The image of the natural homomorphism ${H^2}(X,\mathbb{R})\to {H^2}(X,\mathbb{C})$

Question

I wonder what the image of the natural homomorphism ${H^2}(X,\mathbb{R})\to {H^2}(X,\mathbb{C})$ looks like, in particular, I feel confused why there may exist a differential form of type (1,1) as a representative? Shouldn't all the elements in the image be of type (1,0)?

Answer 1

On any manifold, $H^k(X; \mathbb{C})$ is isomorphic to the de Rham cohomology of complexified differential forms $\mathcal{E}^k(X)\otimes_{\mathbb{R}}\mathbb{C}$. Note that there is a natural inclusion $\mathcal{E}^k(X) \hookrightarrow \mathcal{E}^k(X)\otimes_{\mathbb{R}}\mathbb{C}$ given by $\eta\mapsto \eta\otimes 1$ which induces the map $H^k(X; \mathbb{R}) \to H^k(X; \mathbb{C})$, so the image of $H^k(X; \mathbb{R}) \to H^k(X; \mathbb{C})$ consists of those classes represented by a real differential form.

If $M$ admits an almost complex structure,

$$\mathcal{E}^k(X)\otimes_{\mathbb{R}}\mathbb{C} = \mathcal{E}^{k, 0}(X) \oplus\mathcal{E}^{k-1,1}(X)\oplus\dots\oplus\mathcal{E}^{1,k-1}(X)\oplus\mathcal{E}^{0,k}(X).$$

So if $\eta \in \mathcal{E}^k(X)\otimes_{\mathbb{R}}\mathbb{C}$, then $\eta = \eta^{k,0} + \eta^{k-1,1} + \dots + \eta^{1,k-1} + \eta^{0,k}$ where $\eta^{p,q} \in \mathcal{E}^{p,q}(X)\otimes_{\mathbb{R}}\mathbb{C}$. The conjugation action on $\mathbb{C}$ extends to $\mathcal{E}^k(X)\otimes_{\mathbb{R}}\mathbb{C}$, and $\eta \in \mathcal{E}^k(X)$ if and only if $\eta = \overline{\eta}$ which is equivalent to $\eta^{p,q} = \overline{\eta^{q,p}}$ for all $p$ and $q$.

If $k = 2$, $\eta \in \mathcal{E}(X)$ if and only if $\eta = \alpha + \beta + \overline{\alpha}$ with $\overline{\beta} = \beta$ where $\alpha \in \mathcal{E}^{2,0}$ and $\beta \in \mathcal{E}^{1,1}(X)$. There are examples where $\alpha$ is zero, for example $\eta = dx\wedge dy = \frac{i}{2}dz\wedge d\bar{z}$ on a complex torus $\mathbb{C}/\Lambda$.