Let $A$ be a unital $C^*$ algebra which is equipped with a faithfull trace $T$. In particular we may consider $A=C^*_{\text{red}} (G)$ for some discrete group $G$. We consider the following differential equation on $A$: $$(*)\;\;\;Z'=Z^2-Z$$ (For $A=M_n(\mathbb{C})$ one can easilly check that $$(**)\;\;\;D'=D(T-n)$$ where $D,T$ are standard determinant and trace respectively and $D'$ is the derivative of $D$ along solutions of $(*)$.Note that $n$ in $(**)$ can be regarded as $\text{trace}(I_n)$. We will modify this $n$ to $1$ in the case of normalized trace. In fact "Determinant" is the unique analytic function on $M_n(\mathbb{C})$ satisfying the equation $(**)$ with initial condition $D(I_n)=1$). We try to generalize this situation of matrix algebra to a $C^*$ algebra $A$ with a faithful normalized trace $T$. So we consider the following modified differential equation:$$(***)\;\;\; D'=D(T-1)$$ where the unknow $D$ is a function on $A$ and $T$ is a normalized trace. Moreover $D'$ is the derivative of $D$ along solution of $(*)$. What can be said about existence of a global solution $D$ for $(***)$ with initial condition $D(1)=1$? Does such a solution $D$ satisfy multiplicative condition $D(ab)=D(a)D(b)$? Is $D^{-1}(0)$ equal to the set of all non invertible elements? As a motivation for the later question, we note that the group of invertible elements is flow-invariant under system $(*)$ If there are no some complete answers to the above questions for an arbitrary algebra with a faithful normalized trace, what would be the answer of those questions in the particular case $A=C^*_{\text{red}} (G)$? For which kind of groups the answer to the above questions are known? Remark: We conclude that "Determinant" as a function on matrix algebra can be dynamically and uniquely extracted from "Trace", at least in a neighborhood of the identity matrix since the identity matrix, as a singularity of $(*)$, attracts all nearby orbits,as time goes to $-\infty$. This is somewhat compatible with the classical fact that "Determinant" of a matrix $B$,as an invariant polynomial, can be generated by "Trace" of powers of $B$, that is $\text{trace}(B^k),\;k\in \mathbb{N}$. But this dynamical interpretation we provided, needs merely trace of power $1$ not higher powers. More precisely, if we denote by $\phi$ the flow of $(*)$, then for $B$ sufficiently close to identity matrix we have $$Det(B)=exp(\int_{-\infty}^0 (n-\text{trace})(\phi_t(B))dt$$ So knowing "Trace" leads us to knowing "Detrminant". Login To add answer/comment