Unitarity of the complementary series

In spherical parity put $a_0(\nu )=1$. The Gamma formula in [F5], using $\mathrm{\Gamma }(z+1)=z\mathrm{\Gamma }(z)$, gives the cross-multiplied recurrence as a meromorphic identity for every $\nu$, so it may be used even at its zero denominators. For $\nu \mathrm{\notin }\{-1,-3,\dots \}$, successive use at $n=0,2,\dots ,2j-2$ gives $a_{2j}={\prod }_{l=1}^j(2l-1-\nu )\mathrm{/}(2l-1+\nu )$. The meromorphic symmetry $c_{-n}=(-1{)}^n{c}_n$ gives $a_{-2j}=a_{2j}$. At a positive odd $\nu =2m+1$ with $m\ge 0$, the numerator $2l-1-\nu$ vanishes at $l=m+1$, so every weight with $j\ge m+1$ is zero; a denominator vanishes exactly at a negative odd integer. Thus the displayed weights are finite precisely on the real parameter set in the Statement.

Let $B$ be any positive-definite invariant Hermitian form on the full spherical K-finite module, and put $b_n=B(f_n,f_n)>0$. K-invariance makes distinct K-types orthogonal. By the infinitesimal invariance relation [F6], $B(L_{E_{+}}{f}_0,f_2)=-B(f_0,L_{E_{-}}{f}_2)$, so the ladder formulas give $(1+\nu )b_2=(1-\nu )b_0$. If $\nu =-1$, the left side is zero and the right side positive; if $\nu <-1$, the left coefficient is negative and the right positive; if $\nu =1$, the left is positive while the right is zero; and if $\nu >1$, the right side is negative. Hence such a form can exist only when $\mathrm{\mid }\nu \mathrm{\mid }<1$.

In odd parity, any invariant Hermitian form on the K-finite module is K-diagonal with real weights $b_n=B(f_n,f_n)$. Applying [F6] to each adjacent pair gives $(1+\nu +n)b_{n+2}=(n+1-\nu )b_n$; at $n=-1$ this is $\nu b_1=-\nu b_{-1}$. For real $\nu \mathrm{\ne }0$, a positive-definite form would have $b_1,b_{-1}>0$, contradicting this identity. If $\nu$ is nonexceptional and the form is nonzero, K-diagonality gives some nonzero $b_n$; no ladder coefficient vanishes, so the recurrence propagates this to $b_1\mathrm{\ne }0$ and then to every odd weight. Since $b_{-1}=-b_1$, the form is nondegenerate and indefinite.

At $\epsilon =1$, $\nu =0$, the K-finite odd module splits into the two unitary limit chains by [F8]; the recurrence at $n=-1$ is $0=0$ and leaves their invariant weights independent. This is the separate zero-parameter unitary splitting, not an odd complementary series.

Fix a real $\nu \mathrm{\notin }\{-1,-3,\dots \}$. For all sufficiently large $l$, $\mathrm{\mid }(2l-1-\nu )\mathrm{/}(2l-1+\nu )\mathrm{\mid }\le 1+C\mathrm{/}l$ for a constant depending on $\nu$; bound the finitely many earlier factors separately. Choose an integer $M\ge C$; then ${\prod }_{l=1}^j(1+M\mathrm{/}l)=\left(\genfrac{}{}{0px}{}{j+M}{M}\right)\le (j+M{)}^M$, so $\mathrm{\mid }{a}_n(\nu )\mathrm{\mid }\le C^{\mathrm{\prime }}(1+\mathrm{\mid }n\mathrm{\mid }{)}^M$. If an initial factor is zero the subsequent weights are zero and the same estimate holds. By [F9], $\mathrm{\mid }\widehat{f}(n)\mathrm{\mid }\le C_N{p}_N(f)\mathrm{\mid }n{\mathrm{\mid }}^{-N}$ for $n\mathrm{\ne }0$. Choose $N$ with $2N-M>1$; the p-series bound then gives $\mathrm{\mid }{B}_{\nu }(f,h)\mathrm{\mid }\le C{p}_N(f)p_N(h)$, so the Fourier form is finite and continuous. Its weights are real and symmetric, hence it is Hermitian.

Define $R_{\nu }$ as the smooth K-diagonal multiplier with coefficients $a_n(\nu )$. If $\nu$ is not a common scalar pole and $c_0(\nu )\mathrm{\ne }0$, it equals $A(\nu )\mathrm{/}{c}_0(\nu )$; the smooth continuity follows from [F4]. At ${\nu }_0=-2m$, $m\ge 0$, the Gamma formula in [F5] has a simple pole with nonzero residue in its numerator, while both denominator arguments for every even $n$ are half-integers and finite. Thus all $c_n$, including $c_0$, have the same simple pole and $(\nu -{\nu }_0)c_0(\nu )$ is holomorphic and nonzero at ${\nu }_0$. In the local Laurent expansion of the meromorphic K-diagonal family [F4], every coefficient below degree $-1$ has zero multiplier on each K-type because each scalar $c_n$ has at most a simple pole by [F5]. Such a continuous K-diagonal coefficient sends every smooth vector to a smooth function with all Fourier coefficients zero, hence is zero by completeness in [F2]. Therefore $(\nu -{\nu }_0)A(\nu )$ is holomorphic in the smooth operator topology, and dividing by $(\nu -{\nu }_0)c_0(\nu )$ defines the continuous extension of $R_{\nu }$ there; its K-type multipliers are exactly those in step 1.1. At every real parameter under consideration, [F4] gives the intertwining identity after division when $c_0\mathrm{\ne }0$; at the common poles pass to the limit from neighboring regular parameters, using [F11] for continuity of the compact-picture action in $\nu$. Therefore $R_{\nu }$ intertwines ${\mathrm{\Pi }}_{\nu }$ with ${\mathrm{\Pi }}_{-\nu }$ for every $\nu$ in the stated real domain. By [F2], $B_{\nu }(f,h)=⟨R_{\nu }f,h{⟩}_0$; [F3] and the intertwining identity give $B_{\nu }({\mathrm{\Pi }}_{\nu }(g)f,{\mathrm{\Pi }}_{\nu }(g)h)=B_{\nu }(f,h)$ for every $g\in G$.

If $\mathrm{\mid }\nu \mathrm{\mid }<1$, every numerator and denominator in the spherical weight product is positive, so all $a_n(\nu )>0$. Fourier completeness [F2] then makes $B_{\nu }$ positive definite; at $\nu =0$ all weights equal one and $B_0$ is the ordinary $L^2$ inner product.

At a regular real parameter with $\mathrm{\mid }\nu \mathrm{\mid }>1$, the displayed normalized form has $a_0=1$ and $a_2=(1-\nu )\mathrm{/}(1+\nu )<0$, so it is indefinite. Negative odd parameters are excluded from this assertion because the normalized weights have poles there; the recurrence argument in step 1.2 still rules out any positive-definite full-module form at those parameters.

Let $\nu =2m>0$. The recurrence from [F6] at $n=2m-1$ forces $b_{2m+1}=0$ and then all upper tail weights vanish; at $n=-2m-1$ it forces $b_{-2m-1}=0$ and then all lower tail weights vanish. If $\nu =-2m<0$, the same two recurrence equations force $b_{2m-1}=b_{-2m+1}=0$, propagating through every central odd weight $\mathrm{\mid }n\mathrm{\mid }\le 2m-1$. Thus every invariant form at either nonzero even exceptional parameter is degenerate. These equations force zeros, not signs on the remaining chains, so no blanket indefiniteness conclusion is asserted.

At $\nu =1$, the product in step 1.1 has $a_0=1$ and $a_{\pm 2j}=0$ for every $j\ge 1$. Hence $B_1(f,h)=\widehat{f}(0)\overline{\widehat{h}(0)}$. It is a finite nonzero degenerate invariant form by steps 2.1–2.2. Its radical in $C_0^{\mathrm{\infty }}(K)$ is $\{f:\widehat{f}(0)=0\}$, whose K-finite part is the algebraic span of the nonzero even K-types; the quotient is one-dimensional, and [F7] identifies its K-finite quotient with the trivial module $L_0$.

For $-1<\nu \le 0$ and $j\ge 1$, rewrite $(1+\nu )a_{2j}(\nu )=(1-\nu ){\prod }_{l=2}^j(2l-1-\nu )\mathrm{/}(2l-1+\nu )$. Each factor in the product is at most $l\mathrm{/}(l-1)$, so $0<(1+\nu )a_{2j}(\nu )\le 2j$. The $j=0$ weight $(1+\nu )a_0$ tends to zero, while for $j\ge 1$ the product tends to $2j$ as $\nu \to -1^{+}$. By [F9] and this bound, the rescaled Fourier forms converge on smooth vectors to the finite nonzero form with weights $b_0=0$ and $b_{\pm 2j}=2j$. To pass invariance to the rescaled limit, [F12] bounds the Fourier-decay seminorms of ${\mathrm{\Pi }}_{\nu }(g)f$ and ${\mathrm{\Pi }}_{\nu }(g)h$ uniformly for $\nu \in [-1,0]$; hence the same summable majorant applies to both sides of the invariance identity. The limit is invariant, its radical is exactly $\mathbb{C}{f}_0$, and [F7] identifies that line as the trivial submodule.

Suppose $0<\mathrm{\mid }\nu \mathrm{\mid }<1$. The polynomial bound in step 2.1 and Fourier decay in [F9] give a finite $r$ and $C$ with $\mathrm{\parallel }f{\mathrm{\parallel }}_{B_{\nu }}\le C{p}_r(f)$ for every smooth $f$. The smooth compact action is continuous in each $C^r$ seminorm by [F1, F11], so every smooth vector has a continuous orbit in the $B_{\nu }$ norm. Invariance makes ${\mathrm{\Pi }}_{\nu }(g)$ an isometry with inverse ${\mathrm{\Pi }}_{\nu }(g^{-1})$; it extends to a unitary on the completion, and density plus the isometry bound extends strong continuity to every completed vector. For irreducibility, the completion is the weighted ${\mathrm{\ell }}^2$ completion of the even Fourier modes. If $W$ is a nonzero closed invariant subspace, choose $\xi ={\sum }_{n\in 2\mathbb{Z}}{x}_n{f}_n\mathrm{\ne }0$ in $W$ and an $n$ with $x_n\mathrm{\ne }0$. For integers $N>\mathrm{\mid }n\mathrm{\mid }$, the finite average $Q_N\xi =N^{-1}{\sum }_{j=0}^{N-1}{e}^{-2\pi inj\mathrm{/}N}{\mathrm{\Pi }}_{\nu }(k_{2\pi j\mathrm{/}N})\xi$ lies in $W$ and retains exactly the modes $r\equiv n(\mathrm{m}\mathrm{o}\mathrm{d}N)$. As $N\to \mathrm{\infty }$, all retained modes other than $n$ lie in the weighted ${\mathrm{\ell }}^2$ tail $\mathrm{\mid }r\mathrm{\mid }\ge N-\mathrm{\mid }n\mathrm{\mid }$, so $Q_N\xi \to x_n{f}_n$. Hence $f_n\in W$. Smooth difference quotients in the $B_{\nu }$ norm put both ladder images in $W$; for $\mathrm{\mid }\nu \mathrm{\mid }<1$ all even ladder coefficients are nonzero, so $W$ contains every even K-type. Their finite span is dense by construction of the completion, giving $W$ equal to the full Hilbert space.