The following statement is from Lübke's The Kobayashi-Hitchin Correspondence pp.222:
If $a\in A^2(X)$ is harmonic, and $a=a^++a^-$ with $a^{\pm}\in A^2_{\pm}(X)$, then $a^+$ and $a^-$ are also harmonic. This induces a decomposition $H^2(X,\mathbb{C})=H^2_+(X)\oplus H^2_-(X)$ (and similarly for $H^2(X,\mathbb{R})$).
Here $X$ is a complex surface. $A^p(X),A^{s,t}(X)$ are spaces of smooth complex $p$-forms and forms of type $(s,t)$ respectively. $A^2_+(X),A^2_-(X)\subset A^2(X)$ correspond to the eigenvalues $\pm 1$ of the Hodge star operator $*:A^2(X)\to A^2(X)$. Then $$A^2_+(X)=A^{2,0}(X)\oplus A^{0,2}(X)\oplus A^0(X)\cdot \omega_g, \quad A^2_-(X)=A^{1,1}_0(X),$$ where $g$ is a Hermitian metric in $X$ and $\omega_g$ is the Kähler form. How to see that $a^+$ and $a^-$ are simultaneously harmonic when $a$ is?
Note that $a^{\pm} = \frac{1}{2}(a \pm \ast a)$. As $\Delta$ and $\ast$ commute, we see that
$$\Delta a^{\pm} = \frac{1}{2}(\Delta a \pm \Delta(\ast a)) = \frac{1}{2}(\Delta a \pm \ast\Delta a) = 0.$$