1
$\begingroup$

Let $g$ be a group element, let $G$ be the group generated by $g$, and let $\mathbb C G$ be the group algebra on $G$.

If we define $\tau:G\to\mathbb C$ as \begin{align*}\tag{1} \tau[g^n]=\begin{cases} 1&\text{if }n=0;\text{ and}\\ \alpha_n&\text{otherwise}, \end{cases} \end{align*} where $(\alpha_n:n\in\mathbb N)\subset\mathbb C$ are completely arbitrary, then $\tau$ can be extended to a normalized (i.e., $\tau[1]=1$) linear functional on $\mathbb C G$.

If, however, we would like $\tau$ to be positive (i.e., $\tau[pp^*]\geq0$ for all $p\in\mathbb C G$), there are certain restrictions on how we can choose the $\alpha_n$ in $(1)$ (for instance, we have to make sure that $\tau[g^n]=\overline{\tau[g^{-n}]}$ for all $n$).

In this context, the question I'm interested in is the following:

Question. How many of the $\alpha_n$ can be different from zero (and under what conditions)?

If $g$ has finite order $m\in\mathbb N$, then we can choose as many of $\alpha_1,\ldots,\alpha_{\lfloor m/2\rfloor}$ nonzero as we want, provided $|\alpha_1|,\ldots,|\alpha_{\lfloor m/2\rfloor}|$ are bounded above by a small enough quantity (this is mainly because, in this case, any element of $\mathbb C G$ can be written as $\beta_0+\beta_1g+\beta_2 g^2+\cdots+\beta_{m-1}g^{m-1}$).

However, if $g$ is of infinite order, I can't find an example of a positive $\tau$ with at least one $\alpha_n$ nonzero, but I can't show that all of the $\alpha_n$ need to be zero either. Is there a known characterization of positive linear functionals on such algebras?

$\endgroup$

1 Answer 1

2
$\begingroup$

You are asking for the condition that a function $\tau$ on a cyclic group $G$ be (normalized by $\tau(e)=1$ and) positive-definite, i.e. $$ \tau(p^*p) = \sum_{i,j} \overline{p_i}p_j\,\tau(g^{-i}g^j)\geqslant 0 $$ for all $p = \sum_ip_ig^i\in\mathbf CG$ (finite sum). The answer is given by Bochner's theorem (due in this case to Herglotz, I believe) which says that such $\tau$'s are exactly the Fourier transforms of probability measures on the dual group $\smash{\hat G}$.

  • If $G$ has infinite order, we have $G \simeq \mathbf Z$ and $\hat G$ is the unit circle group $\mathbf T=\mathbf U(1)$, so there must be an (otherwise arbitrary) probability measure $\mu$ on $\mathbf T$ such that $$ \alpha_n=\tau(g^n) =\int_{\mathbf T}z^n\,d\mu(z). $$

  • If $G$ has order $m$, we have $G\simeq\mathbf Z_m$ and $\hat G$ is the subgroup $\simeq\mathbf Z_m\subset \mathbf T$ consisting of all $m$-th roots of 1, i.e. all $m$ powers of $\omega = e^{2\pi i/m}$; and there must be a probability measure $\mu$ on it, or in other words arbitrary nonnegative weights $\mu(\{\omega^k\})_{k=0}^{m-1}$ adding up to 1, such that $$ \alpha_n = \tau(g^n) =\sum_{k=0}^{m-1}\omega^{kn}\,\mu(\{\omega^k\}). $$

With that, it's easy to cook up examples where all $\alpha_n\ne0$. E.g. you can, in both cases, make all $\alpha_n=1$ by taking $\mu =$ Dirac measure at 1.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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