Let $D$ be an elliptic operator of a compact Riemannian manifold and $G(x_0,x_1)$ the Green's function of $D$. Is $G$ always symmetric in variables $x_0$ and $x_1$, i.e. $G(x_0,x_1)=G(x_1,x_0)$? If yes, is it true that $G(x_0,x_1)$ is a function of the distance $d(x_0,x_1)$ just like the Euclidean case? Any reference will be appreciated.
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5$\begingroup$ Even when the elliptic operator is the Laplacian of a Riemannian metric (for which the Green's function is symmetric), you generally don't have that $G(x_0,x_1)$ is a function of $d(x_0,x_1)$. In fact, this property characterizes the so-called 'harmonic Riemannian manifolds', about which there is an extensive literature, even though they are very rare. $\endgroup$– Robert BryantCommented May 13, 2014 at 15:00
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In general it is not. If $L$ is the operator in question, then $G$ satisfies $L_{x_0}G=0$, $L_{x_1}^*G=0$, where $L^*$ is the adjoint operatior to $L$.
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$\begingroup$ Shouldn't it be $L_{x_{0}}G = \delta_{x_{1}}$? Also a formal argument suggests $L^{*}_{x_{1}}G = \delta_{x_{0}}$, but I'm not sure that's true. If it is, then we ought to have $G^{*}(x_{1},x_{0}) = G(x_{0},x_{1})$, which is the familiar symmetry when $L = L^{*}$. $\endgroup$ Commented Jun 26, 2018 at 19:33
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