Area of metric spheres in Riemannian manifolds

I am trying to estimate the integral $\int \mathbb{e} ^{-d(x_0,x)^2} \mathbb{d}x$ on a Riemann manifold $(M,g)$, for some arbitrary fixed $x_0 \in M$ and $d$ the usual distance. The only thing that I can think of is to use some coarea theorem, leading to $\int _0 ^\infty \mathbb{e} ^{-r^2} A(x_0, r) \mathbb{d}r$, where $A(x_0, r)$ is the area of the metric sphere of center $x_0$ and radius $r$. The issue now is to have some estimate of $A$. Surprisingly, even though there are plenty of comparison theorems for the volume of metric balls, I haven't been able to find anything usable on the area of spheres. Therefore, the questions:

• how would you approach this integral (if not by the coarea formula)?
• is it finite? (I assume so)
• is $A$ a polynomial in $r$ (of degree $n-1$)? If so, I would expect it not to have a free term, but what about its leading coefficient?
• can $A$ be computed explicitly in space forms (other than Euclidian, of course)?
• are there comparison theorems for $A$?

Thank you.

• A possible approach: If $\exp_{x_0}$ is a bijection $T_{x_0}M\to M$, then you can turn your integral into some integral over the tangent space (or get an estimate if it is not bijective). You have $d(x_0,\exp_{x_0}v)^2\leq|v|^2$ (with equality if $\exp$ is injective) for $v\in T_{x_0}M$, so you have a Gaussian integral in a Euclidean space with weight coming from pulling the metric to the tangent space. – Joonas Ilmavirta Mar 10 '15 at 20:00
• @JoonasIlmavirta: My manifold is not necessarily complete, so $\mathbb{exp _{x_0} }$ cannot even be defined on the whole $T _{x_0} M$. – Alex M. Mar 10 '15 at 20:15
• @JoonasIlmavirta: The inequality that you suggest would make my integral "greater or equal" than some Gaussian integral, while I'm looking to (hopefully) get "less or equal". – Alex M. Mar 10 '15 at 20:26
• I guess my method would work if $\exp_{x_0}$ is bijective on its maximal domain of definition (which need not be the entire tangent space). If you don't want to assume this much, this approach might be useless. – Joonas Ilmavirta Mar 10 '15 at 20:36
• It seems like things could go quite badly if your manifold is not complete, even in constant curvature. Puncture a round 2-sphere at antipodal points and take its universal cover. Now as $x_0$ heads towards one of the missing points, $A$ grows without bound. – John Harvey Mar 10 '15 at 20:38

The integral might be infinite. Indeed, the function $A(r)$ can grow as fast as you want, but in this case the Ricci curvature in some of the radial direction have to go to $-\infty$ as $r\to\infty$. (Needless to say $A$ can grow faster than any polynomial.)

If Ricci curvature is bounded below then $A(r)$ has at most exponential growth --- this follows from the Bishop--Gromov inequality. In this case you integral has to be finite.

• P.S. if $n=\dim M$ and $\mathrm{Ric}_M\ge -(n-1)$ then Bishop--Gromov inequality says that the function $$r\mapsto A(r)/(\sinh r)^{n-1}$$ is nonincreasing. – Anton Petrunin Mar 11 '15 at 1:36
• Let me come back to your answer, please. You assume lower-boundedness of the Ricci curvature, but not completeness. All the proofs of Bishop-Gromov that I have read require completeness. Do you know of any version of it without this requirement? (I had asked a related question a few days ago, and only now did I remember your answer.) – Alex M. Sep 16 '15 at 20:16
• @AlexM. One needs completeness, without one you can do pretty much anything, in particular there is no Bishop--Gromov. – Anton Petrunin Sep 16 '15 at 21:25

I personally do not think that you can compute that integral explicitly in general just because the geodesic spheres could become quite ugly. $\newcommand{\bx}{\boldsymbol{x}}$ However, when $M$ has finite volume, you can obtain pretty good asymptotic estimates for the integral

$$I(\lambda,\bx_0)=\int_M e^{-\lambda d(\bx_0,\bx)^2} dV_g(x),$$

as $\lambda\to \infty$. To obtain these asymptotic estimates you write $dV_g(\bx)$ in normal coordinates $(t^1,\dotsc, t^n)$ at $\bx_0$ in the form

$$dV_g(x) = \rho(t) dt^1\wedge\cdots \wedge dt^n.$$

The Taylor expansion of $\rho(t)$ at $t=0$ can be described explicitly in terms of geometric invariants at $\bx_0$. The degree 4 Taylor polynomial of $\rho$ at $t=0$ is described in Corollary 9.9 of A. Gray's book Tubes, 2nd Edition, Birhauser, 2006. The Taylor polynomials of degree $>4$ are very complicated.

The integral you are interested could be computed on spaces with rich symmetries, but even in those cases it is not easy.

• Agreed, but normal coordinates cover only a small patch of the whole manifold. What am I going to do with the rest? (And no, I do not hope for an explicit result, just for a "sharp" estimate.) – Alex M. Mar 10 '15 at 20:59
• In any case, this makes me think of the formula of steepest descent in Euclidean spaces. Does a similar thing exist for Riemannian manifolds? With Google I can only find articles about optimizations. – Alex M. Mar 10 '15 at 21:10
• @AlexM. If the injectivity radius at $x_0$ is small then there is very little you can do. However, if you allow for the large parameter $\lambda$, anything that is outside a small geodesic ball is negligible for large $\lambda$. – Liviu Nicolaescu Mar 10 '15 at 21:34

You could integrate the co-area formula by parts, to get: $$\int_0^\infty 2re^{-r^2}V(x_0,r)dr$$ where $V(x_0,r)$ is the volume of the ball at $x_0$ of radius $r$, and then use comparison results for $V$.

• Yes, I realised this after having posted the question. Still, I'm not sure whether this helps: are explicit formulae for the volume of metric balls in space forms known? Or, at least, is $V(x_0, r)$ a polynomial function of $r$? – Alex M. Mar 10 '15 at 23:06
• My point is just that you can use comparison results for $V$ in place of comparison results for $A$. As for evaluating your integral explicitly, the integration by parts I mention is probably not very helpful. – John Pardon Mar 11 '15 at 0:07
• @AlexM.: $V(x_0,r)$ is not generally polynomial in $r$ in non-flat space forms. For example, on the $2$-sphere with Gauss curvature $K\equiv 1$, one finds $$V(x_0,r) = 2\pi \,(1{-}\cos r), \qquad 0\le r\le \pi.$$ The general formula involves polynomials in trig (or hyperbolic trig) functions. – Robert Bryant Mar 12 '15 at 15:20
• @RobertBryant: I am very interested in the general formula for space forms (at least for small balls). Do you know where I could find it? – Alex M. Mar 12 '15 at 21:39
• I don't know a reference off the top of my head, but straightforward calculation shows that $$V(x_0,r) = V_{n-1}\sum_{\ell=0}^{n-1}\frac{(-1)^\ell}{2^{n-1}(\sqrt{-K})^n}{{n{-}1}\choose{\ell}}{{\bigl({e}^{(n-1-2\ell)\sqrt{-K}\,r}-1\bigr)}\over{(n-1-2\ell)}}$$ when $M$ is an $n$-dimensional space form of sectional curvature $K\not=0$ and $V_{n-1}$ is the volume of the unit $(n{-}1)$-sphere in Euclidean $n$-space. (Note that, even when $K>0$, this is a real-valued function with leading order term $V_{n-1}r^n/n$.) Perhaps this formula simplifies? – Robert Bryant Mar 12 '15 at 22:21