I was reading Professor Huybrechts's Lectures on K3 surfaces, there is a statement about Hodge class that I can't figure out.
Let's consider the (integral or rational) Hodge structure $V$ with the Hodge decomposition $$V_{\mathbb{C}} = \bigoplus_{p+q = n}V^{p,q}$$we define the Hodge class to be $\text{Hdg}^{2k}(V) = V\cap V^{k,k}$, the book says in general we have proper inclusion $$(V\cap V^{k,k})_{\mathbb{C}} \subset V^{k,k}$$however $(V_{\mathbb{R}}\cap V^{k,k})_{\mathbb{C}} = V^{k,k}$. I found it confusing, why is there a difference here?
(I found a post https://math.stackexchange.com/a/4192210/360262 which shows it is related to the complex version of Hodge conjecture)
The question is How can I prove that $(V_{\mathbb{R}}\cap V^{k,k})_{\mathbb{C}} = V^{k,k}$ why it does not apply to $\mathbb{Q}$-coefficient? (I guess it's related to the conjugation, but don't know how to write it down clearly?)
(An elementary proof, I guess there may have more advanced proof.)
First, an element $\alpha =\sum \alpha^{p,q} \in V_\mathbb{C}$ lies in $V_{\mathbb{R}}\cap V^{k,k}$ iff $\alpha = \alpha^{k,k}$ such that $\bar{\alpha}^{k,k} = \alpha^{k,k}$.
On the other hand pick an arbitrary element $\alpha \in V^{k,k}$ then since the Hodge type $(k,k)$ is stable under conjugation we have $\bar{\alpha} \in V^{k,k}$ thus $\alpha + \bar{\alpha} \in (V_{\mathbb{R}}\cap V^{k,k})$. Similarily we have $\sqrt{-1}(\alpha -\bar{\alpha}) \in (V_{\mathbb{R}}\cap V^{k,k})$ thus taking complexification we have $$\alpha = 1/2\cdot ((\alpha + \bar{\alpha})- (\sqrt{-1})(\sqrt{-1}(\alpha -\bar{\alpha})) \in (V_{\mathbb{R}}\cap V^{k,k})_{\mathbb{C}}$$
which shows $(V_{\mathbb{R}}\cap V^{k,k})_{\mathbb{C}} = V^{k,k}$.
On the other hand, we have for rational Hodge structure, there is no such action as conjugation to preserve it.