Gauss Lemma, Chapter 3 - Do Carmo's differential geometry.

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.

• 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. – user8469759 Jul 14 at 9:22
• I added a thorough explanation of the first point. – Laz Jul 15 at 7:26
• 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? – user8469759 Jul 15 at 9:00
• Exactly that one. – Laz Jul 15 at 14:03
• Ok, lemme read through again. Thank you (my apologies I'm going back and forth with the questions). – user8469759 Jul 15 at 14:54