Let $n \geq 3$. The ring $H^\bullet(\overline{M}_{0,n},\mathbf Q)$ was determined by Sean Keel. It is generated by the cohomology classes of boundary divisors $D_{A,B}$ corresponding to partitions $A \sqcup B = \{1,\ldots,n\}$ of the marked points with $|A|, |B| \geq 2$, and where $D_{A,B} = D_{B,A}$. All relations are given by: (i) demanding that the product $D_{A,B} \cdot D_{A',B'}$ vanishes if the two divisors are disjoint (i.e. if there are no containments between the four sets $A,A',B$ and $B'$); (ii) the relation $$ \sum_{\substack{\{i,j\} \subseteq A \\ \{k,l\} \subseteq B}} D_{A,B} = \sum_{\substack{\{i,k\} \subseteq A \\ \{j,l\} \subseteq B}} D_{A,B} $$ which follows by pulling back the WDVV relation on $\overline{M}_{0,4}$ to $\overline M_{0,n}$.

It follows in particular that $H^{2\bullet}(\overline{M}_{0,n},\mathbf Q)$ is a quadratic algebra. I was asked during a seminar today whether this algebra is Koszul, but I had no idea what to answer. So, is it Koszul? If so, is its Koszul dual interesting?

  • $\begingroup$ For $n=3$ none of the $\sum$ relations amongst generators occur, so you have a quadratic commutative monomial algebra. Such things are always Koszul. $\endgroup$
    – M T
    Jun 14 '12 at 15:44
  • 3
    $\begingroup$ @mt -- The moduli space $\overline{M}_{0,3}$ is not particularly interesting. $\endgroup$ Jun 14 '12 at 16:33
  • $\begingroup$ Moreover, Dan surely meant to say that $|A|,|B|\ge 2$ so for $n=3$ the set of generators is empty, and the algebra is just $\mathbb{Q}$, as it should be. $\endgroup$ Jun 15 '12 at 12:23
  • $\begingroup$ I gave a more precise definition now. Thanks, Vladimir. $\endgroup$ Jun 15 '12 at 14:06
  • 1
    $\begingroup$ One quick remark: even the somewhat simpler cohomology algebras of the real loci of moduli spaces are only conjectured to be Koszul, see arxiv.org/abs/math/0507514 - so I would be very surprised if this actually is known. It's a very nice question though. $\endgroup$ Jun 16 '12 at 9:44

It is: https://arxiv.org/abs/1902.06318 - this paper also explains how to use the Koszul dual algebra for something, where something is estimating Betti numbers of the free loop spaces of $\overline{M}_{0,n}$; those, by Gromov & Ballmann-Ziller, allow one to estimate the number of closed geodesics of bounded length.


Here is an alternative presentation of the cohomology, taken from the unpublished PhD thesis of my student Daniel Singh. It has the disadvantage that one marked point is treated specially, so some symmetry is lost, but otherwise has many pleasant properties.

Put $S=\{1,\dotsc,n-1\}$.

  • For each subset $T\subseteq S$ with $|T|>1$ we have a generator $x_T$ in degree two.
  • For each pair of sets $T,U$ with $T\cap U\neq\emptyset$ we have $(x_{T\cup U}-x_T)(x_{T\cup U}-x_U)=0$.
  • Now consider a set $T$ as before, and disjoint subsets $U_1,\dotsc,U_r\subseteq T$, again with $|U_i|>1$. Put $m=(|T|-1)-\sum_i(|U_i|-1)$. Then $x_T^m\prod_i(x_T-x_{U_i})=0$.
  • Moreover, there are no more generators or relations.

One can also give a basis for the cohomology consisting of monomials in the generators $x_T$.

  • Consider a monomial $y=\prod_Tx_T^{n_T}$. The shape of $y$ is $\{T : n_T>0\}$.
  • We say that a collection $\mathcal{F}$ of subsets of $S$ is a forest if all elements have size at least two, and any two elements are either disjoint or nested.
  • Given a forest $\mathcal{F}$ and a set $T\in\mathcal{F}$, let $U_1,\dotsc,U_r$ be the maximal elements of $\{V\in\mathcal{F}:V\subset T\}$, and then put $m(\mathcal{F},T)=(|T|-1)-\sum_i(|U_i|-1)$.
  • We say that our monomial $y$ is admissible if $\text{shape}(y)$ is a forest and $n_T\lt m(\text{shape}(y),T)$ for all $T\in\text{shape}(y)$.

It can be shown that the admissible monomials form a basis for the cohomology.

I do not know whether the algebra is Koszul, but I think that this presentation is well-adapted for investigating that question.

  • 1
    $\begingroup$ Unless I am misunderstanding you, this is no longer a quadratic algebra? $\endgroup$ Jun 16 '12 at 7:48
  • $\begingroup$ I have the same question as Dan. This algebra seems to be manifestly non-quadratic, so something must have gone wrong. $\endgroup$ Jun 16 '12 at 9:37
  • $\begingroup$ Indeed, something is funny here. I have not thought about this for a few years, I will have to read it properly again. $\endgroup$ Jun 16 '12 at 10:39

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.