6
$\begingroup$

I have two questions (somewhat related) regarding local geometry on a SMOOTH, COMPACT Riemannian manifold. I still have a hard time getting a "good" understanding of local geometry.

Question 1:

It is true that there exists $\epsilon>0$ such that for all $r < \epsilon$, there exists $c_g >0$ (indep. of $x\in M$) such that

$$ c_g r^n \leq Vol_g (B_x(r)), \forall x \in M, $$

where $B_x(r)$ is the geodesic ball centered at $x$. In other words, is it true that small geodesic ball are "comparable" to Euclidean balls.

Question 2:

I am trying to compute an integral on a "small" geodesic ball, namely the following:

$$ I = \int_{(exp_{x_0}(B_{0}(R))} \rho(x_0,x)^{2-n} dV_g$$

Using normal coordinates at $x_0$, we should have that

\begin{eqnarray*} I & =& \int_{(exp_{x_0}(B_{0}(R))} \rho(x_0,x)^{2-n} dV_g. \\ & \leq & C_g \int_0^{R} \rho^{2-n} \rho^{n-1} (1 + O(\rho^2)) d\rho \\ \end{eqnarray*}

I really need to get that last inequality but I am very unsure about it. I think it should be true, at least for small enough $R$ (hopefully for $R$ smaller than the $\epsilon$ defined in question 1.

Is that possible? Any feedback would be appreciated.

$\endgroup$

1 Answer 1

10
$\begingroup$

The answer to both questions is 'yes'.

To see this, just consider the exponential map $\exp:TM\to M$ and look at the pullback of the Riemannian volume form $dV$, say $\Omega = \exp^*(dV)$ on $TM$. By the usual expansion in normal coordinates, there will be a smooth function $\phi$ on $TM$ that vanishes to order $2$ along the zero section such that, for each $x\in M$, the pullback of $\Omega$ to $T_xM$, say $\iota_x^*\Omega$ satisfies $\iota_x^*\Omega = (1+\iota_x^*\phi) dV_x$ where $dV_x$ is the ordinary Euclidean volume form on $T_xM$ (considered as a Euclidean vector space). This immediately proves what you want, since you can now use compactness to show that, when $\epsilon$ is sufficiently small, one has $|\phi(v)|\le C|v|^2$ for some constant $C>0$ and all $v\in TM$ with $|v|\le \epsilon$.

$\endgroup$
4
  • $\begingroup$ The compactness of $M$ is not needed here, right? See for example: math.stackexchange.com/questions/9768/… $\endgroup$
    – Shaq155
    Mar 25 at 8:59
  • $\begingroup$ @Shaq155: The compactness is needed for the existence of the constant $c_g$. Consider the case of a surface with an infinitely long spike that gets very thin quickly as you go out to infinity, such as, for example, the usual pseudosphere of revolution in $3$-space. There is no $c_g>0$ making the inequality true in this case. $\endgroup$ Mar 25 at 9:26
  • $\begingroup$ @Shaq155: As I explained in my answer, that's always true and for the reason that I gave: It follows from the formula for the volume form in geodesic normal coordinates. The reason you still can't conclude that $c_g$ exists in the noncompact setting is that the injectivity radius can go to zero, so that $B(x,r)$ is covered many times by the $r$-ball in $T_xM$ by the exponential map. For example, consider the metric $g = dx^2 + e^{-2x} d\theta^2$ on the cylinder $\mathbb{R}\times S^1$ (i.e, $\theta$ is periodic with period $2\pi$). As $x$ increases, the injectivity radius rapidly goes to 0. $\endgroup$ Mar 25 at 9:43
  • $\begingroup$ Ah sorry I missed the point that the constant is independent of $x$. Thank you. $\endgroup$
    – Shaq155
    Mar 25 at 9:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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