7
$\begingroup$

Let $A$ be a unital $C^*$-algebra which is equipped with a faithful trace $T$. In particular we may consider $A=C^*_{\text{red}} (G)$ for some discrete group $G$. We consider the following differential equation on $A$:

$$Z'=Z^2-Z.\tag{*}\label{star1}$$

(For $A=M_n(\mathbb{C})$ one can easily check that $$D'=D(T-n)\tag{**}\label{star2}$$ where $D$, $T$ are the standard determinant and trace respectively and $D'$ is the derivative of $D$ along solutions of \eqref{star1}. Note that $n$ in \eqref{star2} can be regarded as $\operatorname{trace}(I_n)$. 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 \eqref{star2} 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:$$D'=D(T-1)\tag{***}\label{star3}$$ where the unknown $D$ is a function on $A$ and $T$ is a normalized trace. Moreover $D'$ is the derivative of $D$ along solution of \eqref{star1}.

What can be said about existence of a global solution $D$ for \eqref{star3} with initial condition $D(1)=1$? Does such a solution $D$ satisfy the multiplicativity 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 \eqref{star1} (see On differential equation $Z'=Z^2-Z$ on a $C^*$ algebra).

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 \eqref{star1}, 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 $\operatorname{trace}(B^k),\;k\in \mathbb{N}$. 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 $$\operatorname{Det}(B)=\exp\left(\int_{-\infty}^0 (n-\operatorname{trace})(\phi_t(B))dt\right).$$ So knowing "trace" leads us to knowing "determinant".

$\endgroup$
7
  • 4
    $\begingroup$ You should have a look at the Fuglede-Kadison determinant. $\endgroup$
    – Epsilon
    Dec 4, 2019 at 16:11
  • 1
    $\begingroup$ 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. $\endgroup$
    – Epsilon
    Dec 6, 2019 at 19:40
  • 1
    $\begingroup$ 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? $\endgroup$
    – LSpice
    Dec 12, 2019 at 21:24
  • 1
    $\begingroup$ Your final equation $\operatorname{Det}(B) = \exp(\int_{-\infty}^0 (n - \operatorname{trace})(\phi_t(B))dt$ was missing a parenthesis. I guessed where it should be, I hope correctly. $\endgroup$
    – LSpice
    Apr 21 at 15:13
  • 1
    $\begingroup$ @LSpice Thank you very much for your edit. $\endgroup$ Apr 24 at 12:02

1 Answer 1

5
$\begingroup$

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 $$D(e^h)=e^{T(h)}.\tag{$\dagger$}\label{dagger}$$ I have tried to prove that the OP's condition $\eqref{star3}$ implies \eqref{dagger} but, although I still think this is true, I cannot come up with a proof (any ideas?).

Regarding the existence of solutions for \eqref{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 Rotation numbers for automorphisms of $C^*$ algebras 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.

$\endgroup$
1
  • $\begingroup$ Dear Prof. Exel Thank you very much for your very interesting answer. $\endgroup$ Dec 13, 2019 at 15:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.