## Lecture 8 (20.10.2015)

Last time we discussed the determinant of a link, which turned out to be the strongest invariant we have so far. Today we will briefly consider a related construction, known as the Alexander polynomial.

To formulate it, we will consider oriented diagrams (the determinant did not require a choice of orientation.)

The main idea is to replace the “usual crossing condition” $2a-b-c=0\mod p$ from Def. 2.4 by a parameter-dependent version. Namely, we let $t$ be a real number, and ask that at each crossing of a given diagram $D$, the equation

$$(1-t)\cdot a-b+t\cdot c=0$$

holds, where $a$ labels the overarc and $b,c$ the underarcs. It is now important to distinguish between $b$ and $c$, and this is done with the help of the orientation:

We have not given the understrand an arrow here, because this is not relevant at the moment: Just look at the crossing from such a direction that the overstrand goes from the left to the right. Then the overstrand variable, $a$, gets a factor $(1-t)$, the lower underarc variable, $b$, gets a factor $-1$, and the upper underarc variable, $c$, gets a factor $t$. Clearly, if we put $t=-1$, we are in the old situation which we already studied in the context of $p$-colourability.

However, now we are not working modulo some prime, and $t$ is a real (not necessarily integer) free parameter.

Definition 2.10: Let $D$ be an oriented link diagram which has no closed loops as arcs. Label the arcs and crossings of $D$ in some arbitrary manner, and write down the $t$-dependent $|D|\times|D|$-matrix $\tilde{A}(t)$ collecting all the equations of the form $(1-t)a-b+tc=0$ as above. Delete one arbitrarily chosen row and one arbitrarily chosen column from $\tilde{A}(t)$. Call the resulting $|D|\times|D|$-matrix $A(t)$. The Alexander polynomial of $D$ is $$\Delta_D(t):=\det A(t).$$

$\Delta_D(t)$ is indeed a polynomial in $t$, hence the name.

Examples:

• The Alexander polynomial of the unknot is $\Delta_{0}(t)=1$ for all $t\in\mathbb R$.
• The Alexander polynomial of a trefoil diagram is $\Delta_{\rm trefoil}(t)=t^2+1-t$, as was calculated in the lecture.

We made several choices in computing the Alexander polynomial. In comparison to the determinant, things are not quite as nice here: The choices made do have a more significant influence on $\Delta_D(t)$, not just a factor $\pm1$ that can be removed by taking the absolute value.

Theorem 2.11: Let $L_1\cong L_2$ be two equivalent oriented links, and $D_1$, $D_2$ diagrams of $L_1$, $L_2$ (without closed loops as arcs). Then there exists an integer $m\in\mathbb Z$ and a choice of $\pm$ such that $$\Delta_{D_1}(t)=\pm t^m\cdot\Delta_{D_2}(t)\,,\qquad t\in\mathbb R.$$

This theorem holds of course in particular $L_1=L_2$ are one and the same link. Due to the choices we have in labeling and deleting rows/columns, there is no unique Alexander polynomial attached to a link, or even diagram. But since the effect is only by a factor $\pm t^m$, one can still tell if two given Alexander polynomials can belong to equivalent links, and in this sense, the Alexander polynomial is “almost” a link invariant. For example, the polynomial $t^2+t-1$ is not a product of $\pm t^m$ and $1$ for any $m\in\mathbb Z$. Hence the trefoil is non-trivial. But the polynomials $1$ and $-t^{10}$ are related by just a factor $-t^{10}$, which is of the form specified in Theorem 2.11. Hence, if we have a knot with Alexander polynomial $-t^{10}$, we do not know if it is trivial or not.

We agree to define the Alexander polynomial of an oriented link $L$ as $\Delta_D(t)$ for some of its diagrams, with some labeling, and some choice of deleted row/column. This is not unique, but can differ only by $\pm t^m$ when different a different diagram or labeling is used.

• In the situation of Thm. 2.11, we write, as a shorthand, $$\Delta_{D}(t)\sim\Delta_{D’}(t).$$ To make this an equivalence relation, it is a good idea to consider the Alexander polynomial as a Laurent polynomial, which means that we allow for both positive and negative powers of the variable $t$. For example, $t^5+2t^{-1}+3$ is a Laurent polynomial.
• The proof of Thm. 2.11 will not be given here, see the book of Livingston, page 50, for a sketch of it. Later we will revisit the Alexander polynomial from a completely different point of view, from which we will be able to extract its properties easily.

The Alexander polynomial is “an almost link invariant” (more precisely, the equivalence class of the Alexander polynomial under the equivalence $\sim$ defined above is a true invariant in the sense of Def. 1.7) which is stronger than the determinant – it can distinguish certain knots that the determinant cannot distinguish, see Problem 6c) for an exercise in this direction. Moreover, any two links that can be distinguished by the determinant can also be distinguished by the Alexander polynomial.

Before we show that, let us collect some properties of the Alexander polynomial. In its formulation, the mirror image $L^*$ of a link $L$ appears (cf. Problem 3) and also the reverse, written $rL$, appears: $rL$ is defined as the same link, but with all orientations opposite to those of $L$.

Proposition 2.12: Let $L$ be an oriented link. Then

• $|\Delta_L(-1)|=\det(L)$.
• $\Delta_{rL}(t)\sim\Delta_L(t^{-1}).$
• $\Delta_{L^*}(t)\sim\Delta_L(t^{-1}).$

Proof: The first part is clear by the definition of $\Delta_L$ and $\det L$. For the second part, consider a crossing

and compare it with the reversed crossing

In the first case, we have the equation $$(1-t)a-b+tc=0,$$ and in the second case, we have the equation $$(1-t)a-c+tb=0.$$ If we multiply the first equation by $-t^{-1}$, and replace $t$ by $t^{-1}$ in the second equation, both coincide. Hence the corresponding $|D|\times|D|$-matrices are related by$$-t^{-1}\tilde{A}_L(t)=\tilde{A}_{rL}(t^{-1}).$$ If we now delete the same rows and columns, we also obtain $$-t^{-1}A_L(t)=A_{rL}(t^{-1}).$$ From here, we compute $$\Delta_{rL}(t^{-1})=\det A_{rL}(t^{-1})=\det (-t^{-1}A_L(t))=(-t^{-1})^{|D|-1}\det A_L(t)=(-t^{-1})^{|D|-1}\Delta_L(t)\sim\Delta_L(t).$$

The proof of the third part is closely related to Problem 7), and therefore not given here.

• We now give a proof of the claim we made earlier: Any two links that can be distinguished by the determinant can also be distinguished by the Alexander polynomial. Equivalently, this means that in case $L_1,L_2$ are two links with $\Delta_{L_1}\sim\Delta_{L_2}$ (i.e. they can not be distinguished by the Alexander polynomial), then $\det L_1=\det L_2$. This follows from the following computation: $$\det L_1=|\Delta_{L_1}(-1)|=|\pm(-1)^m\Delta_{L_2}(-1)|=|\Delta_{L_2}(-1)|=\det L_2.$$
• The third statement of Prop. 2.12 essentially says that the Alexander polynomial cannot distinguish mirror images – if we flip the orientation of a the mirror image, the Alexander polynomials of $L$ and $L^*$ are equivalent.
• Another problem of the Alexander polynomial is that it can not detect the unknot. That means, there exist knots that are not equivalent to the unknot, but still have the same (or: equivalent) Alexander polynmial, which is $1$. An example of such a knot is this one:To check that this diagram has Alexander polynomial $\pm 1$, give it an orientation, write down the $10\times 10$ matrix (this diagram has 11 crossings) according to the rules set up in Def. 2.10, and compute its determinant.

For the last two remarks made here, see also the the slides presented in the lecture (pdf).

Computing invariants like determinants or Alexander polynomials can be done systematically and is therefore a task a computer is good at. This should be compared with the computation of eigenvalues of a matrix — you have to be able to do it on your own (in particular, in an exam), but to check the results, it is easy to work with a computer.

For example, you can use the knot explorer, already shown in the first lecture:

Knot Explorer from the Wolfram Demonstrations Project by Tom Verhoeff

The Knot Explorer works interactively if you have the corresponding browser plugins installed. Alternatively, you can use the knot explorer (and all other applets from the Wolfram Demonstrations Project) if you download the free CDF Player.