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  1. One can show explicitly and easily that the function $G(x,y) = \frac 1 2 |x-y|$ is a positive Green function for the Laplacian $\frac {\mathrm d ^2} {\mathrm d x ^2}$ on $\mathbb R$ (endowed with the usual Riemannian structure).

  2. At the very beginning of "Green's functions, harmonic functions, and volume comparison" by P. Li and L.-F. Tam (J. Differential Geom. 41(2): 277-318 (1995)) one can read the following:
    "A sharp necessary condition was later proved by Varopoulos in [4], which states that if a complete manifold $M$ has a positive Green's function, then $$\int _1 ^\infty \frac t {V_p (t)} \ \mathrm d t < \infty \ ,$$ where $V_p (t)$ is the volume of the geodesic ball of radius $t$ with center at $p$."
    The reference [4], which I do not have access to, is: N. Varopoulos, "Potential theory and diffusion on Riemannian manifolds", Conf. on Harmonic Analysis in Honor of Antoni Zygmund, Vols. I, II, Wadsworth Math. Ser., Wadsworth, Belmont, CA, 1983, 821-837.
    And this is my first problem, because in $\mathbb R$ it is obvious that $V_p (t) = 2t$ and the above integral is $\infty$:

were there any constraints on $\dim M$ in Varopoulos' article, that Li and Tam have forgotten to mention?

  1. Furthermore, it is a deep geometric fact that, on many Riemannian manifolds, the (minimal, positive) Green function of the Laplace-Beltrami operator is the integral in time of the heat kernel, i.e. $$G(x,y) = \int _0 ^\infty h(t,x,y) \ \mathrm d t \ ,$$ where $h$ is the heat kernel on $M$.
    This, though, does not happen on $\mathbb R$, because $$\int _0 ^\infty \frac 1 {\sqrt{4 \pi t}} \mathrm e ^{- \frac {|x-y|^2} {4t}} \ \mathrm d t = \infty \ .$$

Can one recognize those Riemannian manifolds that admit a positive Green function which is not equal to the integral in time of the heat kernel?

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  • $\begingroup$ (1), (2): This is just a guess on my part, but perhaps these authors consider $-\Delta$, and then (1) ceases to be a counterexample (since now $G=-(1/2)|x-y|$). $\endgroup$ Commented Nov 27, 2023 at 15:04
  • $\begingroup$ @ChristianRemling: The cited articles and my own question explicitly ask about positive Green functions. $\endgroup$
    – Alex M.
    Commented Nov 27, 2023 at 15:14
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    $\begingroup$ I was able to look at a later paper on this topic, and indeed the Green function is defined as satisfying $\Delta G=-\delta$. $\endgroup$ Commented Nov 27, 2023 at 15:34
  • $\begingroup$ @MichaelRenardy: This discussion about the sign misses the point of my question, which is the fact that both those integrals are infinite. This has nothing to do with the sign. $\endgroup$
    – Alex M.
    Commented Nov 27, 2023 at 15:55
  • $\begingroup$ @ChristianRemling: Yes, I understand now what you are saying. $\endgroup$
    – Alex M.
    Commented Nov 27, 2023 at 20:00

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