after some calculations I obtain the next expression

$$\lambda^{c}\nabla_{a}g_{bc}=\lambda^{c}{}(\lambda_{c}\nabla_{a}\lambda_{b}-g_{be} \Gamma^{e}_{ad}\lambda^{d}\lambda_{c}) $$ So my question is if I can eliminate $\lambda^{c}$ of this equation, and get something like this

$$\nabla_{a}g_{bc}=\lambda_{c}\nabla_{a}\lambda_{b}-g_{be} \Gamma^{e}_{ad}\lambda^{d}\lambda_{c} $$


1 Answer 1


It depends on what $\lambda$ is. If the first equation is true for all $\lambda\in\mathbb{R}^n$, then the second equation follows. However, if it is true for a single $\lambda$, then from $\lambda^a A_a = \lambda^a B_a$ you can only conclude that $$A_a = B_a + k_a$$ for some $k\in\text{ker}(\lambda):=\{v\in\mathbb{R}^n \mid \lambda^a v_a = 0\}$. Thus, you can only specify the equation up to a term sent to zero when "applying" $\lambda$.



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