torus knots (and, quite possibly, all knots) have even signature?

  • $\begingroup$ I haven't thought about this sort of thing in some time but if I were to try to prove the signature is always even, the most enticing direction would be to express the signature in terms of the Milnor signatures, decomposing the Alexander module along the quadratic factorization of the Alexander polynomial. But I haven't thought through the details. $\endgroup$ – Ryan Budney May 13 '14 at 23:08
  • $\begingroup$ So do you know if the signature really is always even? $\endgroup$ – zygund May 13 '14 at 23:11
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    $\begingroup$ Off the top of my head, no, I don't. But it wouldn't surprise me if it is. I suspect it's known and probably a pretty easy argument, one way or the other. Someone like Danny Ruberman should come along soon with an answer. I'm too absorbed in something else at the moment. $\endgroup$ – Ryan Budney May 13 '14 at 23:22

This is exercise 3.4 in Livingston's book (Knot Theory, Carus math. monographs, vol 24, page 123).

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    $\begingroup$ … and an extension of that exercise would prove that the Tristram-Levine signatures, which can be computed as the signature of a Hermitianized version of the Seifert matrix, are even, so long as that matrix is invertible. This happens as long as you stay away from roots of the Alexander polynomial. $\endgroup$ – Danny Ruberman May 14 '14 at 1:51
  • $\begingroup$ Once we know that the ordinary signature of a knot is even, it follows easily that all Tristram-Levine signatures are even as well (they can only jump by +/- 2 when a root changes sign). $\endgroup$ – zygund May 14 '14 at 2:47
  • $\begingroup$ Signatures can jump by arbitrary even amounts if you are at a multiple root. But we are really just kicking around the same argument here; see mathoverflow.net/questions/85976/… $\endgroup$ – Danny Ruberman May 14 '14 at 13:27
  • $\begingroup$ Fair enough about multiple roots, but the jump would always be even. $\endgroup$ – zygund May 19 '14 at 16:08

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