6.02 Braids: Artin action
Below the video you will some pre-class questions and notes to accompany the video. Apologies for the gurgling noises in the background of the video: it's my office radiator.
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Notes
The Artin action
(0.00) Recall from Braids 1 that the braid group on n strands is the fundamental group of the unordered configuration space UCn. This group acts on the group Z⋆n via automorphisms; this action is called the Artin action of the braid group on the free group. In this section we will outline the reason for this; in the next section, we will compute explicitly the Artin action for some simple braids.
Monodromy
(2.30) Inside C×UCn we have a tautological subspace Tn={(x,c) : x∈c}
In a former version of this module, I had already introduced covering spaces before talking about the Artin action, so the following paragraph made ϵ-more sense. If I were you, I would skip this paragraph on a first reading, read the section below where we do an example and then come back and re-read this paragraph when you've seen the videos on covering spaces and monodromy.
The universal family has a nice property: while it is not a covering space of the configuration space, it is a fibration over the configuration space. This means that if F:X×[0,1]→UCn is a map and ˜F0:X→Un is a lift of F|X×{0} then there exists a (not necessarily unique) lift ˜F:X×[0,1]→Un of F. In particular, given a path γ in UCn, we get a monodromy map from the fibre Fγ(0) to the fibre Fγ(1), which is well-defined up to homotopy. In particular, we get an action of π1(UCn,c) on π1(Fc,[z1,…,zn])≅Z⋆n. This is called the Artin action of the braid group on the free group.
Explicit action
(5.40) Rather than proving the fibration property of the universal family, let us see how some explicit braids act in the Artin action. Hopefully the idea will become clear.
Let n=2 and consider the elementary braid σ1. The fundamental group of C∖{z1,z2} is Z⋆Z, with generators α and β as drawn in the figure below.
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The braid σ1 moves the points z1 and z2 around one another until they switch places. The result on the loops α and β is as follows:
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(9.04) We can see that σ1(β)=α, and, after a homotopy, we can see that σ1(α)=αβα−1:
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(11.16) Therefore we have σ1(α)=αβα−1σ1(β)=α
(11.37) What about when we add more strands? In fact, if we focus just on elementary braids (which generate the braid group), we already know the answer: an elementary braid only affects two of the strands, and we can take the monodromy to be the identity away from these two strands. In other words, the action of σi on α1,…,αn is: α1↦α1α2↦α2⋮⋮αi↦αiαi+1α−1iαi+1↦αi⋮⋮αn↦αn.
In the next video, we will use the Artin action to compute the fundamental group of any knot complement.
Pre-class questions
- What is the Artin action of σ−11?
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