Elementary tensors of tensor product of C*algebras

Question

When $\alpha$ is a $C^*$-norm on $A \times B$, we denote the $C^*$ completion of $A \otimes B$ with respect to $\alpha$ by $A\otimes_{\alpha}B$. I feel a little confused about the elementary tensors of $A\otimes_{\alpha}B$. They have the form $a\otimes b, \ a\in A, \ b\in B$. But the completion of $M$ should be the equivalence classes of Cauchy sequences in $M$.

Answer 1

Recall that in the completion $\tilde X$ of a metric space $X$, the space $X$ isometrically embeds into $\tilde X$ via constant sequences (or more appropriately, the equivalence class containing a constant sequence):

$$X\ni x\mapsto[(x,x,x,\ldots)]\in\tilde X.$$

More often than not, we ignore regarding elements of $x$ as (equivalence classes of) sequences in $\tilde X$, and identify an element of $x\in X$ with this corresponding element in $\tilde X$.

So when considering a completion $A\otimes_\alpha B$, you will more frequently see people refer to $A\otimes B$ as being "inside" of $A\otimes_\alpha B$, rather than be completely formal and consider an isometric embedding of $A\otimes B$ into $A\otimes_\alpha B$.

The elementary tensors of $A\otimes_\alpha B$ are really the images of elementary tensors of $A\otimes B$ under this isometric embedding. You don't get more elementary tensors in the completion.

Answer 2

Not entirely sure if this is what you are asking, but $$ A\otimes B=\left\{\sum_j a_j\otimes b_j:\ a_j\in A,\ b_j\in B\right\}. $$ As you say, the completion can be seen as the set of equivalence classes of Cauchy sequences in $A\otimes B$.