Let $E$ be a Hilbert module over a $C^*$-algebra $A$. Let $T\colon E\to E$ be a densely defined, unbounded $A$-linear operator. (In particular, the initial domain of $T$ is an $A$-submodule of $E$.) If the operators $T\pm i$ each has dense range, then $T$ is an essentially self-adjoint, regular operator on $E$, and there is a continuous functional calculus

$$F_T\colon C_b(\mathbb{R})\to\mathcal{B}_A(E),$$

where $C_b(\mathbb{R})$ denotes the bounded continuous $\mathbb{C}$-valued functions on $\mathbb{R}$, and $\mathcal{B}_A(E)$ denotes the $C^*$-algebra of bounded adjointable $A$-linear operators on $E$.

Question: Suppose $D(T)$ is not an $A$-submodule of $E$, but only an $A_0$-submodule, where $A_0$ is a (metrically) dense $*$-subalgebra of $A$. Then, still assuming that $T\pm i$ each has dense range, is it possible to construct a continuous functional calculus for $T$?

Comment: It seems to me that even though in this situation $T$ is only $A_0$-linear, the operator $1+T^*T$ is injective, and that $(1+T^*T)^{-1}$ extends to an $A$-linear operator. Similarly, it seems possible to make sense of the operator $T(1+T^*T)^{-1/2}$ as being $A$-linear.

  • $\begingroup$ This is Woronowicz's "z transform", right? What source are you following for this material? $\endgroup$ Feb 4, 2021 at 9:47
  • 2
    $\begingroup$ @MatthewDaws I'm looking at Lance's book, which assumes that $D(T)$ is closed under the action of $A$. I think the operator $T(1+T^*T)^{-1/2}$ is the $z$-transform of $T$. I guess what I was speculating in the comment to the question was that even if $T$ is only $A_0$-linear, then it has an $A$-linear $z$-transform, so its functional calculus can be developed. $\endgroup$
    – geometricK
    Feb 4, 2021 at 10:10

1 Answer 1


I think you need to be very careful about hypotheses. Looking in Lance's book, we have

Lemma 9.8: Suppose $t:E\rightarrow E$ is densely-defined and self-adjoint. Then $t$ is regular if and only if the operators $t\pm i$ are surjective.

(Lance's notation $t:E\rightarrow F$ means that $t$ is defined on $D(t)$ a submodule of $E$.) Notice here that $t$ needs to be self-adjoint, which you don't mention in your question. I am afraid I cannot find anything in Lance where $t\pm i$ merely have dense range is sufficient.

Suppose we have $t$ where $D(t)$ is only a dense $A_0$-submodule. What is $t^\ast$? Well, the definition is $$ D(t^\ast) = \{ y\in E : \exists\, z\in E, \ (tx|y) = (x|z) \ (x\in D(t) \} $$ with $t^\ast(y) = z$. Suppose $y\in D(t^\ast)$ with $t^\ast(y) = z$. For $a\in A$ we have $$ (x|za) = (x|z)a = (tx|y)a = (tx|ya) \qquad(x\in D(t), $$ just using $A$-linearity of the inner-product. So $ya\in D(t^\ast)$ and $t^\ast(ya) = za = t^\ast(y)a$. So $D(t^\ast)$ is an $A$-submodule, and $t^\ast$ is $A$-linear.

Thus, if $t=t^*$, then $D(t)$ is already an $A$-submodule! We could go back to the definition of $t$ being "regular", but that requires knowing that $1+t^*t$ has dense range. I suspect in any application, having the weaker hypothesis that $D(t)$ is only an $A_0$-submodule will not be an aid in showing $1+t^*t$ has dense range.

  • $\begingroup$ Thanks for the answer! I should have included that $t$ is symmetric in the hypotheses. Let me ponder this a little more, but I think you are right that knowing $D(t)$ is only an $A_0$-submodule won't help show $1+t^*t$ has dense range. However, if one happened to know that $1+t^*t$ has dense range, then this should be sufficient to define functional calculus of $t$, even in the $A_0$-submodule case. Here one won't expect to recover $t$ from its $z$-transform like in the usual case. I think the functional calculus formed in this way should just be the functional calculus of $t^{**}$. $\endgroup$
    – geometricK
    Feb 4, 2021 at 11:29
  • $\begingroup$ Be careful with the difference between "symmetric" and "self-adjoint". We definitely need self-adjoint here. $\endgroup$ Feb 4, 2021 at 11:50
  • $\begingroup$ Right, I should have said closed and symmetric, which together with the density-of-range assumption implies self-adjoint. On the other hand, all symmetric operators are closable. $\endgroup$
    – geometricK
    Feb 4, 2021 at 12:11

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