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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.

http://en.wikipedia.org/wiki/Normal_coordinates#Geodesic_normal_coordinates

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.

Thanks!

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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.

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  • $\begingroup$ 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

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