8
$\begingroup$

In Aubin's book (nonlinear problems in Riemannian Geometry), starting from p. 106, it is shown that a Green's function of a compact manifold without boundary satisfies

$$G(P,Q) \leq k \rho(P,Q)^{2-n},$$

where $\rho(P,Q)$ denotes the geodesic distance between $P$ and $Q$.

a) I do not understand their proof. In fact, I think they proved that near the diagonal, such estimate holds. However, from what he stated, this looks global (which, for my purposes, is great). Anyone knows that proof and can help me get it?

b) When one considers a manifold with boundary, it seems that the same estimate holds (p. 112). However, their constant $k$ now depends on the distance between P and the boundary of the manifold. Is there any way to avoid that ? Is there any way to get the explicit dependence ?

$\endgroup$
1

2 Answers 2

9
$\begingroup$

Your have two nice texts of Frederic Robert (a descendant of Aubin) about the construction and estimate of Green function

one in french: http://www.iecl.univ-lorraine.fr/~Frederic.Robert/ConstrucGreen.pdf

an other in english where you will find an answer to (b) when assuming Neumann boundary condition: http://www.iecl.univ-lorraine.fr/~Frederic.Robert/NotesGreenNeumannRobert.pdf

Else you can also have a look to appendix A of the book: Blow-up theory for elliptic PDEs in Riemannian geometry, Olivier Druet, Emmanuel Hebey and Frederic Robert, Mathematical Notes, Princeton University Press, Volume 45.

$\endgroup$
9
$\begingroup$

The answer to (a) is easier than you think. As you remark, Aubin shows the estimate holds near the diagonal, so for some $\epsilon$, for all $P$, $Q$ with $\rho(P,Q)<\epsilon$,

$G(P,Q) \leq k \rho(P,Q)^{2-n}.$

However, since the manifold is compact, the continuous function $G$ is bounded (say, by the real number $A$) on the complement of this epsilon-neighbourhood of the diagonal. Thus for all $P$, $Q$

$G(P,Q) \leq max(k, A \rho(P,Q)^{n-2})\rho(P,Q)^{2-n}\leq \max(k, A \ diam(M)^{n-2})\rho(P,Q)^{2-n}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.