Derivative of a Dual Frame Vector

[BOAS]

Homework Statement

Use the relation $\langle \vec e^a, \vec e_b \rangle = \delta^a_b$ and the Leibniz rule to give an expression for the derivative of a dual frame vector $\frac{\partial \vec e_b}{\partial x^a}$ in terms of the connexion components.

Homework Equations

The Attempt at a Solution

I'm not sure how to do this, but this is what I've got so far:

$\partial_c \langle \vec e^a, \vec e_b \rangle = 0$

$\langle \partial_c \vec e^a, \vec e_b \rangle + \langle \vec e^a, \partial_c \vec e_b \rangle = 0$

$\langle -\Gamma^b_{dc} \vec e^d, \vec e_b \rangle + \langle \vec e^a, \Gamma^d_{bc} \vec e_d \rangle = 0$

$\Gamma^d_{bc}\langle \vec e^a, \vec e_d \rangle - \Gamma^b_{dc} \langle \vec e^d, \vec e_b \rangle = 0$

 

[Orodruin]

$\langle \partial_c \vec e^a, \vec e_b \rangle + \langle \vec e^a, \partial_c \vec e_b \rangle = 0$

$\langle -\Gamma^b_{dc} \vec e^d, \vec e_b \rangle + \langle \vec e^a, \Gamma^d_{bc} \vec e_d \rangle = 0$

Between these lines you are using what you are supposed to show ...

 

[BOAS]

Between these lines you are using what you are supposed to show ...

$\frac{\partial \vec e_b}{\partial x^a} = \Gamma^c_{ba} \vec e_c$ was presented as a definition to me, so I'm not sure how I can bring in the connection components without using that fact. (Though I do appreciate your point)

 

[Orodruin]

That is not the derivative of the dual basis vector, it is the derivative of the tangent basis vector.