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I am reading the proof of the proposition $3.6$ of Riemannian Geometry by Do Carmo, which can be read here, and I would like to understand why the following statement is true:

If $l(c) = l(\gamma)$, then $\left| \frac{\partial f}{\partial t} \right| = 0$, that is, $v(t) = \ \text{const.}$, and $|r'(t)| = r'(t) > 0$.

Specifically, my doubts are

  1. Why $l(c) = l(\gamma)$ imply $\left| \frac{\partial f}{\partial t} \right| = 0$?

$l(c) = l(\gamma)$ imply $c'(t) = r'(t)$?

  1. Why $\left| \frac{\partial f}{\partial t} \right| = 0$ imply $v(t) = \ \text{const.}$, and $|r'(t)| = r'(t) > 0$?

Thanks in advance!

George
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1 Answers1

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First, I would like to make a short comment. Do Carmo's book is really great and covers a large part of basics in Riemannian geometry. But I personnally do not really like all this part concerning the exponential map, normal coordinates etc. It is always a pain to me to understand the author's notation and ideas. I would highly recommand to look at the corresponding parts in J. Lee's Riemannian manifolds: an introduction to curvature, or S. Gallot, D. Hulin and J. Lafontaine's Riemannian Geometry, which provide (in my opinion) a bit clearer proofs.

Note that $ |c'|^2 = |r'|^2 + |\partial_tf|^2 \geqslant |r'|^2, $ which yields $$ |c'| \geqslant |r'|. $$ Integrating this between $\varepsilon>0$ and $1$ together with the triangular inequality yields $$ l(c) = \int_0^1|c'| \geqslant \int_{\varepsilon}^1|c'| \geqslant \int_{\varepsilon}^{1} |r'| \underset{(1)}{\geqslant} \left| \int_{\varepsilon}^1 r' \right| = |r(1)-r(\varepsilon)| = r(1)-r(\varepsilon) \to_{\varepsilon \to 0} r(1) =l (\gamma), $$ and therefore, we have $l(c) \geqslant l(\gamma)$. Suppose this last inequality is an equality.

  1. This implies that $|c'|-|r'|$ is a non-negative function with vanishing integral, and it follows that $|c'|-|r'|=0$ and thus $|c'| = |r'|$. Equality $|c'|^2 = |r'|^2 + |\partial_tf|^2$ gives $|\partial_tf|^2 = 0$.

  2. First: note that $\partial_tf = \mathrm{d}\exp_p\left(r(t)v(t)\right)\cdot \left(r(t) v'(t)\right)$. Moreover, the exponential map is supposed to be a diffeomorphism (we are considering a normal geodesic ball!), hence $\mathrm{d}\exp_p\left(r(t)v(t)\right)$ is a linear isomorphism. It follows from $|\partial_t f|=0$ that $r(t)v'(t) = 0$, and therefore, $v'(t) = 0$ (recall that $r>0$). It follows that $v(t) = v_0$. Second: in inequality $(1)$, there is equality in the triangle inequality (when $\varepsilon \to 0$): $$ \left|\int_{0}^1 r' \right| = \int_0^1|r'|, $$ which implies (by an integration-theory result) that $r'$ has constant sign. As $r(\varepsilon) \to 0$ and $r(1) = l(\gamma)>0$, it follows that $r(1) - r(\varepsilon) >0$ for $\varepsilon >0$ small enough.By the MVT, there is $t \in (\varepsilon,1)$ with $r'(t) = \frac{r(1)-r(\varepsilon)}{1-\varepsilon} >0$. The result follows.

Didier
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  • thanks for the answer, but what is this integration-theory result that you mentioned? – George Jul 20 '21 at 13:06
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    It is the equality case in the triangle inequality for integrals, namely: if $f \colon \Omega \to \mathbb{C}$ is integrable such that $\left| \int_{\Omega} f \right| = \int_{\Omega} |f|$, then there exists $\theta \in \mathbb{R}$ such that $f = e^{i\theta}|f|$. In fact here I only use the real case, which is simpler to show – Didier Jul 20 '21 at 13:14
  • I think $\left|\int_{0}^1 r' \right| = \int_0^1|r'|$ does not mean $r'$ has constant sign, since it can be $r'(t)=0$ for some $t$. If so, we can't get $r'(t)>0$. How should I understand it? – Enhao Lan Aug 03 '21 at 11:53
  • By constant sign, I meant $r' \geqslant 0$ everywhere, or $r' \leqslant 0$ everywhere, which follows from the mentioned result of integration. – Didier Aug 03 '21 at 12:07
  • @Didier Thanks, but I still can't get $r'(t)>0$. I agree that there exists $t_0$ such that $r'(t_0)>0$ by MVT. Besides, I also agree $r'\ge 0$. But in my view, the two things do not mean $r'(t)>0$. In fact, I feel $r'$ can be zero at some $t_0$, since in this case, the images of $c$ and $\gamma$ can be same. But if so, the proof of do Carmo is wrong. – Enhao Lan Aug 03 '21 at 12:09
  • That is why I mentioned at first in my answer two other references that are clearer to me. Do Carmo's proof is not "wrong", but it is too short in my opinion – Didier Aug 03 '21 at 12:14
  • @Didier Very thanks your hint, in Lee's book, I find the problem of do Carmo's proof. The do Carmo does not assume $c(t)$ is parametrized by arc length, so, there is not $|\frac{dc}{dt}|>0$. Beacuse of this, we have $r'(t)= |r'(t)| = |\frac{dc}{dt}|>0$. Thanks again. – Enhao Lan Aug 03 '21 at 12:35