Koszulness of the cohomology ring of moduli of stable genus zero curves - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T21:15:18Z http://mathoverflow.net/feeds/question/99613 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/99613/koszulness-of-the-cohomology-ring-of-moduli-of-stable-genus-zero-curves Koszulness of the cohomology ring of moduli of stable genus zero curves Dan Petersen 2012-06-14T14:47:47Z 2012-06-15T17:17:56Z <p>Let $n \geq 3$. The ring <code>$H^\bullet(\overline{M}_{0,n},\mathbf Q)$</code> was determined by Sean Keel. It is generated by the cohomology classes of boundary divisors <code>$D_{A,B}$</code> corresponding to partitions <code>$A \sqcup B = \{1,\ldots,n\}$</code> 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 <code>$D_{A,B} \cdot D_{A',B'}$</code> 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 <code>$$\sum_{\substack{\{i,j\} \subseteq A \\ \{k,l\} \subseteq B}} D_{A,B} = \sum_{\substack{\{i,k\} \subseteq A \\ \{j,l\} \subseteq B}} D_{A,B}$$</code> which follows by pulling back the WDVV relation on <code>$\overline{M}_{0,4}$</code> to <code>$\overline M_{0,n}$</code>. </p> <p>It follows in particular that <code>$H^{2\bullet}(\overline{M}_{0,n},\mathbf Q)$</code> 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?</p> http://mathoverflow.net/questions/99613/koszulness-of-the-cohomology-ring-of-moduli-of-stable-genus-zero-curves/99726#99726 Answer by Neil Strickland for Koszulness of the cohomology ring of moduli of stable genus zero curves Neil Strickland 2012-06-15T17:17:56Z 2012-06-15T17:17:56Z <p>Here is an alternative presentation of the cohomology, taken from the unpublished PhD thesis of my student Daniel Singh.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.</p> <p>Put <code>$S=\{1,\dotsc,n-1\}$</code>. </p> <ul> <li>For each subset $T\subseteq S$ with $|T|>1$ we have a generator $x_T$ in degree two. </li> <li>For each pair of sets $T,U$ with $T\cap U\neq\emptyset$ we have $(x_{T\cup U}-x_T)(x_{T\cup U}-x_U)=0$.</li> <li>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 $x_T^m\prod_i(x_T-x_{U_i})=0$.</li> <li>Moreover, there are no more generators or relations.</li> </ul> <p>One can also give a basis for the cohomology consisting of monomials in the generators $x_T$. </p> <ul> <li>Consider a monomial $y=\prod_Tx_T^{n_T}$. The <em>shape</em> of $y$ is <code>$\{T : n_T&gt;0\}$</code>. </li> <li>We say that a collection $\mathcal{F}$ of subsets of $S$ is a <em>forest</em> if all elements have size at least two, and any two elements are either disjoint or nested. </li> <li>Given a forest $\mathcal{F}$ and a set $T\in\mathcal{F}$, let $U_1,\dotsc,U_r$ be the maximal elements of <code>$\{V\in\mathcal{F}:V\subset T\}$</code>, and then put $m(\mathcal{F},T)=(|T|-1)-\sum_i(|U_i|-1)$. </li> <li>We say that our monomial $y$ is <em>admissible</em> if $\text{shape}(y)$ is a forest and $n_T\lt m(\text{shape}(y),T)$ for all $T\in\text{shape}(y)$. </li> </ul> <p>It can be shown that the admissible monomials form a basis for the cohomology.</p> <p>I do not know whether the algebra is Koszul, but I think that this presentation is well-adapted for investigating that question.</p>