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?

• I don't quite understand the question. How are $l$ and $k$ meant to be related? Where is $x$ taken from in the displayed equation? Is this meant to be for all $k$? Aug 15 '19 at 15:24
• Yes this is for all $k$, and no there is no relation between $k$ and $l$. Of course, we must have that $x$ has degree $k-l$. Aug 15 '19 at 15:29
• I have edited the question to make it clearer, IMHO. Please edit again if it is wrong... Aug 15 '19 at 17:54
• Sure! I guess we could also adopt the convention that $\mathbb{H}$ is $\mathbb{Z}$-graded, with $\mathbb{H}_k = 0$, for $k \leq 0$. But the way you have written is fine too. Aug 15 '19 at 18:04

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.