Given $\delta\in\mathbb C$, let $A(\delta)$ denote the complex unital $*$-algebra generated by an identity $1$ and selfadjoint elements $e_k$, $k\in\mathbb N$, satisfying $e_k^2=\delta e_k$, $e_ke_l=e_le_k$ for $|k-l|\geq2$ and $e_ke_{k\pm1}e_k=e_k$ (This puts well-known constraints on the possible values of $\delta$). I wonder if it is known what all the extremal traces (factor traces) of $A(\delta)$ are, i.e. the positive normalized functionals $\tau:A(\delta)\to\mathbb C$ such that $\tau(ab)=\tau(ba)$ for all $a,b\in A(\delta)$ and having the property that $\tau$ cannot be written as a non-trivial convex combination of other traces.

Markov traces are known to be factor traces. But are there additional factor traces on $A(\delta)$, and can they be specified explicitly? Any reference to the literature would be welcome.


For $\delta < 2$, I think you can read this off from Jones's "index for subfactors" paper that the "finite depth"-ness forces that the Markov trace is the only trace for each allowed $\delta$.

For $\delta \ge 2$, again from Jones's paper the C$^*$-envelope of $A(\delta)$ is independent of $\delta$, and is in fact a quotient of $C^*(S_\infty)$ (which is easy to see for $\delta=2$ from the Schur-Weyl duality). The factor traces on the latter are classified by the Thoma parameter $0 \le \alpha_n$, $0 \le \beta_n$, $\sum_n \alpha_n + \beta_n \le 1$ ($1 \le n < \infty$), and those induced from $A(\delta)$ correspond to $\alpha_1 + \alpha_2 = 1$ (Wassermann's PhD thesis is the original reference I think?). These correspond to the Markov traces again.

So at the end of the day, the Markov traces are the only factor traces if I'm not mistaken.

  • $\begingroup$ For $\delta<2$, your statement is what I am hoping for, I'll check Jones' paper for that - many thanks for the reference. But for $\delta\geq2$, and in particular for $\delta=2$, I think not all the extremal traces of $S_\infty$ that factor through the TL algebra (i.e., those with $\alpha_1+\alpha_2=1$) are Markov. A Markov trace $\tau$ must satisfy $\tau(e_1e_2)=\tau(e_1)^2$, but on the other hand, $\tau(e_1e_2)=\delta^{-1}\tau(e_1^2e_2)=\delta^{-1}\tau(e_1)$, and therefore either $\tau(e_1)=0$ or $\tau(e_1)=\delta^{-1}$. $\endgroup$ – Gandalf Lechner Mar 17 '18 at 15:04
  • 1
    $\begingroup$ The trace corresponding to $\alpha_1 + \alpha_2 = 1$ is given by the restriction of ${\rm Tr}_{M_2(\mathbb{C})}(\cdot \operatorname{diag}(\alpha_1, \alpha_2))^{\otimes \infty}$ to $(M_2(\mathbb{C})^{\otimes \infty})^{{\rm SU}(2)} \simeq A(2)$. This does give you the Markov trace for $\delta = (\alpha_1 \alpha_2)^{-1/2}$. To see this, first you can describe the induced trace in terms of the Bratteli diagram (which Wassermann does), then check that it agrees with the Markov trace (like Jones mentions in p. 19). $\endgroup$ – Makoto Yamashita Mar 18 '18 at 1:33
  • $\begingroup$ I completely agree that the trace with $\alpha_1+\alpha_2=1$ is Markov on the algebra with $\delta=(\alpha_1\alpha_2)^{-1/2}$. But for fixed $\delta$, say $\delta=2$, there is a one-parameter family of factor traces (given by Thoma parameters with $\alpha_1+\alpha_2=1$) most of which are not Markov because $\delta\neq(\alpha_1\alpha_2)^{-1/2}$. $\endgroup$ – Gandalf Lechner Mar 18 '18 at 20:19
  • $\begingroup$ Sure, my point being that those other factor traces just come from the Markov traces up to the identification of C*-envelopes of $A(\delta)$ for different $\delta \ge 2$. (Even purely algebraically I think $A(2) \simeq A(\delta)$, but the concrete formula/argument would be cumbersome.) $\endgroup$ – Makoto Yamashita Mar 19 '18 at 1:02

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.