- #1
rbwang1225
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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.