I was wondering if someone knows of a reference in which $\mathbb{Z}_2$-graded $C^*$-algebra theory is developed using the sign convention $(ab)^* = (-1)^{|a||b|}b^* a^*$. I would be most enthusiastic if someone already performed the exercise of converting powerful machinery like Hilbert modules and Kasparov KK-theory to this language. If that is the case, I would be interested to know whether it simplifies the theory in any way or just makes it even harder.
Let me give some context: a $\mathbb{Z}_2$-graded $C^*$-algebra is usually defined as an ordinary $C^*$-algebra together with a direct sum decomposition $A = A_0 \oplus A_1$ respected by multiplication and $*$. In particular, it satisfies $(ab)^* = b^* a^*$. This expression violates the Koszul sign rule, which instead would require such an antilinear involution $\dagger$ to satisfy $(ab)^\dagger = (-1)^{|a||b|} b^\dagger a^\dagger$. On the algebraic level, there is a correspondence between anti-linear even involutions $*$ on $\mathbb{Z}_2$-graded algebras such that $(ab)^* = b^* a^*$ and ones that satisfy $(ab)^\dagger = (-1)^{|a||b|} b^\dagger a^\dagger$. One such bijection is given by $$ a^\dagger = \begin{cases} a^* & a \text{ even} \\ ia^* & a \text{ odd}. \end{cases} $$ Transferring the $C^*$-property $\|a^* a\| = \|a\|^2$ along this bijection yields an equivalent definition of a $C^*$-algebra which is much harder to work with, but has the 'correct' Koszul sign in the definition. For example, if $a$ is odd, $\operatorname{Spec}a^\dagger a \subseteq i \mathbb{R}_{\geq 0}$. I should also remark that over the real numbers, the two sign conventions no longer agree and many real superalgebras don't admit any of these '$\dagger$-structures' at all.
So why bother? Often in mathematics, it turns out that using Koszul signs from the start yields the correct notions later on. For example, if $A,B$ are $\mathbb{Z}_2$-graded $C^*$-algebras, then the $C^*$-structure on the graded tensor product algebra $A \otimes B$ is defined as $(a \otimes b)^* = (-1)^{|a||b|} a^* \otimes b^*$. Why does the weird sign appear? Using the other sign convention, the correct definition is $(a \otimes b)^\dagger = a^\dagger \otimes b^\dagger$. Translating this definition back under the bijection above yields the definition with the strange sign for $*$.
Main question: Has this perspective on graded $C^*$-algebras been developed?
Any thoughts on this are very welcome.
