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In Algebraic geometry we have a Lefschetz type theorem for Picard groups; Let $X \subset \mathbb{P}^N$ be a smooth and projective variety (over $\mathbb{C}$), with $dim(X) \geq 4$. Then the map $$Pic(X) \to Pic(X \cap H )$$ is an isomorphism, where $H$ is a hyperplane.

This is a special case of Ex 3.1.25 in "Positivity in Algebraic Geometry Vol 1" by Lazarsfeld.

So via the Veronese embedding, we get, for example, if $V \subset \mathbb{P}^N$ is a hypersurface of dimension at least 4, that $Pic(\mathbb{P}^N) \cong Pic(V) $.

My supervisior tells me that there is an extension for higher codimensions: that is,

$$ A^i(V) \cong A^i(\mathbb{P}^N) $$

when $i < \frac{1}{2} dim(V) $, $A^i(X)$ is the ring of codimension $i$ cycles modulo numerical equivalence.

My supervisor says this is a consequence of of the usual Lefschetz theorem for singular homology, but my knowledge of singular homology is not great and I don't see how it follows. My main problem is I don't know how to intepret $A^i(X)$ in terms of homology.

Has anyone here come across this? I haven't been able to find any reference to it. Could anyone provide a reference?

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    $\begingroup$ Suppose you have a subvariety of complex dimension $d$ in $V$. By triangulating it, say, you can think of it as a $2d$-chain, so it has a homology class in $H_{2d}(V)$. Assuming $V$ is smooth, applying Poincare duality gives a corresponding cohomology class in $H^{2n-2d}(V)$ where $n$ is the dimension of $V$. Now homologically equivalent cycles are numerically equivalent, so there is a quotient map $H^i \rightarrow A^i$. Lefeschetz gives an isomorphism $H^i(V) \rightarrow H^i(\mathbf P^n)$ in the appropriate ranges, and this implies the same for $A^i$. $\endgroup$
    – Schemer
    Sep 15 '15 at 16:14
  • $\begingroup$ Sounds convincing enough to me :). If you put this as an answer I can mark it as such. $\endgroup$ Sep 16 '15 at 11:03
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(From the comments)

"Suppose you have a subvariety of complex dimension $d$ in $V$. By triangulating it, say, you can think of it as a $2d$-chain, so it has a homology class in $H_{2d}(V)$. Assuming $V$ is smooth, applying Poincare duality gives a corresponding cohomology class in $H^{2n−2d}(V)$ where $n$ is the dimension of $V$. Now homologically equivalent cycles are numerically equivalent, so there is a quotient map $H^i \to A^i$. Lefeschetz gives an isomorphism $H^i(V) \to H^i(\mathbb{P}^n)$ in the appropriate ranges, and this implies the same for $A^i$."

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