Let $\bar{X}$ be a compact Riemann surface of genus $g>0$. Let $X$ be $\bar{X}$ minus a finite set of points $\{a_1,\ldots,a_n\}$ ($n\geq 1$). Let $X^{(r)}$ be the configuration space of $r$ ordered distinct points on $X$: $$X^{(r)}=X^r-\bigcup_{i\neq j}\{(x_1,\ldots,x_r): x_i=x_j\}$$

(1) Is there an explicit description of the Lie algebra associated to the lower central series of the fundamental group of $X^{(r)}$ (=pure braid group of $r$ strands on $X$) in terms of generators and relations?

(2) Are there cases (i.e. specific curves) for which an explicit description of $H^1_{dR}(X^{(r)})$ is known?

Any relevant references will be appreciated. For (1), a paper of Bezrukavnikov addresses $X=\bar{X}$ case. A paper of Nakamura, Takao and Ueno studies the non-compact situation but works with the weight central series (as opposed to the lower central series). In (2), by an explicit description I mean an explicit basis of $H^1_{dR}(X^{(r)})$.


1 Answer 1


I can at least answer your second question. I'll be a bit brief, but let me know if you need more details. Let $M$ be an oriented manifold and $M^{(r)}$ the configuration space as in your question. Then it follows from the Totaro spectral sequence, i.e. the Leray spectral sequence for $M^{(r)} \to M^r$, that the map $M^{(r)} \to M^r$ induces an isomorphism on $H^1(-,\mathbf Z)$ whenever $\dim(M) >2$ or when $M$ is a surface of positive genus with $n\geq 0$ punctures.

To see this use the exact sequence of low degree terms: $$ 0 \to H^1(M^r,\mathbf Z) \to H^1(M^{(r)},\mathbf Z) \to E_2^{0,1} \to E_2^{2,0}$$ If $\dim(M)>2$ then $E_2^{0,1}=0$. When $\dim(M)=2$ we have $E_2^{0,1} = \mathbf Z^{n(n-1)/2}$ -- a copy of $\mathbf Z$ for each small diagonal -- which injects into $E_2^{2,0} = H^2(M^r,\mathbf Z)$ because each diagonal maps to a different Künneth component of the form $H^1(M,\mathbf Z) \otimes H^1(M,\mathbf Z)$ (the class of the diagonal in $H^1(M,\mathbf Z) \otimes H^1(M,\mathbf Z)$ is nonzero precisely when $g>0$). The result follows.

  • $\begingroup$ Thanks. I read parts of Totaro's paper after this answer and found your explanation clear and helpful. $\endgroup$
    – P.E.
    Mar 7, 2017 at 18:29

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