- #1

Geofleur

Science Advisor

Gold Member

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"Let ## \mathcal{A} ## and ## \mathcal{B} ## be algebras. The the vector space tensor product ## \mathcal{A} \otimes \mathcal{B} ## becomes an algebra tensor product if we define the product

## (\mathbf{a}_1 \otimes \mathbf{b}_1)(\mathbf{a}_2\otimes\mathbf{b_2})=\mathbf{a}_1\mathbf{a}_2\otimes\mathbf{b}_1\mathbf{b}_2 ##

on ## \mathcal{A}\otimes\mathcal{B} ##."

He goes on to say that, because the spaces ##\mathcal{A}\otimes\mathcal{B} ## and ## \mathcal{B}\otimes\mathcal{A} ## are isomorphic, we require that ## \mathbf{a}\otimes\mathbf{b} = \mathbf{b}\otimes\mathbf{a} ## for all ## \mathbf{a}\in\mathcal{A} ## and ## \mathbf{b}\in\mathcal{B}##.

So far, so good. But then he goes on to say that this last requirement is important when an algebra ## \mathcal{A} ## is written as the tensor product of two of its subalgebras ##\mathcal{B} ## and ##\mathcal{C}##; also, that ## \otimes ## in such a case is identified with the multiplication in ## \mathcal{A}##.

I have been trying for a week now to come up with an example to help me make sense of these remarks. Firstly, does anyone know of an example of an algebra that can be written as the tensor product of two of its own subalgebras? Secondly, I have always thought that ## \mathbf{a}\otimes\mathbf{b} ## is just shorthand for an element, ## (|a\rangle, |b\rangle) ##, of the Cartesian product of the underlying vector spaces. So how can ## \otimes ## be the same as the multiplication in ## \mathcal{A} ##?