The identity $x_i = g^{ij}x_j$ in geodesic normal coordinates

Question

Suppose we have a Riemannian manifold $(M^n,g)$ and consider geodesic normal coordinates $x_1,\dots,x_n$ about $p$ in some neighbourhood $U.$ By shrinking $U$ as necessary, we will moreover assume it is geodesically convex.

To fix notation, write $e_i = \partial_{x_i}|_p$ which forms an orthonormal basis of $T_pM.$ Then if $x \in B_r$ for $r$ sufficiently small, we can identify this as a point in $M$ by $\exp_p(\sum x_ie_i)$ (this is really the local coordinates written out explicitly). The metric $g$ with respect to these coordinates will be denoted by $g_{ij}$ and its inverse by $g^{ij}.$

Question: Is it true that $x_i = g^{ij}(x)x_j$ for all $x \in B_r$?

Note in the above I am using the summation convention, so $j$ is summed over from $1, \dots, n.$ Also I am identifying $B_R$ with a neighbourhood of $p$ via the expoential map, as is the usual convention when working with local coordinates.

I am reading a paper which seems to suggest this, appealing to Gauss' lemma. From the perspective of the paper this is a minor point and hence was glossed over, but it would be useful for me if it's true.

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Some ideas: If I let $y_i = g^{ij}x_j,$ I can view $y$ as an element of $T_pM$ similarly to $x.$ Then I can consider how $y$ transforms with respect to geodesic polar coordinates. I suspect you can then show by Gauss' lemma that $|x|^2 = |y|^2$ and combine it with the calculation $\langle x, y\rangle_g = g_{ij}x_ig^{jk}x_k = |x|^2,$ but I can't see the change of coordinates giving anything nice.

Answer 1

Recall Gauss lemma : If $g(0)(V,W)=0$, then $g(tV)(V,W)=0$.

Hence If $V=\sum_i\ x_ie_i,\ W=x_2e_1-x_1e_2$, then $$ x_2g_{1j}x_j =x_1g_{2j}x_j$$

so that $g_{ij}x_j=cx_i$ for some $c$.

Further, $g(V,V)=\sum_i\ x_i^2$ so that $c=1$.

Answer 2

As you assume that $\left\{\partial_{x_j}\right\}_{j=1}^n$ is orthonormal, and that $g=g_{jk}{\rm d}x^j\otimes{\rm d}x^k$, the proposition seems correct, because $$ \delta_{rs}=g(\partial_{x_r},\partial_{x_s})=g_{jk}{\rm d}x^j(\partial_{x_r})\otimes{\rm d}x^k(\partial_{x_s})=g_{rs}, $$ meaning that the representation $g_{rs}$ is an identity matrix, hence so is $g^{rs}$. Therefore, $x_j=g^{ij}x_i$ becomes an immediate consequence.

Yet in general, I am afraid the proposition does not hold, because $\left\{\partial_{x_j}\right\}_{j=1}^n$ may very well not be an orthonormal basis. For example, consider the representation of $g$ under the polar coordinate $\left(r,\theta\right)$, when $M=\mathbb{R}^2$ has been equipped with the standard Euclidean metric. In this case, $$ \left(g_{ij}\right)=\text{diag}\left\{1,r^2\right\}\iff\left(g^{ij}\right)=\text{diag}\left\{1,1/r^2\right\}. $$ Now, for any $p=\left(r_0,\theta_0\right)\in M$, the proposition seems to indicate that $$ \left( \begin{array}{c} r_0\\ \theta_0 \end{array} \right)=\left( \begin{array}{cc} 1&0\\ 0&1/r_0^2 \end{array} \right)\left( \begin{array}{c} r_0\\ \theta_0 \end{array} \right), $$ which might lead to some obvious contradiction...