Well defined Tensoring of spectral triples

2012-10-19T19:35:42Z

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.

Answer by Branimir Ćaćić for Well defined Tensoring of spectral triples

2012-11-16T02:26:44Z

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 $C^\infty(M,\mathbb{R}) \otimes_{\mathbb{R}} A_F$.

Well defined Tensoring of spectral triples - MathOverflow

Well defined Tensoring of spectral triples

Answer by Branimir Ćaćić for Well defined Tensoring of spectral triples