Graded adjointable operators on a graded Hilbert space Given a graded Hilbert space $\mathbf{H} = \bigoplus_{k \in \mathbb{N}} \mathbf{H}_k$, one might extend the notion of adjoint to a "graded adjoint" defined as follows: $L \in B(\mathbf{H})$ is said to be graded adjointable of degree $l$, if there exists an operator $L^{*}$ such that, for each $k\geq l$ and for each $y \in \mathbf{H}_k$, 
$$
\langle L(x),y\rangle = (-1)^{l.k} \langle x,L^*(y)\rangle
\qquad (x\in H_{k-l}).
$$
Do such things appear in the literature, and it they do, is the theory of graded adjointable operators significantly different from the usual case?
 A: I do not know any references.  However, if the following calculation is correct, then $L^*$ always exists and can easily be calculated from the usual adjoint of $L$.
As $H = \bigoplus_{n\geq 1} H_k$ is an orthogonal sum, we can think of $L\in B(H)$ as a matrix of operators, $L=(L_{ij})$ say, where $L_{ij} : H_j \rightarrow H_i$. Thus, for $y\in H_k$ and $x\in H_{k-l}$ we have
$$ \langle L(x), y \rangle = \langle L_{k, k-l}(x), y \rangle
= \langle x, L_{k,k-l}^\star(y) \rangle $$
where I write $\star$ for the usual Hilbert space adjoint.  You want this to be equal to
$$ (-1)^{l.k} \langle x, L^\ast(y)\rangle = (-1)^{l.k} \langle x, L^\ast_{k-l,k}(y) \rangle. $$
Thus, you need $L^\ast_{k-l,k} = (-1)^{l.k} L_{k,k-l}^\star = (-1)^{l.k} (L^\star)_{k-l,k}$ for each $k>l$.  That is, $L^\ast_{j,k} = (-1)^{l^2} (-1)^{jl} (L^\star)_{j,k}$ for all $j$ and all $k>l$.  There is no constraint on the $(j,k)$ component of $L^*$ when $k\leq l$.
As $l$ is fixed, we can obtain $L^*$ from $L^\star$ just by multiplying the appropriate rows by $-1$.  This is a bounded operation, and so $L^*$ always exists.
