4
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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 $k$? $\endgroup$ Aug 15 '19 at 15:24
  • $\begingroup$ 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 $k-l$. $\endgroup$ Aug 15 '19 at 15:29
  • $\begingroup$ I have edited the question to make it clearer, IMHO. Please edit again if it is wrong... $\endgroup$ Aug 15 '19 at 17:54
  • $\begingroup$ 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. $\endgroup$ Aug 15 '19 at 18:04
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.