Holomorphic and antiholomorphic discrete-series models

If $cz+d=0$ for $z\in \mathfrak{H}$, then $c=0$ would force $d=0$, contrary to $ad-bc=1$, while $c\mathrm{\ne }0$ would make $z=-d\mathrm{/}c$ real. Thus $j(g,z)\mathrm{\ne }0$. Expanding numerator times the conjugate denominator gives $\mathrm{Im}((az+b)\mathrm{/}(cz+d))=\mathrm{Im}z\mathrm{/}\mathrm{\mid }cz+d{\mathrm{\mid }}^2>0$. The inverse fractional map is that of $g^{-1}$, and multiplication of matrices proves both the action law and $j(g_1{g}_2,z)=j(g_1,g_2\cdot z)j(g_2,z)$. The maps are holomorphic by [F1]. These identities prove the group law for ${\pi }_n$ and ${\pi }_n^{-}$ as stated, and their identity elements act as the identity.

For $z=x+iy\in \mathfrak{H}$, $\mathrm{\mid }z+i{\mathrm{\mid }}^2-\mathrm{\mid }z-i{\mathrm{\mid }}^2=4y>0$, so $w=(z-i)\mathrm{/}(z+i)$ lies in $\mathbb{D}$. Solving gives the holomorphic inverse $z=i(1+w)\mathrm{/}(1-w)$, and direct substitution gives $\mathrm{Im}z=(1-\mathrm{\mid }w{\mathrm{\mid }}^2)\mathrm{/}\mathrm{\mid }1-w{\mathrm{\mid }}^2>0$; hence the Cayley map is a biholomorphism. The identities $z+i=2i\mathrm{/}(1-w)$ and $\mathrm{\mid }dz\mathrm{/}dw{\mathrm{\mid }}^2=4\mathrm{/}\mathrm{\mid }1-w{\mathrm{\mid }}^4$ now give $\mathrm{\parallel }{f}_{n,j}{\mathrm{\parallel }}_n^2=2^{2-2n}{\int }_{\mathbb{D}}\mathrm{\mid }w{\mathrm{\mid }}^{2j}(1-\mathrm{\mid }w{\mathrm{\mid }}^2{)}^{n-2}dudv$. In polar coordinates, justified by [F2], this is $2^{2-2n}2\pi {\int }_0^1{\rho }^{2j+1}(1-{\rho }^2{)}^{n-2}d\rho \le 2^{2-2n}\pi <\mathrm{\infty }$; the integral is positive because its integrand is positive on a nonempty open set. Thus every $f_{n,j}$ is a nonzero member of ${\mathcal{H}}_n^{+}$, and its complex conjugate belongs to ${\mathcal{H}}_n^{-}$.

Fix $g\in G$ and set $w=g^{-1}\cdot z$, so $z=g\cdot w$. The cocycle from step 1.1 gives $j(g^{-1},g\cdot w)=j(g,w{)}^{-1}$. Under $z=g\cdot w$, step 1.1 and the complex derivative $d(g\cdot w)\mathrm{/}dw=j(g,w{)}^{-2}$ give $d{x}_{z}d{y}_z=\mathrm{\mid }j(g,w){\mathrm{\mid }}^{-4}d{x}_{w}d{y}_w$. Thus the integrand $\mathrm{\mid }{\pi }_n(g)f(z){\mathrm{\mid }}^2(\mathrm{Im}z{)}^{n-2}d{x}_{z}d{y}_z$ becomes $\mathrm{\mid }f(w){\mathrm{\mid }}^2(\mathrm{Im}w{)}^{n-2}\mathrm{\mid }j(g,w){\mathrm{\mid }}^{2n-2(n-2)-4}d{x}_{w}d{y}_w=\mathrm{\mid }f(w){\mathrm{\mid }}^2(\mathrm{Im}w{)}^{n-2}d{x}_{w}d{y}_w$. By [F2] this change of variables holds for the nonnegative measurable integrand, even if the integral is infinite; hence $\mathrm{\parallel }{\pi }_n(g)f{\mathrm{\parallel }}_n=\mathrm{\parallel }f{\mathrm{\parallel }}_n$. The group law makes ${\pi }_n(g^{-1})$ its inverse. Complex conjugation gives the same norm-preserving action for ${\pi }_n^{-}$.

For $k_{\theta }$ as in [F3], $k_{-\theta }\cdot z-i=e^{-i\theta }(z-i)\mathrm{/}j(k_{-\theta },z)$ and $k_{-\theta }\cdot z+i=e^{i\theta }(z+i)\mathrm{/}j(k_{-\theta },z)$, with $j(k_{-\theta },z)=\sin \theta z+\cos \theta$. Substitution into the inverse-action formula cancels the denominator powers and gives ${\pi }_n(k_{\theta })f_{n,j}=e^{-i(n+2j)\theta }{f}_{n,j}$. Taking complex conjugates gives ${\pi }_n^{-}(k_{\theta }){\tilde{f}}_{n,j}=e^{i(n+2j)\theta }{\tilde{f}}_{n,j}$; hence their algebraic spans are K-finite.

Each displayed vector is smooth for the group action. Under the Cayley map, $G$ acts by disk automorphisms; the inverse automorphy factor in the transformed coordinate is a smooth scalar times $(c_{g}w+d_g{)}^{-n}$, with $\mathrm{\mid }{d}_g\mathrm{\mid }>\mathrm{\mid }{c}_g\mathrm{\mid }$. Thus the transform of $f_{n,j}$, whose disk-coordinate function is $w^j$, is a rational function with denominator $(c_{g}w+d_g{)}^{n+j}$. Near any fixed $g_0$, smooth dependence of the disk-automorphism coefficients and the strict bound $\mathrm{\mid }{c}_{g}w+d_g\mathrm{\mid }\ge \mathrm{\mid }{d}_g\mathrm{\mid }-\mathrm{\mid }{c}_g\mathrm{\mid }>0$ uniformly for $\mathrm{\mid }w\mathrm{\mid }\le 1$ show that every parameter derivative is uniformly bounded on the closed disk. The weighted measure $(1-\mathrm{\mid }w{\mathrm{\mid }}^2{)}^{n-2}dudv$ is finite for $n\ge 2$, so the parameter difference quotients and all their derivatives converge in its $L^2$ norm by uniform convergence. Hence each orbit map is $C^{\mathrm{\infty }}$ in the Hilbert norm. Complex conjugation gives the same conclusion for ${\tilde{f}}_{n,j}$.

For a real matrix $X=\left(\begin{array}{cc}a& b\\ c& -a\end{array}\right)$, differentiating $\exp (-sX)$ at $s=0$ in the inverse-action formula gives the derived operator $L_{X}f=(c{z}^2-2az-b)f^{\mathrm{\prime }}(z)+(-na+ncz)f(z)$ on these smooth vectors. Hence $L_H=-2z{\mathrm{\partial }}_z-n$, $L_S=(z^2-1){\mathrm{\partial }}_z+nz$, and $L_{E_{+}}=\frac{i}{2}((z+i{)}^2{\mathrm{\partial }}_z+n(z+i))$, $L_{E_{-}}=-\frac{i}{2}((z-i{)}^2{\mathrm{\partial }}_z+n(z-i))$. Substitution of $f_{n,j}$ yields $L_{E_{+}}{f}_{n,j}=-j{f}_{n,j-1}$ (zero for $j=0$) and $L_{E_{-}}{f}_{n,j}=(n+j)f_{n,j+1}$. Complex conjugation gives $L_{E_{+}}{\tilde{f}}_{n,j}=(n+j){\tilde{f}}_{n,j+1}$ and $L_{E_{-}}{\tilde{f}}_{n,j}=-j{\tilde{f}}_{n,j-1}$.

Choose $\epsilon \in \{0,1\}$ with $\epsilon \equiv n(\mathrm{m}\mathrm{o}\mathrm{d}2)$; then $\nu =n-1$ lies in ${\mathcal{W}}_{\epsilon }$ of Highest- and lowest-weight submodules at the exceptional parameters. Its target module has $L_{E_{+}}{f}_{-n-2j}=-j{f}_{-n-2(j-1)}$ and $L_{E_{-}}{f}_{-n-2j}=(n+j)f_{-n-2(j+1)}$; these match step 3.1 under $f_{n,j}\mapsto f_{-n-2j}$. On the positive tail, $L_{E_{+}}{f}_{n+2j}=(n+j)f_{n+2(j+1)}$ and $L_{E_{-}}{f}_{n+2j}=-j{f}_{n+2(j-1)}$, matching the antiholomorphic formulas under ${\tilde{f}}_{n,j}\mapsto f_{n+2j}$. The K-weights match by step 2.2, so these maps identify the displayed algebraic K-finite spans with $M_{-n}^{+}$ and $M_n^{-}$.