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Let $\Delta_{hyp}=\Delta_{hyp,1}=-y^2(\partial_x^2+\partial_y^2)$ be the hyperbolic Laplacian acting on functions of $\mathfrak{h}$ (the Poincare upper half-plane) and consider its resolvent $$ R(s)=(\Delta_{hyp}-s(1-s))^{-1}, $$ where $s$ is in a suitable open subset of $\mathbf{C}$. It is known that $R(s)$ is an integral operator which has a kernel which equals to: $$ G_s(z,w):=-\frac{\Gamma(s)^2}{4\pi\cdot\Gamma(2s)}\cdot(1-\tanh^2 \frac{r}{2}))^s\cdot {}_{2}F_1\left(s,s,2s;1-\tanh^2 \frac{r}{2}\right), $$ where $r:=d_{hyp}(z,w)$, $z=x+iy,w\in\mathfrak{h}$, and $d_{hyp}$ is the hyperbolic distance. Here one may think of $w$ as the kernel variable over which one integrates. In particular, if we set $s=1$ in $G_s(z,w)$, then we obtain the classical hyperbolic Green function on $\mathfrak{h}$ which reduces in this case to $-\frac{1}{2\pi}\cdot\log |\frac{z-w}{z-\overline{w}}|$. Since the operator $\Delta_{hyp}$ is invariant under the group of hyperbolic rotations (isomorphic to $SO(2)$) it follows that the kernel of the resolvent (for a fixed $s$) must satisfy an ODE of order 2 in the variable $r$, where $r$ corresponds to the hyperbolic radius. The fact that we are also looking at a function which has a logarithmic singularity at $z=w$ (so it should behave like $-\frac{1}{2\pi}\log r$ when $r$ is small) singles out a special line in the two dimensional solution space of this ODE.

Replace now $\mathfrak{h}$ by $\mathfrak{h}^2=\mathfrak{h}\times \mathfrak{h}$ (more generally we could consider say $g$ copies of $\mathfrak{h}$) with its product metric and let $\Delta_{hyp,2}$ be the corresponding Laplacian acting on $\mathfrak{h}^2$.

Question: Do we have an explicit expression for the kernel of the resolvent of $\Delta_{hyp,2}$ on $\mathfrak{h}^2$ ?

If we try to follow the same strategy as in the case where $g=1$, I guess that the invariance of the Laplacian under $SO(2)\times SO(2)$ would reduce the problem to solving a certain PDE in TWO variables, namely $r_1$ and $r_2$, the two radii corresponding to each copy of $\mathfrak{h}$. I don't know much about PDE's and in general they seem to be underdetermined which frightens me a little bit... Any help or advice would be more than appreciated.

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    $\begingroup$ I would rather start with the heat kernel: an explicit expression is known for the hyperbolic space (see, for example, here), and the heat kernel for the product of two spaces is just the product of the corresponding heat kernels. Now try to integrate $e^{-\lambda t}$ times the heat kernel to get the resolvent kernel of $(\lambda-L)^{-1}$. $\endgroup$ – Mateusz Kwaśnicki Oct 1 '17 at 18:38
  • $\begingroup$ The heat kernel $H(z,z',t)$ of $\mathfrak{h}$ has a continuous part since it is not compact. Using this approach do you at least recover $G_s(z,w)$ when you integrate $H$ with respect to $t$ from $0$ to $\infty$ ? In fact, because of the continuous part I'm not sure if this integral will converge... $\endgroup$ – Hugo Chapdelaine Oct 1 '17 at 23:23
  • $\begingroup$ Not sure if we are speaking about the same thing: do you mean the continuous part of the spectrum? The expression for the resolvent in terms of the heat kernel works for any strongly cont. semigroup. The heat kernel is given by (1.7) in the linked article, integration with respect to $e^{-\lambda t}dt$ gives an integral expression for the resolvent kernel: $2^{-3/2}\pi^{-1}\int_\rho^\infty e^{-\sqrt{4\lambda+1}s}(\cosh s-\cosh \rho)^{-1/2}ds$. For the bi-variate case, one gets an expression as a double integral that involves the Bessel function $K_2$ (if I typed it correctly to Mathematica). $\endgroup$ – Mateusz Kwaśnicki Oct 2 '17 at 8:46
  • $\begingroup$ Dear Mateusz, yes I meant the continuous part of the spectrum of the Laplacian. I knew how to write down the heat kernel when the spectrum is discrete and I naively thought that for non-compact Riemannian manifolds one should also to take into account the continuous part (I imagined a kind of of integral over the continous part). In any case, thanks a lot for the reference. Using the explicit formula in (1.7) I shall try at least to recover the Green function, that will be a first test. $\endgroup$ – Hugo Chapdelaine Oct 2 '17 at 18:58
  • $\begingroup$ Oh, I see. Indeed, there are (generalised) eigenfunction expansions of the heat kernel when the spectrum is continuous, and for the hyperbolic space (or, more generally, symmetric spaces) this reduces to writing the heat kernel as the inverse Fourier transform in the underlying Lie group. But someone has already gone through this, and we already have a nice expression for the heat kernel. Now it remains to do the integrals – and let me stress that I have no idea whether the double integral that I mentioned above can be simplified in any reasonable way. $\endgroup$ – Mateusz Kwaśnicki Oct 2 '17 at 19:26
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The answer depends on what you call "explicit", and the passage from the heat kernel to the Green function is not that straightforward from this point of view. There is a quite efficient uniform asymptotic formula for the Green function on the product of two hyperbolic spaces (of arbitrary dimension) given in Theorem 1 of this paper by Giulini - Woess.

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  • $\begingroup$ Thanks @RW for the reference. I'll look at the paper but do you know if we can dig from their paper an aymptotic for the Green function of $X=\mathfrak{h}\times\mathfrak{h}$ as $r\rightarrow \infty$ where $r=\sqrt{r_1^2+r_2^2}$ stands for the hyperbolic radius on $X$ ? $\endgroup$ – Hugo Chapdelaine May 25 at 17:09
  • $\begingroup$ As far as I remember their asymptotic formula is quite precise as they manage to use it for an identiciation of the Martin boundary (which requires knowing the behaviour of ratios of Green functions). The main diffuculty is in understanding what happens near the walls of Weyl chambers (i.e., when $r_1/r_2$ goes to 0 or to infinity in your notation). $\endgroup$ – R W May 25 at 19:12
  • $\begingroup$ Thanks I'll look it up. $\endgroup$ – Hugo Chapdelaine May 26 at 1:12

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