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Question on algebraic Algebraic cycles

I have a very basic knowledge of algebraic geometry. Could you please let me know if the following holds:

Let $X$ be a smooth complex surface. Assume that $2$-dimensional cohomology class $[A] \in H^{2}(X, \mathbb{C})$ belongs to $H^2(X,\mathbb{Z}) \cap H^{1,1}(X)$. Then the class $[A]$ can be represented by a complex-algebraic submanifold of $X$. Does this follow from Lefschetz theorem on $(1,1)$-classes?

Thanks

Question on algebraic cycles

I have a very basic knowledge of algebraic geometry. Could you please let me know if the following holds:

Let $X$ be a smooth complex surface. Assume that $2$-dimensional cohomology class $[A] \in H^{2}(X, \mathbb{C})$ belongs to $H^2(X,\mathbb{Z}) \cap H^{1,1}(X)$. Then the class $[A]$ can be represented by a complex-algebraic submanifold of $X$. Does this follow from Lefschetz theorem on $(1,1)$-classes?

Thanks

Algebraic cycles

Let $X$ be a smooth complex surface. Assume that $2$-dimensional cohomology class $[A] \in H^{2}(X, \mathbb{C})$ belongs to $H^2(X,\mathbb{Z}) \cap H^{1,1}(X)$. Then the class $[A]$ can be represented by a complex-algebraic submanifold of $X$. Does this follow from Lefschetz theorem on $(1,1)$-classes?

Thanks

Source Link
upd
upd
  • 33
  • 4

Question on algebraic cycles

I have a very basic knowledge of algebraic geometry. Could you please let me know if the following holds:

Let $X$ be a smooth complex surface. Assume that $2$-dimensional cohomology class $[A] \in H^{2}(X, \mathbb{C})$ belongs to $H^2(X,\mathbb{Z}) \cap H^{1,1}(X)$. Then the class $[A]$ can be represented by a complex-algebraic submanifold of $X$. Does this follow from Lefschetz theorem on $(1,1)$-classes?

Thanks