I'm having a hard time trying to understand something that I'm suspicious is pretty stupid. I'll refer to Wikipedia to settle the term's I'll refer to.


That said, having $V=(X_1,...,X_n)\in T_p M$, how can the element $\gamma_V(t)=(tX_1,...,tX_n)$ be on the manifold $M$? Isn't the geodesic defined by $\gamma_V(t)=exp_p (t.V)$? This to me looks like just a parameterization of a straight line which does not need to be on the manifold, necessarily.

I need some help. I'll also link this lecture note which I'm trying to understand: http://www-personal.umich.edu/~wangzuoq/635W12/Notes/Lec%2023.pdf.



Linear coordinates $X_{i}$ on $T_{p}M$ are identified, via the exponential map, with normal coordinates in a sufficiently small neighborhood of $p$ in $M$. When one writes $\gamma_{V}(t) = (tX_{1},\dots tX_{n})$ in normal coordinates (as on the wikipedia page you linked), this identification is implicit.

  • $\begingroup$ So, the coefficients $X_i$ are defined as functions of $p$? $\endgroup$ – Marra May 16 '14 at 1:08
  • $\begingroup$ Well...$p$ is a specific point of $M$, but yes, if $V \subset T_{p}M$ is mapped diffeomorphically to $U \subset M$ by the exponential map, then the $X_{i}$ (which strictly speaking are functions on $T_{p}M$) may be viewed as local coordinates in $U$. $\endgroup$ – Andrew D. Hwang May 16 '14 at 1:12
  • $\begingroup$ Perfection! Thank you for your time :) $\endgroup$ – Marra May 16 '14 at 1:13
  • $\begingroup$ You're very welcome. :) $\endgroup$ – Andrew D. Hwang May 16 '14 at 1:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.