Let $k\in\mathbb{N}$ be odd and $N\in\mathbb{N}$. You may assume that $N>k^2/4$ although I don't think that is relevant.

Let $\zeta:=\exp(2\pi i/k)$ and $\alpha_v:=\zeta^v+\zeta^{-v}+\zeta^{-1}$.

As it comes from the trace of a positive matrix I know that the following is real:

$$\sum_{v=1}^{\frac{k-1}{2}}\sec^2\left(\frac{2\pi v}{k}\right)\frac{(\overline{\alpha_v}^N+\alpha_v^N\zeta^{2N})(\zeta^{2v}-1)^2}{\zeta^N\zeta^{2v}}.$$

I am guessing, and numerical evidence suggests, that in fact


is real for (at least) each $v=1...(k-1)/2$. Therefore I am assuming that there is some nice simplification of it.

Can anyone simplify this expression?

Summing or even bounding the series would go above and beyond.


I need to calculate or rather bound traces to calculate a distance to random for the convolution powers of a $\nu\in M_p(\mathbb{G}_k)$ for $\mathbb{G}_k$ a series of quantum groups of order $2k^2$ ($k$ odd).


Following mercio's answer below I am now dealing with:

$$\frac{2}{4^{2N+1}}\sum_{v=1}^{\frac{k-1}{2}}\sec^2\left(\frac{2\pi v}{k}\right)\left(8+8|\alpha_v|^{2N}-8\sin^2\left(\frac{2\pi v}{k}\right)\Re\left((\alpha_v\zeta)^N\right)\right).$$

I can handle the first term (it is $2(k^2-1)/4^{2N}$) and am now looking at the other two terms.


Since $|\zeta|=1$, $\zeta^{-1} = \overline {\zeta}$, and so :

$\frac{(\overline{\alpha_v}^N+\alpha_v^N\zeta^{2N})}{\zeta^N} = (\overline{\alpha_v\zeta})^N + (\alpha_v\zeta)^N \in \Bbb R$

and $\frac{(\zeta^{2v}-1)^2}{\zeta^{2v}} = (\zeta^{2v}-1)^2\overline {\zeta^{2v}} = (\zeta^v - \overline {\zeta^v})^2 \in \Bbb R$

  • $\begingroup$ Excellent... thank you for that. $\endgroup$ – JP McCarthy Apr 3 '16 at 13:31

Just one more step towards a symmetrical representation.

Since \begin{align*} \sec^2\left(\frac{2\pi v}{k}\right)&=\frac{1}{\cos^2\left(\frac{2\pi v}{k}\right)} =\frac{4}{\left(\zeta^v+\overline{\zeta^v}\right)^2} \end{align*} we obtain using the representation from @mercio: \begin{align*} \sum_{v=1}^{\frac{k-1}{2}}&\sec^2\left(\frac{2\pi v}{k}\right)\frac{(\overline{\alpha_v}^N+\alpha_v^N\zeta^{2N})(\zeta^{2v}-1)^2}{\zeta^N\zeta^{2v}}\\ &=\sum_{v=1}^{\frac{k-1}{2}}\frac{4}{\left(\zeta^v+\overline{\zeta^v}\right)^2}\left((\overline{\alpha_v\zeta})^N + (\alpha_v\zeta)^N\right) (\zeta^v - \overline {\zeta^v})^2 \\ &=4\sum_{v=1}^{\frac{k-1}{2}}\left(\frac{ \zeta^v - \overline {\zeta^v}}{\zeta^v+\overline{\zeta^v}}\right)^2 \left((\overline{\alpha_v\zeta})^N + (\alpha_v\zeta)^N\right) \end{align*}

If this calculation is correct I have the feeling, that some further simplification might be possible.

  • $\begingroup$ Thank you... the $\sec^2$ term actually came from what you have at the end here. $\endgroup$ – JP McCarthy Apr 8 '16 at 9:31
  • $\begingroup$ @JpMcCarthy: You're welcome! $\endgroup$ – Markus Scheuer Apr 8 '16 at 10:16

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