Given a basis of $T_pM (e_i)$. Extend this base to a local orthonormal frame $(E_i)$ with $\nabla E_i (p) = 0 $

Question

I have been working this problem almost 3 days, I would appreciate any help or idea:

Let (M,h) be a Riemannian manifold. For every $ p \in M $ and $ (e_1 , ... , e_n)$ basis of $T_pM$. There exists an orthonormal frame in an neighborhood of p $(E_1, ... , E_n)$ , with $E_i (p) = e_i$ and $\nabla E_i (p) = 0 $

Hint: Fix an orthonormal frame $(\overline E_i)$ near p with $\overline E_i (p) = e_i$ and define $E_i = \alpha_i^j\overline E_i $ with $(\alpha_i^j(x))_{ij} \in SO(n)$ and $\alpha_i^j(p)= \delta_i^j$.

What I have got:

The construction of an orthonormal frame $(\overline E_i)$ near p with $\overline E_i (p) = e_i$. Follows from Gram-Schmidt process with no problem.

Defining $E_i= \sum_j \alpha_i^j\overline E_i$ the frame is still orthonormal, that follows from a direct calculation and the fact that $(\alpha_i^j(x))_{ij} \in SO(n)$ and $\overline E_i (p) = e_i$ because $\alpha_i^j(p)= \delta_i^j$.

Since $h(E_i,E_j)= \delta_i^j$ then: $h(\nabla_X E_i, E_j) + h(E_i,\nabla_X E_j) = 0 $.

So all I have to see is that:

$h(\nabla_X E_i, E_j) = h(E_i,\nabla_X E_j) $

Writing the above equation with the koszul formula and doing the calculations, Using the antisymmetry of Lie bracket and the fact that the inner product is commutative:

$h(\nabla_X E_i, E_j) - h(E_i,\nabla_X E_j) = -h([E_i,E_j],X) $

So all at have to see is that in p: $[E_i,E_j](p) = 0$

That the Lie bracket its 0 at p for the basis constructed in this way! I have been trying to justify this from lie bracket proprieties with no luck

Answer 1

Choose a coordinate system $\varphi = (x^1, \dots, x^n)$ around $p$ with $p$ corresponding to $(0,\dots,0)$ and define the Christoffel symbols of the frame $\overline{E}_j$ with respect to the frame $\frac{\partial}{\partial x^i}$ by

$$ \nabla_i \overline{E}_j = \nabla_{\frac{\partial}{\partial x^i}} \overline{E}_j = \Gamma_{ij}^k \overline{E}_k $$

(summation convention is in place). We want to define a new frame $E_i = \alpha_i^j \overline{E}_j$ so that the following conditions hold:

1. The frame $E_i$ is an orthonormal frame.
2. $E_i(p) = \overline{E}_i(p)$.
3. $(\nabla_i E_j)(p) = \left( \nabla_{\frac{\partial}{\partial x^i}} E_j \right)(p) = 0$ for all $1 \leq i, j \leq n$.

Let us write explicitly the conditions on the matrix $\alpha = (\alpha_i^j)_{i,j=1}^n$ must satisfy for each of the following conditions to hold:

1. The matrix $\alpha(q) = (\alpha_i^j(q))_{i,j=1}^n$ must be an orthonormal matrix for each $q \in U$ (where $U$ is the relevant neighborhood of $p \in M$).
2. $\alpha(p) = I_n$ (the identity matrix).
3. We must have $$ (\nabla_i E_j)(p) = (\nabla_i (\alpha_j^k \overline{E}_k))(p) = \left( \frac{\partial \alpha_{j}^k}{\partial x^i} \overline{E}_k + \alpha_j^k \Gamma_{ik}^l \overline{E}_l \right)(p) = \\ \left( \frac{\partial \alpha_j^l}{\partial x^i}(p) + \alpha_j^k(p) \Gamma_{ik}^l(p) \right) \overline{E}_l(p) = 0$$ which implies that $$ \frac{\partial \alpha_j^l}{\partial x^i}(p) + \alpha_j^k(p) \Gamma_{ik}^l(p) = \frac{\partial \alpha_j^l}{\partial x^i}(p) + \delta_j^k \Gamma_{ik}^l(p) = \frac{\partial \alpha_j^l}{\partial x^i}(p) + \Gamma_{ij}^l(p) = 0 $$ for $ 1 \leq i,j,l \leq n$. If we set $\Gamma_i = (\Gamma_{ij}^k)_{j,k=1}^n$, this can be written as $$ \frac{\partial \alpha}{\partial x^i}(p) = -\Gamma_i(p) $$ for $1 \leq i \leq n$.

Note that the matrix $\Gamma_i := (\Gamma_{ij}^k)_{j,k=1}^n$ is anti-symmetric (since our connection is metric).

If we set

$$ \alpha(\varphi^{-1}(x^1, \dots, x^n)) := e^{-x^i \Gamma_i(p)} $$

then since the exponential of an anti-symmetric matrix is orthogonal, we have $(1)$ and by direct calculation we also have $(2)$ and $(3)$.

Answer 2

Hint: The actual task is to show that you can choose the functions $\alpha^i_j$ in such a way that the conditions are satisfied, which mainly amounts to fixing the derivatives at $x$ without destroying the fact that the matrix $(\alpha^i_j)$ is orthogonal.

My preferred way to solve the problem would not be by choosing an orthonormal frame and then adapting via the $\alpha^i_j$ but rather use parallel transport to directly construct an appropriate frame.