Homological Representations of the Braid Groups
3.4 The Alexander–Conway polynomial of links
3.4.2 The Alexander–Conway polynomial
Therefore,
(1 +t+· · ·+tn−1) det(ψn+1r (β+)−In)
= det
⎛
⎝X−In−2 Y 0
Z T−1 t
0 0 −(1 +t+· · ·+tn)
⎞
⎠.
This implies (3.11). Hence
fn+1(σnι(β)) =fn+1(ι(β)σn) =fn+1(β+) =fn(β).
A similar argument shows thatfn+1(σ−n1ι(β)) =fn(β). This verifies the sec- ond condition in the definition of a Markov function.
For an oriented link L ⊂R3, set f(L) =fn(β), where β is an arbitrary braid on n strings whose closure is isotopic to L. By Section 2.5.2 and the previous lemma,f(L) is an isotopy invariant of Lindependent of the choice ofβ. We study this invariant in the next subsection.
L+ L− L0
Fig. 3.4.A Conway triple
L+ L− L0
Fig. 3.5.Example of a Conway triple
Theorem 3.13.The Alexander–Conway polynomial of links exists and is unique. The invariantfof links in R3 constructed in Section 3.4.1 coincides with the Alexander–Conway polynomial.
Proof. We first prove the uniqueness: there is at most one mapping from the set of oriented links in R3 to Z[s, s−1] satisfying axioms (i)–(iii). The proof requires the notion of an ascending link diagram, which we now introduce.
An oriented link diagramDonR2isascending if it satisfies the following two conditions:
(a) the components ofDcan be indexed by 1, . . . , m(wheremis the number of the components) so that at every crossing of distinct components, the component with smaller index lies below the component with larger index;
(b) each component ofD can be provided with a base point (not a crossing) such that starting from this point and moving along the component in the positive direction, we always reach the self-crossings of this component for the first time along the undergoing branch and for the second time along the overgoing branch.
An example of an ascending link diagram is given in Figure 3.6. It is a simple geometric exercise to see that the link presented by an ascending diagram is necessarily trivial.
Suppose now that there are two mappings from the set of oriented links in R3 to Z[s, s−1] satisfying axioms (i)–(iii) of the Alexander–Conway poly- nomial. Let∇ be their difference. We have to prove that∇= 0.
1 2
Fig. 3.6.An ascending link diagram
It is clear from the axioms and the computation before the statement of the theorem that ∇ is an isotopy invariant of links annihilating trivial knots and links and satisfying the Alexander–Conway skein relation. We prove by induction on N that ∇ annihilates all oriented links presented by link diagrams withN crossings. ForN = 0, this is obvious, since a link presented by a diagram without crossings is trivial. Suppose that our claim holds for a certain N. Let L be an oriented link presented by a link diagram with N+ 1 crossings. Exchanging over/undergoing branches at a single crossing, we obtain a diagram of another link, L. The links L, L together with the linkL0obtained by smoothing the crossing in question form a Conway triple as in Figure 3.4. The linkL0is presented by a link diagram withN crossings. By the induction assumption,∇(L0) = 0. The skein relation gives∇(L) =∇(L).
Thus, the value of∇ onL is not changed when overcrossings are traded for undercrossings. However, these operations can transform our diagram into an ascending one. Since∇annihilates the links presented by ascending diagrams,
∇(L) = 0. This completes the induction step. Hence∇= 0.
To prove the remaining claims of the theorem, it is enough to show that the link invariantfconstructed in the previous subsection satisfies the axioms of the Alexander–Conway polynomial. By Corollary 2.9 and the results above,f is a well-defined isotopy invariant of links. IfLis a trivial knot, thenLis the closure of a trivial braid on one string and thereforef(L) = 1. We verify now thatfsatisfies the Alexander–Conway skein relation.
Givenn≥2,i∈ {1, . . . , n−1}, and two braidsα, β∈Bn, we see directly from the definitions that the closures of the braidsασiβ, ασi−1β, andαβform a Conway triple of links inR3. The proof of Alexander’s theorem (Theorem 2.3) shows that conversely, an arbitrary Conway triple of links inR3arises in this way from certainn, i, α, β. Thus, we need to prove the identity
fn(ασiβ)−fn(ασ−i 1β) = (s−1−s)fn(αβ).
Since fn is invariant under conjugation in Bn and σi is a conjugate of σ1 inBn(see Exercise 1.1.4), we can assume without loss of generality thati= 1.
Further conjugating byα, we can assume thatα= 1. Thus we need to prove that for anyβ∈Bn,
fn(σ1β)−fn(σ−11β) = (s−1−s)fn(β).
This reduces to the equality
s−1g(D+)−s g(D−) = (s−1−s)g(D0), (3.13) where
D±= det(ψnr(σ±11β)−In−1) and D0= det(ψrn(β)−In−1). Multiplying both sides of (3.13) bys, we reduce (3.13) to the equality
D+−tD−= (1−t)D0. To verify the latter, we expand
ψrn(β) =
⎛
⎝a b x
c d y
p q M
⎞
⎠,
wherea, b, c, d∈Λ,x, yare rows overΛof lengthn−3,p, qare columns overΛ of heightn−3, andM is an (n−3)×(n−3) matrix overΛ. By definition, D0= detA0, where
A0=
⎛
⎝a−1 b x
c d−1 y
p q M−In−3
⎞
⎠.
Also
ψrn(σ1β) =
⎛
⎝−t 0 0
1 1 0
0 0 In−3
⎞
⎠
⎛
⎝a b x
c d y
p q M
⎞
⎠=
⎛
⎝−ta −tb −tx a+c b+d x+y
p q M
⎞
⎠.
SubtractingIn−1, then multiplying the first row by−t−1, and finally sub- tracting the first row from the second one, we obtainD+=−tdetA+, where
A+=
⎛
⎝a+t−1 b x c−t−1 d−1 y
p q M −In−3
⎞
⎠.
Similarly,
ψrn(σ1−1β) =
⎛
⎝−t−1 0 0 t−1 1 0 0 0 In−3
⎞
⎠
⎛
⎝a b x
c d y
p q M
⎞
⎠
=
⎛
⎝ −t−1a −t−1b −t−1x t−1a+c t−1b+d t−1x+y
p q M −In−3
⎞
⎠.
Subtracting In−1, then adding the first row to the second one, and finally multiplying the first row by−t, we obtainD− =−t−1detA−, where
A−=
⎛
⎝a+t b x c−1 d−1 y
p q M−In−3
⎞
⎠.
The matricesA0,A+,A− differ only in the first columns, which we denote by A10,A1+,A1−, respectively. Clearly,
−tA1++A1−= (1−t)A10. We conclude thatD+−tD−= (1−t)D0.
The functionL →f(L) satisfies all conditions of the Alexander–Conway polynomial except that a priori it takes values in the field of rational functions in s rather than in its subring of Laurent polynomials Z[s, s−1]. However, applying the skein relation and an induction on the number of crossings of a link diagram as at the beginning of the proof, one observes that all the values offare integral polynomials ins−s−1. In particular, all the values offare
Laurent polynomials ins.