Chirality and the Conway Polynomial

I'd like to relay a simple conjecture in knot theory, which, despite its simplicity to state, remains unproven. It has to do with the notion of chirality and its opposite amphicheirality. A knot $K\subset S^3$ is said to be amphicheiral if it is isotopic to its mirror image. There are two types. If the strand orientation is preserved, it is said to be $+$ amphicheiral, while if the strand orientation is reversed, it is said to be $-$ amphicheiral. Equivalently in both cases, there exists an orientation-reversing homeomorphism $h\colon S^3\to S^3$ which fixes the knot setwise and which either preserves of reverses the knot's strand orientation. If the knot is hyperbolic, then $h$ can be realized by an isometry, and must be of finite order. One particularly interesting case is when $h$ is an involution. In that case, whether or not $K$ is hyperbolic, we say the knot is strongly $\pm$ amphicheiral. It is possible that a knot can be both $+$ and $-$ amphicheiral simultaneously. The figure $8$ knot has this property.

The Jones Polynomial $V_K(t)\in \mathbb Z[t,t^{-1}]$ is a knot invariant which does not see strand orientation and which satisfies $V_{K^*}(t)=V_{K}(t^{-1})$ where $K^*$ is the mirror image of $K$. Thus the Jones Polynomial of an amphicheiral knot must be symmetric with respect to $t$, and indeed, this is a good way to detect chirality of many knots, e.g. the trefoil.

The Conway Polynomial $C_K(z)\in \mathbb Z[z^2]$ on the other hand is a knot invariant which does not distinguish between $K$ and $K^*$, so it cannot detect chirality in the same way as the Jones. However, we have

Conjecture:
Let $\overline{C_K(z)}\in\mathbb Z_4[z^2]$ be the image of $C_K(z)$ under the natural projection $\mathbb Z[z^2]\twoheadrightarrow \mathbb Z_4[z^2]$. If $K$ is amphicheiral, then
$$\exists f(z)\in \mathbb Z_4[z] \text{ such that }\overline{C_K(z)}=f(z)f(-z)\in \mathbb Z_4[z].$$

(Like a true topologist, I am using $\mathbb Z_k$ to denote the cyclic group of order $k$.)

In this blog post, I would like to explain several confluent lines of evidence for this conjecture. I made an equivalent conjecture in 2006, in a paper with the same name as my blog post. Later, my student Vajira Manathunga showed that my conjecture is equivalent to the one just stated.

The case of strongly amphicheiral knots:
Kawauchi and Hartley proved some time ago that if $K$ is strongly amphicheiral, then $C_K(z)=f(z)f(-z)$ for some $f(z)\in \mathbb Z[z]$. Indeed in the $+$ case, they showed that $C_K(z)=f(z^2)^2$ for some $f$. (They phrased their results in terms of Alexander Polynomials.)

Later Hartley extended this result to the case of all $-$ amphicheiral knots. At this point, the reader may wonder whether the real conjecture ought to be over the integers and not over $\mathbb Z_4$, but it turns out to be false over $\mathbb Z$. Hartley and Kawauchi did not consider the general case in their papers. It may be that they didn't see a reason why it should hold, or perhaps they had a counterexample which they didn't publish. In any event the first published counterexample was due to Ermotti et al, who found a $+$ amphicheiral knot with an orientation reversing symmetry of order $4$, which does not split over $\mathbb Z$.

Independently of the Kawauchi and Hartley story, I was led to conjecture the $\mathbb Z_4$ version after looking at certain properties of Vassiliev invariants related to the Conway polynomial. Gratifyingly the Ermotti et all example does split over $\mathbb Z_4$. Manathunga has since come up with lots of examples which don't split over $\mathbb Z$. See our joint paper on the arXiv. Indeed, in that paper, we prove the conjecture for the case of a positive amphicheiral symmetry that preserves a braid axis, using the Burau representation to calculate the Alexander (and hence Conway) polynomial.

I would like to sketch the argument given by Hartley and Kawauchi for the strongly positive case as well as explain a little bit the Vassiliev invariant background which led to this conjecture.

Review of the Alexander Polynomial:
The Alexander Polynomial is an invariant which is strongly related to the Conway polynomial, and has a very pleasant topological definition. The complement of any knot in $S^3$ has $H_1(S^3\setminus K)=\mathbb Z$, so the covering space $X_K$ of $S^3\setminus K$ associated with the commutator subgroup has deck transformation group $\mathbb Z$. This means that $H_1(S^3\setminus K)$ is a $\mathbb Z[\mathbb Z]$-module. We will think of the group ring $\mathbb Z[\mathbb Z]$ as Laurent polynomials in $t$: $\mathbb Z[t,t^{-1}]$. Moreover, we will switch to rational coefficients, since $\mathbb Q[t^{\pm 1}]$ is a PID. Let $M_K$ be the first homology of $M_K$ with rational coefficients, looked upon as a $\Lambda=\mathbb Q[t^{\pm 1}]$ module. This is called the Alexander Module. One can show, for example by finding an explicit presentation for $M_K$, that $M_K$ is a finitely generated torsion module over $\Lambda$. By the classification of modules over a PID, we have
$$M_K\cong \bigoplus \Lambda/ f_i \Lambda$$
for some $f_i\in \Lambda$.

The Alexander polynomial $\Delta_K(t)$ is defined to be the product of all $p_i$: $\Delta_K(t)=\prod_i f_i$. I.e. it is the order of the Alexander module $M_K$. The polynomials $f_i$ are only well-defined up to units in $\Lambda$. So in addition to multiplying by rational numbers, one can also multiply by powers of $t$. The Alexander polynomial is therefore only well-defined up to these operations. However,  there is a satisfying normal form. It turns out that $\Delta_K(t)$ is equivalent up to multiplying by powers of $t$ to $\Delta_K(t^{-1})$. So one can assume that it is symmetric $\Delta_K(t)=\Delta_K(t^{-1})$. Moreover one can multiply by a constant so that $\Delta_K(1)=1$. With these conventions, though not obviously, $\Delta_K(t)$ has integer coefficients. In fact, Levine showed that each $f_i$ has these properties.

The Conway Polynomial is a repackaging of the Alexander polynomial under the change of variables $z=t^{1/2}-t^{-1/2}$.

Hartley and Kawauchi's Argument for strongly positive amphicheiral knots:
Theorem: If $K$ is strongly positive amphicheiral, then the Alexander module is a direct double $M_k\cong N \oplus N$ for some $\Lambda$-module $N$. This implies that the Conway polynomial is a square $f(z^2)^2$.

Proof: Let $\alpha\colon S^3\to S^3$ be an orientation-reversing involution that preserves the knot's strand orientation. It turns out that it must have two fixed points in the complement of the knot. This implies that $\alpha$'s restriction to $S^3\setminus K$ lifts to an involution $\tilde{\alpha}$ of the universal abelian cover $X_K$. We will use that $M_K$ has a nondegenerate sesquilinear pairing, called the Blanchfield form:
$$B\colon M_K\otimes M_K\to \mathbb Q(t)/\mathbb Q[t,t^{-1}].$$
By definition, $B(x,y)$ is defined as follows. Since $M_K$ is torsion, there is some $\Delta\in \mathbb Q[t,t^{-1}]$ such that $\Delta\cdot y=\partial \sigma$. Indeed one could simply let $\Delta$ be the Alexander polynomial. Then we let
$$B(x,y)=\frac{1}{\Delta}\sum t^{-n}(t^nx)\cdot \sigma,$$
where $\cdot$ denotes the algebraic intersection number.

$B$ is sesquilinear: $B(fx,gy)=f\bar g B(x,y)$ and $B(x,y)=\overline{B(y,x)}$.

Moreover, because $\alpha$ preserves the knot's orientation, it must reverse a meridian. Hence $\tilde{\alpha} t\tilde\alpha=t^{-1}$, which implies that $B(\tilde\alpha x,\tilde\alpha y)=-\overline{B(x,y)}.$ Now the trick is to define $B'(x,y)=B(x,\tilde{\alpha} y)$. It is not hard to see that this is an antisymmetric nondegenerate bilinear form (not sesquilinear) on $M_K$. Now $M_K$ is a finitely generated module over a PID; so it decomposes as a direct sum of pieces $\Lambda/f\Lambda$. Moreover we may assume that the $f$'s are powers of primes $p\in \Lambda$. We can therefore decompose $M_K$ into a direct sum of $p$-primary components: $$M_K=\oplus_p M_K^{(p)}$$ where each $M_K^{(p)}$ is a direct sum of cyclic modules of order $p^k$ for some $k$. It is not hard to see that this must be an orthogonal decomposition with respect to $B'$. Now, each $M_K^{(p)}$ decomposes into a direct sum of cyclic modules of order $p$, plus those of order $p^2$ and so forth. This decomposition is not necessarily orthogonal, but can be fixed up to be so, using an upper triangular argument. So we get summands $\oplus_{i=1}^{j(k,p)} \Lambda/p^k\Lambda$ supporting a nondegenerate antisymmetric bilinear form. Tensoring with the field $F=\Lambda/p\Lambda$, we get a symplectic vector space of dimension $j(k,p)$, which must therefore be even! This implies that $M_K$ is a direct double, as desired. $\Box$



So, now that we understand this classical result of Hartley and Kawauchi, how did Vassiliev invariants enter the picture? For the purposes of this blog post it is not that important to know what exactly a Vassiliev invariant is. It is a type of knot invariant. Some are "Vassiliev" and some are not. The name is used as an adjective to describe some knot invariants in much the way the proper name "Hausdorff" is used as an adjective to describe certain topological spaces. The first thing we want to bring attention to is that the coefficients of the Conway polynomial are Vassiliev invariants of knots:
$$C_K(z)=1+\sum_{i\geq 1}c_{2i}(K)z^{2i},$$
and their Vassiliev degree is $2i$. These invariants are not additive under connected sum, which can be desirable. A standard trick to make them additive is to apply the formal logarithm of power series $\log\colon 1+z\mathbb Z[[z]]\to z\mathbb Z[[z]]$ to $C_K(z)$ and then consider the coefficients. However, this trick does not preserve integrality of the coefficients. This leads to the following definitions:

Definition: Define $$\exp_{\mathbb Z}\colon z\mathbb Z[[z]]\to 1+z\mathbb Z[[z]]$$ by the formula $$\exp_{\mathbb Z}(a_1 z+a_2 z^2+\cdots)= (1-z)^{a_1}(1+z^2)^{a_2}(1-z^3)^{a_3}\cdots=\prod (1+(-z)^n)^{a_n}$$

It is not hard to see that $\exp_{\mathbb Z}$ is bijective and sends addition to multiplication. We define $\log_{\mathbb Z}\colon 1+z\mathbb Z[[z]]\to z\mathbb Z[[z]]$ as its inverse.

Primitive Vassiliev Invariants from the Conway Polynomial: We now define primitive (i.e. additive under connected sum) Vassiliev invariants of degree $2k$ coming from the Conway polynomial, $pc_{2k}\colon \{\text{Knots}\}\to \mathbb Z$, by
$$\log_{\mathbb Z}(C_K(z))=\sum_{i=1}^\infty pc_{2i}(K) z^{2i},$$
where the formal logarithm $\log_{\mathbb Z}$ is taken with respect to the variable $z^2$.

These invariants have an interesting property. Modulo $2$, they drop in degree. $pc_{2n}\pmod 2$ is of degree $2n-1$. On the other hand, no torsion is known among Vassiliev knot invariants. It is thus natural to conjecture

Conjecture: There exist integer-valued Vassiliev knot invariants $v_{2n-1}$ of degree $2n-1$ such that $v_{2n-1}\cong pc_{2n}\pmod 2$.

Such invariants have been explicitly identified for $n=1,2$.

A knot invariant taking values in an abelian group is said to be odd if it changes sign under mirror image. Thus odd $\mathbb Z$-valued knot invariants vanish on amphicheiral knots.

Somewhat unmotivated conjecture: The invariants $v_{4n-1}$ can be chosen to be odd.

This appears to be untrue for general $v_{2n-1}$, but experimental evidence suggested to me that it could be true in the stated half of cases! If this is true, it implies that $pc_{4n}\cong 0\pmod 2$ on amphicheiral knots.

The following proposition then tells us how these considerations on Vassiliev invariants lead to the mod $4$ splitting conjecture.

Proposition: The invariants $pc_{4n}\cong 0\pmod 2$ iff $C_K(z)$ factors as $f(z)f(-z)$ modulo $4$.

So, one approach to proving the mod 4 splitting conjecture is to explicitly identify degree $4n-1$ Vassiliev invariants $v_{4n-1}$ which have the same parity as $pc_{4n}$. This appears to be a difficult problem. Already for  $n=2$ the answer is not known. In degree $7$, the space of all Vassiliev invariants is spanned by polynomial invariants coming from the HOMFLY and Kauffman polynomials, together with a $2$-fold cable. Computer calculations show that $v_{4n-1}$ must involve this cabled invariant, though we have not explicitly identified it.