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I was self-learning Do Carmo's Riemannian Geometry, there is a step in the proof of Gauss's Lemma what I can't quite figure out.

Since $d\,\exp_p$ is linear and, by the definition of $\exp_p$, $ \langle (d\,\exp_p)_v(v),(d\,\exp_p)_v(w_T)\rangle=\langle v,w_T\rangle. $

So I went on wikipedia hoping to find something that can help me figure this out.

I did find something. HERE. It says that $(d\,\exp_p)_v(v)=v$. In order to do that, it constructs that curve $\alpha(t)$ with $\alpha(0)=v$, $\alpha'(0)=v$. And it gives $\alpha(t)=(t+1)v$. I agree with all these. Then it argues that you can view $\alpha (t)=vt$ since it's just a shift of parametrization. Okay. I am okay with that. But in this case, should $(d\,\exp_p)_v(v)={d\over dt}(\exp_po \alpha(t))|_{t=1}$, instead of evaluating the derivative at $t=0$ as argued in wiki?

Also, I found myself really uncomfortable with all the abuse of notations in Differential Geometry. Like here, the identification of the $T_p M$ and $T_v(T_p M)$ freaks me out. Does this mean that when a local coordinate system is picked, $T_p M$ under the natural basis will be the same as $R^n$, and so does $T_v (T_p M)$?

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    @ZhenLin Thank you for the clarification!2012-07-27

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The equation you are citing in combination with the reasoning you are citing, too, is used in do Carmo only for $w_T$ parallel to $v$. Otherwise the equation would be still correct (that is the content of the Gauss lemma) but requires a different justification (as presented later on in Do Carmo).

If $w_T$ is parallel to $v$ then $\exp_p(tw)$ is just a parametrization of the geodesic through $p$ in direction $v$ with constant speed $|w|$ (this is where the definition of $\exp$ is used) and the derivative in your formula can be computed as the derivative of this geodesic w.r.t to $t$, so it is simply the tangent vector to that geodesic of length $|w|$ in the point under examination.

'Abuse' of notation in the way you describe it is quite common in Differential Geometry whenever possible, cause otherwise equations are often quite clumsy. A (slightly) more formal notation can be found, e.g., in Klingenbergs Riemannian Geometry.

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    Now that you confirmed what I wrote was right, there's definitely some mistake in Wiki. Thanks a lot. For beginners like me, some times it's hard to tell right from wrong.2012-07-27