-\alpha )}\mathcal{S}$, where $0\leq \alpha \leq \pi /M$, have the same distance profiles. We have $$\begin{aligned}
\nonumber
\mathcal{S}+\gamma e^{j (\frac{\pi}{M} -\alpha )}\mathcal{S}&=\mathcal{S}+\gamma e^{j (\frac{\pi}{M} -\alpha )}(\mathcal{S}e^ {\frac{-2\pi}{M}})\\
\label{Lemma2_1}
&=\mathcal{S}+\gamma e^{-j (\frac{\pi}{M} +\alpha )}\mathcal{S}.\end{aligned}$$ The first equality is because $\mathcal{S}=\mathcal{S}e^{j k 2\pi /M},$ *i.e.* rotating $\mathcal{S}$ by $2k\pi /M$ gives the same constellation. Thus for $k=-1$, $\mathcal{S} = \mathcal{S}e^ {\frac{-2\pi}{M}}$. Also due to the symmetric nature of $M$-PSK constellation, the distance distribution of the sum constellation depends only on the relative angle of rotation between the input constellations. Thus $\mathcal{S}+\gamma e^ \beta \mathcal{S}$ and $\mathcal{S}+\gamma e^ {-\beta} \mathcal{S}$ have same distance profiles, for any $\beta \in [0,\pi]$. This together with proves the second part of the lemma.

From Lemma \[lemmaangle\], it is clear that when both users use $M$-PSK signal sets, if $ (\gamma',\theta') $ is a singular fade state, then there exists singular fade states at $ (\gamma',\theta' + p \frac{2\pi}{M}) $, where $1\leq p \leq M-1$ because distance distribution in $\mathcal{S}_{\text{eff}}$ is periodic with period $\frac{2\pi}{M}$. The distance distribution of $\mathcal{S}_{\text{eff}}$ is the basis for channel quantization. Also from Lemma \[lemmaangle\], it suffices to obtain such a quantization only for the wedge $ [0, \pi /M]$. This can then 