# How can one define a kind of “Determinant” on reduced group $C^*$ algebra?

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".

• You should have a look at the Fuglede-Kadison determinant. – Epsilon Dec 4 '19 at 16:11
• @Epsilon Thank you for introducing me this kind of determinant. – Ali Taghavi Dec 6 '19 at 19:09
• You can also have a look at arxiv.org/pdf/1608.07395.pdf Section 5. In this paper we derive properties of the determinant from algebraic K theory. – Epsilon Dec 6 '19 at 19:40
• I think that maybe I don't understand the normalisation. With $\tau$ the normalised trace (so $\tau(I_n) = 1$), then $D' = n D(\tau - 1)$, not $D' = D(\tau - 1)$; the solutions $D_1$ to the latter equation seem to be related to the solutions $D_0$ of the former by $D_1(A) = D_0(A/n)$ (right?), so the $n$ is just moved around, not disposed of. Where does the $n$ go in the $\mathrm C^*$-algebra picture? – LSpice Dec 12 '19 at 21:24
• @LSpice yes you are right. Maybe the terminology I used was not appropriate. I just noticed that "T-n" vanish at identity so I thought that an appropriate modification is T-1 which vanishes at identity. So I am curious whether this modified equation has some meaningfull interpretation? For example is its solution compatible to what Prof Exel sugfests $D(e^A)=e^{tr(}A)$?(in reduced group C^* algebra or some other infinite domensional cases with a trace with trace(1)=1 – Ali Taghavi Dec 13 '19 at 15:13

This is a long comment, rather than an answer, so it perhaps fits here.

Another way in which the trace determines the determinant, at least locally near the identity, is via the equation $$D(e^h)=e^{T(h)}.\qquad (\dagger)$$ I have tried to prove that the OP's condition $$(***)$$ implies $$(\dagger)$$ but, although I still think this is true, I cannot come up with a proof (any ideas?)

Regarding the existence of solutions for $$(\dagger)$$, there is a clear obstruction: if $$p$$ is an idempotent element, then $$e^{2\pi i p}=1$$, so one would need $$e^{2\pi i T(p)}=1$$, which is equivalent to saying that $$T(p)$$ is an integer.

Conversely, it is a consequence of Theorem (II.10) in my PhD thesis that the existence of projections with non-integer trace is the only possible obstruction to the existence of a determinant.

Regarding (reduced) group C*-algebras the question becomes very interesting. If the group $$G$$ has torsion, then the spectral projections of any nontrivial torsion element will have non-integer trace (for the standard normalized trace, of course), so no determinant function exists.

On the other hand, if $$G$$ has no torsion one would hope to prove that all projections have integer trace, but doing so would solve the Kadison–Kaplansky conjecture which has been proved for many groups, including all hyperbolic groups, but I think the most general case is still open.

• Dear Prof. Exel Thank you very much for your very interesting answer. – Ali Taghavi Dec 13 '19 at 15:06