# (1,1)-form that does not come from a divisor

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 $$\operatorname{Pic}(M)_{\mathbb{C}}$$?

• 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"? Commented Feb 18, 2021 at 23:28
• Let's please not rush to close this, before giving this relative newcomer a chance to respond. This may yet elicit a good answer. Commented Feb 19, 2021 at 14:28
• 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 $\{ \mathrm{point} \} \times E_2$. Commented Feb 19, 2021 at 14:31
• 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 $E_i$. Commented Feb 20, 2021 at 14:43