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In the Laplacian comparision theorem by Yau and Schoen, the proof assume a $n$ dimensional space form $N$ with sectional curvature $-k^{2}$. And we have a separate $M$ with $Ric<-(n-1)k^{2}$. But after reading the proof several times and scratching my head for hours, I do not understand what is the role of $N$ at here. I tried to find online sources on the same theorem, and the ones I found did not state $N$ at all. Further in the proof I noticed the authors assumed $M$ has curvature $-k^{2}$ instead. Is there a typo here or somewhere? I am really confused. This is in page 5 of Yau and Schoen (sorry I do not know where to find an online version).

Their original theorem statement is the following:

Let $M$ be an $n$-dimensional complete Riemannian manifold with $Ric(M)\ge -(n-1)k^{2},(k\ge 0)$. Let $N$ be the $n$-dimensional simply connected space of constant sectional curvature $-k^{2}$. Let $p_{M}$ and $p_{N}$ be distance functions with respect to some fixed point $x$ on $M$ and $N$ respectively. If $x\in M$ and $p_{M}$ is differentiable at $x$, then for any $y\in N$ with $p_{N}(y)=p_{M}(x)$ we have $$\Delta p_{M}(x)\le \Delta p_{N}(y)$$

But reading their proof I could not find where did they use the assumption $Ric(M)\ge -(n-1)k^{2}$ or $N$ is simply connected at all.

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    $\begingroup$ To estimate the Laplacian, you will need to play with Jacoby or Riccati equation (depending on your religion). That is the place where curvature comes into game. Simply connectedness is not needed, but is natural to assume that is to compare with the best possible space. $\endgroup$
    – Anton Petrunin
    Commented May 1, 2013 at 4:14
  • $\begingroup$ @Anton Petrunin: Thanks for the help. But this does not address the issue of $M$ having curvature $-k^{2}$ in proof. I do not know if this is a typo or something. $\endgroup$
    – Bombyx mori
    Commented May 1, 2013 at 4:51
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    $\begingroup$ Try to work out the proof yourself step-by-step, and you'll see how curvature is used. As Anton says, it appears because the Hessian of each $p$ satisfies a matrix Riccati equation, which allows you to show that one is greater than the other. Or you can write the Hessian in terms of a frame of Jacobi fields and their radial derivatives and use the Sturm comparison theorem. If you have trouble reading Schoen and Yau, try other sources. The book by Gallot, Hulin, Lafontaine probably has it all. I also like papers by Karcher, including ones with Heintze and with Jost. $\endgroup$
    – Deane Yang
    Commented May 1, 2013 at 13:48
  • $\begingroup$ @Deane Yang: I typed quite a few pages of latex trying to figure out the proof in detail. But I think the $M$ in page 5 should really be $N$ instead. In the theorem statement $M$ is not assumed to have constant sectional curvature, and that is essential in producing the final result. Here is my problem: I cannot use Hessian comparision theorem in the Laplacian comparison theorem with the stated curvature requirements, maybe I am confused with something really elementary. $\endgroup$
    – Bombyx mori
    Commented May 1, 2013 at 14:07
  • $\begingroup$ I can't comment on what typos the paper might have. If you understand the overall approach of the proof (compare the solution of the matrix Jacobi or Riccati equation with variable Ricci curvature against a corresponding solution with constant sectional curvature), then, perhaps with some struggle, you should be able to figure out what should be $M$ and what should be $N$ and write out your own proof, so you don't have to rely on the one by Schoen and Yau. $\endgroup$
    – Deane Yang
    Commented May 1, 2013 at 14:22

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The form $-k^2$ is just a way of saying that the sectional curvature is nonpositive. There is a nice, modern and detailed discussion of this Laplacian comparison result in P. Petersen, "Riemannian Geometry" (second edition), Springer-Verlag, 2006. The result you want is the first inequality in Lemma 34, page 268. There, one only assumes a lower bound on the Ricci curvature of the form $Ric(M)\geq (n-1)k$, with $k\in\mathbb{R}$, and no simple connectedness assumption is made, as Anton Petrunin commented above. The proof is entirely based on Proposition 39, pages 266-267, which relates the volume form of $(M,g)$ to the Laplacian of $p_M$. The needed inequality in that Proposition is obtained from direct calculations and the Cauchy-Schwarz inequality for the Hilbert-Schmidt scalar product of matrices.

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