Tensoring Spectral triples that are composed from Real algebras. I have a misunderstanding that I am hoping is really quite trivial. 
In connes standard Non-commutative geometry model of electroweak interactions he takes the algebra input in his finite spectral triple to be $A_F=\mathbb{C}⊕\mathbb{H}$. He then tensors this finite algebra with the algebra of complex functions over a smooth manifold $C^\infty(M)$.
As far as I am aware this tensor product must be over the complex numbers, but the quaternions $\mathbb{H}$ are a real algebra. It is impossible to centralize the complex numbers as a sub algebra of the quaternions. For this reason I am wondering how his tensor product over $\mathbb{C}$ is well defined.
Question: How is tensoring two spectral triples over the complex numbers well defined when a real algebra is chosen? 

There's just a convenient abuse of notation in play. When dealing with almost-commutative spectral triples, $C^\infty(M)$ means $C^\infty(M,\mathbb{C})$ except when forming $C^\infty(M) \otimes A_F$ for $A_F$ real, in which case $C^\infty(M) \otimes A_F$ really means $C^\infty(M,\mathbb{R}) \otimes_{\mathbb{R}} A_F$.
EDIT: This is the same answer I gave at MathOverflow.
