Geometric Topology – Geometric Interpretation of Second Derivative of Alexander Polynomial at 1

alexander-polynomialfinite-type-invariantsgt.geometric-topologyknot-theoryqa.quantum-algebra

For an (oriented) knot in $S^3$ the number $\Gamma(K) := \Delta_K’’(1)$ shows up in a number of places in knot theory, for example the Casson-Walker-Lescop invariant. Here $\Delta_K(t)$ is the Alexander-Conway polynomial.

One explanation is that it’s the unique nontrivial order $2$ finite-type invariant, so it’s going to show up anywhere there’s something finite-type of small degree. Is there a more geometric interpretation than this? What does $\Gamma$ really mean?

This question discusses some related invariants but not $\Gamma$.

Best Answer

Given a knot in $S^3$, think of it as an embedding

$$f : S^1 \to S^3.$$

The configuration space of $5$ distinct points in $S^3$ is denoted $C_5(S^3)$, this is a $15$-dimensional manifold and it consists of all $5$-tuples of distinct points in $S^3$.

Similarly, we can talk about $C_5(S^1)$, but since $S^1$ has a cyclic order, this space has $5!/5 = 4!$ diffeomorphic path-components, so consider the subspace where the points are in cyclic order.

There is a subspace of $C_5(S^3)$ consisting of $5$ points that sit on a round circle. I.e. these are points that sit on an affine $2$-dimensional subspace of $\Bbb R^4$, and they also happen to be in $S^3 \subset \Bbb R^4$.

Since any three points in $S^3$ sit on a unique round circle, the subspace of $5$ points on a circle has co-dimension 4 in $C_5(S^3)$.

So our map

$$ f : S^1 \to S^3 $$

has an induced map

$$ f_* : C_5(S^1) \to C_5(S^3)$$

the domain is $5$-dimensional. Consider the pre-image of the circular subspace under $f_*$. Generically, it is a $1$-dimensional submanifold of $C_5(S^1)$. We consider only the components where when you compare the circular ordering of the $5$ points in $C_5(S^1)$ with the circular ordering of the circle in $S^3$, you get a pentagram. i.e. if any pair of points are adjacent in $S^1$, they are not adjacent in the circle in $S^3$.

This manifold is canonically an oriented manifold, inherited by the normal orientation induced by $f_*$. So we can consider its projection to $S^1$, i.e. forget $4$ of the $5$ coordinates in this manifold. This gives us a map from a $1$-manifold (the pentagrammic 5-tuples in the preimage of $f_*$) to $S^1$, so we can take the degree.

This degree is the invariant you are discussing.

So this is perhaps a "very geometric" interpretation of what you are looking for. Is this more or less what you want, or are you looking for something else?

This was written up by Garrett Flowers in JKTR, 2013.

https://doi.org/10.1142/S021821651350017X