Coordinate-independence of equation for the parallel transport

[rbwang1225]

Homework Statement

Please show that the defining equation for the parallel transport of a contravariant vector along a curve $\dot{\lambda}^a+\Gamma^a_{bc}\lambda^b\dot{x}^c=0$ is coordinate-independent, given that the transformation formula for the christoffel symbol being $\Gamma^{a'}_{b'c'}=(\Gamma^{d}_{ef}X^{a'}_d-X^{a'}_{ef})X^{e}_{b'}X^{f}_{c'}$.

The Attempt at a Solution

I have stuck by the following derivation $X^{a'}_{ab}\dot x^b\lambda^a+X^{a'}_a\dot\lambda^a+(\Gamma^{d}_{ef}X^{a'}_d-X^{a'}_{ef})X^{e}_{b'}X^{f}_{c'}(X^{c'}_{ef}\dot x^fx^e+X^{c'}_e\dot x^e)$, where I can't simplify it to the unprimed equation $\dot{\lambda}^a+\Gamma^a_{bc}\lambda^b\dot{x}^c=0$.
Any advice will be appreciated!

 

[rbwang1225]

Hello!
After think twice, I met some mistakes in the derivation.
First, dummy indexes were repeated in the same term.
Second, tangent vectors are transformed by $\dot{x} ^{a'}=X^{a'}_a\dot x^a$

Now, I describe my derivation as follows.

The parallel transport equation for contravariant vector $\lambda ^{a'}$ along a curve $\gamma$ parametrized by t
$\dot\lambda^{a'}+\Gamma^{a'}_{b'c'}\lambda^{b'} \dot{x}^{c'}=0$
was transformed as
$\frac{dX^{a'}_a\lambda^a}{dt}+(\Gamma^{d}_{ef}X^{a'}_dX^{e}_{b'}X^f_{c'}-X^e_{b'}X^f_{c'}X^{a'}_{ef})(X^{b'}_b\lambda^bX^{c'}_c\dot x^c)$
$=X^{a'}_{ab}\lambda^a\dot x^b+X^{a'}_a\dot\lambda^a+\Gamma^d_{ef}X^{a'}_d \lambda ^e \dot x^f-\lambda^e \dot x^fX^{a'}_{ef}$
$=X^{a'}_a(\dot\lambda^a+\Gamma^a_{bc}\lambda^b \dot x^b)=0$
Since $X^{a'}_a$ are arbitrary coefficients of transformation, $\dot\lambda^a+\Gamma^a_{bc}\lambda^b \dot x^c)=0$ in $x^a$ coordinate frame, as desired.

If I have fault, please kindly inform me.