Let $ A $ be a $ C^{\ast} $-algebra, $ I $ a closed two-sided ideal of $ A $, and $ \mathcal{E} $ a Hilbert $ A $-module. Let
$$
\mathcal{E}\_{I}
\stackrel{\text{df}}{=}
\{ x \in \mathcal{E} \mid \langle x,x \rangle\_{\mathcal{E}} \in I \}.
$$
Using the Cauchy-Schwarz Inequality for Hilbert $ C^{\ast} $-modules, it is not difficult to show that
$$
\mathcal{E}\_{I} =
\{
x \in \mathcal{E} \mid
(\forall y \in \mathcal{E})(\langle x,y \rangle\_{\mathcal{E}} \in I)
\}.
$$
This implies that $ \mathcal{E}\_{I} $ is a linear subspace of $ \mathcal{E} $.


Furthermore, as the $ A $-valued inner product on $ \mathcal{E} $ is continuous, $ \mathcal{E}\_{I} $ is a $ \| \cdot \|\_{\mathcal{E}} $-closed subset of $ \mathcal{E} $.


All of this implies that $ \mathcal{E}\_{I} $ is a Hilbert $ I $-module.


Now, the quotient space $ \mathcal{E} / \mathcal{E}\_{I} $ is a Banach space w.r.t. the quotient norm $ \| \cdot \|\_{\text{q}} $ defined by
$$
\forall x \in \mathcal{E}: \qquad
\| x + \mathcal{E}\_{I} \|\_{\text{q}}
\stackrel{\text{df}}{=}
\inf\_{y \in \mathcal{E}\_{I}} \| x + y \|\_{\mathcal{E}}.
$$
It is also a right $ (A / I) $-module equipped with an $ (A / I) $-valued pre-inner product $ [\cdot,\cdot] $ defined by
$$
\forall x\_{1},x\_{2} \in \mathcal{E}: \qquad
[x\_{1} + \mathcal{E}\_{I},x\_{2} + \mathcal{E}\_{I}]
\stackrel{\text{df}}{=}
\langle x\_{1},x\_{2} \rangle\_{\mathcal{E}} + I.
$$




---



> 
> **Question.** Is the norm on $ \mathcal{E} / \mathcal{E}\_{I} $ that is induced by $ [\cdot,\cdot] $ the same as $ \| \cdot \|\_{\text{q}} $, i.e.,
>  $$
> \forall x \in \mathcal{E}: \qquad
>  \| x + \mathcal{E}\_{I} \|\_{\text{q}}
> = \sqrt{\| \langle x,x \rangle\_{\mathcal{E}} + I \|\_{A / I}}?
> $$
> 
> 
> 


Thank you for your help!



On the coincidence (or non-coincidence) of two norms defined on the quotient of a given Hilbert $ C^{\ast} $-module by a certain linear subspace
Yes, I think this is true. Here is an argument to show why. First for $E=A$, then this is just the standard theorem saying that the quotient norm on $A/I$ is the $C^\*$-algebra norm. (any ref on $C^\*$-algebras has a proof)


Then you argue directly that the case $E=A$ implies that it is also true for $E=l^2(\mathbb{N})\otimes A$.


Then you could use Kasparov stabilization theorem to prove it for coutably generated $C^\*$-modules.


To prove it for all modules, you might argue that both norms calcuated at a fixed element, only needed a countably generated submodule to be computed on. (Both are infimum of something so you take the submodules generated by a sequence reaching the infimum)


Edit: Here some more details,


I will use $E'$ to denote the module $E\oplus l^2(\mathbb{N})\otimes A$. We directly see that $E'\_I=E\_I\oplus l^2(\mathbb{N})\otimes I$, hence $E'/E'\_I=E/E\_I\oplus l^2(\mathbb{N})\otimes A/I.$ Hence the inclusion map $i:E/E\_I\to E'/E'\_I$ preserves the two norms defined in the question. 


Theorem(Kasparov's Specialization theorem)(reference in a comment) Let $E$ be a countably generated $A$-module, then $E\oplus l^2(\mathbb{N})\otimes A\simeq l^2(\mathbb{N})\otimes A$. Here the isomorphism is a unitary isomorphism.


Since the module $E'$ is untarly equivalent to $l^2(\mathbb{N})\otimes A$ it follows that the two norms are equal on $E'/E'\_I$ and hence on $E/E\_I$.


