# Différences between KKO and KKR in Kasparov theory

In Kasparov article : The operator K functor and extensions of $C^*$algebras there is the definition of the two bifunctors $KKO : ralg^{op} \times ralg \to Ab$ and $KKR : Ralg^{op}_r \times Ralg_r \to Ab$. Where $ralg$ is the category of $C^*$ algebras over the real numbers, and $Ralg_r$ the category of complex $C^*$ algebras together with an antilinear involution $\bar \cdot$ (with $\bar{ab} = \bar a \cdot \bar b$.) The two functors take values in the category of abelian groups $Ab$

There is an equivalence of category $Ralg \to ralg_r$ wich take an algebra to its complexification with involution given by conjugation on complexe numbers.

The two functors of Kasparaov commutes with this identification.

Now, I don't understand why in the chapter 5 of his article Kasparov defines Bott elements $\alpha_{p,q}$ in $KR_{p-q}(\mathbb R^{p,q}) = KKR(C_0(\mathbb R^{p,q})\otimes Cl_{p,q}, \mathbb C)$ and only $\alpha_n$ in $KO_{n}(\mathbb R^{n}) = KKO(C_0(\mathbb R^{n})\otimes Cl_{n,0}, \mathbb R)$

Is there a legitimate reason to do such a restriction ?

• This is just because when working with complex clifford algebra $Cl_{p,q} \simeq Cl_{p+q,0}$ so there is no point in keeping the double indices while in the real case it is not the case. – Simon Henry Nov 30 '17 at 16:00

I think the difference is because one has real Clifford algebras $Cl_{p,q}(\mathbb{R})$ (indexed by two variables $p,q$), but only complex Clifford algebras $Cl_n(\mathbb{C})$.
My main concern is what you said first about the equivalence between real $C^*$-algebras and complex $C^*$-algebras:
If you have a complex $C^*$-algebra $A$, then it is not simply the complexification of the self-adjoint part $A_h$, because the product $xy$ of self-adjoint elements $x,y \in A_h$ is usually not self-adjoint, that is $xy \notin A_h$.
• I see the confusion. I was speaking of the category of real algebras (in the usual sense, a $C^*$ algebra over $\mathbb R$ and Real algebras : complex $C^*$ algebras with an antilinear involution. This involution is not the $*$. in a Real algebras you have the conjugation $\bar \cdot$ and the star $*$ – InfiniteLooper Dec 1 '17 at 9:50