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Sorry, I posted a related question last week on here, but I'm still having trouble and this is a little different, I hope it's OK. Thank you! ( proof that this is an isometric map (on a $C^*$-module) )


Notation: ${}_A\langle \cdot ,\cdot \rangle $ and $ \langle \cdot ,\cdot \rangle _B$ denotes the left and right inner product.

Notation: $\theta_{x,y}(z) = x\langle y,z \rangle _B $

Notation: $\mathcal{K}(X) = \{ \theta _{x,y} : x,y \in X\} $

Definition: A (right/left) inner product A-module $V$ is called a (right/left Hilbert $A$-module if $E$ is complete with respect to the norm defined by $\| x \| = \| \langle x,x \rangle \| ^{1/2} $ $(x \in V) $.

Definition: Let $A$ and $B$ be $C^*$-algebras. A Hilbert $A$-$B$-module V is both a left Hilbert $A$-module and a right Hilbert $B$-module such that $$ (ax)b=a(xb) \ \text{ and } \ _A \langle x,y \rangle z = x\langle y, z \rangle _B $$ for all $a\in A$, $b\in B$, and $x,y,z\in V$.

Proposition: If $X$ is a Hilbert $A$-$B$-module, then $$ \| _A \langle x,x \rangle \| = \| \langle x,x \rangle _B \| \ \text{ for all } x\in X. $$

Proof: Let $x\in X$ and $a= {}_A \langle x,x \rangle $. In the following, note that $\| x \| = \| \langle x,x \rangle _B \| ^{1/2}_B $. So, when one considers $\| \lambda _a \| _{\mathcal{K}(X)}$, one has to consider $\| az \| = \| az \| _B $ with $ \| z \| = \| z \| _B \leq 1$ (i.e., the norm on $X$ is taken from the inner product $\langle \cdot , \cdot \rangle _B $). Now we have $$ \theta _{x,x} (z) = x\langle x,z \rangle _B = {}_A\langle x,x \rangle z = az = \lambda _a (z) \ (z\in X), $$ where $\lambda _a$ is the map $z\mapsto az$ on $X$.

The following part of the proof is causing me trouble, but here is my attempt

Moreover, $ \lambda $ is bounded since for all $z\in X$ such that $\| z \| _A \leq 1 $, we have $$ \| a z \| _A \leq \| a \| _A \| z \| _A \leq \| a \| _A . $$ Therefore, to show that $\| \lambda _a \| _{\mathcal{K}(X)} = \| a \| _A$, it suffices to show that $\lambda $ has trivial kernel.

Assume $\lambda _a = 0 $. Let $e_\alpha = {}_A\langle u_\alpha , v_\alpha \rangle $ be an approximate identity of ${}_A\langle X,X \rangle $. Then, \begin{align*} ae_\alpha a^* &= a{}_A \langle u_\alpha , v_\alpha \rangle a^* \\ &= {}_A \langle au_\alpha , au_\alpha \rangle \\ &= {}_A \langle \lambda _a (u_\alpha ), \lambda _a (v_\alpha ) \rangle \\ &= 0. \end{align*} Therefore, $aa^*=0$ and hence $a=0$. So, $\| \lambda _a \| _{\mathcal{K}(X)} = \| a \| _A$.

It follows that $$ \| _A\langle x,x \rangle \| _A = \| \lambda _a \| _{\mathcal{K}(X)} = \| \theta _{x,x} \| _{\mathcal{K}(X)} = \| \langle x,x \rangle _B \| _B. $$


I also found a proof of this proposition, which I am having a hard time understanding...

http://books.google.ca/books?id=8LddfdeBbHgC&pg=PA1154&lpg=PA1154&dq=Quasi-multipliers+and+Embeddings+of+Hilbert+C*-bimodules&source=bl&ots=QGa0CfW5Rh&sig=QEmke3JZLCZe392l998-Pw5NiHo&hl=en&sa=X&ei=V0lGUdyxHaOg2gXd5IGADQ&ved=0CE0Q6AEwAw

(in particular, they don't show why $\lambda $ is bounded... and I don't understand why $\sum _i {}_A \langle \lambda _a x_i,x_i \rangle = 0 $ if $\lambda _a = 0 $)

(proposition 1.10 and cor 1.11)

so... my attempt is a little different, so I don't know if it is correct or not...

Thank you.

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1 Answer 1

up vote 2 down vote accepted

First of all, $ {_{A}} \langle X,X \rangle $ is defined as follows: $$ {_{A}} \langle X,X \rangle \stackrel{\text{def}}{=} ~ \overline{\text{Span}(\{ {_{A}} \langle x_{1},x_{2} \rangle ~|~ x_{1},x_{2} \in X \})}^ {\| \cdot \|_{A}}. $$ Notice that this is a C$ ^{*} $-subalgebra of $ A $. Notice also that $ \mathcal{K}(X) $ is a closed two-sided ideal of $ \mathcal{L}(X) $, which is the C$ ^{*} $-algebra of adjointable operators on $ X $. Hence, $ \mathcal{K}(X) $ is a C$ ^{*} $-subalgebra of $ \mathcal{L}(X) $.

Next, define a continuous $ ^{*} $-homomorphism $ \Theta: {_{A}} \langle X,X \rangle \to \mathcal{K}(X) $ by first declaring $$ \forall x_{1},x_{2} \in X: \quad \Theta({_{A}} \langle x_{1},x_{2} \rangle) \stackrel{\text{def}}{=} \theta_{x_{1},x_{2}} := (z \longmapsto {_{A}} \langle x_{1},x_{2} \rangle z) $$ and then extending this initial definition by

  • linearity and involution and

  • finally by continuity, which is possible because $$ \Theta: \text{Span}(\{ {_{A}} \langle x_{1},x_{2} \rangle ~|~ x_{1},x_{2} \in X \}) \to \text{Span}(\{ \theta_{x_{1},x_{2}} ~|~ x_{1},x_{2} \in X \}) $$ is a bounded mapping. More precisely, $$ \forall {_{A}} \langle x_{1,1},x_{2,1} \rangle,\ldots,{_{A}} \langle x_{1,n},x_{2,n} \rangle: \quad \left\| \theta_{\sum_{i=1}^{n} {_{A}} \langle x_{1,i},x_{2,i} \rangle} \right\|_ {\mathcal{K}(X)} \leq \left\| \sum_{i=1}^{n} {_{A}} \langle x_{1,i},x_{2,i} \rangle \right\|_{A}. $$

Now, one error that you have made about how approximate identities of $ {_{A}} \langle X,X \rangle $ look like is in assuming that $ e^{\alpha} = {_{A}} \langle u^{\alpha},v^{\alpha} \rangle $. In general, $ e^{\alpha} $ is of the form $$ e^{\alpha} = \sum_{i=1}^{n} {_{A}} \langle u_{i}^{\alpha},v_{i}^{\alpha} \rangle, $$ as explained in the linked article (using an argument of Dixmier). This is only a minor issue however, and once you have established that $ \text{Ker}(\Theta) $ is trivial, then you have obtained an injective $ ^{*} $-homomorphism from $ {_{A}} \langle X,X \rangle $ to $ \mathcal{K}(X) $. As an injective $ ^{*} $-homomorphism from one C$ ^{*} $-algebra to another is necessarily an isometry, and as $ \text{Range}(\Theta) $ is clearly dense in $ \mathcal{K}(X) $, we conclude that $$ {_{A}} \langle X,X \rangle \cong \mathcal{K}(X). $$ Therefore, $$ \forall x \in X: \quad \| {_{A}} \langle x,x \rangle \|_{A} = \| \theta_{x,x} \|_{\mathcal{K}(X)} = \| \langle x,x \rangle_{B} \|_{B}. $$

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@Euthenia: Do you understand the explanation given here? I hope that I’ve managed to clear your doubts. Wegge-Olsen’s book K-Theory and C$ ^{\ast} $-Algebras also contains a proof of this result. –  Haskell Curry Mar 19 '13 at 1:57
    
I understand now, thank you –  Euthenia Mar 19 '13 at 2:25

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