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Part of the definition of the Casson invariant is that if you have an integer homology sphere $\Sigma$ and a knot $k,$ then $$\lambda(\Sigma + \frac{1}{m} k) - \lambda(\Sigma + \frac{1}{m+1} k)$$ does not depend on $m,$ where adding a multiple of a knot means doing a Dehn surgery with the indicated slope. My question is: with the standard normalization, when is the difference actually equal to $\pm 1?.$

EDIT In the case I was specifically looking at, the complement of the knot was fibered, so the Alexander polynomial was the (normalized to be a symmetric Laurent series) characteristic polynomial of the monodromy action on homology. However, it is not clear what the deep significance (from the algebraic standpoint) is of taking the second derivative of this object and evaluating it at one (the evaluating at $1$ part is natural, since the homology of a mapping torus is equal to the cokernel of $f_* - I,$ but the second derivative is a little mysterious).

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  • $\begingroup$ I think the relation to the Alexander polynomial (and in particular the second derivative) is just a distraction. Better to think of it (as Dylan Thurston suggests below) as one of the simplest finite type knot invariants. $\endgroup$ – Kevin Walker Apr 26 '14 at 15:18
  • $\begingroup$ Dear Igor, thank you for an interesting question. I can't believe, that the Alexander polynomial is just a distraction. I guess that this definition of the Casson invariant makes some sense. Have you seen this paper of Donaldson? $\endgroup$ – nikitamarkarian May 3 '14 at 12:19
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The expression in your question is equal to $\frac{1}{2}\Delta''_k(1)$, where $\Delta_k(t)$ is the Alexander polynomial of the knot k with symmetric normalization. This is Theorem 8.7 of Akbulut and McCarthy.


[edit]

Here are some further remarks in response to Igor's comments.

In addition to the first derivative of the Casson invariant, $$\lambda'(K) = \lambda(K_{1/(m+1)}) - \lambda(K_{1/m}) = \frac{1}{2}\Delta''_K(1) ,$$ there are analogously defined second and third derivatives $\lambda''(L_2)$ and $\lambda'''(L_3)$. Here $L_i$ is an $i$-component link in a homology 3-sphere with all pairwise linking numbers zero. (The fourth and higher derivatives of the Casson invariant are zero.)

$\lambda''(L_2)$ and $\lambda'''(L_3)$ have geometric interpretations. $\lambda''(L_2)$ is (up to a scalar which I don't recall) the Sato-Levine invariant -- the self-linking number of the intersection of Seifert surfaces for the two components of $L_2$. In particular, it is zero if $L_2$ is a boundary link.

$\lambda'''(L_3)$ is the Milnor $\mu_{123}$ invariant of $L_3$ (again up to a multiplicative constant which I don't recall). It too can be defined in terms of intersections of Seifert surfaces and linking numbers.

The above observations about $\lambda''$ and $\lambda'''$ allow one to construct large families of knots $K$ with prescribed $\lambda'(K)$.

Perhaps in the finite type invariant literature there is a similar geometric characterization of $\lambda'$, but I don't recall seeing one.


[edit #2]

When the knot $K$ is fibered with monodromy $f$ and fiber $Y$, $\lambda'(K)$ is equal to (plus or minus) the Lefschetz number of the induced map $f^*$ on the reduced representation variety of $\pi_1(Y)$ into $SU(2)$. When the fiber $Y$ is genus 1, this space is just a point, so in this case $\lambda'(K)$ is $\pm 1$.

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    $\begingroup$ Thanks, Kevin. I had, in the meantime, found references to similar statements (people talk about the Conway polynomial instead). However, to be annoying, this only pushes the question one level back. Is there some criterion for this quantity to be equal to $1?$ $\endgroup$ – Igor Rivin Apr 25 '14 at 16:53
  • $\begingroup$ I'm not aware of any such criterion, but I haven't thought about these things for a long time. $\endgroup$ – Kevin Walker Apr 25 '14 at 17:13
  • $\begingroup$ To me, $\Delta''_k(1)$ is a basic knot invariant: it's the unique finite-type invariant of order 2. Not sure what more you want. $\endgroup$ – Dylan Thurston Apr 25 '14 at 21:42
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    $\begingroup$ @DylanThurston It might be the "unique finite-type invariant of order 2", but this is not how I think about geometry/topology (perhaps my PhD advisor is to blame?), so I am looking for some more geometric criterion for when this thing is/is not zero or one, which, you will admit, is a natural question. $\endgroup$ – Igor Rivin Apr 25 '14 at 22:54
  • $\begingroup$ @KevinWalker Thank, that's really interesting! You can take a look at my edit as well for more confusion :) $\endgroup$ – Igor Rivin Apr 26 '14 at 13:52

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