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?
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Thank you very much.



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


