Inclusion $M(A) \otimes M(B)\subseteq M(A\otimes B)$ of multiplier algebras

Question

Consider the following definitions given in Timmerman's book "An invitation to quantum groups and duality":

m

Further in the book, it is claimed that if $A$ and $B$ are non-degenerate algebras, then we have a canonical inclusion $M(A)\otimes M(B)\hookrightarrow M(A\otimes B)$. How exactly does this inclusion arise?

My try:

Given $S=(S_l, S_r) \in M(A), T=(T_l, T_r) \in M(B)$, we want to associate a multiplier in $M(A\otimes B)$. My guess is that we can take $$(S_l \otimes T_l, S_r\otimes T_r).$$ The universal property of the tensor product will then give the desired inclusion.

Is my idea correct? I'm doubting this because this map does not seem very 'practical' to work with. Any help/insight will be appreciated!

Answer 1

$\newcommand{\bC}{\mathbb{C}}\newcommand{\finite}{\text{fin}}$What you propose is absolutely correct. Note that when multiplier algebras appear in practice, they’re either treated as universal objects (i.e., as maximal unitalisations) or using some special concrete realisation. For example, if $\bC_\finite(X)$ is the commutative algebra of all finitely-supported $\mathbb{C}$-valued functions on some set $X$ with the usual pointwise operations, then $M(\bC_\finite(X))$ can be canonically identified with the commutative algebra of all $\mathbb{C}$-valued functions on $X$; indeed, the multiplier $T = (T_l,T_r)$ corresponding to a function $f : X \to \mathbb{C}$ is given by $$ \forall a \in \bC_\finite(X), \quad T_l(a) := f \cdot a = a \cdot f =: T_r(a). $$ If the business with left and right multipliers seems impractical, it’s because it is—it’s a conceptually meaningful way of constructing multiplier algebras that you can then use to prove their basic general properties; compare, for instance, the tedium you need to invoke to construct tensor products of Hilbert spaces.

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Let me sketch a proof of the following claim, which was cited above:

Let $X$ be a set, let $\mathbb{C}(X)$ be the commutative algebra of all $\mathbb{C}$-valued functions on $X$, let $\bC_\finite(X)$ be the subalgebra of all finitely supported functions. Then the homomorphism $\tau : \bC(X) \to M(\bC_\finite(X))$ given by $$ \forall f \in \bC(X), \, \forall a \in \bC_\finite(X), \quad \tau(f)_l(a) := f \cdot a =: \tau(f)_r $$ is an isomorphism of unital algebras intertwining the inclusion $\bC_\finite(X) \hookrightarrow \bC(X)$ and the canonical injection $\bC_\finite(X) \hookrightarrow M(\bC_\finite(X))$.

By commutativity of $\bC(X) \supseteq \bC_\finite(X)$ and construction of the canonical injection $\bC_\finite(X) \hookrightarrow M(\bC_\finite(X))$, one can quickly check that $\tau$ is a well-defined homomorphism intertwining the inclusion $\bC_\finite(X) \hookrightarrow \bC(X)$ and the canonical injection $\bC_\finite(X) \hookrightarrow M(\bC_\finite(X))$. The only non-trivial point, then, is bijectivity of $\tau$. Now, for $x \in X$, let $\delta_x \in \bC_\finite(X)$ be the characteristic function of the singleton $\{x\}$, so that $\{\delta_x\}_{x\in X}$ gives a basis for $\bC_\finite(X)$ as a $\bC$-vector space. Then, for all $x \in X$, since $\delta_x^2 = \delta_x$, it follows that for every $T \in M(\bC_\finite(X))$, $$ T_l(\delta_x) = T_l(\delta_x^2) = T_l(\delta_x)\delta_x = \delta_xT_r(\delta_x) = T_r(\delta_x^2) = T_r(\delta_x), $$ and hence that $$ T_l(\delta_x) = T_r(\delta_x) = T(x)\delta_x, \quad T(x) := T_l(\delta_x)(x) = T_r(\delta_x)(x). $$ One can now check that $\tau^{-1} : M(\bC_\finite(X)) \to \bC(X)$ is given by $$ M(\bC_\finite(X)) \ni T \mapsto (x \mapsto T(x)) \in \bC(X). $$