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Let $M$ be projective complex manifold. The Lefschetz (1,1)-theorem says that the cycle map $$ \text{cl}:\operatorname{Pic}(M) \to \text{Hod}^1(M) $$ is surjective.

Question. Is there an interesting example of (1,1)-form $\omega \in H^1(M,\Omega_M^1)$ which isn`t spanned by $\operatorname{Pic}(M)_{\mathbb{C}}$?

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    $\begingroup$ As soon as $\operatorname{rk} \operatorname{NS}(M) < \dim H^{1,1}(M)$, you will have many such classes. This happens for example on a K3 surface of Picard rank less than $20$. Can you say a little more what you mean by "interesting"? $\endgroup$
    – R. van Dobben de Bruyn
    Commented Feb 18, 2021 at 23:28
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    $\begingroup$ Let's please not rush to close this, before giving this relative newcomer a chance to respond. This may yet elicit a good answer. $\endgroup$
    – Todd Trimble
    Commented Feb 19, 2021 at 14:28
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    $\begingroup$ A simple example is the product of two non-isogenous elliptic curves, $E_1$ and $E_2$. Then $H^{(1,1)}(E_1 \times E_2)$ is $4$-dimensional. If $E_1$ and $E_2$ are not isogenous, then $\mathrm{Pic}(E_1 \times E_2)$ is only two dimensional, spanned by the classes of $E_1 \times \{ \mathrm{point} \}$ and $\{ \mathrm{point} \} \times E_2$. $\endgroup$
    – David E Speyer
    Commented Feb 19, 2021 at 14:31
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    $\begingroup$ Just for concreteness — in David Speyer’s example, the two-form $dz_1\wedge d\overline{z_2}$ is not in the span of the image of the cycle class map, where $dz_i$ is the pullback of any nonzero invariant differential form on $E_i$. $\endgroup$
    – Daniel Litt
    Commented Feb 20, 2021 at 14:43

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