4
$\begingroup$

Given a smooth projective variety $X$ of dimension $l$, we denote with $F(X,n)$ the configuration space of points $$ F(X,n):=\{(x_{1}, \dots, x_{n})\in X^{n}\: : \: x_{i}\neq x_{j}\text{ for each }i,j \} $$ In https://www.jstor.org/stable/2946581?seq=1#page_scan_tab_contents there is an explicit rational dg algebra $E(n)$ such that $H^{\bullet}(E(n))\cong H^{\bullet}(X, \mathbb{Q})$ (a model). In particular , as an algebra, $E(n)$ is isomorphic to a free $H^{\bullet}(X^{n})$ algebra with generators $G_{ab}$ for $1\leq a,b \leq l$, modulo some easy relations. The same is true for the differential. Now, choose $k$ distinct points $y_{1}, \dots y_{k}$ on $X$. We consider the configuration space of $X-\{y_{1}, \dots , y_{k}\}$

$$F(X, n; y_{1}, \dots , y_{k}):=F(X-\{y_{1}, \dots , y_{k}\},n)$$

Here my question: do you know a model $E(n, k )$ that compute the rational cohomology of $F(X, n; y_{1}, \dots , y_{k})$? What are generators, relations and the differential?

I think that such a model can be obtained from $E(n)$ by the fact that $F(X, n; y_{1}, \dots , y_{k})$ is isomoprhic to the fiber at the point $(y_{1}, \dots, y_{k})$ of the projection $$ F(n+k,X)\rightarrow F(k, X) ,$$ but I wonder if there is something in the literature. I am interested to the case $dimX=1$.

Edit: I don't know if this is an open problem. I am interested to find a 1-minimal model for the case $dimX=1$, $k=1$. Is there a way procedure to compute $H^{1}( F(n, X;y) $ and the subspace $V\subset H^{2}( F(n, X;y)$ generated by $H^{1}( F(n, X;y) $ via the wedge product?

$\endgroup$
4
$\begingroup$

If you read Totaro's paper "Configuration spaces of algebraic varieties" he derives the same cdga calculating the cohomology of $F(X,n)$ as Kriz, but in a different way, via the Leray spectral sequence for $F(X,n) \to X^n$. First he calculates what the first nontrivial page of the spectral sequence looks like for an arbitrary oriented manifold, and he finds that it's given precisely by this expression $$H^\ast(X^n,\mathbf Q)[G_{ab}]/\text{relations.}$$ Then he shows that if $X$ is smooth projective algebraic, the spectral sequence degenerates after the first differential using a weight argument: no differential after the first one can be compatible with the weights in the natural mixed Hodge structure on the spectral sequence. The argument works more generally if $H^i(F(X,n),\mathbf Q)$ is pure of weight $i$ for all $i$ --- in particular, it works for a once-punctured smooth projective variety. So for $k=1$, the exact same cdga works to compute the cohomology of $F(X,n)$. No assumption on the 1-connectedness of $X$ is necessary.

Also. You may find the paper "Koszul dg-algebras arising from configuration spaces" by Bezrukavnikov useful.

Re your last question, let $X$ be a once punctured Riemann surface with first cohomology group $V$. Then $H^1(F(X,n),\mathbf Q) \cong V^{\oplus n}$, i.e. the map $F(X,n) \to X^n$ induces an isomorphism on $H^1$. The image of $H^1(F(X,n)) \otimes H^1(F(X,n)) \to H^2(F(X,n))$ is the direct sum of $\binom n 2$ copies of $V_{2,0} \oplus V_{1,1}$, where these denote representations of $\mathrm{Sp}(V)$. Specifically, $V_{2,0}$ is the symmetric square, and $V_{1,1}$ is the "primitive part" of the exterior square (the orthogonal complement of the class of the symplectic form).

$\endgroup$
7
  • $\begingroup$ Thanks, I have some difficult to visualize $V_{1,1}$. Here an example. assume that $X$ is a punctured elliptic curve (i.e. genus 1). Then $V$ is generated by $a,b$; $V_{0,2}$ is generated by $a^2, b^2, ab$ and $V_{1,1}=a \wedge b$. Is that correct? $\endgroup$
    – Cepu
    Jul 15 '16 at 12:14
  • $\begingroup$ ups $a\wedge b$ is symplectic, hence $V_{1,1}=0$ in this case. $\endgroup$
    – Cepu
    Jul 15 '16 at 13:02
  • $\begingroup$ Not exactly. If $V$ is generated by $a,b$ then $V \otimes V$ has the subspace $V_{0,2}$ (the symmetric square) gen'd by $a \otimes a$, $b \otimes b$, and $a \otimes b + b \otimes a$. The exterior square is spanned by $a \otimes b - b \otimes a$, which is also the class of the symplectic form, so $V_{1,1}=0$. $\endgroup$ Jul 15 '16 at 16:27
  • $\begingroup$ BTW - note that $H^2(X^n)$ is isomorphic to $\binom n 2$ copies of $V \otimes V$, and it's easy to see that the class of the symplectic form in each of these copies of $V \otimes V$ goes to zero in $H^2(F(X,n))$. Indeed, the "$(i,j)$th" copy of the symplectic form is exactly the class of the diagonal $\Delta_{ij}$ in $X^n$, so of course it goes to zero in the cohomology of $F(X,n)$ where we have removed all the diagonals. $\endgroup$ Jul 15 '16 at 16:29
  • $\begingroup$ A useful reference for you could be my paper "The structure of the tautological ring in genus one", specifically the final section, which does some very explicit computations about the cohomology of $F(X,n)$ where $X$ is a punctured elliptic curve. $\endgroup$ Jul 15 '16 at 16:31
2
$\begingroup$

The question is addressed by Berceanu, Markl, and Papadima in Multiplicative models for configuration spaces of algebraic varieties, Topology 44 (2005), no. 2, 415–440.

$\endgroup$
1
  • $\begingroup$ Thanks! The paper contain a solution to the problem when $X$ is simply connected and $k=1$ . Do you know if the general answer is an open problem? $\endgroup$
    – Cepu
    Jul 14 '16 at 17:22

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.