Rankin–Selberg periods for spherical principal series

Abstract

By the unfolding method, Rankin–Selberg L-functions for ${{\,\mathrm{GL}\,}}(n)\times {{\,\mathrm{GL}\,}}(n^{\prime })$ can be expressed in terms of period integrals. These period integrals actually define invariant forms on tensor products of the relevant automorphic representations. By the multiplicity-one theorems due to Sun–Zhu and Chen–Sun such invariant forms are unique up to scalar multiples and can therefore be related to invariant forms on equivalent principal series representations. We construct meromorphic families of such invariant forms for spherical principal series representations of ${{\,\mathrm{GL}\,}}(n,{\mathbb {R}})$ and conjecture that their special values at the spherical vectors agree in absolute value with the archimedean local L-factors of the corresponding L-functions. We verify this conjecture in several cases. This work can be viewed as the first of two steps in a technique due to Bernstein–Reznikov for estimating L-functions using their period integral expressions.

Appendix A. Integral formulas

We collect some integral formulas for the hypergeometric function and Meijer’s G-function.

1.1 A.1. Hypergeometric function

By [14, equation 7.512 (10)] we have for ${\text {Re}}\gamma ,{\text {Re}}(\alpha -\gamma +\sigma ),{\text {Re}}(\beta -\gamma +\sigma )>0$ and $|\arg z|<\frac{\pi }{2}$:

$$\begin{aligned}&\int _0^\infty x^{\gamma -1}(x+z)^{-\sigma }{_2F_1}(\alpha ,\beta ;\gamma ;-x)\,dx = \frac{\Gamma (\gamma )\Gamma (\alpha -\gamma +\sigma )\Gamma (\beta -\gamma +\sigma )}{\Gamma (\sigma )\Gamma (\alpha +\beta -\gamma +\sigma )}\nonumber \\&\quad \times {_2F_1}(\alpha -\gamma +\sigma ,\beta -\gamma +\sigma ;\alpha +\beta -\gamma +\sigma ;1-z). \end{aligned}$$

(A.1)

The following integral formula holds for $|u|>|\beta |$ and $0<{\text {Re}}\mu <-2{\text {Re}}\nu$ (see [14, equation 3.254 (2)] for $\lambda =0$):

$$\begin{aligned} \int _u^\infty (x-u)^{\mu -1}(x^2+\beta ^2)^\nu \,dx= & {} B(\mu ,-\mu -2\nu )u^{\mu +2\nu }\nonumber \\&\times {_2F_1}\left( -\frac{\mu }{2}-\nu ,\frac{1-\mu }{2}-\nu ;\frac{1}{2}-\nu ;-\tfrac{\beta ^2}{u^2}\right) .\nonumber \\ \end{aligned}$$

(A.2)

Using the relation (see [1, Theorem 2.3.2])

$$\begin{aligned} {_2F_1}(a,b;c;x)= & {} \frac{\Gamma (c)\Gamma (b-a)}{\Gamma (c-a)\Gamma (b)}(-x)^{-a}{_2F_1}(a,a-c+1;a-b+1;x^{-1})\\&+\frac{\Gamma (c)\Gamma (a-b)}{\Gamma (c-b)\Gamma (a)}(-x)^{-b}{_2F_1}(b,b-c+1;b-a+1;x^{-1}) \end{aligned}$$

the integral formula in (A.2) can be extended to $u\in {\mathbb {R}}$, $\beta >0$, by analytic continuation:

$$\begin{aligned}&\int _u^\infty (x-u)^{\mu -1}(x^2+\beta ^2)^\nu \,dx = \frac{\Gamma \left( \frac{\mu }{2}\right) \Gamma \left( -\frac{\mu }{2}-\nu \right) }{2\Gamma (-\nu )}\beta ^{\mu +2\nu }{_2F_1}\left( -\tfrac{\mu }{2}-\nu ,\tfrac{1-\mu }{2};\tfrac{1}{2};-\tfrac{u^2}{\beta ^2}\right) \nonumber \\&\quad -\frac{\Gamma \left( \frac{\mu +1}{2}\right) \Gamma \left( \frac{1-\mu }{2}-\nu \right) }{\Gamma (-\nu )}\beta ^{\mu +2\nu -1}u\cdot {_2F_1}\left( \tfrac{1-\mu }{2}-\nu ,\tfrac{2-\mu }{2};\tfrac{3}{2};-\tfrac{u^2}{\beta ^2}\right) . \end{aligned}$$

(A.3)

For ${\text {Re}}c>{\text {Re}}b>0$ the following integral representation holds (see [1, Theorem 2.2.1]):

$$\begin{aligned} {_2F_1}(a,b;c;x) = \frac{\Gamma (c)}{\Gamma (b)\Gamma (c-b)}\int _0^1 t^{b-1}(1-t)^{c-b-1}(1-xt)^{-a}\,dt. \end{aligned}$$

(A.4)

The Euler integral representation holds for ${\text {Re}}(\gamma -\beta ),{\text {Re}}\beta >0$ (see [1, equation (2.3.17)]):

$$\begin{aligned} {_2F_1}(\alpha ,\beta ;\gamma ;1-x) = \frac{\Gamma (\gamma )}{\Gamma (\gamma -\beta )\Gamma (\beta )}\int _0^\infty t^{\beta -1}(1+t)^{\alpha -\gamma }(1+xt)^{-\alpha }\,dt.\nonumber \\ \end{aligned}$$

(A.5)

The following transformation formula holds (see [1, Theorem 2.2.5]):

$$\begin{aligned} {_2F_1}(a,b;c;x) = (1-x)^{c-a-b}{_2F_1}(c-a,c-b;c;x). \end{aligned}$$

(A.6)

The following integral formula holds for $\alpha ,\beta >0$ and $0<{\text {Re}}\lambda <2{\text {Re}}(\mu +\nu )$ (see [14, 3.259 (3)]):

$$\begin{aligned}&\int _0^\infty x^{\lambda -1}(1+\alpha x^2)^{-\mu }(1+\beta x^2)^{-\nu }\,dx\nonumber \\&\quad = \frac{1}{2}\alpha ^{-\frac{\lambda }{2}}B\left( \frac{\lambda }{2},\mu +\nu -\frac{\lambda }{2}\right) {_2F_1}\left( \nu ,\frac{\lambda }{2};\mu +\nu ;1-\frac{\beta }{\alpha }\right) . \end{aligned}$$

(A.7)

The following integral formula holds for ${\text {Re}}\mu ,{\text {Re}}\nu >0$ (see [14, 7.512 (12)]):

$$\begin{aligned}&\int _0^1 t^{\mu -1}(1-t)^{\nu -1}{_pF_q}(a_1,\ldots ,a_p;b_1,\ldots ,b_q;tx)\,dt\nonumber \\&\quad =\frac{\Gamma (\mu )\Gamma (\nu )}{\Gamma (\mu +\nu )}{_{p+1}F_{q+1}}(a_1,\ldots ,a_p,\mu ;b_1,\ldots ,b_q,\mu +\nu ;x). \end{aligned}$$

(A.8)

Lemma A.1

For ${\text {Re}}\rho ,{\text {Re}}(\alpha -\sigma -\rho +1),{\text {Re}}(\beta -\sigma -\rho +1)>0$ and $u>0$ we have

$$\begin{aligned}&\int _0^\infty x^{\rho -1}(x+u)^{\sigma -1}{_2F_1}(\alpha ,\beta ;\gamma ;-x)\,dx \\&\quad = \frac{\Gamma (\rho )\Gamma (1-\sigma -\rho )}{\Gamma (1-\sigma )}u^{\rho +\sigma -1}{_3F_2}(\alpha ,\beta ,\rho ;\gamma ,\sigma +\rho ;u)\\&\quad +\frac{\Gamma (\gamma )\Gamma (\alpha -\sigma -\rho +1)\Gamma (\beta -\sigma -\rho +1)\Gamma (\sigma +\rho -1)}{\Gamma (\beta )\Gamma (\alpha )\Gamma (\gamma -\sigma -\rho +1)}\\&\quad \times {_3F_2}(\alpha -\sigma -\rho +1,\beta -\sigma -\rho +1,1-\sigma ;\gamma -\sigma -\rho +1,2-\sigma -\rho ;u). \end{aligned}$$

Note that the integral in Lemma A.1 is more general than the one in [14, equation 7.512 (10)], which corresponds to $\rho =\gamma$.

Proof

We make use of the integral representation (for ${\text {Re}}c>{\text {Re}}b>0$, see [1, Theorem 2.2.1])

$$\begin{aligned} {_2F_1}(a,b;c;x) = \frac{\Gamma (c)}{\Gamma (b)\Gamma (c-b)}\int _0^1 t^{b-1}(1-t)^{c-b-1}(1-xt)^{-a}\,dx \end{aligned}$$

and the transformation formula (see [1, Theorem 2.3.2])

$$\begin{aligned}&{_2F_1}(a,b;c;x) = \frac{\Gamma (c)\Gamma (c-a-b)}{\Gamma (c-a)\Gamma (c-b)}{_2F_1}(a,b,a+b-c+1;1-x)\\&\quad + \frac{\Gamma (a+b-c)\Gamma (c)}{\Gamma (a)\Gamma (b)}(1-x)^{c-a-b}{_2F_1}(c-a,c-b;c-a-b+1;1-x). \end{aligned}$$

First, substituting $x\mapsto ux$ and using the integral representation we find

$$\begin{aligned}&\int _0^\infty x^{\rho -1}(x+u)^{\sigma -1}{_2F_1}(\alpha ,\beta ;\gamma ;-x)\,dx\\&\quad = \frac{\Gamma (\gamma )}{\Gamma (\gamma -\beta )\Gamma (\beta )}u^{\rho +\sigma -1}\int _0^\infty \int _0^1 t^{\beta -1}(1-t)^{\gamma -\beta -1}x^{\rho -1}(1+x)^{\sigma -1}(1+tux)^{-\alpha }\,dt\,dx. \end{aligned}$$

Next, we compute the integral over x with (A.5):

$$\begin{aligned} = \frac{\Gamma (\gamma )\Gamma (\alpha -\sigma -\rho +1)\Gamma (\rho )}{\Gamma (\gamma -\beta )\Gamma (\beta )\Gamma (\alpha -\sigma +1)}u^{\rho +\sigma -1}\int _0^1 t^{\beta -1}(1-t)^{\gamma -\beta -1}{_2F_1}(\alpha ,\rho ;\alpha -\sigma +1;1-tu)\,dt. \end{aligned}$$

Apply the transformation formula:

$$\begin{aligned}&= \frac{\Gamma (\gamma )\Gamma (\rho )\Gamma (1-\sigma -\rho )}{\Gamma (\gamma -\beta )\Gamma (\beta )\Gamma (1-\sigma )}u^{\rho +\sigma -1}\int _0^1 t^{\beta -1}(1-t)^{\gamma -\beta -1}{_2F_1}(\alpha ,\rho ;\sigma +\rho ;tu)\,dt\\&\quad +\frac{\Gamma (\gamma )\Gamma (\alpha -\sigma -\rho +1)\Gamma (\sigma +\rho -1)}{\Gamma (\gamma -\beta )\Gamma (\beta )\Gamma (\alpha )}\\&\quad \times \int _0^1 t^{\beta -\sigma -\rho }(1-t)^{\gamma -\beta -1}{_2F_1}(1-\sigma ,\alpha -\sigma -\rho +1;2-\sigma -\rho ;tu)\,dt \end{aligned}$$

and finally (A.8):

$$\begin{aligned}&= \frac{\Gamma (\rho )\Gamma (1-\sigma -\rho )}{\Gamma (1-\sigma )}u^{\rho +\sigma -1}{_3F_2}(\alpha ,\beta ,\rho ;\gamma ,\sigma +\rho ;u)\\&\quad +\frac{\Gamma (\gamma )\Gamma (\alpha -\sigma -\rho +1)\Gamma (\beta -\sigma -\rho +1)\Gamma (\sigma +\rho -1)}{\Gamma (\beta )\Gamma (\alpha )\Gamma (\gamma -\sigma -\rho +1)}\\&\quad \times {_3F_2}(\alpha -\sigma -\rho +1,\beta -\sigma -\rho +1,1-\sigma ;\gamma -\sigma -\rho +1,2-\sigma -\rho ;u). \end{aligned}$$

$\square$

For the special value of the generalized hypergeometric function ${_3F_2}$ at $x=1$ we have the following transformation formula, which holds for ${\text {Re}}(d+e-a-b-c),{\text {Re}}(c-d+1)>0$ (see [1, Theorem 2.4.4]):

$$\begin{aligned}&{_3F_2}(a,b,c;d,e;1) = \frac{\Gamma (d)\Gamma (d-a-b)}{\Gamma (d-a)\Gamma (d-b)}{_3F_2}(a,b,e-c;e,a+b-d+1;1)\nonumber \\&\quad +\frac{\Gamma (d)\Gamma (e)\Gamma (d+e-a-b-c)\Gamma (a+b-d)}{\Gamma (a)\Gamma (b)\Gamma (d+e-a-b)\Gamma (e-c)}\nonumber \\&\quad \times {_3F_2}(d-a,d-b,d+e-a-b-c;d+e-a-b,d-a-b+1;1) \end{aligned}$$

(A.9)

For ${\text {Re}}(\gamma -\alpha -\beta )>0$ the special value of ${_2F_1}$ at $x=1$ is given by (see [1, Theorem 2.2.2])

$$\begin{aligned} {_2F_1}(\alpha ,\beta ;\gamma ;1) = \frac{\Gamma (\gamma )\Gamma (\gamma -\alpha -\beta )}{\Gamma (\gamma -\alpha )\Gamma (\gamma -\beta )}. \end{aligned}$$

(A.10)

1.2 Meijer’s G-function

For ${\text {Re}}(a)<1$ and ${\text {Re}}(b_1),{\text {Re}}(b_2),{\text {Re}}(b_3)>0$:

$$\begin{aligned} G^{30}_{13}\Big (z\Big |\begin{array}{l}a_1\\ b_1,b_2,b_3\end{array}\Big ) = \frac{1}{2\pi \sqrt{-1}} \int _{\mathbb {R}}\frac{\Gamma (b_1+is)\Gamma (b_2+is)\Gamma (b_3+is)}{\Gamma (a_1+is)}z^{-is} \,ds. \end{aligned}$$

We have the following integral representation for $-1<{\text {Re}}\nu <2\max ({\text {Re}}\alpha ,{\text {Re}}\beta )-\frac{3}{2}$, $y>0$ (see [10, 8.17 (5)]):

$$\begin{aligned} \int _0^\infty \cos (xy){_2F_1}(\alpha ,\beta ;\gamma ;-x^2)\,dx = \frac{\Gamma (\frac{1}{2})\Gamma (\gamma )}{\Gamma (\alpha )\Gamma (\beta )}y^{-1}G^{30}_{13}\Big (\big (\tfrac{y}{2}\big )^2\Big |\begin{array}{l}\gamma \\ \frac{1}{2},\alpha ,\beta \end{array}\Big ).\nonumber \\ \end{aligned}$$

(A.11)

Lemma A.2

$$\begin{aligned} \int _1^\infty K_\nu (ax)(x^2-1)^\lambda x^\mu \,dx = 2^{-\nu -1}\Gamma (\lambda +1)a^{\nu -1}G^{30}_{13}\Big (\big (\tfrac{a}{2}\big )^2\Big |\begin{array}{l}-\frac{\mu +\nu -2}{2}\\ \frac{1}{2},\tfrac{1}{2}-\nu ,-\lambda -\tfrac{\mu +\nu }{2}\end{array}\Big ). \end{aligned}$$

Proof

By [14, 3.771 (2)] we have for $a>0$, ${\text {Re}}x>0$ and ${\text {Re}}\nu <\frac{1}{2}$

$$\begin{aligned} \int _0^\infty (x^2+y^2)^{\nu -\frac{1}{2}}\cos (ay)\,dy = \frac{1}{\sqrt{\pi }}\Big (\frac{2x}{a}\Big )^\nu \cos (\pi \nu )\Gamma (\nu +\frac{1}{2})K_\nu (ax). \end{aligned}$$

Multiplying with $(x^2-1)^\lambda x^{\mu -\nu }$ and integrating over $(1,\infty )$ gives

$$\begin{aligned}&\int _1^\infty K_\nu (ax)(x^2-1)^\lambda x^\mu \,dx = \frac{2^{-\nu }\sqrt{\pi }}{\cos (\pi \nu )\Gamma (\nu +\frac{1}{2})}a^\nu \\&\quad \times \int _1^\infty (x^2-1)^\lambda x^{\mu -\nu }\int _0^\infty (x^2+y^2)^{\nu -\frac{1}{2}}\cos (ay)\,dy\,dx. \end{aligned}$$

Interchanging the order of integration and substituting $x=t^{-\frac{1}{2}}$ gives

$$\begin{aligned} = \frac{2^{-\nu -1}\sqrt{\pi }}{\cos (\pi \nu )\Gamma (\nu +\frac{1}{2})}a^\nu \int _0^\infty \cos (ay)\int _0^1 t^{-\frac{\nu +\mu +2}{2}-\lambda }(1-t)^\lambda (1+ty^2)^{\nu -\frac{1}{2}}\,dt\,dy. \end{aligned}$$

The inner integral can be computed in terms of the hypergeometric function using (A.4):

$$\begin{aligned}&= \frac{2^{-\nu -1}\sqrt{\pi }\Gamma (\lambda +1)\Gamma (-\lambda -\frac{\mu +\nu }{2})}{\cos (\pi \nu )\Gamma (\nu +\frac{1}{2})\Gamma (-\frac{\mu +\nu -2}{2})}a^\nu \int _0^\infty \cos (ay){_2F_1}\left( \tfrac{1}{2}-\nu ,-\lambda -\tfrac{\mu +\nu }{2};\right. \\&\quad \left. -\tfrac{\mu +\nu -2}{2};-y^2\right) \,dy. \end{aligned}$$

By (A.11) and Euler’s reflection formula this equals the claimed formula. $\square$