Existence of a local geodesic frame

Question

Let $(M,g)$ be a Riemannian manifold of dimension $n$ with Riemannian connection $\nabla,$ and let $p \in M.$ Show that there exists a neighborhood $U \subset M$ of $p$ and $n$ (smooth) vector fields $E_1,...,E_n \in \chi (U),$ orthonormal at each point of $U,$ s.t. at $p,$ $\nabla_{E_i}E_j(p)=0.$

What I've found so far;

Let $V$ be a normal neighborhood at $p.$ Let $(W,x(x_1,...,x_n))$ be a local coordinate system at $p$ s.t. $x(W) \subset V.$ Let $\{X_i:=\frac{\partial}{\partial x_i}\}$ be the standard basis of $T_pM.$ By Gram-Schmidt, there exists an orthonormal basis $\{E_i\}$ of $T_pM.$ Consider the orthonormal frame $E_i(t),$ by parallel transporting $\{E_i(0)=E_i\}$ along a a geodesic $\gamma :I \to M$ starting at $\gamma(0)=p$ and ending in $\gamma(1)=q,$ where $q \in V.$ This gives us an orthonormal frame at each point of $V.$ To satisfy the second condition, let $E_i(t)=\sum_{l}a_{li}X_l$ and $E_j(t)=\sum_{s}b_{sj}(t)X_s.$ Then

$$\nabla_{E_i}E_j=\sum_{k}(\sum_{l,s} a_{li}(t)b_{sj}(t)\Gamma^k_{ls}+E_i(b_{kj}))X_k,$$

where $\Gamma^k_{is}$ are Christoffel symbols.

To have $\nabla_{E_i}E_j(p)=0,$ we must have $\sum_{l,s}a_{li}(0)b_{sj}(0)\Gamma^k_{ls}+E_i(b_{kj})(p)=0$ for each $k.$

How can I conclude the argument?

Answer 1

Let $(U,\phi)$ be a coordinate neighborhood of $p$ and let $g_{ij}$ and $\Gamma_{ij}^{k}$ denote the Riemannian metric tensors and the Christoffel symbols, respectively. If we recall that $g_{ij}=(E_i,E_j)$ and $\Gamma_{ij}^{k}=\nabla_{E_i}E_j(E_k)$ for all $1\leq i,j,k\leq n$ (where $M$ is a smooth $n$-manifold and $E_1,\dots,E_n$ are the coordinate frames on $(U,\phi)$), then we need only have $g_{ij}=\delta_{ij}$ on $U$ and $\Gamma_{ij}^{k}(p)=0$.

You are right in choosing a normal coordinate system at $p$, that is, choosing a normal coordinate neighborhood $(U,\phi)$ at $p$. Let me recall that this means choosing an orthonormal basis $F_1,\dots,F_n$ of the tangent space $T_p(M)$, choosing a star-shaped neighborhood of the origin in $T_p(M)$ which is diffeomorphically mapped onto $U$ under the exponential mapping $\text{exp}_p:D_p\to M$ (where $D_p$ is an open subset of $T_p(M)$ containing $0$) and defining $\phi=\exp_{p}^{-1}$ on $U$ (here we identify $T_p(M)$ with $\mathbb{R}^n$ by the linear isomorphism mapping $F_i$ onto $e_i$, $1\leq i\leq n$, where $e_1,\dots,e_n$ is the standard basis of $\mathbb{R}^n$).

In normal coordinates, the geodesics are of the form $y^i=a^it$ where $a_i$ is a constant for all $1\leq i\leq n$. If you prove this, then you should easily be able to see that $\Gamma_{ij}^{k}(p)=0$ for all $1\leq i,j,k\leq n$ by looking at the second-order ordinary differential equation of geodesics. You can also check that $g_{ij}(p)=\delta_{ij}$ for all $1\leq i,j\leq n$.

I hope this helps!

Answer 2

Recall parallel transport is a isometry, i. e. if $P=P_{\gamma, t_0, t}: T_{c(t_0)}M\rightarrow T_{c(t)}M$ is parallel transport then

$<u, v>_{c(t_0)} = <P(u), P(v)>_{P(c(t_0))}$

then, because you get ${E_i}$ is orthonormal, $<E_i, E_j> = \delta_{i j}$ is the kronecker delta. Therefore $g_{ij} = <E_i, E_j> = \delta_{i j}$ thus $\Gamma_{ij}^k(p) = 0$.

Thereupon $\nabla_{E_i}E_j(p) = \sum_k \Gamma_{ij}^k(p)E_k(p) = 0$.