Let $M$ be a Riemannian manifold and $p\in M$ a point. Let $E_{i}$ be an orthonormal basis of $T_{p}M$ and let $E\colon \mathbb{R}^{n} \to T_{p}M$ be the isomorphism sending $(x_{1},\ldots,x_{n})\mapsto \sum_{i=1}^{n}x_{i}E_{i}$. On a normal neighborhood $U$ of $p$ in $M$ (that is, the image under the exponential map $\exp_{p}$ of a star shaped open neighborhood of $0\in T_{p}M$) we can define a coordinate chart by $\varphi =E^{-1}\circ \exp_{p}^{-1}\colon U\to \mathbb{R}^{n}$. As with any other chart, the basis $\frac{\partial }{\partial x_{i}}$ of $T_{0}\mathbb{R}^{n}$ pulls back to a basis $\frac{\partial }{\partial x_{i} }|_{p}$ of $T_{p}M$.

Is this pullback the orthnormal basis that we started from?

My attempt: let $f\in C^{\infty }(M)$. We have to show that $E_{i}(f)=\frac{\partial }{\partial x_{i} }|_{p}(f)$. By definition:

$$\frac{\partial }{\partial x_{i} }|_{p}(f)=\frac{\partial( f\circ \exp_{p}\circ E)}{\partial x_{i} }(0)$$

I guess now I need to deal with the definition of the exponential map explicitly, but I am not sure on how to do this. So first $E$ sends $(x_{1},\ldots,x_{n})\mapsto \sum_{i=1}^{n}x_{i}E_{i}$. Then the exponential map sends a vector $v\in T_{p}M$ to $\gamma_{v}(1)$, where $\gamma_{v}$ is a geodesic with $\gamma_{v}(0)=p$ and $\dot{\gamma_{v} }(0)=v$. How do I use this now? I don't even know how geodesics look like without knowing the Riemannian metric $g$.

For example, the point $(1,0,\ldots,0)$ is sent first to $E_{1}$ and then we take the geodesic $\gamma$ with $\gamma(0)=p=(0,\ldots,0)$ (using the expression of the point in the normal coordinate chart) and $\dot{ \gamma}(0)=E_{1}$. Now $\gamma(1)$ should be the point in the trajectory of $\gamma $ at time 1, and I guess if we track $\gamma$ on $\mathbb{R}^{n}$ via the normal coordinates then this point should correspond to $(1,0,\ldots,0)$, but I am already not sure of these claims.

This also led me to the following question:

In a chart $(V,\psi)$ with $\psi(p)=0$, is it true that $\frac{\partial }{\partial x_{i} }|_{p}=\dot{ \gamma }(0)$ for any curve $\gamma $ in $M$ with $\gamma(0)=p$ and such that $\psi(\gamma)$ has speed $1$ in the positive direction of the $x_{i}$ axis?

EDIT (current thoughts on the questions):

I suspect now that the answer to the first question is positive and I suspect it will follow from the differential of the exponential map at $p$ being the identity on $T_{p}M$. But I still couldn't manage to write down a formal argument.

Regarding the second question, I still didn't find an answer and I still very interested in finding one. I think that there are at least two ways of defining the tangent space of $M$ at a point $p$, one with this abstract derivations and the other one by considering derivatives of curves in the manifold passing through $p$ at time $0$. I have never worked with this definition, but after thinking about the second question I got to the following conclusion: isn't that second question precisely the relation between these two definitions of the tangent space?


The answer to the first question uses (as you have already noted) that $(\exp_{p})_{\ast p} = \mathrm{id}_{T_p M}$. A proof of this can be found in any textbook on Riemannian Geometry. Now, we obtain $$ (\exp_p \circ\; E)_{\ast p} = E_{\ast p} = E$$ since $E$ is itself linear. Hence, $$ (\exp_p \circ \; E)_{\ast p} \left(\tfrac{\partial}{\partial x^i}\big\vert_{x=0}\right) = E \left(\tfrac{\partial}{\partial x^i}\big\vert_{x=0}\right) = E_i,$$ which affirmatively answers your first question. Subsequently, your second question is easily answered as well, since it is the very same question! You reformulated the problem in the geometric conception of tangent vectors being equivalence classes of curves (where equivalence holds iff the vectors have identical differential at the tangent point).

  • $\begingroup$ Nice, thanks for your answer! $\endgroup$ – Pedro Jan 15 '18 at 21:47

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