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ag.algebraic geometry – Cohomology ring of a hypersurface in toric selection Answer

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ag.algebraic geometry – Cohomology ring of a hypersurface in toric selection

I cerebrate this will breathe finished, not less than for rational cohomology, utilizing the Lefschetz hyperplane theorem and the Hard Lefschetz theorem.

The pullback map $H^i (X, mathbb Q) to H^i ( Z, mathbb Q)$ is an isomorphism for $i < n-1$ and injective for $i=n-1$.

Thus Poincare duality provides an isomorphism $H^i (Z, mathbb Q) = H^{2n-2-i}(X,mathbb Q)^vee$ for $i > n+1$.

For $i = n+1$, the picture of $H^i(X,mathbb Q)$ inside $H^i (Z,mathbb Q)$ is a non-degenerate subspace for the Poincare duality pairing by the Hard Lefschetz theorem. So it has an orthogonal complement $V$, whose dimension you may compute.

We can thus narrate $H^i (Z, mathbb Q)$ as $H^i(X,mathbb Q)$ for $i<n$, $H^i (X,mathbb Q) oplus V$ for $i=n$, and $ H^{2n-2-i}(X,mathbb Q)^vee$ for $i>n$.

Using this, we will secure the ring construction. Consider two courses $alpha,beta$, let’s compute the cup product $alpha cup beta$.

If $alpha in H^i(X,mathbb Q)$ and $beta in H^j(X,mathbb Q)$, then we will take $alpha cup beta in H^{i+j}(X,mathbb Q)$ as a result of the pullback map is appropriate with cup merchandise. If $i+j>n$, we necessity to know tips on how to map to $H^{2n-2-i-j}(X,mathbb Q)^vee$, which is equal to taking a category $gamma$ in $H^{2n-2-i-j}(X,mathbb Q)$ and integrating $alpha cup beta cup gamma$ over $Z$, which is the identical as integrating $alpha cup beta cup gamma cup L$ over $X$, which we will do by describing the ring construction of $X$.

If $alpha in H^i (X,mathbb Q)$ and $beta in V$, then $alpha cup beta$ has diploma $>n-1$ so it suffices to combine $alpha cup beta cup gamma$ over $Z$ for $gamma in H^{n-1-i} (X,mathbb Q)$. But $(alpha cup beta cup gamma) = (alpha cup gamma) cup beta =0$ since $beta$ is within the orthogonal complement of $H^{n-1}(X,mathbb Q)$.

If $alpha in H^i (X,mathbb Q)$ and $beta in H^{2n-2-j}(X,mathbb Q)^vee$, then it suffices to combine $(alpha cup beta cup gamma)$ for $gamma in H^{2n-2-i-j}(X,mathbb Q)$, however that is simply the linear figure $beta$ utilized to $alpha cup gamma$.

If $alpha, beta in V$, then $alpha cup beta$ has diploma $2n-2$ so it suffices to compute the Poincare duality pairing on $V$. This is a nondegenerate symmetric pairing if $n-1$ is plane or a nondegenerate symplectic pairing if $n-1$ is queer. There is a exclusive such as much as isomorphism, so we will at all times select that one.

Any different pairing may have diploma $>2n-2$ and thus vanish.

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