why Hard Lefschetz is stated in the real coefficient in Huybrechts' complex geometry book? In  Huybrechts' complex geometry book , the Hard Lefschetz theorem is stated as follows:
Proposition 3.3.13 (Hard Lefschetz theorem) Let $(X, g)$ be a compact Kähler manifold of dimension $n$. Then for $k \leq n$
$$
L^{n-k}: H^k(X, \mathbb{R}) \cong H^{2 n-k}(X, \mathbb{R})
$$
and for any $k$
$$
H^k(X, \mathbb{R})=\bigoplus_{i \geq 0} L^i H^{k-2 i}(X, \mathbb{R})_{\mathrm{p}}
$$
Moreover, both assertions respect the bidegree decomposition. In particular, $H^k(X, \mathbb{R})_{\mathrm{p}} \otimes \mathbb{C}=\bigoplus_{p+q=k} H^{p, q}(X)_{\mathrm{p}}$
Proof. The first assertion is iii) of Remark 3.2.7. (which used the isomorphism of harmonic forms).

There seem to have many things to talk about the coefficient of the cohomology. In the book, we use that Lefschetz operator induced isomorphism on the harmonic form to prove the statement. However, the harmonic form is defined using the complex coefficient. I was confused why we can use it for the real coefficient case?
 A: For the completeness I write down the proof in detail.
The idea is for any real manifold, we can define the harmonic operator and the space of Harmonic form isomorphic to the deRham group with real coefficient. therefore the problem can be reduced to the problem of harmonic form.
For the Kahler setting we have the commutative relation $[L,\Delta_d]$ holds on the space $\mathcal{A}^{\cdot}_{\Bbb{C}}(X)$. However the key observation is that both $L$ and $d$ are real operator, that is if we restrict to the form with real coefficient $\mathcal{A}_{\Bbb{R}}^\cdot$ the image is still real form. $\Delta_d$ is real operator since $d$ is real operator ,$L$ is real operator since Kahler form is real (1,1) form.
Therefore we have the Kahler identity restrict to the real case. Since $L$ respect $\Delta_d$ (with real coefficient also complex coefficient), we have the isomorphism on the level of differential form $L^{n-k}: \mathcal{A}^k(X, \mathbb{R}) \cong \mathcal{A}^{2n-k}(X, \mathbb{R})
$ induce the isomorphism on the level of harmonic form.
Since harmonic form isomorphic to deRham group, we have the isomorphism on the deRham cohomology with real coefficient.

For the second statement, needs a little lemma: if $\alpha = \sum_i L^i \alpha_i$ being the decomposition such that $\alpha$ is harmonic form, then $\alpha_i$ is also harmonic form. This due to the fact that $\Delta_d$ commute with $L$, and the uniqueness of the decomposition.

For the third statement(which is called Lefschetz decomposition is compatible with the Hodge decomposition)
given $\alpha$ which is premitive k form, it has Hodge decomposition as $\alpha  = \sum_{p+q = k} \alpha^{p,q}$  then apply the $\Lambda \alpha = 0$ then it will decomposition by bidegree, as $\Lambda$ is pure type (-1,-1) therefore all $\alpha^{p,q} \in \ker \Lambda$
(Here we use a little fact that kernel of complexified map is the kernel of original map tensor with $\Bbb{C}$. You can find the result on Conrad's note Theorem 2.9. here used the fact that the free module are flat.)
