I tried to prove example 3.4 from the book Morita Equivalence and Continuous-Trace C$^*$-Algebras by Iain Raeburn and Dana P. Williams, but I get uneasy with notations and ideas. Let me restate my assumption about this example.

Suppose that $T$ is locally compact Hausdorff space and that $H$ is a complex Hilbert space. Then $$X:=C_0(T,H)=\left\{ x:T\rightarrow H: x \:continuous,t\rightarrow \left\| x(t)\right\|\in C_0(T)\right\}$$ complex vector space and $$A:=C_0(T,K(H))=\left\{ f:T\rightarrow K(H): f \:continuous,t\mapsto \left\| x(t)\right\|\in C_0(T) \right\}$$ C*-algebra. Then $C_0(T,H)$ is left Hilbert-$C_0(T,K(H))$ module under this operation: $$f\bullet x(t)=f(t)(x(t))$$ $$_A \left\langle x,y \right\rangle (t)=x(t) \otimes \bar{y(t)} $$ for $x,y\in C_0(T,H)$ and $f\in C_0(T,K(H))$.

My questions are:

  1. Is $\otimes$ here related to tensor product? What kind of tensor product?
  2. Is $t\rightarrow \left\| x(t)\right\|\in C_0(T)$ and $t\mapsto \left\| x(t)\right\|\in C_0(T)$ different?
  3. Is norm for $_A\left\| x(t) \right\|$ a sup-norm?
  • $\begingroup$ I think your definition of $A$ should read $f:T \to K(H)$ continuous such that $t \mapsto \|f(t)\| \in C_0(T)$. $\endgroup$ – Prahlad Vaidyanathan Aug 20 '15 at 6:48
  • $\begingroup$ It fixed, thanks @PrahladVaidyanathan :D $\endgroup$ – Raden Muhammad Hadi Aug 20 '15 at 10:30
  1. For two vectors $x,y\in H$, the notation $x\otimes y$ is frequently used to denote the rank-one operator $z\longmapsto \langle z,y\rangle x$. It is related to tensor products, but not in the obvious way: it is seeing $x\otimes y$ as an element of $H\otimes H^*$ (most often, no one pays attention to this and just uses the notation).

  2. I think it is just a typo, it should have been $\longmapsto $ in both cases.

  3. I'm not sure what you mean by $_A\|x(t)\|$. The elements of the Hilbert module are functions $T\to H$. And $_A\langle x,y\rangle$ is a function $T\to K(H)$. In a Hilbert module you often construct a norm by taking the square root of the norm in $A$, i.e. $$ \|x\|:=\|_A\langle x,x\rangle\|_A^{1/2}. $$ In this case $A$ is a function algebra, so its norm is a sup norm. So $$ \|x\|=\sup\{|_A\langle x,x\rangle (t)|^{1/2}:\ t\in T\} $$

  • $\begingroup$ I use $_A\left\| x \right\|$ (norm on left pre-Hilbert-C* module) just to differ it from $\left\| x \right\| _A$ (norm on right pre-Hilbert-C* module), because the book says nothing about left Hilbert-C* module. I want to ask more about (2). Is $t\mapsto \left\| x(t) \right\| \in C_0(T)$ means that norm on $X$ and $A$ are take values (or similar?) to norm on $C_0(T)$ ? $\endgroup$ – Raden Muhammad Hadi Aug 20 '15 at 21:12
  • $\begingroup$ I write $\|\cdot\|_A$ to denote the norm on $A$. As for $t\longmapsto\|x(t)\|$, I don't understand what you asking. $\endgroup$ – Martin Argerami Aug 21 '15 at 1:50
  • $\begingroup$ I want to ask about what $t\mapsto \left\| x(t) \right\| \in C_0(T)$ means. That part still not clear for me. $\endgroup$ – Raden Muhammad Hadi Aug 21 '15 at 6:35
  • $\begingroup$ That's a (common, and clear when you get used to it) notation for the function $f (t)=\|x (t)\|$. Every time you have a rule that assigns a number to each point in $T $, you can ask yourself whether that assignment is continuous, and whether it vanishes at infinity. $\endgroup$ – Martin Argerami Aug 21 '15 at 13:16

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