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In 1968, Arnol'd proved that the integral cohomology of the pure braid group $P_n$ is isomorphic to the exterior algebra generated by the collection of degree-one classes $\omega_{i,j}\ (1 \le i < j \le n)$, subject to the following relation:

$$ \omega_{k,l} \omega_{l,m} + \omega_{l,m}\omega_{m,k} + \omega_{m,k}\omega_{k,l} = 0. $$

The classes $\omega_{k,l}$ are realizable as differential forms on the pure configuration space of $n$ points in $\mathbb C$ as follows: $$ \omega_{k,l} = \frac{1}{2 \pi i} \frac{dz_k - dz_l}{z_k - z_l}, $$

from which it can be seen that the classes $\omega_{k,l}$ are computing the winding number of the $k^{th}$ point around the $l^{th}$. In his paper (which can be found here), Arnol'd merely remarks that the above relation can be seen to hold for the forms $\omega_{k,l}$ by direct computation.

My question is, does the above relation have an interpretation in the context of winding numbers? How would I know to find this relation if I were trying to compute the cohomology of the pure braid group on a desert island?

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    $\begingroup$ At least you will have a lot of time to draw pictures of braids in the sand. $\endgroup$
    – Lee Mosher
    Commented May 18, 2013 at 19:03
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    $\begingroup$ One you notice that those forms are closed (and considering those forms is very natural once you see that you are working with an arrangement of hyperplanes), the relations of Arnold are the very first that come to mind. Why would one conjecture that these are all relations, I don't know, but that these forms generate the whole thing is a sensible optimistic guess —although Arnold surely had reasons too. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented May 18, 2013 at 19:09

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You know some kind of relation has to hold, since the configuration space of three points in the plane $C_3 \mathbb R^2$ has the homotopy-type of $(S^1 \vee S^1) \times S^1$, so $H^2$ only has rank $2$. The homotopy-equivalence comes from noticing the Faddell-Neuwirth fibration $C_1 (\mathbb R^2 \setminus \{0,1\}) \to C_3 \mathbb R^2 \to C_2 \mathbb R^2$ is trivial. But why a relation of that specific type? You can follow this line of reasoning to its conclusion and derive the relation from a close inspection of this model, say, using cellular cohomology. Because of the action of the symmetric group, there's basically no other relation possible (modulo a small sign issue).

Another way to go about it would be to think of cohomology as dual to homology, but for that you need a compact manifold. So you could compactify the manifold in the Fulton-Macpherson manner. In this model, $\omega_{i,j}$ is dual to the subspace of the compactified configuration space where $j$ is directly above $i$. Cup product corresponds to intersection product, so your relation boils down to saying that the homology class where $m$ is directly over $l$ and $l$ is directly over $k$ (or any cyclic permutation of that) is a boundary. You can write that as a boundary of a class -- the idea is to swing the bottom point of the configuration around to be the top point.

Those are two things that come to mind. I suspect someone like Dev Sinha or Fred Cohen have cuter ways of thinking of this.

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  • $\begingroup$ Ryan, can you direct me to an explicit reference that gives a proof that the above Faddell-Neuwirth fibration is trivial? Their paper doesn't have the statement. I would have thought that there is non-trivial monodromy. $\endgroup$
    – John Klein
    Commented Oct 5, 2021 at 14:06
  • $\begingroup$ Theorem IV.6.1 in Fadell and Husseini's book "Geometry and Topology of Configuration Spaces" states that the fiber bundle $C_k(\mathbb{R}^2)\rightarrow C_r(\mathbb{R}^2)$ is fiber-homotopically trivial if and only if $r\leq 2$. In the case $k=3$ and $r=2$, that's what you asked for, right? $\endgroup$
    – user164898
    Commented Oct 5, 2021 at 15:41
  • $\begingroup$ @A.S. Thanks much. $\endgroup$
    – John Klein
    Commented Oct 5, 2021 at 16:11
  • $\begingroup$ @JohnKlein: My preferred way to think about it is that $C_2 \mathbb R^2$ has the homotopy-type of $S^1$, i.e. a group. Think of that group as $SO_2$ acting on the plane by rotations. That gives you the splitting of the bundle at the fiber (rotate the configuration of three poiints so the first two points have a standard direction vector). $\endgroup$
    – Ryan Budney
    Commented Oct 5, 2021 at 16:55
  • $\begingroup$ @RyanBudney I looked at the Fadell and Husseini book yesterday. Despite their awful notation, the book does indeed prove it. I managed to distill the argument into the same one you made in your comment. Fadell and Husseini work with O(2) rather than SO(2) and that complicates their picture. $\endgroup$
    – John Klein
    Commented Oct 6, 2021 at 15:08

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