3
$\begingroup$

The following doubt came after reading the book "Hilbert C*-modules" by E.C. Lance. Let $A$ be a C*-algebra and $E$ a Hilbert $A$-module, there's a natural structure of Hilbert $A$-module on $E^n$ given by $$\langle (x_1,\dots,x_n),(y_1,\dots,y_n)\rangle_A = \sum \langle x_i,y_i\rangle$$ but also on page 39 of the book we're introduced a Hilbert $M_n(A)$-module structure on $E^n$ given by $$(x_1,\dots,x_n)\cdot (a_{ij})=\left(\sum x_ia_{i1},\dots ,\sum x_ia_{in}\right)$$ and $$\langle (x_1,\dots,x_n),(y_1,\dots,y_n)\rangle_{M_n(A)}=(\langle x_i,y_j\rangle)$$ Later on, on page 58, there's a result that states $\mathcal{L}_{M_n(A)}(E^n)\simeq \mathcal{L}_{A}(E^n) $. The *-homomorphism that establishes this isomorphism according to results on the previous pages seems to be $T\mapsto T$. I'm pretty sure this isn't exactly the isomorphism since $T$ being adjointable in the $M_n(A)$ sense doesn't seem to imply it being adjointable in the $A$ sense.

My question is: explicitly what would be the isomorphism between these two algebras? If $\varphi:\mathcal{L}_{M_n(A)}(E^n)\rightarrow \mathcal{L}_{A}(E^n)$ is the isomorphism then what would $\varphi(T)(x_1,\dots,x_n)$ be?

$\endgroup$
2
$\begingroup$

Realized the morphism $T\mapsto T$ actually works. Take $x,y\in E^n$, $T\in \mathcal{L}_{M_n(A)}(E^n)$ and write $T=(T_1,\dots,T_n)$ and $T^*=(S_1,\dots,S_n)$ where $T_i,S_i:E^n\rightarrow E$ are linear maps. Then $$(\langle T_i(x),y_j \rangle)_{i,j=1}^n=\langle T(x),y \rangle_{M_n(A)}=\langle x , S(y) \rangle_{M_n(A)}=(\langle x_i,S_j(y) \rangle)_{i,j=1}^n$$ meaning $\langle T_i(x),y_j \rangle= \langle x_i,S_j(y) \rangle$ for any $i,j\in \{1,\dots, n \}$. But now $$\langle T(x),y \rangle_{A}=\sum_{i=1}^n \langle T_i(x), y_i\rangle=\sum_{i=1}^n \langle x_i, S(y)\rangle=\langle x , T^*(y) \rangle$$ this means not only that $T\in \mathcal{L}_A(E^n)$ but that it has the same adjoint as in the $M_n(A)$ case and therefore $T\mapsto T$ is a *-isomorphism between $\mathcal{L}_{M_n(A)}(E^n)$ and $\mathcal{L}_A(E^n)$.

$\endgroup$

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.