Parameter-sign equivalence and its exceptional failures for SL2(R)

Let $P_{\epsilon }=\{-m:m\in {\mathbb{Z}}_{\ge 0},m\equiv \epsilon (\mathrm{m}\mathrm{o}\mathrm{d}2)\}$ be the common scalar pole set from [F1]. It is disjoint from ${\mathcal{W}}_{\epsilon }$. For $\nu \mathrm{\notin }{\mathcal{W}}_{\epsilon }\cup P_{\epsilon }$, [F2] shows that $c_{n_0}(\nu )$ is finite and nonzero, so $\widehat{A}(\nu )=A(\nu )\mathrm{/}{c}_{n_0}(\nu )$ is a continuous K-diagonal map. At ${\nu }_0=-m\in P_{\epsilon }$, the Gamma formula in [F2] has one numerator Gamma factor with a simple pole of nonzero residue, while both denominator arguments are half-integers and hence finite and nonzero for every allowed $r$. Thus every $c_r$, including $c_{n_0}$, has the same simple pole and $h(\nu )c_{n_0}(\nu )$ is holomorphic and nonzero at ${\nu }_0$, where $h(\nu )=\nu -{\nu }_0$. By [F1], $hA(\nu )$ is holomorphic in the smooth operator topology there. Therefore $\widehat{A}=(hA)\mathrm{/}(h{c}_{n_0})$ extends continuously through that pole as a K-diagonal map.

Let $\nu \in {\mathcal{W}}_{\epsilon }\setminus \{0\}$ and write $n=\mathrm{\mid }\nu \mathrm{\mid }\ge 1$. The pole set $P_{\epsilon }$ has parity $\epsilon$ and is disjoint from ${\mathcal{W}}_{\epsilon }$, so $A(\nu )$ is regular. If $\nu =n$, [F2] gives $c_r(n)=0$ on every allowed type with $\mathrm{\mid }r\mathrm{\mid }\ge n+1$; if $\nu =-n$, it gives $c_r(-n)=0$ for $\mathrm{\mid }r\mathrm{\mid }\le n-1$. In either case a nonzero K-type is killed, so $A(\nu )$ is not injective.

For every allowed $r$, normalize the recurrence in [F2] to obtain ${\widehat{c}}_{r+2}(\nu )={\widehat{c}}_r(\nu )(r+1-\nu )\mathrm{/}(r+1+\nu )$ and ${\widehat{c}}_{-r}(\nu )=(-1{)}^r{\widehat{c}}_r(\nu )$, with ${\widehat{c}}_{n_0}=1$. If $\nu \mathrm{\notin }{\mathcal{W}}_{\epsilon }$, neither $r+1+\nu$ nor $r+1-\nu$ vanishes for any allowed $r$, since $r+1$ has the opposite parity. Thus every multiplier is finite and nonzero, including at $P_{\epsilon }$ by step 1.1.

Applying the recurrence at $\nu$ and $-\nu$ gives ${\widehat{c}}_{r+2}(\nu ){\widehat{c}}_{r+2}(-\nu )={\widehat{c}}_r(\nu ){\widehat{c}}_r(-\nu )$; the base value is $1$ and the symmetry handles negative indices. Therefore $\widehat{A}(-\nu )\widehat{A}(\nu )$ is the identity on every K-type. Both operators are continuous and K-diagonal by step 1.1; for smooth $f$, every Fourier coefficient of $(\widehat{A}(-\nu )\widehat{A}(\nu )-I)f$ is zero, so [F5] makes that vector zero in $L^2$. A smooth function that is zero almost everywhere is zero everywhere, since a nonzero value would by continuity give a nonempty open arc on which its modulus is bounded below, and every open arc has positive normalized Haar measure. The same argument applies in the reverse order, proving that the maps are continuous inverses on $C_{\epsilon }^{\mathrm{\infty }}(K)$.

For $\nu \mathrm{\notin }{\mathcal{W}}_{\epsilon }\cup P_{\epsilon }$, [F1] gives the G-intertwining identity for the regular operator $A(\nu )$, and [F2] gives $0\mathrm{\ne }{c}_{n_0}(\nu )$; division yields the identity for $\widehat{A}(\nu )$. At a pole ${\nu }_0\in P_{\epsilon }$, take regular ${\nu }_j\to {\nu }_0$ outside ${\mathcal{W}}_{\epsilon }\cup P_{\epsilon }\cup (-P_{\epsilon })$; step 1.1 gives convergence of $\widehat{A}({\nu }_j)$ in the smooth-operator topology. For fixed $g$, the compact formula in [F4] depends continuously on $\nu$ in every smooth seminorm, because $\log \mathrm{\mid }\alpha (p(k,g))\mathrm{\mid }$ is smooth and bounded on compact $K$. Passing to the limit in the intertwining identity gives it at ${\nu }_0$ as well. Consequently the inverse maps of step 3.1 give $I_{\epsilon ,\nu }\cong I_{\epsilon ,-\nu }$ for every $\nu \mathrm{\notin }{\mathcal{W}}_{\epsilon }$. When in addition the unnormalized $A(\nu )$ is regular, [F2] gives finite nonzero $c_{n_0}(\nu )$, so $A(\nu )=c_{n_0}(\nu )\widehat{A}(\nu )$ is an isomorphism.

Suppose $\nu \in i\mathbb{R}\setminus {\mathcal{W}}_{\epsilon }$. Every recurrence factor $(r+1-\nu )\mathrm{/}(r+1+\nu )$ has modulus one, and the base multiplier is ${\widehat{c}}_{n_0}=1$; the symmetry gives $\mathrm{\mid }{\widehat{c}}_r(\nu )\mathrm{\mid }=1$ on every allowed K-type. Thus the Fourier multiplier preserves the $L^2$ norm on finite Fourier sums and its inverse multiplier does too; by density [F5] it extends to a unitary operator on $L_{\epsilon }^2(K)$. Step 4.1 makes it a smooth G-intertwiner, so continuity of the two unitary actions and density extend the intertwining identity to all of $L_{\epsilon }^2(K)$. Separately, for $Cf=\bar{f}$, the compact-picture formula [F4] gives $C{\mathrm{\Pi }}_{\nu }(g)={\mathrm{\Pi }}_{-\nu }(g)C$ because $\mathrm{\mid }\alpha (p(k,g))\mathrm{\mid }$ is positive real and $\bar{\nu }=-\nu$; thus $C$ is an anti-unitary intertwiner.

Let $n\ge 1$ lie in ${\mathcal{W}}_{\epsilon }$. In $I_{\epsilon ,n}^K$ put $T^{+}={⨁}_{j\ge 0}\mathbb{C}{f}_{n+1+2j}$. By [F6], $L_{E_{-}}{f}_{n+1}=0$, while $E_{+}$ preserves these weights and both arrows are nonzero between every adjacent pair above the boundary; $K$ also preserves $T^{+}$. Hence $T^{+}$ is a nonzero $(\mathfrak{g},K)$-submodule. It is simple: for any nonzero submodule choose a nonzero finite Fourier sum; a $k_{\theta }\in K$ can be chosen with distinct eigenvalues on its finite set of K-types, since only finitely many angles cause a collision, and a polynomial in its action isolates one nonzero K-type. The nonzero ladder arrows then generate all of $T^{+}$. If a $(\mathfrak{g},K)$-module isomorphism $T$ existed from $I_{\epsilon ,n}^K$ to $I_{\epsilon ,-n}^K$, it would send $T^{+}$ to an irreducible submodule at $-n$, which [F3] says must be $L_{n-1}$. But $L_{n-1}$ has only K-types $-(n-1),-(n-3),\dots ,n-1$, disjoint from the support of $T^{+}$, contradicting K-equivariance. Thus the K-finite modules are not isomorphic. Any continuous G-equivalence of the smooth compact-picture representations restricts to a K-finite module isomorphism: continuity lets one differentiate $T{\mathrm{\Pi }}_n(g)={\mathrm{\Pi }}_{-n}(g)T$ along real one-parameter subgroups, while K-commutation preserves K-finiteness. Therefore no continuous G-equivalence exists; exchanging $n$ and $-n$ gives the same conclusion in the reverse direction. The odd case $\nu =0$ is excluded: there $\nu \in {\mathcal{W}}_1$ and [F3] gives a direct sum of the two limit chains.