Let $X$ be a comapct Riemann surface of genus $g$ and let $J\: : \: X\to \mathbb{C}^{g}/\Lambda$ be the AbelJacobi map. This map is a smooth embedding. Let $p\in X$ such that $J(p)=\Lambda$ and consider $$ J\: : \: X\setminus p \to (\mathbb{C}\Lambda)/\Lambda. $$ Then $J^*$ induces a surjection in complex de Rham cohomology. $J$ is injective and hence is defined at the level of the (ordered) configuration space $$ \operatorname{Conf}_{l}(J)\: : \: \operatorname{Conf}_{l}(X\setminus p) \to \operatorname{Conf}_{l}((\mathbb{C}\Lambda)/\Lambda). $$ Let $V_{1}\subset H^{2}(\operatorname{Conf}_{l}(X\setminus p), \mathbb{C})$ , resp. $V_{2}\subset H^{2}(\operatorname{Conf}_{l}((\mathbb{C}\Lambda)/\Lambda), \mathbb{C})$ be the space generated by Massey products $(a_{1}, \dots, a_{n})$ between degree $1$ elements, for $n\geq 2$. Does $\operatorname{Conf}_{l}(J)^{*}$ induces a surjection $$ \operatorname{Conf}_{l}(J)^{*}\: : \: \mathbb{C}\oplus H^{1}(\operatorname{Conf}_{l}((\mathbb{C}\Lambda)/\Lambda), \mathbb{C})\oplus V_{2}\to \mathbb{C}\oplus H^{1}(\operatorname{Conf}_{l}(X\setminus p), \mathbb{C})\oplus V_{1}? $$ Notice that $\mathbb{C}$ represents the $0$th cohomology group.
1 Answer
The answer is yes.
First off, the map $\mathrm{Conf}_l M \to M^l$ induces an isomorphism on $H^1$ if $M$ is an oriented manifold such that $\dim M > 2$, or such that $\dim(M)=2$ and $M$ has positive genus. See my answer to a previous question fundamental group of configuration spaces of ordered points on open Riemann surfaces.
Secondly, if $X$ is a smooth projective variety, then $X \setminus \{pt\}$ is a formal topological space. This is because the mixed Hodge structure on the cohomology of $X \setminus \{pt\}$ is pure. See e.g. Dupont's paper https://arxiv.org/abs/1505.00717. So all the triple or higher Massey products of classes in $H^1(\mathrm{Conf}_l (X \setminus \{pt\})) = H^1((X\setminus \{pt\})^l)$ vanish, because the Massey products are functorial and they vanish in the cohomology of $(X\setminus \{pt\})^l$. The same holds for the jacobian.
So the result follows from what you've already said about the case $l=1$ and the Kunneth formula.


$\begingroup$ Thanks for your answer, It seems to me that you are saying that $V_{1}$ is generated by the ordinary product (all the higher product vanish), for any $l$. I'm right? Does this follows from the fact that $X\setminus{p}$ is formal, i.e. if $M$ is formal, then so is $\operatorname{Conf}_{l}(M)$? $\endgroup$– CepuNov 21, 2017 at 13:51

$\begingroup$ You're right, I was very confused. I edited the answer. $\endgroup$ Nov 21, 2017 at 16:29

$\begingroup$ Thanks! Could you explain why an isomorphism between first cohomology group implies an equality between Massey products? They are contained in the second cohomology group. Consider the map induced by $\operatorname{Conf}_{l}(M)\to M^{l}$ in the second cohomology group. I 'don't get why this map does not kill some Massey products. $\endgroup$– CepuNov 21, 2017 at 20:15

$\begingroup$ I don't understand what you mean. The map goes $H^2(M^l) \to H^2(\mathrm{Conf}_l(M))$, and the claim is that the Massey products vanish in $H^2(M^l)$. $\endgroup$ Nov 21, 2017 at 20:23