I am reading the chapter on Griffiths Residues in Arapura's book "Algebraic Geometry over the Complex Numbers" and I find a particular statement very puzzling.

Let $X \subset \mathbb{P}^{n+1}$ be a hypersurface. Then Arapura claims that "by weak Lefschetz" the Gysin homomorphism $H^{n-1}(X) \rightarrow H^{n+1}(\mathbb{P}^{n+1})$ is an isomorphism. Here I mean singular cohomology over $\mathbb{C}$. But I thought that the Lefschetz hyperplane theorem (otherwise known as "weak Lefschetz") relates the cohomology of a variety with that of its hyperplane sections. $X$ need not be a hyperplane section, if could be a hypersurface of arbitrary degree.

  • $\begingroup$ Have you looked at Arapura's statement of the weak Lefschetz (theorem 14.3.1)? This is exactly what that theorem says. $\endgroup$
    – KReiser
    Mar 23 '20 at 18:37
  • $\begingroup$ @KReiser I have done. Theorem 14.3.1 is stated for $Y \subset X$ where $Y=X \cap H$, where $H$ is a hyperplane. I am confused precisely because in my question $X \not = \mathbb{P}^{n+1} \cap H$. $\endgroup$
    – baltazar
    Mar 23 '20 at 18:56

A hypersurface is always a hyperplane section by choosing an "ampler" projective embedding.

Let $d$ be the degree of $X$, and let $\mathbb P^N$ parameterizes degree $d$ monomials in $\mathbb P^{n+1}$. By Veronese embedding $$i: \mathbb P^{n+1}\hookrightarrow\mathbb P^N,$$ the defining equation of $X$ becomes linear in $\mathbb P^N$, so $i(X)$ is a hyperplane section of $i(\mathbb P^{n+1})$.


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