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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
6 days 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 6 days ago by Renat Abugaliev
Alexander Petrov
28 days 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.