Unitarity of the unitary principal series

For every $\nu \in i\mathbb{R}$, [F1] gives the strongly continuous unitary compact-picture action on $L_{\epsilon }^2(K)$; the inversion and ${\rho }^{-1\mathrm{/}2}$ map in that supplier is the model equivalence, so no left/right covariance convention is silently identified. By [F2], the K-types are the one-dimensional mutually orthogonal lines $\mathbb{C}{f}_n$, $n\equiv \epsilon (\mathrm{m}\mathrm{o}\mathrm{d}2)$, and their finite span is dense. By [F7] this is a strongly continuous unitary representation with K-type multiplicity one.

At $\epsilon =0$, ${\mathcal{W}}_0$ consists of odd integers, so $0\mathrm{\notin }{\mathcal{W}}_0$. Part (a) of [F3] therefore says that the Hilbert compact-picture representation $I_{0,0}$ is irreducible.

At $\epsilon =1$ and $\nu =0$, set $z=-e^{2i\theta }$. The odd positive Fourier polynomials have the form $f(k_{\theta })=e^{i\theta }F(z)$ with $F$ a polynomial; their closure $H_{+}$ is the closed span of $f_1,f_3,f_5,\dots$. The negative odd Fourier polynomials have the form $\overline{f(k_{\theta })}$ for such positive polynomials; let their closure be $H_{-}$. By [F2], $H_{+}$ and $H_{-}$ are orthogonal and $L_1^2(K)=H_{+}\oplus H_{-}$.

Put $\mathcal{C}=2^{-1\mathrm{/}2}\left(\begin{array}{cc}1& i\\ 1& -i\end{array}\right)$ and let $v_{\theta }=(-\sin \theta ,\cos \theta {)}^T$. Direct multiplication gives $\mathcal{C}{v}_{\theta }=(i{e}^{i\theta },-i{e}^{-i\theta }{)}^T\mathrm{/}\sqrt{2}$, so the ratio of its coordinates is $z=-e^{2i\theta }$. For $g\in G$, direct multiplication using $\det g=1$ gives $\mathcal{C}{g}^T{\mathcal{C}}^{-1}=\left(\begin{array}{cc}a& b\\ \bar{b}& \bar{a}\end{array}\right)$ with $\mathrm{\mid }a{\mathrm{\mid }}^2-\mathrm{\mid }b{\mathrm{\mid }}^2=1$. The row action $v_{\theta }^{T}g$ therefore sends $z$ to ${\mathrm{\Phi }}_g(z)=(az+b)\mathrm{/}(\bar{b}z+\bar{a})$; its derivative is ${\mathrm{\Phi }}_g^{\mathrm{\prime }}(z)=(\bar{b}z+\bar{a}{)}^{-2}$. Since $\mathrm{\mid }a{\mathrm{\mid }}^2-\mathrm{\mid }b{\mathrm{\mid }}^2=1$ and both moduli are nonnegative, $\mathrm{\mid }a\mathrm{\mid }>\mathrm{\mid }b\mathrm{\mid }$, so the denominator has no zero on the closed disk.

On the boundary, $z=-e^{2i\theta }$ and ${\mathrm{\Phi }}_g(z)=-e^{2i{\kappa }_g(\theta )}$. Differentiating in $\theta$ gives $\mathrm{\mid }{\kappa }_g^{\mathrm{\prime }}(\theta )\mathrm{\mid }=\mathrm{\mid }{\mathrm{\Phi }}_g^{\mathrm{\prime }}(z)\mathrm{\mid }=\mathrm{\mid }\bar{b}z+\bar{a}{\mathrm{\mid }}^{-2}$. To determine the sign, write the bottom row of $k_{\theta }g$ as the column $u(\theta )=g^T{v}_{\theta }$, where $v_{\theta }=(-\sin \theta ,\cos \theta {)}^T$. For $k_{\theta }g=a_{t(\theta )}{n}_{x(\theta )}{k}_{{\kappa }_g(\theta )}$ one has $u=e^{-t\mathrm{/}2}{v}_{{\kappa }_g(\theta )}$. Since $\det g=1$ and $\det (v_{\theta },v_{\theta }^{\mathrm{\prime }})=1$, we have $\det (u,u^{\mathrm{\prime }})=1$; the factorized expression gives $\det (u,u^{\mathrm{\prime }})=e^{-t}{\kappa }_g^{\mathrm{\prime }}$. Therefore ${\kappa }_g^{\mathrm{\prime }}=e^t>0$, and the boundary derivative identity gives ${\kappa }_g^{\mathrm{\prime }}=\mathrm{\mid }{\mathrm{\Phi }}_g^{\mathrm{\prime }}(z)\mathrm{\mid }=\mathrm{\mid }\bar{b}z+\bar{a}{\mathrm{\mid }}^{-2}=\mathrm{\mid }\alpha (p(k_{\theta },g)){\mathrm{\mid }}^2$. On $\mathrm{\mid }z\mathrm{\mid }=1$, $az+b=z\overline{(\stackrel{‾}{b}z+\stackrel{‾}{a})}$, so $e^{2i({\kappa }_g(\theta )-\theta )}=\bar{d}\mathrm{/}d$ for $d=\bar{b}z+\bar{a}$ and $e^{i{\kappa }_g(\theta )}=\pm e^{i\theta }\bar{d}\mathrm{/}\mathrm{\mid }d\mathrm{\mid }$ with a constant sign on the circle. Using $\mathrm{\mid }\alpha (p(k_{\theta },g))\mathrm{\mid }=e^{t\mathrm{/}2}=\mathrm{\mid }d{\mathrm{\mid }}^{-1}$, if $f(k_{\theta })=e^{i\theta }F(z)$ then [F4] gives $({\mathrm{\Pi }}_0(g)f)(k_{\theta })=\pm e^{i\theta }(\bar{b}z+\bar{a}{)}^{-1}F({\mathrm{\Phi }}_g(z))$.

For polynomial $F$, each term of $(\bar{b}z+\bar{a}{)}^{-1}F({\mathrm{\Phi }}_g(z))$ is a polynomial divided by a positive integer power of $\bar{a}+\bar{b}z$. Since $\mathrm{\mid }\bar{b}\mathrm{/}\bar{a}\mathrm{\mid }<1$, [F5] expands each reciprocal power uniformly on $\mathrm{\mid }z\mathrm{\mid }\le 1$ as a series with only nonnegative powers of $z$. Thus ${\mathrm{\Pi }}_0(g)$ maps positive odd Fourier polynomials into $H_{+}$. The action is unitary by [F1], so approximation by these polynomials and closedness give ${\mathrm{\Pi }}_0(g)H_{+}\subseteq H_{+}$; applying the same argument to $g^{-1}$ gives equality. At $\nu =0$ the cocycle and the odd inducing sign are real, so complex conjugation commutes with ${\mathrm{\Pi }}_0(g)$; consequently $H_{-}$ is also invariant.

By [F3], the K-finite parts of $H_{+}$ and $H_{-}$ are exactly $M_1^{-}$ and $M_{-1}^{+}$. Each is algebraically irreducible by the chain argument in [F3]. If a closed invariant subspace of either summand is nonzero, [F6] puts a nonzero K-finite vector in it; irreducibility then gives the whole corresponding chain, which is dense in that summand. Hence both invariant Hilbert summands are irreducible unitary limits of discrete series, and their orthogonal sum is $I_{1,0}$.