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Let $G$ be a (discrete) torsion free group with identity $e$. Recall that for an element $\alpha=\sum a_gg$ in $\mathbb C[G]$ (complex functions on $G$ with compact support), $\alpha^*$ is defined to be $\sum\overline{a_g}g^{-1}$ and for $\beta=\sum b_gg\in\mathbb C[G]$, we have the (convolution) product $\alpha\beta:=\sum a_gb_hgh$.

We call a selfadjoint elemet $\alpha\in\mathbb C[G]$ (i.e. $\alpha=\alpha^*$) golden if $a_e\in\mathbb R$ and $a_e-\sum_{g\neq e}|a_g|\geq0$. For $\beta\in\mathbb C[G]$, if $\beta^*\beta$ (or some of it's powers, $(\beta^*\beta)^n,(\beta^*\beta)^\frac 1n$, $n\in\mathbb N$) is golden?

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    $\begingroup$ It is obvious that if $\alpha$ is "golden" then so is $r\alpha$ for any $r\geq 0$. The triangle inequality for $\mathbb C$ shows that if $\alpha_1, \alpha_2$ are golden then so are $\alpha_1 + \alpha_2$. Thus the collection of golden elements is already a convex cone, and you don't need to take combinations. I have tried some simple examples: you can come up with cases when $\beta^*\beta$ is golden, and when it is not.. $\endgroup$
    – Matthew Daws
    Commented Jun 19, 2018 at 8:47

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Consider $G=\mathbb Z$ and $\chi = \delta_{-1} + \delta_1 \in \mathbb C[\mathbb Z]$. Then $\chi=\chi^*$. One can check that $$(\chi^{2n})_k= \binom{2n}{n+k}.$$ Hence, with your definition (and the remark by Matthew Daws) the element $\chi^{2n}$ is golden if $$\binom{2n}{n} \geq \sum_{k \neq 0} \binom{n+k}{2n} = 2^{2n} - \binom{2n}{n}.$$ However, by https://en.wikipedia.org/wiki/Central_binomial_coefficient $$ \binom{2n}{n} \leq \frac{1}{\sqrt{3n+1}}2^{2n},$$ so that $\chi^{2n}$ is golden if and only if $n=0$ or $n=1$. We conclude that $\chi^2$ has the property that none of its proper powers is golden.

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  • $\begingroup$ But $\chi^2=2+\delta_2+\delta_{-2}$ is golden so this is not the counterexample! $\endgroup$
    – MSMalekan
    Commented Jun 27, 2018 at 8:44
  • $\begingroup$ Doesn't Andreas say that $\chi^{2n}$ is golden if and only if $n=0$ or $1$. So $\chi^4$ is your counter-example... I think? $\endgroup$
    – Matthew Daws
    Commented Jun 27, 2018 at 9:41
  • $\begingroup$ Yes, exactly. No proper power of $\chi^2$ is golden. $\endgroup$
    – Andreas Thom
    Commented Jun 27, 2018 at 13:10
  • $\begingroup$ I edited the last sentence to clarify the meaning. $\endgroup$
    – Andreas Thom
    Commented Jun 27, 2018 at 14:53

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