Hilbert C*-modules

An increasingly prominent tool in operator theory is the Hilbert C*-module, which are (loosely speaking) Hilbert spaces where the inner product takes values in a C*-algebra. The next level of generalization is that of Hilbert modules over locally C*-algebras (we briefly mentioned locally C*-algebras in this post), and much of the following theory extends to this setting as well.
Here I give the definition of a Hilbert C*-module and collect some of it’s properties, mostly as a reference for personal use. I will likely update this post with new material later on, hopefully without making it too bloated. The theory is now well developed in the literature so the proofs will kept to a bare minimum. For references I will mostly use [1] and [2].

Definition 1 – Hilbert C*-module
A (right) module $M$ over a C*-algebra $\mathcal{A}$ equipped with the a sesquilinear form (over $\mathbb{C}$!), $\langle \cdot, \cdot \rangle: M \times M \to \mathcal{A}$, satisfying

• $\langle x, x \rangle \geq 0$ with equality if and only if $x = 0$.
• $\langle x, y \rangle = \langle y, x \rangle^*$
• $\langle x, y \cdot a \rangle = \langle x, y \rangle a$

is called a pre-Hilbert C*-module over $\mathcal{A}$. The completion of this module with respect to the norm
$$||x||^2 = ||\langle x, x \rangle||^{1/2}$$
is called a right Hilbert C*-module over $\mathcal{A}$ or just a Hilbert $\mathcal{A}$-module.

By convention the above sesquilinear form is linear in the second variable and conjugate linear in the first, this is not what I’m used to, but you can’t argue with conventions, so let’s stick with this for now.

Here are some norm-inner product identities which will come in handy. They can all be found in the first chapters of [1] and [2].

Proposition 1
For any $x,y \in \mathcal{M}$ and any $a\in \mathcal{A}$, the above norm/inner product satisfy the following identities,

1. $||x \cdot a || \leq ||x|| ||a||$
2. $||\langle x, y \rangle || \leq ||x|| ||y||$
3. $\langle x, y \rangle \langle y, x \rangle \leq ||y||^2 \langle x, x \rangle$
4. $||x|| = \sup \{ ||\langle x, y \rangle|| ~:~ y \in \mathcal{M}, ||y|| \leq 1 \}$

Property (2) resembles Cauchy-Schwartz inequality, and property (4) is also very similar to the Hilbert space setting.

1. Operators on Hilbert C*-modules

Let $\mathcal{M}$ and $\mathcal{N}$ be $\mathcal{A}$-modules. A morphism $f$ between $\mathcal{M}$ and $\mathcal{N}$ is a continuous $\mathcal{A}$-module homomorphism, that is, a continuous $\mathbb{C}$-linear map $f: \mathcal{M}\to \mathcal{N}$ satisfying

$$f(xa) = f(x) a \qquad \text{for all } x\in \mathcal{M} \text{ and for all } a \in \mathcal{A}.$$

The collection of all such morphisms from $\mathcal{M}$ to itself will be denoted $Hom_\mathcal{A}(\mathcal{M})$ and is a Banach algebra with respect to the “usual” operator norm, i.e.
$$||f|| = \sup\{ ||f(x)|| ~: ~ x\in \mathcal{M}, ||x|| = 1 \}.$$

This follows from the fact that $Hom_\mathcal{A}(\mathcal{M})$ is a closed subalgebra of the algebra of bounded $\mathbb{C}$-linear maps on $\mathcal{M}$, know from Banach space theory to be complete. We donete by $\mathcal{L}(\mathcal{M})$ the collection of maps $f\in Hom_\mathcal{A}(\mathcal{M})$ for which there exists a map $s \in Hom_\mathcal{A}(\mathcal{M})$ such that
$$\langle f(x), y \rangle = \langle x, s(y) \rangle \qquad \text{for all } x, y \in \mathcal{M}.$$
The map $s$ is uniquely determined and will be denoted $f^\star$, imitating the adjoint operator for Hilbert spaces. $\mathcal{L}(\mathcal{M})$ is easily seen to be a closed subalgebra of $Hom_\mathcal{A}(\mathcal{M})$, hence a Banach algebra in its own right. Elements of $\mathcal{L}(\mathcal{M})$ are called adjointable operators. Remarkably the collection $\mathcal{L}(\mathcal{M})$ can also be characterized simply as the set of all maps $f$ on $\mathcal{M}$ for which there exists an adjoint map $f^\star$ such that
$$\langle f(x), y \rangle = \langle x, f^\star(y) \rangle$$
the point being that continuity and $\mathcal{A}$-linearity of both $f$ and $f^*$, and uniqueness of $f^*$ are all implied (see Lemma 2.1.1 [2]).

$\mathcal{L}(\mathcal{M})$ is actually a C*-algebra since the involution $f\mapsto f^*$ is an isometry, and

\begin{align} || f^*f || & := \sup\{ ||f^*f x|| ~ : ~ ||x|| \leq 1 \} \\ & = \sup\{ ||\langle f^*f x, y \rangle || ~ : ~ ||x||, ||y|| \leq 1 \} \\ & \geq \sup \{ ||\langle f^*fx, x \rangle|| ~ :~ ||x|| \leq 1 \} \\ & = \sup \{ ||\langle fx, fx \rangle||~ :~ ||x|| \leq 1\} = ||f||^2 \end{align} where the second equality follows from Proposition 1 (4).

Another interesting subalgebra of $Hom_\mathcal{A}(\mathcal{M})$, denoted $\mathcal{K}(\mathcal{M})$, is the closed linear span of the collection of all operators of the form
$$\theta_{x,y}: \mathcal{M} \to \mathcal{M} \qquad \text{given by } \theta_{x,y}(z) = x\langle y, z \rangle$$
where $x, y$ are fixed elements in $\mathcal{M}$. Elements of $\mathcal{K}(\mathcal{M})$ are called compact operators, since they mirror the characterization of compact operators on Hilbert spaces as the closure of the span of the rank-1 operators, and as such if $\mathcal{A} = \mathbb{C}$, they are precisely the usual compact operators. Keep in mind though that these operators need not in general be compact operators when viewed as maps between the underlying Banach space (see [2] page 10 for a counterexample).Since $\theta_{x,y}^* = \theta_{y,x}$ one has that $\mathcal{K}(\mathcal{M}) \subset \mathcal{L}(\mathcal{M})$, but equality does not hold in general. Recently it was shown by E. Troitsky that for an adjointable map $T:\mathcal{M} \to \mathcal{N}$ with $\mathcal{N}$ countably generated, the two notions of compactness actually coincide.

2. Basic Operations on Hilbert C*-moduels

Let’s check out some of the common operations that can be done with the Hilbert C*-modules. Given a countable collection of Hilbert $\mathcal{A}$-modules $\mathcal{M}_i$, their direct sum is defined to be the the set
$$\bigoplus \mathcal{M}_i := \left\{ (x_i) \subset \prod \mathcal{M}_i ~~ : ~~ \sum \langle x_i, x_i \rangle \text{ is (norm) convergent in } \mathcal{A} \right\}$$
with inner produce
$$\langle (x_i), (y_i) \rangle := \sum \langle x_i, y_i \rangle.$$
It follows easily from Proposition 1 (2) that the inner product is well defined on $\bigoplus \mathcal{M}_i$, and the proof of completeness with respect to the induced norm is similar to the usual proof of completeness of $l^\infty$ (see for instance [1] Example 1.3.5).

The tensor product is not as straight forward. Let $\mathcal{M}, \mathcal{N}$ be Hilbert C*-modules over $\mathcal{A}$ and $\mathcal{B}$ respectively. The natural starting point is the tensor product $\mathcal{M} \otimes_{alg} \mathcal{N}$ of $\mathcal{M}$ and $\mathcal{N}$ (treated as linear spaces over $\mathbb{C}$). We are going to define an inner product and a $\mathcal{A} \otimes_{min} \mathcal{B}$-module structure on $\mathcal{M} \otimes_{alg} ,\mathcal{N},$ where $\mathcal{A} \otimes_{min} \mathcal{B}$ is the minimal tensor product of the C*-algebras $\mathcal{A}$ and $\mathcal{B}$, henceforth simply denoted $\mathcal{A} \otimes \mathcal{B}$. This is commonly done in the following two ways

• The exterior tensor product of the modules $\mathcal{M}$ and $\mathcal{N}$, which we denote $\mathcal{M} \otimes \mathcal{N}$, is defined to be the completion of $\mathcal{M} \otimes_{alg} \mathcal{N}$ with respect to the norm induced by the inner product $$\langle m_1 \otimes n_1, m_2 \otimes n_2 \rangle := \langle m_1, n_1 \rangle \otimes \langle m_2, n_2 \rangle.$$
• The Interior tensor product of $\mathcal{M}$ and $\mathcal{N}$, with respect to a morphism (of C*-algebras) $\rho: \mathcal{A} \to \mathcal{L}(\mathcal{N})$, which we denote $\mathcal{M} \otimes_\rho \mathcal{N}$ is defined as follows. First define the (possibly degenerate) inner product
$$\langle m_1 \otimes m_2, n_1 \otimes n_2 \rangle_\rho := \langle m_2, \rho(\langle m_1, n_1 \rangle ) m_2 \rangle.$$
Let $K = \{ x\otimes y \in \mathcal{M} \otimes_{alg} \mathcal{N} ~:~ \langle x\otimes y, x\otimes y\rangle = 0 \}.$ The form $\langle \cdot, \cdot \rangle_\rho$ lifts to an inner product on the quotient space $(\mathcal{M} \otimes_{alg} \mathcal{N})/ K$, satisfying the conditions of Definition 1, and the completion of the quotient $\mathcal{M} \otimes_{alg} \mathcal{N}/K$ with respect to the corresponding induced norm is called the interior tensor product of $\mathcal{N}$ and $\mathcal{M}$ with respect to $\rho$.

In both cases the action of $\mathcal{A} \otimes \mathcal{B}$ is defined in the natural way, by its action on decomposable tensors; $m\otimes n \cdot a \otimes b := ma \otimes nb$. There is a lot to check here, which I will not be covering in this post, as it would get way to bloated, but it is all covered well in both [2] and [1] or any introductory book to Hilbert C*-modules.

Limits

For projective and injective limits of Hilbert C*-modules over over fixed C*-algebras, one needs to be a little careful as to the type of directed system and inverse systems we take the limit over, since the category of Hilbert C*-modules over a fixed C*-algebra seems not to be complete. See this post for more on projective and injective limits. Here is one example where an injective (direct) limit does exists:

If $(\mathcal{M}_i, T_{ij})$ is a directed system of Hilbert $\mathcal{A}$-modules, with $T_{ij}:\mathcal{M}_i \mapsto \mathcal{M}_j$ (whenever $i \prec j$) are $\mathcal{A}$-linear module homomorphisms which preserve the inner product (but not necessarily adjointable) , we can define an inner product on the algebraic direct limit of the modules;
$$\bigsqcup \mathcal{M}_i/\sim$$

(recall that $m_i \sim m_j$ if there exist a $k$ such that $i, j \prec k$, and $T_{ik}(m_i) = T_{jk}(m_j)$). One can do this by picking arbitrary representatives, that is $\langle [m_i], [m_j] \rangle := \langle T_{ik}m_i, T_{jk} m_j \rangle$, since the module maps in the directed system preserves inner products. This turns out to be a Hilbert C*-module over $\mathcal{A}$ with the desired universal properties as is shown in Proposition 1.3 of this article.

3. Can I bring my tools?

Every Hilbert C*-module has the structure of a Banach space over $\mathbb{C}$ (if we neglect the module structure) and morphisms of Hilbert C*-modules as defined above are continuous linear maps of the underlying Banach spaces. Hence many of the results that hold for Banach spaces carry over to the setting of Hilbert C*-modules. Off the top of my head, among these are the usual suspects from Banach space theory; the open mapping theorem, the closed graph theorem and the uniform boundedness principle. Additionally, as we have seen, the collection of adjointable operators forms a C*-algebra, on which we may use the spectral theory and functional calculus as usual without any hassle.

Here is a list of things that do carry over, and some known pathologies for Hilbert C*-modules. The list is by no means exhaustive, but hopefully it gets you quickly up to speed. In what follows $H$ will be a Hilbert space over $\mathbb{C}$ and $\mathcal{M}$ will be a Hilbert $\mathcal{A}$-module. The inner products of both $H$ and $\mathcal{M}$ will be denoted by $\langle \cdot, \cdot \rangle$, as this is unlikely to cause serious confusion.

Riesz representation theorem and sesquilinear forms. On Hilbert spaces the Riesz representation theorem asserts that every bounded linear functional $\phi$ on $H$ is of the form $x \mapsto \langle x, y \rangle$ for some uniquely determined $y\in H$, and that $||\phi|| = ||y||.$
For Hilbert C*-modules we have a similar statement, namely that the C*-algebra $\mathcal{K}(\mathcal{M}, \mathcal{A})$, where $\mathcal{M}$ is an $\mathcal{A}$-module, consists entirely of functions of the form $f_x(y) = \langle x, y \rangle$, and it’s easy to check uniqueness and that $||f_x|| = ||x||$.

On Hilbert spaces there is is also a Riesz representation theorem for continuous sesquilinear forms, which says that there is a 1-1 correspondence between contunuous (or bounded) sesquilinear forms and bounded operators on a Hilbert space. The correspondence is give by
$$[\cdot, \cdot ] \mapsto T\qquad \text{where} \qquad [x, y ] = \langle Tx, y \rangle.$$
It would be nice to find some similar statement (or counterexamples) for Hilbert C*-modules.

Polarization Identity: The identity $$\langle x, y \rangle = \frac{1}{4} \sum_{n=0}^3 i^n \langle x + i^ny, x + i^ny \rangle$$ holds for any sesquilinear form, so it clearly also holds for inner products on Hilbert C*-modules.

Polar decomposition The polar decomposition from Hilbert space theory, decomposing operators on Hilbert spaces into a product of a partial isometry and a positive operator does not in general work on Hilbert C*-modules. That is, if $T$ is an adjointable operator on a Hilbert $\mathcal{A}$-module $\mathcal{M}$, then it does not follow that the partial isometry $U$ in the polar decomposition $T = U|T|$ of $T$ is adjointable. The operator $|T|$ is found using functional calculus on the C*-subalgebra of $\mathcal{L}(\mathcal{M})$ generated by $T$ and $T^*$, so this is an adjointable module morphism, but recall that the partial isometry $U$ is contained in the von Neumann algebra generated by $T$, which may not be contained in the C*-algebra of adjointable operators. There are situations where the decomposition works though. One such situation is when both $T$ and $T^*$ have dense ranges (see Proposition 3.8 [2]).

Orthogonal complementing subspace: The most striking pathology of Hilbert C*-modules is the non-existence of an orthogonal complementing subspace. On Hilbert spaces any closed subspace has a natural complementing subspace, namely the orthogonal complement. This does not hold in the setting of Hilbert C*-modules. That is, if $K \subset \mathcal{M}$ is a closed submodule of a Hilbert C*-module $\mathcal{M}$, then $$K^\perp = \{ x\in \mathcal{M} ~:~ \langle x, y\rangle = 0 ~\text{for all } y\in K \}$$ is indeed also a closed submodule of $\mathcal{M}$, but $K$ and $K^\perp$ need not be complementary, that is $K\oplus K^\perp$ may not be isomorphic to $\mathcal{M}$. In fact, there may be no complementary submodule at all. However, if either of the following conditions holds;

• $K$ is a finitely generated projective $\mathcal{A}$ module, (that is, if there exists a Hilbert $\mathcal{A}$-module $\mathcal{N}$ such that $K \oplus \mathcal{N} \simeq \mathcal{M} \oplus … \oplus \mathcal{M}$)
• $K$ is finitely generated
• $K$ is the kernel or image of an adjonable module morphism with closed range
• $K$ is self-dual (explained below)

then $K \oplus K^\perp \simeq \mathcal{M}$ (section 2 of [1] and Th 1.4.6 [2]).

Existence of a minimal distance : On Hilbert spaces, for any closed subspace $K\subset H$ and any $x\in H\backslash K$ there exists an element $y \in K$ such that $$||x – y|| = \inf_{z\in K} || x – z|| \tag{(\star)}.$$
This does not hold for Hilbert C*-modules in general. Try to find a counterexample, or look at Exercise 3 below.

Dual space: The dual space of a Hilbert space $H$ has a natural Hilbert space structure, with respect to the inner product $\langle \phi_x, \phi_y \rangle = \langle x, y \rangle$ where $\phi_x, \phi_y \in H^\star$ are the functionals corresponding to $x$ and $y$ in $H$ by Riesz. The Riesz representation theorem is an (antilinear) isomorphism of these Hilbert spaces, hence all Hilbert spaces are isomorphic to their bidual. Such spaces are called reflexive. For Hilbert C*-modules the dual space of a $\mathcal{A}$-module $\mathcal{M}$, denoted $\mathcal{M}^*$, is defined as
$$M^* : = Hom_\mathcal{A}(\mathcal{M}, \mathcal{A})$$
and is a complete banach module over $\mathcal{A}$ with respect to the usual operator. The action of $\mathbb{C}$ and $\mathcal{A}$ on $\mathcal{M}^*$ is given by $(\lambda \cdot f)(x) = \overline{\lambda} f(x)$ and $(f \cdot a)(x) = a^*f(x)$ respectively, which makes the natural inclusion of $\mathcal{M}$ into $\mathcal{M}^\star$ given by
$$x \mapsto \langle x, \cdot \rangle =: \hat{x}$$
an imbedding of Banach spaces. When it is surjective, the module $\mathcal{M}$ is called self-dual. Two important property of self-dual Hilbert C*-modules are

• $\mathcal{L}(\mathcal{M}) \simeq Hom_\mathcal{A}(\mathcal{M})$ (see [1] Proposition 2.5.2)
• every closed submodule is orthogonally complementable (Corollary 2.5.4 of [1]).

When the C*-algebra $\mathcal{A}$ is a vN-algebra the dual module $\mathcal{M}^*$ can be given a Hilbert $\mathcal{A}$-module structure with an inner product for which $\langle f, \hat{x} \rangle := f(x)$ for all $f,\hat{x}\in M^*$ (see Theorem 3.2.1 of [1]).

Bidual and Reflexivity
Luckily the bidual $\mathcal{M}^{**} = Hom_\mathcal{A}(\mathcal{M}^*, \mathcal{A})$ of a Hilbert C*-module always admits a Hilbert C*-module structure, determined by the inner product
$$\langle F, G \rangle = F(\tilde{G})$$
where $F, G\in \mathcal{M}^{**}$ and $\tilde{G} \in \mathcal{M}^*$ is given by $\tilde{G}(x) := G(\hat{x})$ (see “dual space” above for the notation). The norm induced by this inner product is the same as the usual operator norm (Theorem 4.1.4 [1]).

Existence of a basis and dimensionality : This may be silly, but for correctness let’s assert that C*-algebras are not division rings, which makes it difficult to talk about dimensions and basis in general. There are non-trivial examples where a basis does exist, like $H_\mathcal{A}$, where $\mathcal{A}$ is (say) a finite dimensional C*-algebra with basis $\{ a_j\}_{j=1}^n$. The $\mathcal{A}$-module $H_\mathcal{A}$ has basis $\{(e_i, a_j) ~ :~ j= 1, …, n,~ i = 1, … \}$. One often has to make due with generating sets rather than a basis. Many results rely on the existence of a countable generating set, and as a sidenote, a useful characterization of countably generated $\mathcal{A}$ modules $\mathcal{M}$ is the following; $\mathcal{M}$ is countably generated if and only if $\mathcal{K}(\mathcal{M})$ is $\sigma$-unital (if and only if $\mathcal{K}(\mathcal{M})$ posesses a strictly positive element).

Spectral theory and functional calculus: As previously mentioned, the theory carries over to the adjointable operators without issues, since $\mathcal{L}(\mathcal{M})$ is a C*-algebra.

Characterization of Positive elements: Just as in the case of operators on (complex) Hilbert spaces, a map T is a positive element of the C*-algebra $\mathcal{L}(\mathcal{N})$ if and only if
$$\langle T x, x \rangle \geq 0 \qquad \text{for all } x\in \mathcal{M}.$$ (see [1] Prop. 2.1.3).

Unitary operators: An operator $u: \mathcal{M} \to \mathcal{N}$ is unitary if and only if it is surjective isometric and $\mathcal{A}$-linear (Th. 2.3.5 [2])

Boundedness Let T be any map on a Hilbert $\mathcal{A}$-module $\mathcal{M}$. Then $T\in Hom_\mathcal{A}(\mathcal{M})$ if and only if there exists a $K \geq 0$ such that $\langle Tx, Tx \rangle \leq K \langle x, x \rangle$ (see Theorem 2.1.4 of [1]).

Invertibility For Hilbert spaces an operator $T: H \to H$ is invertible if and only if $Im(T)$ is dense in $H$ and $T$ is bounded away from zero, that is, there exists a real constant $k>0$ such that $||Tx|| \geq k||x||$ for all non-zero x.
For Hilbert C*-modules a similar statement holds, at least for self adjoint operators. Let $t^* = t\in \mathcal{L}(\mathcal{M})$ be self adjoint, then $t$ is invertible if and only if there exists a $k>0$ such that $||tx|| \geq k||x||$ for all $x\in \mathcal{M}$. (See Lem. 3.1 [2])

4 Common Examples

Here are some common and useful examples to keep in mind when working with Hilbert C*-modules. First of which, any C*-algebra $\mathcal{A}$, (or any right ideal of $\mathcal{A}$) can be treated as a Hilbert $\mathcal{A}$-module with respect to the inner product
$$\langle x, y \rangle = x^*y.$$
(note again the linearity in the second term!). I will denote this module by $\mathcal{M}_\mathcal{A}$.

The standard Hilbert C*-module over $\mathcal{A}$ is defined to be the Hilbert $\mathcal{A}$-module
$$H_\mathcal{A} := \bigoplus^\infty \mathcal{M}_\mathcal{A},$$
that is, the set of all sequences $(a_i)\subset \mathcal{A}$ such that $\sum a_i^*a_i$ is norm convergent, with inner product $$\langle (a_i), (b_i) \rangle = \sum a_i^*b_i.$$

Next, the continuous section of bundles of Hilbert subspaces of a common Hilbert space $H$, over a compact Hausdorff space $X$, can be endowed with a Hilbert $C(X)$-module structure by pointwise operations, that is if $\xi$ is a continuous section, and $f\in C(X)$, then define the $C(X)$-module action by $(f\cdot \xi)(x) = f(x)\xi(x)$, and if $\nu$ is another continuous section, define the $C(X)$-valued inner product $(\langle \xi, \nu \rangle)(x) = \langle \xi(x), \nu(x) \rangle$ which is clearly a continuous function, being the composition of continuous functions. One can check this inner product satisfies the conditions of Definition 1.

The following identities are worth memorizing as they are often tacitly employed in the literature. Let $\mathcal{M}$ and $\mathcal{E}$ be Hilbert $\mathcal{A}$-modules, and $\mathcal{N}$ be a Hilbert $\mathcal{B}$ module, then

• $\mathcal{K}(\mathcal{A}) \simeq \mathcal{A}$
• $\mathcal{K}(M) \otimes \mathcal{K}(\mathcal{N}) \simeq \mathcal{K}(\mathcal{M}\otimes \mathcal{N})$
• $\mathcal{L}(\bigoplus_{i=1}^m \mathcal{M}, \bigoplus_{i=1}^n \mathcal{E} ) \simeq M_{n\times m}(\mathcal{L}(\mathcal{M}, \mathcal{E}))$
• $\mathcal{K}(\bigoplus_{i=1}^m \mathcal{M}, \bigoplus_{i=1}^n \mathcal{E} ) \simeq M_{n\times m}(\mathcal{K}(\mathcal{M}, \mathcal{E}))$
• $\mathcal{L}(\mathcal{E}) = M(\mathcal{K}(\mathcal{E}))$ where $M$ is the multiplier algebra
Proof
• Prove that the subalgebra $\mathcal{A}^2 = \{a \in \mathcal{A} ~ : ~ a = bc, \text{ for some } b,c \in \mathcal{A}\}$ is dense in $\mathcal{A}$ using an approximate unit. The map sending $\mathcal{A}^2 \ni xy \mapsto \theta_{x,y}$ hence extends to an isomorphism of $\mathcal{A}$ and $\mathcal{K}(\mathcal{A})$
• The map that establishes the isomorphism is the extension of the map $\theta_{u, v}\otimes \theta_{x, y} \mapsto \theta_{u\otimes x, v\otimes y}$.
• Let $e_j$ be the j’th coordinate projection on $\bigoplus_{i=1}^m \mathcal{M}$, and let $e^i$ be the i’th coordinate inclusion on $\bigoplus_{i=1}^n \mathcal{E}$, i.e. $e^i(r) = (0, …, r, …,0)$ with r in the i’th index. Then the map $t\mapsto [t_{ij}]$ where $t_{ij} = e_j \circ t\circ e^i$, determines the needed isomorphism of C*-algebras.
• Repeat the above proof mutatis mutandis
• For a proof of this fact, consult Theorem 2.4 in [2]

Note how the first and fifth equality together implies that $\mathcal{L}(\mathcal{A})$ is isomorphic to the multiplier algebra of $\mathcal{A}$ and that $\mathcal{A}$ imbeds into $\mathcal{L}(\mathcal{A})$ as the C*-subalgebra of compact operators.

In general we only have an imbedding $\mathcal{L}(\mathcal{M}) \otimes \mathcal{L} (\mathcal{N}) \rightarrow \mathcal{L}(\mathcal{M} \otimes \mathcal{N})$ given in the most natural way as the unique lift of the bilinear map from $\mathcal{L}(\mathcal{M}) \times \mathcal{L}(\mathcal{N}) \to \mathcal{L}(\mathcal{M} \otimes \mathcal{N}),$ determined by sending $(s, t) \mapsto s\otimes t,$ where $(s\otimes t)(x\otimes y) = sx\otimes ty,$ to the space $\mathcal{L}(\mathcal{M}) \otimes \mathcal{L} (\mathcal{N}).$

Exercises

Here are some useful facts which I have left as exercises, most of which are taken from the cited references where they pop-up in various proofs and remarks.

Ex. 1
Show that $H_\mathcal{A} \simeq H \otimes \mathcal{A}$ for a separable Hilbert space $H$.
Proof
Let $\{ e_i\}$ be the countably infinite orthonormal basis of $H$. The map in question is the one from $H_\mathcal{A}$ to $H \otimes \mathcal{A}$ sending $(0, …,0, a_i, 0, …. ) \mapsto e_i \otimes a_i$, where $a_i$ is in the i’th index.
Ex. 2
Show that the if $\mathcal{A}$ is a non-unital C*-algebra, and $\mathcal{A}^1$ is it’s unitization, then there is a natural isomorphism
$$\overline{H_{\mathcal{A}^1}\mathcal{A}} \simeq H_{\mathcal{A}}.$$
where $H_{\mathcal{A}^1}\mathcal{A} = \text{span}\{ xa ~ :~ x \in H_{\mathcal{A}^1}, ~ a\in \mathcal{A} \}.$
Proof
First use the identities $H_{\mathcal{A}^1} \simeq H \otimes \mathcal{A}^1$ and $H_{\mathcal{A}} \simeq H \otimes \mathcal{A}$, where $H$ is a separable infinite dimensional Hilbert space (module over $\mathbb{C}$), and the C*-algebras are treated as modules over themselves with the inner product $\langle a, b \rangle = a^* b$ (as above), and action given (as defined earlier) by $(x\otimes b) \cdot a := x \otimes ba$. Now check that the map $(x \otimes 1) a \mapsto (x \otimes a)$ determines an isomorphism of Hilbert C*-modules.
Ex. 3
Find a Hilbert C*-module for which the minimal distance property of $(\star)$ fails to hold. Hint, start with a C*-algebra which is not strictly convex, like the C*-algebra $l^\infty(\mathbb{C})$ of all bounded complex sequences (with pointwise operations and sup norm).

Proof
Let $\mathcal{A} = l^\infty(\mathbb{C})$ be the afformentioned C*-algebra, treated as a Hilbert C*-module over itself. $K_1 = \{ f \in C(\mathbb{N}) ~:~ f(n) = 0 ~ \text{for all } n \geq 2 \}$ is a submodule (since it is an ideal). Let $y = \delta_2$ where $\delta_2(m) = \begin{cases} 1 & m=2 \\ 0 & \text{else} \end{cases}$. Then any $f\in K_1$ with $||f|| \leq 1$ has the property that $||y – f|| = \inf_{z\in K_1} ||y- z||$.
[1] V. M. Manuilov, E. V. Troitsky, Hilbert C*-modules AMS Rhode Island, Mathematical Monographs, 226 (2005)
[2] E.C. Lance Hilbert C*-modules: A toolkit for operator algebraists, Cambridge university press, 210 (1995)

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