I am trying to learn about the exponential map $\mathrm{exp}$ and read about it in Lee's book Riemannian Manifolds: An Introduction to Curvature. The Lemma I am confused about can be found on page $76$. It wants to show that for any $p \in M$ there exists an open neighbourhood $V$ of $0$ in $T_pM$ and $U$ of $p$ in $M$ such that $\mathrm{exp}_p:V \to U$ is a diffeomorphism. The proof goes as follows.
First identify $T_pM \cong T_0(T_pM)$ and compute $\mathrm{exp}_p(V)$ for a tangent vector $V \in T_pM$. Show that this is equal to $V$ under this identification and use the inverse function theorem.
So as far as I can see it, one first has to note that in general $\mathrm{exp}_p$ is only defined on $$\{v \in T_pM \ | \ c_v \ \text{is defined on an interval containing} \ [0,1]\},$$ where $c_v$ denotes the geodesic with $\dot{c}_v(0)=v.$ However, one wants to compute the composition $$T_pM \to T_0(T_pM) \to T_pM$$ where the first map is given by the identification and the second map is given by $d(\mathrm{exp}_p)_p.$ Now, since $\mathrm{exp}_p$ is differentiable one can define the differential of that map, but why is it defined on $T_0(T_pM)$ even though $\mathrm{exp}_p$ is not defined on the entire tangent space $T_pM$? Furthermore, I fear that this is rather trivial, but what exactly is the identification map $T_pM \cong T_0(T_pM)$? And why does $$(\mathrm{exp}_p)_*V=\left.\frac{d}{dt}\right\vert_{t=0}(\mathrm{exp}_p \circ \tau)(t)$$ hold, where $(\mathrm{exp}_p)_*$ is defined as the composition of the exponential map with the identification above and $\tau(t)=tV.$ Is $V$ viewed as a derivation or a geometric tangent vector here? Thanks in advance for any help.
Best Answer
1: Domain of exp and identification map $T_pM \simeq T_0T_pM$
Even though $\exp_p$ is not entirely defined on $T_pM$, it is defined on an open neighbourhood of $0$ in $T_pM$, and that is all you need to calculate differential of maps between manifolds. It might be a little bit harder to prove that this domain of definition of $\exp_p$ is open in $T_pM$, but I will assume this is done in the book.
The identification map $f:T_pM\simeq T_0(T_pM)$ goes as follows: identify each $V\in T_pM$ with the derivative of a curve through $V$ given by $$ (-\varepsilon,\varepsilon)\ni t\mapsto \tau_V(t):=tV $$
Then, the identification is explicitly written as
$$ f(V) := \frac{d}{dt}|_{t=0}\left( \tau_V(t)\right)=\frac{d}{dt}|_{t=0}\left( tV\right) $$ where the right-hand side is a vector in $T_0(T_pM)$, since it is the derivative $\tau_V'(0)$ of a curve passing thru $0=\tau_V(0)\in T_pM$.
It is easy to prove$^\dagger$ that this identification is a linear isomorphism of vector spaces. The smallness of $t$ will be required so as not to exit the open nbd on which $\exp_p$ is defined.
Note this argument can be extended to show that for any manifold $N$ which happens to be a finite dimensional vector space (with natural topology $\simeq \mathbb{R}^N$) the tangent space at any point $x\in N$ is naturally isomorphic to the space itself $T_xN \simeq N$.
2: Yoga with exponential map
The result will follow from the following two results.
Lemma: let $\gamma_V:(-\varepsilon,\varepsilon)\rightarrow M$ be a geodesic satisfying $\gamma_V(0)=p\in M$ and $\frac{d}{dt}|_{t=0} \gamma_V(t) = V\in T_pM$.
Then, the curve $\beta_t(s):=\gamma_V(ts)$ is a geodesic satisfying $\beta(0)=p$ and $\frac{d}{ds}|_{s=0}\beta_t(s)=tV$.
Proof: we have that $\frac{d\beta_t}{ds} = t\frac{d\gamma_V}{dt}(ts)$. Then, the covariant derivative thru the curve $\beta_t$ of the field $\frac{d\beta_t}{ds}$ is via chain-rule: $$ \frac{D}{ds}\frac{d\beta_t}{ds} = \frac{D}{ds}\left( t \frac{d\gamma_V}{dt}(ts) \right) = t \frac{D}{ds}\left(\frac{d\gamma_V}{dt}(ts)\right) = t \left(\frac{D}{dt} \frac{d\gamma_V}{dt}\right)|_{ts} \frac{d(ts)}{ds} $$ and the lest term vanishes because $\gamma_V$ is a geodesic. Thus $\beta$ is a geodesic, and it is obvious it satisfies the above.
Corollary: The exponential map satisfies:
$$ \frac{d}{dt}|_{t=0} \exp_p(tV) = V $$
for all $V\in T_pM$ that are also in the open domain of $\exp_p$.
Proof: the exponential $\exp_p(tV)$ is defined as $\gamma_{tV}(1)$ where $\gamma_{tV}(s)$ is the unique geodesic (w.r.t. s) that has initial velocity $tV$ through $p$. By the above lemma, we know that $\gamma_V(ts)$ is also a geodesic through $p$ with initial velocity $t\cdot V$. But geodesics are unique, and thus $\gamma_{tV}(s)=\gamma_V(ts)$ for small enough $t$ ($t<1$ actually suffices). Evaluating at $s=1$ yields $$\gamma_{tV}(1) = \gamma_V(t)$$ or $$\exp_p(tV) = \gamma_V(t).$$ Differentiating at $t=0$ you get $$\frac{d}{dt}|_{t=0} \exp_p(tV) = V$$
3: Endgame
You want to show that $$ T_pM \overset{f}{\rightarrow} T_0(T_pM)\overset{(\exp_p)_{*0}}{\rightarrow}T_pM $$ is an isomorphism. From the Section 1 we know that for any $V\in T_pM$ $$ f(V) = \frac{d}{dt}|_{t=0} (tV) $$ therefore the composition with the differential at $0$ of the exponential $\exp_p$ gives $$ (\exp_p)_{*0} \circ f (V) = (\exp_p)_{*0})\left[ \frac{d}{dt}|_{t=0} (tV) \right] = \frac{d}{dt}|_{t=0} \exp_p(tV) = V $$ due to the corollary above and how the differential of a map acts on tangent vectors. Thus the composition coincides with the identity map $I_{T_pM}$ and it is an isomorphism. This implies that both $f$ and $(\exp_p)_{*0}$ are linear isomorphisms, and this, by the inverse function theorem, implies that $\exp_p$ is a local diffeomorphism from a nbd of $0$ in $T_pM$ to a nbd of $p$ in $M$.
Comments:
$^\dagger$: in order to show this is an isomorphism, you can take a coordinate chart $\{x^i\}$ around $p$, and use this to induce a basis of vectors $\{\frac{\partial}{\partial x^i}\}$ of $T_pM$. This induces global coordinates $\tilde{x}^i$ (w.r.t. this basis) on $T_pM$ which in turn induces a basis of vectors on $T_0(T_pM)$. Prove that the identification map $f$ sends $\frac{\partial}{\partial x^i}$ to $\frac{\partial}{\partial \tilde{x}^i}$.