Rank-one PUE singular-plane calculation

Put $s=\sqrt{1-{\sigma }^2}$. Direct multiplication gives $U_{\sigma }^{\ast }{U}_{\sigma }=U_{\sigma }^2=\left(\begin{array}{cc}{\sigma }^2+s^2& \sigma s-\sigma s\\ \sigma s-\sigma s& s^2+{\sigma }^2\end{array}\right)=I$, so $U_{\sigma }$ is unitary. The projection onto $\mathrm{\mid }0⟩$ gives $A\mathrm{\mid }0⟩={\mathrm{\Pi }}_L{U}_{\sigma }\mathrm{\mid }0⟩=\sigma \mathrm{\mid }0⟩$, hence the PUE block is $[\sigma ]$. Its complementary matrix entry is $s$, so the ordered basis realizes $W_R(\sigma )$ on the full ambient space: for $0<\sigma <1$ this is the singular plane, at $\sigma =0$ it is the zero-kernel/complement block, and at $\sigma =1$ it splits into the singular line $\mathbb{C}\mathrm{\mid }0⟩$ and residual complement $\mathbb{C}\mathrm{\mid }1⟩$.

The zero phases make both endpoint phase matrices the identity, so the degree-one sequence is $U_{\sigma }$ itself. Its compressed block is ${\mathrm{\Pi }}_L{U}_{\sigma }{\mathrm{\Pi }}_R{\mathrm{\mid }}_{H_R}=[\sigma ]$. The scalar polynomial is the upper-left entry of $W_R(x)$, namely $P(x)=x$; therefore for $\sigma >0$ the odd transform is $P_{\mathrm{o}\mathrm{d}\mathrm{d}}^{\mathrm{S}\mathrm{V}}(A)=[P(\sigma )]=[\sigma ]$, and for $\sigma =0$ the positive singular sum is empty and both blocks are zero. The same formulas give $[1]$ at $\sigma =1$.