We know that the Kälher identity $\Delta=2\Delta_{\partial}=2\Delta_{\bar{\partial}}$ on a Kähler manifold $(X,g)$ implies that the usual Laplacian $\Delta:=dd^*+d^*d$ respects the bidegree, i.e. for any $(p,q)$-form $\alpha$ on $(X,g)$, the form $\Delta(\alpha)$ is still a $(p,q)$-form.

But for general hermitian manifolds, $\Delta$ does not respect bidegree. Are there any famous concrete examples for this?

Another question is: if $\Delta$ respects bidegrees, can we show that $(X,g)$ is Kälher? Or is there any non-Kälher hermitian manifold, whose Laplacian respects bidegrees?

We know that the Kähler identity $\Delta=2\Delta_{\partial}=2\Delta_{\bar{\partial}}$ on a Kähler manifold $(X,g)$ implies that the usual Laplacian $\Delta:=dd^*+d^*d$ respects the bidegree, i.e. for any $(p,q)$-form $\alpha$ on $(X,g)$, the form $\Delta(\alpha)$ is still a $(p,q)$-form.

But for general hermitian manifolds, $\Delta$ does not respect bidegree. Are there any famous concrete examples for this?

Another question is: if $\Delta$ respects bidegrees, can we show that $(X,g)$ is Kähler? Or is there any non-Kähler hermitian manifold, whose Laplacian respects bidegrees?