KAK integration formula for K-bi-invariant functions on SL2(R)

Let $g\in G$ and put $A=g^{T}g$. This is a positive definite symmetric endomorphism of ${\mathbb{R}}^2$, so [F1] gives an orthonormal eigenbasis with eigenvalues ${\lambda }_{+}\ge {\lambda }_{-}>0$. Since ${\lambda }_{+}{\lambda }_{-}=\det (g^{T}g)=1$, one has ${\lambda }_{+}\ge 1$ and ${\lambda }_{-}={\lambda }_{+}^{-1}$. Choose the eigenbasis matrix $U\in \mathrm{S}\mathrm{O}(2)$, changing the sign of one basis vector if needed, and set $t=\log {\lambda }_{+}\ge 0$ and $S=U{a}_t{U}^T$. Then $S^2=A$ and $\det S=1$. The matrix $k=g{S}^{-1}$ satisfies $k^{T}k=S^{-1}A{S}^{-1}=I$ and $\det k=1$, so $k\in K$ and $g=kU{a}_t{U}^T$ is a KAK factorization. Moreover $\mathrm{tr}(g^{T}g)=e^t+e^{-t}=2\cosh t$, so this nonnegative parameter $t$ is uniquely determined by $g$.

For a nonnegative continuous $K$-bi-invariant $\psi$, [F2] and ${\int }_{K}dk=1$ give ${\int }_G\psi (g)dg={\int }_{\mathbb{R}}{\int }_{\mathbb{R}}\psi (n_x{a}_s)e^{-s}dxds$. Set $y=e^s>0$ and $z=x+iy$. Direct multiplication gives $\mathrm{tr}((n_x{a}_s{)}^T(n_x{a}_s))=(x^2+y^2+1)\mathrm{/}y$. Define $w=(z-i)\mathrm{/}(z+i)$ and $\rho =\mathrm{\mid }w\mathrm{\mid }<1$. Then $1-{\rho }^2=4y\mathrm{/}(x^2+(y+1{)}^2)$ and $1+{\rho }^2=2(x^2+y^2+1)\mathrm{/}(x^2+(y+1{)}^2)$. For $r=2\mathrm{artanh}\rho \ge 0$, this gives $\cosh r=(1+{\rho }^2)\mathrm{/}(1-{\rho }^2)=(x^2+y^2+1)\mathrm{/}(2y)$. The unique KAK parameter of $n_x{a}_s$ therefore equals $r$ by step 1.1, so $\psi (n_x{a}_s)=\psi (a_r)$. Since $e^{-s}ds=dy\mathrm{/}{y}^2$, the integral reduces to ${\int }_{y>0}\psi (a_{r(x,y)})dxdy\mathrm{/}{y}^2$.

The inverse Cayley map is $z=i(1+w)\mathrm{/}(1-w)$; it satisfies $\mathrm{Im}z=(1-\mathrm{\mid }w{\mathrm{\mid }}^2)\mathrm{/}\mathrm{\mid }1-w{\mathrm{\mid }}^2$ and $\mathrm{\mid }dz\mathrm{/}dw{\mathrm{\mid }}^2=4\mathrm{/}\mathrm{\mid }1-w{\mathrm{\mid }}^4$. Hence $dxdy\mathrm{/}{y}^2=4dudv\mathrm{/}(1-\mathrm{\mid }w{\mathrm{\mid }}^2{)}^2$ for $w=u+iv$. On the disk with the nonnegative real radius removed, $w=\rho e^{i\theta }$ is a $C^1$ diffeomorphism from $(0,1)\times (0,2\pi )$ with Jacobian $\rho$; the omitted radius is a countable union of compact subsegments on which the weight is bounded, and the origin is a singleton, so both have zero weighted measure. By [F3]–[F4], and with $r=2\mathrm{artanh}\rho$ so $d\rho =(1-{\rho }^2)dr\mathrm{/}2$, one has $4\rho d\rho d\theta \mathrm{/}(1-{\rho }^2{)}^2=2\rho drd\theta \mathrm{/}(1-{\rho }^2)=\sinh (r)drd\theta$. The integrand is independent of $\theta$, whose interval has length $2\pi$. This proves the extended radial identity, including the zero function and the endpoint $r=0$, which contributes no atom.

Apply the nonnegative identity to $\mathrm{\mid }\phi {\mathrm{\mid }}^p$ for $p=1,2$ to obtain both extended $L^p$ formulas and their finiteness criteria by [F5]. When $\phi \in L^1$, its real and imaginary positive and negative parts are continuous nonnegative $K$-bi-invariant functions; applying the identity to those four parts and recombining gives the absolutely convergent formula for $\phi$. For the character-equivariant case, $\mathrm{\mid }{\chi }_i(k)\mathrm{\mid }=1$, hence $\mathrm{\mid }\phi (k_{1}g{k}_2)\mathrm{\mid }=\mathrm{\mid }\phi (g)\mathrm{\mid }$, and the same $L^p$ conclusions follow.