SL2(R): Discrete Series and the Unitary Dual — Examples

Since $y^{2-2}=1$ and $\mathrm{\mid }z+i{\mathrm{\mid }}^2=x^2+(y+1{)}^2$ for $z=x+iy$, [F4] gives $\mathrm{\parallel }{f}_{2,0}{\mathrm{\parallel }}_2^2={\int }_0^{\mathrm{\infty }}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(x^2+(y+1{)}^2{)}^{-2}dxdy$. For $a>0$, the substitution $x=a\tan t$ yields ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}(x^2+a^2{)}^{-2}dx=a^{-3}{\int }_{-\pi \mathrm{/}2}^{\pi \mathrm{/}2}{\cos }^{2}tdt=\pi \mathrm{/}(2a^3)$. Taking $a=y+1$ therefore gives $\mathrm{\parallel }{f}_{2,0}{\mathrm{\parallel }}_2^2=(\pi \mathrm{/}2){\int }_0^{\mathrm{\infty }}(y+1{)}^{-3}dy=\pi \mathrm{/}4$, and $\mathrm{\parallel }(2\mathrm{/}\sqrt{\pi })f_{2,0}{\mathrm{\parallel }}_2=1$.

Setting $n=2$ in [F1] gives $f_{2,j}(z)=(z-i{)}^j(z+i{)}^{-2-j}$, K-character $e^{-i(2+2j)\theta }$, and $L_{E_{+}}{f}_{2,0}=0$, $L_{E_{+}}{f}_{2,j}=-j{f}_{2,j-1}$ for $j\ge 1$, and $L_{E_{-}}{f}_{2,j}=(2+j)f_{2,j+1}$ for $j\ge 0$. Thus every step from $f_{2,j}$ to $f_{2,j+1}$ and every return step for $j>0$ has nonzero coefficient. By [F2], these lines give the full multiplicity-one K-type decomposition of $D_2^{-}$, and [F3] gives its irreducibility.

Direct matrix multiplication gives $w^2=-I$ and $w^{-1}=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$, so $w\cdot z=-1\mathrm{/}z$ and $j(w^{-1},z)=z$. Since $w^{-1}\cdot z=-1\mathrm{/}z$ and $j(w^{-1},z)=z$, the model action is ${\pi }_2(w)f(z)=z^{-2}f(-1\mathrm{/}z)$. The map $z\mapsto -1\mathrm{/}z$ is an involutive C1 diffeomorphism of $\mathfrak{H}$, so this formula defines a holomorphic function there.

Apply [F3] to $z=-1\mathrm{/}u$. By [F2], $d{A}_z=\mathrm{\mid }u{\mathrm{\mid }}^{-4}d{A}_u$, while $\mathrm{\mid }{z}^{-2}{\mathrm{\mid }}^2=\mathrm{\mid }u{\mathrm{\mid }}^4$; hence $\mathrm{\parallel }{\pi }_2(w)f{\mathrm{\parallel }}_2^2={\int }_{\mathfrak{H}}\mathrm{\mid }u{\mathrm{\mid }}^4\mathrm{\mid }f(u){\mathrm{\mid }}^2\mathrm{\mid }u{\mathrm{\mid }}^{-4}d{A}_u=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2$. The integrand is nonnegative, so the change-of-variables identity also holds as an extended integral; for $f\in {\mathcal{H}}_2^{+}$ it is finite.

Since $w^2=-I$ and the scalar automorphy factor of $-I$ at weight $2$ is $(-1{)}^{-2}=1$, the group law in [F4] gives ${\pi }_2(w{)}^2=I$. Therefore the norm-preserving map of step 1.2 is onto and is a bijective linear isometry. The cancellation is the special $n=2$ instance of [F5], where the density weight is $y^0=1$.

Let $c_2(g)=⟨{\pi }_2(g)u_2,u_2⟩$. By [F1], $c_2(a_{\tau })={\mathrm{sech}}^2(\tau \mathrm{/}2)$. The vector $u_2$ is a $K$-eigenvector, so unitarity gives $c_2(k_{1}g{k}_2)=\chi (k_1)\chi (k_2)c_2(g)$ for its character $\chi$; hence $\mathrm{\mid }{c}_2{\mathrm{\mid }}^2$ is continuous and $K$-bi-invariant by [F2].

Applying [F3] gives ${\int }_G\mathrm{\mid }{c}_2(g){\mathrm{\mid }}^{2}dg=2\pi {\int }_0^{\mathrm{\infty }}{\mathrm{sech}}^4(\tau \mathrm{/}2)\sinh \tau d\tau$. Set $u=\tau \mathrm{/}2$; the radial integrand times $d\tau$ becomes $4\sinh u{\cosh }^{-3}udu$. For $M>0$, its integral on $[0,M]$ is $2(1-{\cosh }^{-2}M)$, which tends to $2$. The truncated integrands increase to the full nonnegative integrand, so [F6] gives the full radial integral $2$ and therefore $\mathrm{\parallel }{c}_2{\mathrm{\parallel }}_2^2=4\pi$.

By [F4], this explicit coefficient lies within the K-finite square-integrable coefficient family of $D_2^{-}$. At the $n=1$ endpoint, [F5] and [F3] instead give ${\int }_G\mathrm{\mid }{c}_1(g){\mathrm{\mid }}^{2}dg=2\pi {\int }_0^{\mathrm{\infty }}2\tanh (\tau \mathrm{/}2)d\tau =+\mathrm{\infty }$: for $\tau \ge \log 3$ the integrand is at least $1$, and ${\mathbf{1}}_{[\log 3,\log 3+m]}\uparrow {\mathbf{1}}_{[\log 3,\mathrm{\infty })}$ has integral $m$, so [F6] forces divergence. Thus the concrete $n=2$ coefficient is square-integrable while the displayed limit coefficient is not.

Let $\pi$ be the restriction of $I_{1,0}$ to $D_1^{+}$ and set $c(g)=⟨\pi (g)f_1,f_1⟩$. By [F1], this is a matrix coefficient of an irreducible unitary limit representation; by [F2] it is continuous, and [F3] gives $c(a_{\tau })=\mathrm{sech}(\tau \mathrm{/}2)$. If $\chi (k_{\theta })=e^{i\theta }$, unitarity and $\pi (k)f_1=\chi (k)f_1$ give $c(k_{1}g{k}_2)=\chi (k_1)\chi (k_2)c(g)$. Thus [F4] applies to $\mathrm{\mid }c{\mathrm{\mid }}^2$.

The KAK formula and $\sinh \tau =2\sinh (\tau \mathrm{/}2)\cosh (\tau \mathrm{/}2)$ give ${\int }_G\mathrm{\mid }c(g){\mathrm{\mid }}^{2}dg=2\pi {\int }_0^{\mathrm{\infty }}{\mathrm{sech}}^2(\tau \mathrm{/}2)\sinh \tau d\tau =2\pi {\int }_0^{\mathrm{\infty }}2\tanh (\tau \mathrm{/}2)d\tau$. For $\tau \ge \log 3$, $2\tanh (\tau \mathrm{/}2)\ge 1$. The indicators ${\mathbf{1}}_{[\log 3,\log 3+m]}$ increase to ${\mathbf{1}}_{[\log 3,\mathrm{\infty })}$ and their integrals are $m$, so [F7] shows the last nonnegative integral is infinite.

By [F6], the irreducible unitary representation $D_1^{+}$ is not square-integrable because the coefficient in step 2.1 is not in $L^2(G)$. For $n\ge 2$, [F4] and [F5] give the corresponding extremal coefficient integral $2\pi {\int }_0^{\mathrm{\infty }}{\cosh }^{-2n}(\tau \mathrm{/}2)\sinh \tau d\tau$. With $u=\tanh (\tau \mathrm{/}2)$ this equals $8\pi {\int }_0^{1}u(1-u^2{)}^{n-2}du=4\pi \mathrm{/}(n-1)<\mathrm{\infty }$, confirming the endpoint contrast and refuting the claim in the Statement.

By [F1], ${\mathcal{W}}_0=2\mathbb{Z}+1$ and ${\mathcal{W}}_1=2\mathbb{Z}$. The generic parameter set is their complement, so the sign-equivalence claim in the Statement is restricted to irreducible induced modules; nonzero lattice points are reducible.

The compact-picture theorem [F6] supplies the principal unitary axis; [F3] supplies irreducibility at its even zero and [F5] supplies the odd zero split. The positive weighted Hilbert completion and odd-parity invariant-form obstruction in [F7] give precisely the spherical complementary interval and absence of an odd complementary family.

For $\nu \mathrm{\notin }{\mathcal{W}}_{\epsilon }$, [F3] gives $I_{\epsilon ,\nu }\cong I_{\epsilon ,-\nu }$. For nonzero $\nu \in {\mathcal{W}}_{\epsilon }$, [F3] gives the opposite positions of $L_{\mathrm{\mid }\nu \mathrm{\mid }-1}$ in the two modules, so they are not isomorphic. For spherical real $0<\mathrm{\mid }\nu \mathrm{\mid }<1$, the normalized smooth intertwiner $R_{\nu }$ of [F3] has multipliers $a_n(\nu )$. By [F7], $B_{-\nu }(R_{\nu }f,R_{\nu }f)={\sum }_n{a}_n(-\nu )\mathrm{\mid }{a}_n(\nu )\widehat{f}(n){\mathrm{\mid }}^2=B_{\nu }(f,f)$ on finite Fourier sums. Its inverse is $R_{-\nu }$, so density extends it to an onto unitary intertwiner of the completed complementary representations. At $\epsilon =1,\nu =0$, [F5] gives the direct sum of the two limits.

The K-type parity and multiplicity statement is [F2]. At a positive exceptional integer $n$, [F4] gives the finite quotient and two extremal submodules; at $-n$ it gives the reversed submodule/quotient orientation. When $n=1$ and $\epsilon =0$, $L_0$ is the trivial representation, so the two endpoints $\nu =\pm 1$ have the stated trivial subquotient. For $n\ge 2$, [F5] identifies the extremal modules at $\nu =n-1$ with the algebraic K-finite spans of $D_n^{-}$ and $D_n^{+}$, whose weighted Hilbert completions give the group representations. Thus an exceptional full induced module is not a second irreducible class to be counted alongside its irreducible constituents.

For two generic parameters of the same parity, [F4] gives the central scalar $c(\nu )=({\nu }^2-1)\mathrm{/}8$; equivalent representations must have equal central scalars, hence their parameters differ only by sign. Different parities have disjoint K-type supports by [F2]. On the principal unitary axis $\nu =is$, the scalar is $-(s^2+1)\mathrm{/}8$, so it determines $\mathrm{\mid }s\mathrm{\mid }$; on the spherical complementary interval $0<r<1$, it is $(r^2-1)\mathrm{/}8$, so it determines $r$. These ranges are disjoint, and the sign equivalence [F3] therefore leaves no further identification between distinct absolute parameter values in either family.