In abstract Hodge theory there is the following lemma:
Let $H$ be a Hilbert space and $A \in \mathcal{C}(H)$ a densely defined, closed operator (so possibly unbounded) and $A^*$ its adjoint operator. Now given the condition that $A^2 = 0$, i.e. $\mathop{\mathrm{im}} A \subseteq \ker A$, the Hodge operator $D := A + A^*$ defined on $\mathop{\mathrm{dom}} D = \mathop{\mathrm{dom}} A ∩ \mathop{\mathrm{dom}} A^*$ is again densely defined, closed and also self-adjoint.
The question is now, whether this lemma still holds without requiring that $A^2 = 0$. That $D$ is symmetric is obvious, but I couldn't show that $\mathop{\mathrm{dom}} D = \mathop{\mathrm{dom}} D^*$. Is there maybe a good counter example?