Sl2 R Principal and Complementary Series — Examples

The first column $(p,q{)}^T$ is nonzero because $ps-qr=1$. Put $a=\sqrt{p^2+q^2}>0$, $t=2\log a$, $k(g)=a^{-1}\left(\begin{array}{cc}p& -q\\ q& p\end{array}\right)$, and $x=(pr+qs)\mathrm{/}{a}^2$. Then $k(g)\in K$, $a(g)=\mathrm{diag}(a,a^{-1})=a_t$, and $n(g)=n_x$. Direct multiplication gives $k(g)a(g)n(g)=\left(\begin{array}{cc}p& px-q\mathrm{/}{a}^2\\ q& qx+p\mathrm{/}{a}^2\end{array}\right)$. Its top-right entry is $r$ because $px-q\mathrm{/}{a}^2-r=q(ps-qr-1)\mathrm{/}{a}^2=0$; its bottom-right entry is $s$ because $qx+p\mathrm{/}{a}^2-s=p(1-ps+qr)\mathrm{/}{a}^2=0$. Thus the factors multiply to $g$, and uniqueness in [F1] makes them its Iwasawa coordinates. For $g_{\ast }=\left(\begin{array}{cc}2& 1\\ 1& 1\end{array}\right)$, these formulas give $a=\sqrt{5}$, $t=\log 5$, $x=3\mathrm{/}5$, and $k=5^{-1\mathrm{/}2}\left(\begin{array}{cc}2& -1\\ 1& 2\end{array}\right)$, whose product is $g_{\ast }$.

For left multiplication by $a_u$, the first column of $a_u{k}_{\theta }$ has squared norm $D_u(\theta )=e^u{\cos }^2\theta +e^{-u}{\sin }^2\theta$. Its Iwasawa factors therefore have $t_1=u_1(\theta )=\log D_u(\theta )$, $x_1=v_u(\theta )=\frac{(e^u-e^{-u})\sin \theta \cos \theta }{D_u(\theta )}$, and $k_{{\theta }_1}$ with $\cos {\theta }_1=e^{u\mathrm{/}2}\cos \theta \mathrm{/}\sqrt{D_u(\theta )}$ and $\sin {\theta }_1=e^{-u\mathrm{/}2}\sin \theta \mathrm{/}\sqrt{D_u(\theta )}$. Hence $a_u{k}_{\theta }=k_{{\theta }_1}{a}_{u_1(\theta )}{n}_{v_u(\theta )}$. Differentiating this circle map gives $\frac{d{\theta }_1}{d\theta }=\frac{e^{-u}}{{\cos }^2\theta +e^{-2u}{\sin }^2\theta }=D_u(\theta {)}^{-1}=e^{-u_1(\theta )}>0$, including at $\cos \theta =0$.

For $g_0=k_{\varphi }$, one has $k_{\varphi }{k}_{\theta }=k_{\theta +\varphi }$, so the $AN$ factor is the identity ($t_1=x_1=0$) and the $K$ coordinate is translated by $\varphi$. Thus the two requested left-translation cocycles are $a_{u_1(\theta )}{n}_{v_u(\theta )}$ and $I$, respectively.

Fix $0<\eta <1\mathrm{/}2$ and put ${\psi }_{\eta }(z)=\max (1-\mathrm{\mid }z\mathrm{\mid }\mathrm{/}\eta ,0)$. Using the representative $z\in (-1\mathrm{/}2,1\mathrm{/}2]$ of the torus coordinate in [F3], define $f_{\eta }(k_{2\pi z}{a}_t{n}_x)={\psi }_{\eta }(z){\psi }_{\eta }(t){\psi }_{\eta }(x)$. The function vanishes near the angular coordinate cut and has compact support in an arbitrarily small coordinate neighborhood of the identity as $\eta \downarrow 0$. By [F2] and [F4], its Haar integral factors as $\left({\int }_{\mathbb{T}}{\psi }_{\eta }d{m}_{\mathbb{T}}\right)({\int }_{-\eta }^{\eta }{e}^t{\psi }_{\eta }(t)dt)({\int }_{-\eta }^{\eta }{\psi }_{\eta }(x)dx)$. The torus factor is ${\int }_0^{\eta }(1-z\mathrm{/}\eta )dz+{\int }_{1-\eta }^1(1-(1-z)\mathrm{/}\eta )dz=\eta$, and the $x$ factor is $\eta$. The middle factor is $2{\int }_0^{\eta }(1-t\mathrm{/}\eta )\cosh tdt=2(\sinh \eta -\frac{\eta \sinh \eta -\cosh \eta +1}{\eta })=2(\cosh \eta -1)\mathrm{/}\eta$. Therefore ${\int }_G{f}_{\eta }(g)dg=2\eta (\cosh \eta -1)$.

Write $\xi =[\theta \mathrm{/}(2\pi )]\in \mathbb{T}$, so $dk=d{m}_{\mathbb{T}}$ by [F4]. For a general coordinate $k_{2\pi \xi }{a}_t{n}_x$, left translation by $a_u$ has map $(\xi ,t,x)\mapsto ({\xi }_1,t+u_1(2\pi \xi ),x+e^{-t}{v}_u(2\pi \xi ))$, since $n_v{a}_t=a_t{n}_{e^{-t}v}$, where ${\xi }_1=[{\theta }_1(2\pi \xi )\mathrm{/}(2\pi )]$. Its Jacobian is triangular with determinant $d{\xi }_1\mathrm{/}d\xi =d{\theta }_1\mathrm{/}d\theta =e^{-u_1(2\pi \xi )}>0$; the Haar weight changes to $e^{t+u_1(2\pi \xi )}$, so the density $e^{t}dkdtdx$ is preserved. The pullback calculation and [F5], applied on circle coordinate charts containing the compact support of $f_{\eta }$ and its translate, show that its integral is unchanged. Left translation by $k_{\varphi }$ sends $\xi$ to $\xi +\varphi \mathrm{/}(2\pi )$ and leaves $t,x$ fixed, which preserves $dk$ and the density by [F4]; [F5] gives the same integral identity. Thus both computed cocycles agree with the left invariance of the Haar formula [F2].

For $\epsilon =0$, the only indices in $\{-3,-2,\dots ,3\}$ with the required parity are $-2,0,2$; for $\epsilon =1$ they are $-3,-1,1,3$. Applying the three formulas in [F2] to these indices gives every entry of the table, and [F1] shows that the table omits no K-type in the requested range.

Let $m\in {\mathcal{W}}_{\epsilon }$ with $m\ge 0$; $m=0$ occurs only for odd parity. At $\nu =m$, [F2] gives $L_{E_{-}}{f}_{m+1}=0$ and $L_{E_{+}}{f}_{-m-1}=0$, since their coefficients are respectively $(1+m-(m+1))\mathrm{/}2$ and $(1+m+(-m-1))\mathrm{/}2$. At $\nu =-m$, it gives $L_{E_{-}}{f}_{1-m}=0$ and $L_{E_{+}}{f}_{m-1}=0$, since both coefficients are zero. When $m=0$, these parameter cases coincide and the table shows $L_{E_{-}}{f}_1=L_{E_{+}}{f}_{-1}=0$ at $\epsilon =1,\nu =0$. For the other exceptional values visible in the table, $m=1$ in even parity and $m=2$ in odd parity: the positive parameter zeros occur at $f_{\pm 2}$ and $f_{\pm 3}$, respectively, and the negative parameter zeros occur at $f_0$ and $f_{\pm 1}$. These are exactly the boundary arrows expected from the exceptional K-type strings; no parity class is identified with the other.

Normalize the meromorphic recurrence in [F1] by its base scalar wherever the quotient is initially regular. The $j=0$ and $j=1$ instances give ${\widehat{c}}_2=(1-\nu )\mathrm{/}(1+\nu )$ and ${\widehat{c}}_4={\widehat{c}}_2(3-\nu )\mathrm{/}(3+\nu )=(1-\nu )(3-\nu )\mathrm{/}((1+\nu )(3+\nu ))$. The negative-index symmetry gives ${\widehat{c}}_{-2}={\widehat{c}}_2$ and ${\widehat{c}}_{-4}={\widehat{c}}_4$. These ratios extend meromorphically; [F2] identifies their regular values at $\nu =0$ and agrees with the displayed products.

If $\mathrm{\mid }\nu \mathrm{\mid }<1$, then for every $l\ge 1$ both $2l-1-\nu$ and $2l-1+\nu$ are positive. Thus ${\widehat{c}}_0=1$, ${\widehat{c}}_{\pm 2}>0$, and ${\widehat{c}}_{\pm 4}>0$ throughout the open interval, including $\nu =0$. More generally every factor in the finite product for any fixed even K-type is positive there.

On $1<\nu <3$, the numerator $1-\nu$ is negative and the denominator $1+\nu$ positive, so ${\widehat{c}}_2(\nu )<0$; at the midpoint $\nu =(1+3)\mathrm{/}2=2$ its value is $-1\mathrm{/}3\mathrm{\ne }0$. On $-1<\nu <1$ the coefficient is positive, and it vanishes at $\nu =1$, so its sign changes at that endpoint.

AC enters through the normalized Haar construction and the countable-choice hypotheses of the cited suppliers described in [A1]; this finite computation makes no additional selection.

At $\nu =3$, the normalized weights are finite. By [F1], $a_0(3)=1$ and $4a_2(3)=-2a_0(3)$, so $a_2(3)=-1\mathrm{/}2$. Thus $B_3(f_0,f_0)=1>0$ and $B_3(f_2,f_2)=-1\mathrm{/}2<0$: this regular invariant form is indefinite and refutes positive definiteness. The parameter $3$ is a reducibility point, but regularity of the normalized form does not require irreducibility.

At $\nu =1$, [F2] gives $a_0=1$ and $a_2=0$. Hence $B_1(f_0,f_0)=1$, while $B_1(f_2,h)=0$ for every smooth $h$ by the Fourier-diagonal formula; $f_2\mathrm{\ne }0$, so the endpoint form is nonzero and degenerate. This also contradicts the refuted claim at its boundary and confirms that the positive-definite range $\mathrm{\mid }\nu \mathrm{\mid }<1$ is strict.