Alexander polynomial or Reidemeister torsion for knotted surfaces? - MathOverflow

Alexander polynomial or Reidemeister torsion for knotted surfaces?

2010-07-22T07:13:53Z

An important invariant of a knot in $S^3$ is its Alexander polynomial, related also to Reidemeister torsion. Is there something like that for knotted surfaces in $S^4$? If not, what are the difficulties?

Answer by Scott Carter for Alexander polynomial or Reidemeister torsion for knotted surfaces?

2010-07-22T12:38:45Z

The ideals that define the homology of the infinite cyclic covering are not principal. So the Alexander polynomial, as a single polynomial, is not defined. Fox's example 12 from "A quick trick in knot theory" is the 2-twist spun trefoil. The ideal is generated by a pair of polynomials. I haven't thought about Reidemeister torsion.

Answer by Daniel Moskovich for Alexander polynomial or Reidemeister torsion for knotted surfaces?

2010-07-22T19:44:29Z

For ribbon 2-knots, there is an "Alexander polynomial", which is the universal finite-type invariant for such knots. The original references are
K. Habiro, T. Kanenobu, and A. Shima, Finite type invariants of ribbon 2-knots. In: Low-dimensional topology (Funchal, 1998), 187-196, Contemp. Math., 233, Amer. Math. Soc., Providence, RI, (1999).
and this paper by Habiro and Shima. Dror Bar-Natan is also working on such things- see this talk for instance. If one takes the "quantum" view of the world, this is the "ultimate" Alexander polynomial, because it behaves well with respect to cabling, is calculable in linear time, induces the Alexander polynomial for knots, etc. I'll have more to say on such topics presently...

Answer by Paul for Alexander polynomial or Reidemeister torsion for knotted surfaces?

2010-07-27T23:41:23Z

Yes for $S^2$, and more generally, depending on what you are after.

Given any torsion module $M$ over the PID $Q[t,t^{-1}]$, the order of $M$ is well defined in $Q[t, t^{-1}]$ up to units, in the usual way. Moreover $M\otimes Q(t)=0$. These two facts are at the heart of why the Alexander polynomial is related to Reidemeister torsion. The way it works is that if $C$ is a f.g. chain complex over $Q[t, t^{-1}]$ its homology is torsion iff $C\otimes Q(t)$ is acyclic. Then the two notions of torsion are related by

$$\prod_k order(H_k(C))^{(-1)^k}=\tau(C)$$

If $S^n\subset S^{n+2}$ and $X$ its complement, then $X$ has a $Z$ cover $\tilde{X}$ and hence an exact sequence of chain complexes $0\to C(\tilde{X})\to C(\tilde{X})\to C(X)\to 0.$ The corresponding long exact sequence shows that $H_k(\tilde{X})$ is torsion for all $k$, since $H_k(X)=H_k(S^1)$.

Setting $\Delta_k$ to be the order of $H_k(\tilde{X})$ gives the $k$-th Alexander polynomial. You get Reidemeister torsion equals the multiplicative Euler characteristic:

$$\tau= \prod \Delta_k^{(-1)^k}$$

In the case of $S^1\subset S^3$ Poincare duality relates $\Delta_2$ and $\Delta_1$, and $\Delta_0$ is independent of the knot, so you recover Alexander poly "=" Reidemeister torsion. But in general $\tau$ combines all the $\Delta_k$.

All this (including how to pick the basis of the acyclic complex) is explained in Milnor's article "infinite cyclic covers."

As far as what happens more generally, all this extends, but you have to be careful when the homology is not torsion, which can happen for surfaces in $S^4$. Then you have to view Reidemeister torsion as a function of the homology , etc. Milnor's other Torsion articles are worthwhile reading for these topics.