Timeline for What is an interpretation of the relation in the cohomology of the pure braid groups?

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Oct 6, 2021 at 15:08	comment	added	John Klein		@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.
Oct 5, 2021 at 16:55	comment	added	Ryan Budney		@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).
Oct 5, 2021 at 16:11	comment	added	John Klein		@A.S. Thanks much.
Oct 5, 2021 at 15:41	comment	added	user164898		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?
Oct 5, 2021 at 14:06	comment	added	John Klein		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.
May 18, 2013 at 20:19	history	answered	Ryan Budney	CC BY-SA 3.0