Left m-invertibility by the adjoint of Drazin inverse and m-selfadjointness of Hilbert space operators

Abstract: A Hilbert space operator $A\in\B$ is left $(X,m)$-invertible by $B\in\B$ (resp., $B\in\B$ is an $(X,m)$-adjoint of $A\in\B$) for some operator $X\in\B$ if $\triangle_{B,A}^m(X)=\sum_{j=0}^m(-1)^j\left(\begin{array}{clcr}m\j\end{array}\right)B^{m-j}XA^{m-j}=0$ (resp., $\delta_{B,A}^m(X)=\sum_{j=0}^m(-1)^j\left(\begin{array}{clcr}m\j\end{array}\right)B^{(m-j)}XA^j=0$). No Drazin invertible operator $A\in\B$, with Drazin inverse $A_d$, can be left $(I,m)$-invertible (equivalently, $m$-invertible) by its adjoint or its Drazin inverse or the adjoint of its Drazin inverse. For Drazin inverrtible operators $A$, it is seen that the existence of an $X$ acts as a conduit for implications $\triangle_{B,A}(X)=0\Longrightarrow \delta^m_{C,A}(X)=0$, where the pair $(B,C)=$ either $(A,A_d)$ or $(A_d,A)$ or $(A^,A^_d)$ or $(A^_d,A^)$. Reverse implications fail. Assuming certain commutativity conditions, it is seen that $\triangle_{A^d,A}^m(X)=0=\triangle^n{B^_d,B}(Y)$ implies $\delta^{m+n-1}{A^B^,AB}(XY)=0=\delta^{m+n-1}{A^+B^,A+B}(XY)$.