Let $ X $ be a smooth complex projective algebraic variety of complex dimension $ n $.
Then, there exists the following Hodge decomposition:
$$ H^{2k} (X, \mathbb{C}) = \displaystyle \bigoplus_{p + q = 2k} H^{p, q} (X).
$$
Since $ X $ is a complex manifold, then $ X $ has a real differential manifold structure of real dimension $ 2n $.
My question is the following,
Since $ X $ has a real differential manifold structure, does,
$$ H^{2k} (X, \mathbb{C}) = H_{\mathrm{dR}}^{2k} (X, \mathbb{R}) \otimes \mathbb{C} \qquad ?
$$
- $ H^{2k} (X, \mathbb{C}) $ is the space in the Hodge decomposition formula above.
- $ H_ {\mathrm{dR}}^{2k} (X, \mathbb{R}) $ is the real deRham cohomology such that any element of this space is a linear combination of real differential forms for the coordinate system $ (x_1, y_1, \dots, x_n, y_n) $, with real coefficients.
Thanks in advance for your help.
Best Answer
The answer is yes. Let's denote by $\mathcal{E}^k(X,\mathbb{C})$ and $\mathcal{E}^k(X,\mathbb{R})$ the spaces of complex- and real-valued $k$-forms respectively. The crucial observation is that $\mathcal{E}^k(X,\mathbb{C})\cong\mathcal{E}^k(X,\mathbb{R})\otimes_{\mathbb{R}}\mathbb{C}$, i.e. every complex-valued can be uniquely written as the sum of a real-valued differential form and $i$ times a real-valued differential form (to see this, pick a local frame, expand everything out in terms of the frame and collect the terms containing $i$). Furthermore, the complex exterior differential $d_{\mathbb{C}}\colon\mathcal{E}^k(X,\mathbb{C})\rightarrow\mathcal{E}^{k+1}(X,\mathbb{C})$ corresponds to the complexification $d_{\mathbb{R}}\otimes_{\mathbb{R}}1_{\mathbb{C}}$ of the real exterior differential $d_{\mathbb{R}}\colon\mathcal{E}^k(X,\mathbb{R})\rightarrow\mathcal{E}^{k+1}(X,\mathbb{R})$; indeed, the complex exterior differential is $\mathbb{C}$-linear and agrees with the real exterior differential when restricted to $\mathcal{E}^k(X,\mathbb{R})\subset\mathcal{E}^k(X,\mathbb{C})$. Thus, the chain complex $(\mathcal{E}^{\bullet}(X,\mathbb{C}),d_{\mathbb{C}})$ is just the complexification of the chain complex $(\mathcal{E}^{\bullet}(X,\mathbb{R}),d_{\mathbb{R}})$. It is then a purely algebraic observation that the cohomology of the complexification of a chain complex of real vector spaces is just the complexification of its cohomology.