Hi,


I have a misunderstanding that I am hoping is really quite trivial. I will give my question directly and context below for those that need/want it.


Question: In connes standard model he takes the finite algebra input $\mathcal{A}\_F = \mathbb{C}\oplus\mathbb{H}\oplus M\_3(\mathbb{C})$. I am a little confused how connes takes his spectral tensor products over $\mathbb{C}$ to form an almost commutative triple. Connes represents the quaternions $\mathbb{H}$ as $2\times2$ complex matrices, but the quaternions 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 ('prior' to representation).


Context:


In Connes almost commutative models he forms his almost commutative algebra as $\mathcal{A}\_{ac}= C^\infty(M)\otimes\_C \mathcal{A}\_F$,


and what I am going to call his almost commutative 'cyclic coboundary operator' is given as $\delta\_{ac} = \delta\otimes\_C\mathbb{I} + c\otimes\_C \delta\_F$, where c is a Hochschild cycle. These are both represented faithfully on some hilbert space $\mathcal{H}\_{ac} = \mathcal{H}\otimes\mathcal{H}\_F$.


The representation of the 'cyclic coboundary operator' is then given by the unfluctuated almost commutative dirac operator $D\_{ac} = D\otimes\_C\mathbb{I} + \gamma\_5\otimes\_C = i\gamma^\mu \otimes\_C\partial\_\mu\mathbb{I} + \gamma\_5\otimes\_C D\_F$ (notice that here we really require the tensor product to be over C and not R).


The fluctuated dirac operator is given by $\tilde{D}\_{ac} = D\_{ac} + A + JAJ^\*$ 


Where A is a representation of a one form given by $A = \sum\_i \pi\_{\alpha\_i} [D\_{ac},\pi\_{\beta\_i}] $. A general 'one form' is given by given by a sumation of elements $\alpha\delta\_{ac}\beta = (a\otimes X)(\delta\_{ac} = \delta\otimes\_C\mathbb{I} + c\otimes\_C \delta\_F)(b\otimes Y)$, with $\alpha,\beta\in\mathcal{A}\_{ac}$. 


It seems that forming well defined fluctuated operators in an for an almost commutative geometry requires tensor products to be over C but I am confused how this is done when a real finite algebra is chosen.



Well defined Tensoring of spectral triples 
There's a reasonably 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 $C^\infty(M,\mathbb{R}) \otimes\_{\mathbb{R}} A\_F$.


