# 6.02 Braids: Artin action

## Video

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
\(UC_n\). This group acts on the group \(\mathbf{Z}^{\star n}\) 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 \(\mathbf{C}\times UC_n\) we have a tautological
subspace \[T_n=\{(x,c)\ :\ x\in c\}\] which meets the disc
\(\mathbf{C}\times\{c\}\) over a configuration \(c\in UC_n\) in the
\(n\) points defining the configuration \(c\). Consider the complement
of \(T_n\), that is \[U_n:=\left(\mathbf{C}\times UC_n\right)\setminus
T_n.\] This is called the *universal family over configuration
space*. The space \(U_n\) has a natural projection \[p\colon U_n\to
UC_n\] whose fibre \(F_c=p^{-1}(c)\) over \(c\) is precisely the plane
punctured along the configuration \(c\).

*In a former version of this module, I had already introduced
covering* *spaces before talking about the Artin action, so the
following* *paragraph made \(\epsilon\)-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 \(F\colon X\times [0,1]\to
UC_n\) is a map and \(\tilde{F}_0\colon X\to U_n\) is a lift of
\(F|_{X\times\{0\}}\) then there exists a (not necessarily unique) lift
\(\tilde{F}\colon X\times [0,1]\to U_n\) of \(F\). In particular, given a
path \(\gamma\) in \(UC_n\), we get a monodromy map from the fibre
\(F_{\gamma(0)}\) to the fibre \(F_{\gamma(1)}\), which is well-defined up
to homotopy. In particular, we get an action of \(\pi_1(UC_n,c)\) on
\(\pi_1(F_c,[z_1,\ldots,z_n])\cong\mathbf{Z}^{\star n}\). 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 \(\sigma_1\). The fundamental group of \(\mathbf{C}\setminus\{z_1,z_2\}\) is \(\mathbf{Z}\star\mathbf{Z}\), with generators \(\alpha\) and \(\beta\) as drawn in the figure below.

The braid \(\sigma_1\) moves the points \(z_1\) and \(z_2\) around one another until they switch places. The result on the loops \(\alpha\) and \(\beta\) is as follows:

*(9.04)* We can see that \(\sigma_1(\beta)=\alpha\), and, after a
homotopy, we can see that \(\sigma_1(\alpha)=\alpha\beta\alpha^{-1}\):

*(11.16)* Therefore we have

*(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 \(\sigma_i\) on
\(\alpha_1,\ldots,\alpha_n\) is:

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 \(\sigma_1^{-1}\)?

## Navigation

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**Braids: the Wirtinger presentation**. - Index of all lectures.