A couple of questions about the Lemma 3.5.

Lemma 3.5. (Gauss) : Let $p \in M$ and $v \in T_p M$ such that $\exp_p v$ is defined. Let $w \in T_p M \approx T_v(T_p M)$. Then $$\left\langle (d \exp_p)_v (v), (d \exp_p)_v (w) \right\rangle = \left\langle v, w\right\rangle \;\;\;\;\; (2)$$ Proof. Let $w = w_T + w_N$ is parallel to $v$ and $w_N$ is normal to $v$. Since $d \exp_p$ is linear and, by the definition $\exp_p$, $$\left\langle (d \exp_p)_v (v), (d \exp_p)_v (w_T) \right\rangle = \left\langle v, w_T\right\rangle$$ it suffices to prove (2) for $w = w_N$. It is clear that we can assume $w_N \neq 0$.

The very first bit I don't get:

Since $\exp_p v$ is defined, there exists $\epsilon > 0$ such that $\exp_p u$ is defined for $$ u = tv(s), \; 0 \leq t \leq 1, \; -\epsilon < s < -\epsilon $$ where $v(s)$ is a curve in $T_p M$ with $v(0) = v, v'(0) = w_N$, and $\left| v(s) \right| = const$.

Why there's such an $\epsilon$ that defines $\exp_p$ for $u = tv(s)$?

And continuing

We can, therefore, consider the parametrized surface $$ f : A \to M, \;\;\;\; A = \left\{ (t,s) ; 0 \leq t \leq 1, -\epsilon < s < \epsilon \right\} $$ given by $$f(t,s) = \exp_p tv(s)$$ Observe the curves $t \to f(t,s_o)$ are geodesics. To prove (2) for $w = w_N$, observe first that: $$ \left\langle \frac{\partial f}{\partial s}, \frac{\partial f}{\partial t} \right\rangle(1,0) = \left\langle (d \exp_p)_v (w_N), (d \exp_p)_v (v) \right\rangle = \left\langle v, w\right\rangle \;\;\; (3) $$

Where does (3) come from?

In addition, for all $(t,s)$, we have $$ \frac{\partial}{\partial t}\left\langle \frac{\partial f}{\partial s}, \frac{\partial f}{\partial t} \right\rangle = \left\langle \frac{D}{\partial t}\frac{\partial f}{\partial s}, \frac{\partial f}{\partial t} \right\rangle + \left\langle \frac{\partial f}{\partial s}, \frac{D}{\partial t} \frac{\partial f}{\partial t} \right\rangle $$ The last term of the expression above is zero, since $\frac{\partial f}{\partial t}$ is the tangent vector of a geodesic. From the symmetry of the connection, the first term of the sum is transformed in $$ \left\langle \frac{D}{\partial t} \frac{\partial f}{\partial s}, \frac{\partial f}{\partial t}\right\rangle = \left\langle \frac{D}{\partial s} \frac{\partial f}{\partial t}, \frac{\partial f}{\partial t}\right\rangle = \frac{1}{2} \frac{\partial}{\partial s} \left\langle \frac{\partial f}{\partial t} , \frac{\partial f}{\partial t}\right\rangle = 0 $$ It follows that $\left\langle \frac{\partial f}{\partial s}, \frac{\partial f}{\partial t}\right\rangle$ is independent of t. Since $$ \lim_{t\to 0} \frac{\partial f}{\partial s}(1,0) = \lim_{t\to 0} (d \exp_p)_{tv} t w_N = 0 $$ we conclude $\left\langle \frac{\partial f}{\partial s}, \frac{\partial f}{\partial t}\right\rangle(1,0) = 0$, which together with (3) proves the lemma.

Why is $\left\langle \frac{\partial f}{\partial s}, \frac{\partial f}{\partial t}\right\rangle$ indipendent from $t$? and why is the computed limit $0$?

There are still a couple of questions, but they might get clarified once I understood the once I've asked.

Thank you so much.


For the first thing you do not get:
That $\exp_p$ is defined at $v\in T_pM$, means that there is some $\delta>0$, with $|v|<\delta$ such that $\exp_p$ is defined on $B(0_p,\delta)\subset T_pM$. Just check the domain of the generalized $\exp$ in Proposition 2.7. So, the sphere $S$ of all $w$s with $|w|=|v|$ is contained in the domain of $\exp_p$. Now, since $T_pM$ is a vector space, we identify $T_v(T_pM)$ with $T_pM$ itself. As $w_N\perp v$, $w_N$ is tangent at $v$ to $S$, so that there is a curve $v:(-\epsilon, \epsilon)\rightarrow S$ with $v(0)=v$, $v'(0)=w_N$. Observe that for any $t\in [0,1]$ and $s\in (-\epsilon, \epsilon), |tv(s)|\leq |v|<\delta$, thus $tv(s)$ is in the domain of $\exp_p$. Geometrically, $tv(s)$ is the circular sector portrayed in Fig. 2 of the proof. Its image via the exponential map is a parametrized surface in $M$, and this is important, as it is used later on.
Where does (3) come from?
$f(t,s)=\exp_ptv(s) \implies \frac{\partial f}{\partial s}(1,0)= \frac{d}{ds}|_{s=0}f(1,s)=\frac{d}{ds}|_{s=0}\exp_pv(s)=d(\exp_p)_{v(0)}(v'(0))=d(\exp_p)_{v}(w_N)$.
Do the same for the second term.
Why is $\langle \frac{\partial f}{\partial s}, \frac{\partial f}{\partial t}\rangle$ independent from $t$?
Because you differentiate it along any geodesic $t\to f(t,s)$ and it gives you $0$. Indeed, $\frac{\partial}{\partial t}\langle \frac{\partial f}{\partial s}, \frac{\partial f}{\partial t}\rangle = 0$, as written in your post, right after "Where does (3) come from?".
And why is the computed limit $0$?
$d(\exp_p)_{tv}$ is linear $\forall$ $t$ as the differential of a smooth map. So the $t$ in $tw_N$ gets out.

  • $\begingroup$ Going through your answer one bit at the time. Where does the expression u =tv(s) come from? (talking about the first bit of your answer). I get the existence, but I don't get the expressions that come afterwards. $\endgroup$ – user8469759 Jul 14 at 9:22
  • $\begingroup$ I added a thorough explanation of the first point. $\endgroup$ – Laz Jul 15 at 7:26
  • $\begingroup$ Hi again, I'm reading through the proposition 2.7, the term generalized $\exp$ is not used by author. Do you mean by generalized $\exp$ the map $\gamma : (-2,2) \times \mathcal{U} \to M$ defined by the proposition? $\endgroup$ – user8469759 Jul 15 at 9:00
  • $\begingroup$ Exactly that one. $\endgroup$ – Laz Jul 15 at 14:03
  • $\begingroup$ Ok, lemme read through again. Thank you (my apologies I'm going back and forth with the questions). $\endgroup$ – user8469759 Jul 15 at 14:54

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