Let $X$ be a comapct Riemann surface of genus $g$ and let $J\: : \: X\to \mathbb{C}^{g}/\Lambda$ be the Abel-Jacobi 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 1


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.

  • $\begingroup$ I edit the question because of too many $n$ $\endgroup$
    – Cepu
    Nov 21, 2017 at 13:46
  • $\begingroup$ 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 $\operatorname{Conf}_{l}(M)$? $\endgroup$
    – Cepu
    Nov 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. $\endgroup$
    – Cepu
    Nov 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.