Definition of $C^*$-algebraic quantum group

Question

A compact quantum group is a pair $(A, \Delta)$ where $A$ is a unital $C^*$-algebra and $\Delta: A \to A \otimes A$ is a $C^*$-morphism such that

(1) $(\Delta \otimes \operatorname{id}_A) \circ \Delta = (\operatorname{id}_A \otimes \Delta) \circ \Delta$

(2) $\Delta(A) (1 \otimes A)$ and $\Delta(A)(A \otimes 1)$ are dense subspaces of $A \otimes A$.

Here, the tensor products are the tensor products of $C^*$-algebras with respect to the minimal tensor norm.

Question: How is $1 \otimes A$ and $\Delta(A)(1 \otimes A)$ defined?

My guess would be $$1 \otimes A = \operatorname{span}\{1 \otimes a: a \in A\}$$ and $$\Delta(A) (1 \otimes A) = \{\Delta(a)x : a \in A, x \in 1 \otimes A\}$$

Is the above correct? I'm not even sure if $\Delta(A)(1 \otimes A)$ is a "multiplication" of two sets.

Answer 1

OK, so here is an expansion of my comments above:

My guess would be $$1 \otimes A = \operatorname{span}\{1 \otimes a: a \in A\}$$

and

$$\Delta(A) (1 \otimes A) = \{\Delta(a)x : a \in A, \ x \in 1 \otimes A\}$$

Regading the first guess, the map $a\in A\mapsto 1\otimes a\in A\otimes A$ is linear so its range is a linear subspace, and hence the "span" above is innocuous.

As for the second guess, let me first say that, for two subsets $X$ and $Y$ of the same $C^*$-algebra $A$, the notation $XY$ is often used in the literature with different and incompatible meanings. To my knowledge the most common uses are:

1. $XY = \{xy: x\in X, \ y\in Y\}$,

2. $XY = \text{span}\{xy: x\in X, \ y\in Y\}$,

3. $XY = \overline{\text{span}}\{xy: x\in X, \ y\in Y\}$.

The fact that this is not a standard notation obviouly makes life hard for the reader, unless of course the author makes it clear which convention they are using.

In the case in point, namely in the definition of a quantum group, the apropriate choice is (2). In other words, one should consider the definition

$$\Delta(A) (1 \otimes A) := \text{span}\{\Delta(a)x : a \in A, \ x \in 1 \otimes A\}.$$

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But how is one supposed to know it in case the author does not make it clear? Well, Math is said to be an exact science but reading math isn't! It is actually full of tricks. The sentence:

$Δ()(1⊗)$ and $Δ()(⊗1)$ are dense subspaces of ⊗.

carries two cues: first of all, if something is said to be dense, and one also expects this to be a non-trivial statement, one would guess that whatever we are talking about is not already defined as a closed set, so I would rule out interpretation (3).

On the other hand the word subspaces seem to indicate that the author thinks that $Δ()(1⊗)$ and $Δ()(⊗1)$ are subspaces so, after spending a few minutes and realizing that interpretation (1) does not lead to a subspace, by exclusion the only sensible guess is (2).

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Going slightly off topic (in case I haven't already done so), it might be relevant to mention the Cohen-Hewitt Theorem which asserts that if $A$ is a Banach algebra with an approximate unit, and $M$ is a Banach module (i.e. a Banach space with a left $A$-module structure satsfying $\|am\|\leq \|a\|\|m\|$, for all $a$ in $A$ and $m$ in $M$), then the above three interpretations of $AM$ lead exactly to the same set, an in fact for every element $n\in \overline{\text{span}}\{am: a\in A, \ m\in M\}$, it is possible to write $n=am$ with nice estimates on the norms of $a$ and $m$. In other words, you are sometimes allowed to be vague in your use of $AM$ without risking being imprecise!