Looking for a better counterexample

This post is doubling my old question asked on Mathoverflow.

Let $X$ be a smooth projective complex variety of dimension $d$ , and let $Y\subset X$ be an ample irreducible divisor. Recall that a cohomology class $b\in H^k(X,\mathbf{Q})$ is called $Y$ -primitive, if $b\smile c_1(Y)^{d-k+1} =0$ . Here, $c_1(Y)\in H^{2}(X,\mathbf{Q})$ is the first Chern class of the line bundle associated to $Y$ . We denote by $H^k_{\mathrm{prim}}(X,\mathbf{Q}) \subset H^k(X,\mathbf{Q})$ the subset of primitive elements.

Let denote by $U$ the complement to $Y$ , and by $j\colon U \to X$ the open embedding. I claim that if $Y$ is smooth then the composition

$H^k_{\mathrm{prim}}(X,\mathbf{Q}) \subset H^k(X,\mathbf{Q}) \xrightarrow{j^*} H^k(U,\mathbf{Q})$

is an injection. I skip a proof; this claim is an exercise on the Gysin exact sequence and the Lefschetz theorems. Perhaps, it is more interesting to find counterexamples with a singular $Y$ . In my old question, I came up with something strange, and I hope that now somebody can give me a better example.

My construction goes as follows. Let $V$ be a 3-dimensional vector space. Consider $\mathbb{P}(V)=\mathbb{P}^2$ embedded in $\mathbb{P}(\mathrm{Sym}^2(V)) =\mathbb{P}^5$ by Veronese. Let $X$ be the Grassmann variety parameterizing 3-dimensional planes in $\mathbb{P}^5$ , so $X=\mathrm{Gr}(4,6)$ . Let $Y\subset X$ be the subset of planes which intersect embedded $\mathbb{P}^2$ non-transversally. One can show that $Y$ is irreducible. Moreover, $Y$ is ample because any effective divisor in $Gr(4,6)$ is ample.

I claim that the complement $U=X\setminus Y$ is the configuration space of four non-ordered points in $\mathbb{P}^2$ in general position. Skipping details, it follows from the fact that any four such points in $\mathbb{P}^2$ can be given as an intersection of two quadrics. Nevertheless, the group $PGL(V)$ acts on $U$ , and this action is transitive and with finite stabilizers as it follows from the previous description. Thus,

$H^*(U,\mathbf{Q})\cong H^*(PGL(V),\mathbf{Q}).$

In particular, $H^4(U,\mathbf{Q})=0$ . However, $\dim H_{\mathrm{prim}}^4(X,\mathbf{Q})=8$ .

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Renat Abugaliev

3 years ago

I shall write the simplest example I know but you can generalize it to all even dimensional smooth quadric (except surface) or to any Grassmannian except $\mathbb{P}^n$ . Let $X$ be a smooth quadric in $\mathbb{P}^5$ or the same $Gr(2,4)$ . We take as $Y$ the tangent hypeplane to a point $x \in X$ .
I claim that $H^4_{prim}(X,\mathbb{Z})=\mathbb{Z}$ and $H^4(U,\mathbb{Z})=0$ and it is a counterexample.
To obtain the first equality we need to show that $H^4(X,\mathbb{Z})=\mathbb{Z}^2$ . One can use the fact that even dimensional quadric has $\mathbb{Z}^2$ as middle cohomologies.
The second equality follows from the fact that $U=\mathbb{A}^4$ . The same ( $X=\mathbb{A}^n \sqcup Y$ ) is true for all quadrics and all Grassmannians.

Last edited 3 years ago by Renat Abugaliev

Alexander Petrov

4 years ago

Can’t we simply take $Y$ to be a union of several divisors on a surface $X$ whose classes span all of $H^2(X)$ ? For instance, take $X=\mathbb{P}^1\times \mathbb{P}^1$ with $Y=\mathbb{P}^1\times pt\cup pt\times \mathbb{P}^1$ . This divisor is ample and the class $[\mathbb{P}^1\times pt]-[pt\times\mathbb{P}^1]$ is primitive yet it surely dies in $H^2(U)=0$ .

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Nikolay Konovalov

Reply to Alexander Petrov

Sorry, I said it unclear in the main post. I want an example with a singular but still irreducible $Y$ .