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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?

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But, $\Bbb H$ is also an algebra over $\Bbb C$, at least a right $\Bbb C$-module, using a fix embedding $\Bbb C\hookrightarrow\Bbb H$. Why should they be centralized? – Berci Oct 20 '12 at 20:05
Say I tensor $\mathbb{H}\otimes_C\mathbb{H}\otimes_C\mathbb{H}$ This means I should be able to write $a(A\otimes B\otimes C) = (aA\otimes B\otimes C) = (A\otimes B\otimes aC)$ for all $a \in \mathbb{C}$ and for all $A,B,C \in \mathbb{H}$. But this implies that $a$ commutes with $A,B,C$ does it not? In which case I require my embeding of $\mathbb{C}$ in $\mathbb{H}$ to be central. But this is not the case as $\mathbb{H}$ can not be writen as an algebra over $\mathbb{C}$. – SMF Oct 25 '12 at 14:57

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.

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