# Help with proving: If $X$ is a Hilbert $A$-$B$-module, then $\| _A \langle x,x \rangle \| = \| \langle x,x \rangle _B \|$ for all $x\in X$.

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...

(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.

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
• 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, $$\forall x \in X: \quad \| {_{A}} \langle x,x \rangle \|_{A} = \| \theta_{x,x} \|_{\mathcal{K}(X)} = \| \langle x,x \rangle_{B} \|_{B}.$$
• @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