A Hilbert-space completion of a Hilbert $ C^{*} $-module over a separable $ C^{*} $-algebra

Question

Let $ B $ be a separable $ C^{*} $-algebra and $ \mathcal{E} $ a Hilbert $ B $-module. We know that $ B $ has a faithful state $ \phi $. Using $ \phi $, we can construct a $ \mathbb{C} $-valued pre-inner product $ [\cdot,\cdot] $ on $ B $ by $$ \forall \xi,\eta \in \mathcal{E}: \quad [\xi,\eta] \stackrel{\text{df}}{=} \phi(\langle \xi,\eta \rangle_{B}), $$ where $ \langle \cdot,\cdot \rangle_{B} $ denotes the $ B $-valued inner product on $ \mathcal{E} $.

Question: Is it necessarily true that $ \mathcal{E} $ is already complete with respect to the metric induced by $ [\cdot,\cdot] $? If the answer is ‘no’, then is there a judicious choice of a faithful state $ \phi $ on $ B $ that would make this true?

Thank you very much.

Answer 1

One example of such a bimodule $\mathcal E$ is $B$ itself, with the inner product $\langle b,b'\rangle_B = b^* b'$. Choose $B\supset B_0 = C(X)$ a unital abelian $*$-subalgebra, which can be identified with the algebra of bounded continuous functions on a compact space $X$. If what you are asking were true, you would get that $C(X)\subset B$ is the same as its $L^2$-completion $L^2(X,\mu)\subset L^2(B,\phi)$ for some measure $\mu$ which assigns non-zero values to each non-empty open subset of $X$. From this you see that with the choice $\mathcal E=B$, the state $\phi$ with the property you want exists iff $B$ is finite-dimensional.