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I am studying Riemannian geometry, and there is something stated in my lecture notes that I simply cannot understand:

If $\gamma$ is a geodesic curve on a manifold M so $\gamma(p)=0$, $\gamma(q)=1$, then we have $\gamma(t)=exp_p(t\cdot\dot\gamma(0))$ (on some neighbourhood). Then it is stated that this means that:

$\ (exp_p)_{*\dot\gamma(0)}(\dot\gamma(0))=\dot\gamma(1)$

I have tried to show this, but I fail to do it. Is there anyone who can see why this is true?

Also one more thing that is confusing me a bit, is that when I write out the definition of $\ (exp_p)_{*\dot\gamma(0)}(\dot\gamma(0))$ (who is a vector in $\ T_pM$) acting at a smooth function $\ f \in C^\infty(M)$ I get $\ (\dot\gamma(0))(f\circ\exp_p)$, but $\ f\circ\exp_p $ is not a function from $\ M\rightarrow\mathbb{R}$, so how can the vector $\ \dot\gamma(0)$ act on this?

Thank you for any help.

Christine
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  • You have a few typos at the beginning: $\gamma(0)=p$, $\gamma(1)=q$. Note, next, that $\dot\gamma(1) \in T_qM$. You need to think about what it means for $\gamma$ to be a geodesic. – Ted Shifrin Oct 08 '19 at 19:56
  • It means that it is a straight line in normal coordinates? Or is it something else you are pointing at? I am afraid that I am lost in this. – Christine Oct 08 '19 at 20:12
  • I'm talking about the basic definitions and notions of parallel translation. – Ted Shifrin Oct 08 '19 at 20:24

1 Answers1

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To answer your first question, perhaps a change of notation helps. For any $u\in T_pM$, let $\gamma_u$ denote the geodesic satisfying $\gamma_u(0)=p$ and $\dot{\gamma}_u(0) = u$. Then the expontential map $\exp_p:T_p M \to M$ is defined by $\exp_p(u): = \gamma_u(1)$. Also, for any $c\in\mathbb{R}$, the curve $s\to \gamma_u(cs)$ is also a geodesic through $p$ with initial derivative $cu$, and so $\gamma_{cu}(s) = \gamma_u(cs)$. Taking $c=t$ and $s=1$, we get $\exp_p(tu) = \gamma_{tu}(1) = \gamma_u(t)$. Differentiating this with respect to $t$ at $t=1$ gives $$ (\exp_p)_{*u}(u) = \dot{\gamma}_u(1). $$ Since $u = \dot{\gamma}_u(0)$, this is the result you seek.

To answer your second question, there is an implicit isomorphism going on: we have $$ \exp_p:T_pM\to M \qquad\textrm{and so}\qquad (\exp_p)_{*u}:T_u(T_pM)\to T_{\exp_p(u)}M. $$ However, $T_u(T_pM)$ is canonically isomorphic to $T_pM$, and so we usually just write $$ (\exp_p)_{*u}:T_pM\to T_{\exp_p(u)}M. $$ So in your question, $f\circ \exp_p:T_pM\to\mathbb{R}$, but $\dot{\gamma}(0)$ is really an element of $T_{\dot{\gamma}(0)}(T_pM)$, not $T_pM$. Hence it can act on $f\circ \exp_p$.

user17945
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