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 $T_v (T_p M)$?