Braids, Knots, and Links
2.3 Alexander’s theorem
We establish here a fundamental theorem, due to J. W. Alexander, asserting that all links inR3are isotopic to closed braids.
2.3.1 Closed braids in R3
Pick a Euclidean circle in the planeR2× {0} ⊂R3 with center at the origin O = (0,0,0). We identify a closed cylindrical neighborhood of this circle in R3 with the solid torus V = D ×S1. By a closed n-braid in R3, we shall mean an oriented geometric link in R3 lying in V ⊂ R3 as a closed n-braid with its canonical counterclockwise orientation (cf. Figure 2.3, where the planeR2× {0}is the plane of the picture).
In particular, for anyβ∈Bn, the closed braidβ⊂V yields a closed braid inR3via the inclusionV ⊂R3; cf. Figure 2.4. The latter closed braid is also denoted byβand is called theclosureofβ. A diagram ofβis obtained from a diagram ofβby connecting the bottom endpoints with the top endpoints byn standard arcs; cf. Figure 2.5, where the dotted circles should be disregarded.
We stress that closed braids inR3 areoriented geometric links.
For example, the closure of the trivial braid on n strings is a trivial n- component link. The closure of σ±11 ∈ B2 is a trivial knot. The closure of σ1±2 ∈B2 is an oriented Hopf link. More generally, the closure of σ1m ∈ B2
withm∈Zis a so-calledtorus(2, m)-link. It has two components for evenm and one component for oddm.
We can give an equivalent but more direct definition of closed braids inR3. Consider the coordinate axis={(0,0)}×R⊂R3meeting the planeR2×{0} at the origin O = (0,0,0). The counterclockwise rotation about O in the planeR2×{0}determines a positive direction of rotation about. An oriented geometric link L ⊂ R3− is a closed n-braid if the vector from O to any pointX∈Lrotates in the positive direction about whenX moves along L in the direction determined by the orientation ofL. The equivalence of this definition with the previous one can be seen as follows. Pick a Euclidean diskD lying in an open half-plane bounded byin R3and having its center in R2× {0}. Rotating D around , we sweep a solid torus V = D×S1 as above. TakingDbig enough, we can assume that a given linkL⊂R3−lies in V. It is clear thatL is a closed braid in the sense of the first definition if and only ifLis a closed braid in the sense of the second definition.
Theorem 2.3 (J. W. Alexander). Any oriented link in R3 is isotopic to a closed braid.
Proof. By apolygonal link, we shall mean a geometric link inR3 whose com- ponents are closed broken lines. Byvertices and edges of a polygonal link, we mean the vertices and the edges of its components. It is well known that any geometric link inR3 is isotopic to a polygonal link (cf. the proof of The- orem 1.6). We need only to prove that any oriented polygonal linkL ⊂R3 is isotopic to a closed braid. Moving slightly the vertices ofLin R3, we ob- tain a polygonal link isotopic toL. We use such small deformations to ensure thatL⊂R3− and that the edges ofL do not lie in planes containing the axis={(0,0)} ×R. Let
AC⊂L⊂R3−
be an edge ofL, whereLis oriented fromAtoC. The edgeAC is said to be positive(resp.negative) if the vector from the originO∈to a pointX ∈AC rotates in the positive (resp. negative) direction aboutwhenXmoves fromA to C. The assumption that AC does not lie in a plane containing implies that AC is necessarily positive or negative. The edge AC of L is said to be accessible if there is a pointB ∈ such that the triangleABC meets Lonly alongAC.
If all edges ofLare positive, thenLis a closed braid and there is nothing to prove. Consider a negative edgeAC ofL. We replaceAC with a sequence of positive edges as follows. If AC is accessible, then we pick B ∈ such that the triangleABC meets L only alongAC. In the plane ABC we take a slightly bigger triangleABC containing B in its interior, meeting only at B, and meeting L only along AC; see Figure 2.7. We apply to L the Δ-move Δ(ABC) replacingAC with two positive edgesAB and BC (see Section 1.2.3 for similar moves on geometric braids; in contrast to the setting of braids, we impose here no conditions on the third coordinates of the vertices).
The resulting polygonal link is isotopic toLand has one negative edge fewer thanL.
B
C
A
B
Fig. 2.7.The triangleABC
Suppose that the edgeACis not accessible. Note that every pointP ofAC is contained in an accessible subsegment ofAC. (To see this, pickB∈such that the segmentP BmeetsLonly atP and then slightly “thicken” this seg- ment inside the triangleABC to obtain a triangleP−BP+ meeting Lalong its sideP−P+⊂AC containingP. Then P−P+ is an accessible subsegment of AC.) Since AC is compact, we can split it into a finite number of con- secutive accessible subsegments. We apply to each of them the Δ-move as above choosing the corresponding pointsB∈distinct and choosingB close enough toB to stay away from other edges ofL. Since AC does not lie in a plane containing, the triangles determining these Δ-moves meet only at the common vertices of the consecutive subsegments ofAC (to see this, consider the projections of these segments and triangles to the plane{0} ×R2orthog- onal to ). Therefore these Δ-moves do not hinder each other and may be performed in an arbitrary order. They replaceAC⊂Lwith a finite sequence of positive edges, beginning atA and ending at C. The resulting polygonal link is isotopic toLinR3. Applying this procedure inductively to all negative edges ofL, we obtain a closed braid isotopic toL.
Exercise 2.3.1.Verify that the oriented 2-component links obtained by clos- ingσ12∈B2 and σ1−2 ∈B2 are not isotopic, while the underlying unoriented links are isotopic. (Hint: Consider the linking number of the components.) Exercise 2.3.2.Observe that the closure ofσ13is the trefoil knot shown on the left of Figure 2.1 and endowed with an orientation. Observe that the closure ofσ−11σ2σ−11σ2is the figure-eight knot shown in Figure 2.1 and endowed with an orientation.
Exercise 2.3.3.Verify that the oriented link inR3obtained by inverting the orientation of all components of the closure of a braidσir1
1σir2
2 · · ·σrim
m, where r1, r2,. . . , rm∈Z, is isotopic to the closure ofσirmm· · ·σri22σir11.