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?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ This is related to mathoverflow.net/questions/22252 $\endgroup$– Bruce WestburyCommented Jul 22, 2010 at 7:34
-
$\begingroup$ Perhaps it is. Can you tell me, however: if we calculate the Reidemeister torsion for the complement of a 2-knot, then what happens? For instance: 1) it works OK, 2) it doesn't work, 3) nobody knows? $\endgroup$– John VremCommented Jul 22, 2010 at 12:40
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
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.
$\begingroup$
$\endgroup$
4
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.
answered Jul 22, 2010 at 12:38
-
$\begingroup$ Scott, thanks. I vaguely remember having read something about difficulties with Reidemeister torsion in even dimensions. Is it right? What are they? To calculate Reidemeister torsion, we must have a chain complex with some bases both in vector spaces and homologies - what kinds of such complexes are known? $\endgroup$ Commented Jul 22, 2010 at 12:48
-
1$\begingroup$ Scott, if you use $Q$ coefficients in the infinite cyclic cover, then the ring $Q[Z]$ is a PID, so you bypass this problem. You lose little, since you probably want a field anyway if you consider Reidemeister torsion, and the fraction fields of $Z[Z]$ and $Q[Z]$ are the same. Of course, for more subtle versions of torsion (eg whitehead torsion) you can get more info by sticking to $Z[Z]$. For any knotted $X=S^4-S^2$, $H_1(\tilde{X};Q)$ is torsion as a $Q[Z]$ module, and hence a sum of cyclic modules, so the Alexander poly is defined, but it might have some indeterminacy as a $Z$ poly. $\endgroup$– PaulCommented Jul 27, 2010 at 23:58
-
1$\begingroup$ Thanks Paul. I should have been more careful in my answer. As an addendum to your comment, I suggest that the questioner also cunsult with Milnor's paper about Infinite Cyclic Covering Spaces as well as Kirk's book on Algebraic Topology. (I am in a train station in Korea and don't have the urge to look these things up now). Probably, I too should review these:-) $\endgroup$ Commented Jul 30, 2010 at 9:53
-
$\begingroup$ Scott, by the way, could you please send a copy of your book with Masahico Saito to [email protected] ? Not that I could not buy that book, but I do not support selling such things for money. $\endgroup$ Commented Sep 11, 2010 at 9:11
$\begingroup$
$\endgroup$
1
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...
answered Jul 22, 2010 at 19:44
-
$\begingroup$ Dan, thanks. Let's hope the quizzical posterity will accept your polynomial as "ultimate". I'll try to look through the papers you cite and see whether there is something about Reidemeister torsion in them as well. $\endgroup$ Commented Jul 23, 2010 at 5:46