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!


2 Answers 2


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 $E/E_I$.

  • $\begingroup$ Hello Omar. Could you kindly provide me with more details about the use of the Kasparov Stabilization Theorem to prove the equality of the two norms in the case of countably generated Hilbert $ A $-modules? Thanks! $\endgroup$ May 19, 2017 at 5:44
  • 1
    $\begingroup$ @Transcendental Yes sure, I added some details on the countably generated step. The step to go from countably generated to noncountably generated is not really important since usually people only use countably generated modules. You can find Kasparov stabilization theorem here ro.uow.edu.au/cgi/… $\endgroup$
    – Omar
    May 19, 2017 at 12:41
  • 1
    $\begingroup$ Hi Omar. Thank you for your speedy response. Allow me some time to check the details of your explanation before I award you the bounty. For my work, I do require the result to hold even for Hilbert $ A $-modules that aren’t necessarily countably generated, but as you’ve already mentioned, this appears to be only a minor step from the case for countably generated Hilbert $ A $-modules. $\endgroup$ May 19, 2017 at 20:29

Here are a couple of supplements to Omar's answer.

The first thing to note is that the module $\mathcal{E}_I$ is equal to the module $\mathcal{E}I=\{ei\ |\ e\in \mathcal{E},\ i\in I\}$. The inclusion $\mathcal{E}I\subset \mathcal{E}_I$ follows easily from the linearity of the inner product and from the fact that $I$ is an ideal. For the reverse inclusion, let $i_\lambda$ be an approximate unit for $I$, and then show (by expressing the norm in terms of the inner product) that for each $e\in \mathcal{E}_I$ one has $\| e-ei_\lambda\|\to 0$ as $\lambda\to\infty$. This gives $\mathcal{E}_I\subset \overline{\mathcal{E}I}$, and the latter is equal to $\mathcal{E}I$ by the Cohen factorisation theorem. For full details see Lemma 3.23 in:

Raeburn, Iain; Williams, Dana P., Morita equivalence and continuous-trace $C^*$-algebras, Mathematical Surveys and Monographs. 60. Providence, RI: American Mathematical Society (AMS). xiv, 327 p. (1998). ZBL0922.46050.

Now, turning to the inner product on $\mathcal{E}/\mathcal{E}_I$: here are three arguments showing that the norm defined by the inner product is equal to the quotient norm.

(1) A short direct computation is given in Lemma 3.1 in:

Zettl, Heinrich H., Ideals in Hilbert modules and invariants under strong Morita equivalence of C*-algebras, Arch. Math. 39, 69-77 (1982). ZBL0498.46034.

(2) An argument based on the uniqueness of the $C^*$-norm on the linking algebra is given in Proposition 3.25 in the book of Raeburn-Williams cited above. (The authors attribute the argument to Siegfried Echterhoff.)

(3) Here is a third proof, using ideas around operator modules and the Haagerup tensor product. All of the necessary background can be found in

Blecher, David P.; Le Merdy, Christian, Operator algebras and their modules -- an operator space approach, London Mathematical Society Monographs. New Series 30; Oxford Science Publications. Oxford: Oxford University Press (ISBN 0-19-852659-8/hbk). x, 387~p. (2004). ZBL1061.47002.

Consider $I\to A\to A/I$ as an exact sequence of operator $A$-bimodules. Taking the Haagerup tensor product with $\mathcal E$ (viewed as a right operator module over $A$) gives $$ \mathcal E\otimes_A I \to \mathcal E\otimes_A A \to \mathcal E\otimes_A (A/I). $$ By the exactness property of the Haagerup tensor product over a $C^*$-algebra (a theorem of Anantharaman-Delaroche and Pop), the first map in the display is a (completely) isometric embedding, and the second map induces a (completely) isometric isomorphism $$ (\mathcal E\otimes_A A)/ (\mathcal E\otimes_A I) \cong \mathcal E\otimes_A (A/I)\qquad (*) $$ where the left-hand side carries its canonical quotient operator space structure (and in particular, the usual Banach-space quotient norm).

The module action gives a (completely) isometric isomorphism $\mathcal E\otimes_A A \to \mathcal E$, which restricts to an isomorphism $\mathcal E\otimes_A I \to \mathcal E I=\mathcal E_I$. Making these identifications, the isomorphism $(*)$ is given by the formula $$ \mathcal E/\mathcal E_I\to \mathcal E\otimes_A (A/I),\qquad (ea+\mathcal E_I)\mapsto e\otimes(a+I).\qquad (**) $$ Now, $A/I$ is a (right) Hilbert $C^*$-module over itself, and the quotient map $A\to A/I$ gives a (left) action of $A$ on this $C^*$-module by adjointable operators. We can thus form the Hilbert $C^*$-module tensor product $\mathcal E\otimes^{C^*}_A (A/I)$, which will be a Hilbert $C^*$-module over $A/I$. A theorem of Blecher asserts that the identity map on the algebraic tensor products extends to a (completely) isometric isomorphism between $\mathcal E\otimes^{C^*}_A (A/I)$ and the Haagerup tensor product $\mathcal E\otimes_A (A/I)$. A straightforward computation shows that the map $(**)$ is isometric with respect to the inner product $[\cdot,\cdot]$ on $\mathcal E/\mathcal E_I$, and the canonical inner product on $\mathcal E\otimes^{C^*}_A (A/I)$. Since $(**)$ is also isometric for the quotient norm, the latter must coincide with the norm induced by $[\cdot,\cdot]$.

  • $\begingroup$ Hi t.c. Thank you very much for your very detailed response! $\endgroup$ Jun 9, 2017 at 6:39
  • $\begingroup$ @Transcendental You're welcome! $\endgroup$
    – user85913
    Jun 21, 2017 at 7:43

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