Absolute Derivative: Proving Vector Field Along ##\gamma##

[rbwang1225]

Homework Statement

Please show that the absolute derivative is a vector field along $\gamma$, i.e., $(\frac{d\lambda ^{a'}}{du}+\Gamma^{a'}_{b'c'}\lambda^{b'}\frac{dx^{c'}}{du})=X^{a'}_d(\frac{d\lambda ^{d}}{du}+\Gamma^{d}_{ef}\lambda^{e}\frac{dx^{f}}{du})$

The Attempt at a Solution

I don't know how to reduce the following eq. $(\frac{d\lambda ^{a'}}{du}+\Gamma^{a'}_{b'c'}\lambda^{b'}\frac{dx^{c'}}{du}) = X^{a'}_{bc}\lambda ^b\dot x ^c+X^{a'}_b\dot\lambda ^b-\Gamma ^d_{ef}X^{a'}_d\lambda^eX^{d'}_cX^{e'}_g X^f_{d'e'}\dot x^gx^c+\dot x^f\Gamma^d_{ef}X^{a'}_d \lambda^e$
Any comment would be appreciated.

 

[mark726]

Thank you for your question. To show that the absolute derivative is a vector field along $\gamma$, we can use the definition of the absolute derivative as $\frac{D\lambda^{a'}}{du}=\frac{d\lambda ^{a'}}{du}+\Gamma^{a'}_{b'c'}\lambda^{b'}\frac{dx^{c'}}{du}$. This means that the left-hand side of the given equation can be rewritten as $\frac{D\lambda^{a'}}{du}$.

Next, we can use the properties of the absolute derivative to simplify the equation. For example, we know that the absolute derivative is linear in its first argument, so we can write $\frac{D\lambda^{a'}}{du}=\frac{D(X^{a'}_d\lambda^d)}{du}=\frac{d(X^{a'}_d\lambda^d)}{du}+\Gamma^{a'}_{b'c'}(X^{b'}_d\lambda^d)\frac{dx^{c'}}{du}$.

Using the chain rule, we can expand the first term on the right-hand side as $\frac{d(X^{a'}_d\lambda^d)}{du}=\frac{dX^{a'}_d}{du}\lambda^d+X^{a'}_d\frac{d\lambda^d}{du}$.

Plugging this back into our equation and simplifying, we get $\frac{D\lambda^{a'}}{du}=\frac{dX^{a'}_d}{du}\lambda^d+X^{a'}_d\frac{d\lambda^d}{du}+\Gamma^{a'}_{b'c'}(X^{b'}_d\lambda^d)\frac{dx^{c'}}{du}$.

Now, we can use the definition of the absolute derivative again to rewrite the right-hand side as ##\frac{D\lambda^{a'}}{du}=X^{a'}_d\frac{d\lambda^d}{du}+\frac{dX^{a'}_d}{du}\lambda^d+\Gamma^{a'}_{b'c'}X^{b'}_d\frac{dx^{c'}}{du}\lambda