2
$\begingroup$

I have some trouble on understanding the domain of the "volume density function" on Riemannian manifold. Putting the volume density function in quote means actually I am working on the function defined on the Riemannian manifold $M$, rather than its tangent space $T_pM$ at $p\in M$. The function is defined by (6.75), page 217 of T.J. Willmore's book "Riemannian Geometry". Anyway I will recall the definition.

Let $(M, g)$ be a connected compact $n$-dimensional Riemannian manifold. For simplicity assume $M$ is orientable and is oriented. Let $p\in M$, and $r_p0$ the injectivity radius at $p$, which is positive. Then $\exp_p:B(0; r_p)\to B(p; r_p)$ is a diffeomorphism.

Let $\omega_g\in\Omega^(M)$ be the Riemannian volume form on $M$, and still denote by $\omega_g$ the Riemannian volume form on $B(p; r_p)$ (obtained by pulling it back). Consider $T_pM$ with the inner product $g_p$ as a Riemannian manifold. The orientation on $M$ induces an orienation on $T_pM$. Denote by $\omega_p$ the Riemannian volume form on $(T_pM, \omega_p)$. Then there exists a unique smooth function $\delta_p:T_pM\to\mathbb{R}$ such that $$\exp_p^*\omega_g=\delta_p\omega_p.$$

The function $\delta_p$ is called the volume density function. Define a function $\theta_p:B(p; r_p)\to\mathbb{R}$ by $$\theta_p:=\delta_p\circ\exp_p^{-1}.$$ Thus we do obtain a function $\theta_p$ defined on an open geodesic ball in $M$.

My question is whether the function $\theta_p$ can be defined on all of $M$. I do not think so, while there exist some papers say yes by citing arguments in page 219-221 of T.J. Willmore's book "Riemannian Geometry". On page 154 of Authur Besse's book "Manifolds all of whose Geodesics are Closed" there seems to be a similar argument.

The related argument in T.J. Willmore's book is the function $\theta:TM\to\mathbb{R}$ defined by (6.91) on page 219, and Proposition (6.6.5) on page 221. The related argument in Authur Besse's book is the last paragraph on page 154. My understanding is these two arguments just say the function $\theta:TM\to\mathbb{R}$ is defined on all of $TM$ whenever $M$ is complete (which, under our assumption, is true). However, these two arguments do not say the function $\theta_p:B(p; r_p)\to\mathbb{R}$ can be extended to all of $M$ (otherwise it implies $\exp:T_pM\to M$ would be a diffeomorphism, which is definitely absurd). Is my understanding correct or did I miss something?

Thank you.

$\endgroup$
  • $\begingroup$ You are correct in everything that you say. Could you please point us to those papers which claim that $\theta_p$ can be defined globally? Indeed it can, but it wouldn't necessarily be smooth anymore (not even continuous). $\endgroup$ – Alex M. Jun 2 '18 at 11:15
  • $\begingroup$ Hi. So how do we define $\theta_p$ on all of $M$? Since I think publicly talking a paper may not be a good idea, may I let you know the papers privately by email? $\endgroup$ – GRR Jun 2 '18 at 11:24
  • 1
    $\begingroup$ It is my policy not to disclose information related to my identity on the internet. On the other hand, aren't those papers public? Haven't they been published? If so, then it is both morally and legally correct to discuss them in the open, what the big secrecy? $\endgroup$ – Alex M. Jun 2 '18 at 11:50
  • $\begingroup$ I certainly respect your policy. There is no secret, and I think I may worry too much. One of the papers is "Non-parametric regression estimation on closed Riemannian manifolds" By Prof. Bruno Pelletier. His argument is on page 59. Thank you. $\endgroup$ – GRR Jun 2 '18 at 11:53
  • $\begingroup$ No, Pelletier seems to have a very original understanding of the problem. Notice that he claims that $\theta_p(q) = \theta_q(p)$ if $p$ and $q$ are close to each other, citing p. 221 of Willmore to back this statement. Alas, there is nothing even remotely related to this on that page (and nearby ones). The only smooth thing that is defined globally (for $M$ complete) is $\theta$: if $v \in T_p (M)$, then $\theta(v) = \delta_p(v) = \theta_p(\gamma_v (\|v\|))$. On the other hand, $\theta_p$ is only defined on the maximal neighbourhood of $p$ on which $\exp_p ^{-1}$ is a diffeomorphism. $\endgroup$ – Alex M. Jun 2 '18 at 12:45

Your Answer

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

Browse other questions tagged or ask your own question.