Is Ricci's theorem can be simply deduced using covariant derivatives of fundamental tensors? [duplicate]

Question

Well Ricci's theorem is given by: $$\mathrm{D}g_{ij}=\mathrm{D}g^{ij}=0$$ I was wondering that if the theorem can be proved using covariant derivatives of $\delta_i^k$, $g_{ij}$ and $g^{ik}$.

I really need this proof.

Answer 1

I will give you a hint :

The components of a tensor $\mathbf{U}$ are $u_{st}^r$ the covariant derivative is $\nabla_k u_{st}^r$ it's given by $$\nabla_k u_{st}^r=\partial_k u_{st}^r+u_{st}^i \Gamma_{ki}^r-u_{it}^r\Gamma_{ks}^i-u_{si}^r \Gamma_{kt}^i$$

The covariant derivative of $\delta_i^k$ : $$\nabla_k \delta_s^r=\partial_k \delta_{s}^r+\delta_{s}^i \Gamma_{ki}^r-\delta_{i}^r\Gamma_{ks}^i=0+\Gamma_{ks}^r-\Gamma_{ks}^r=0$$ the same for $g_{ij}$

$$\nabla_{k}g_{ij}=\partial_k g_{ij}-g_{lj}\Gamma_{ki}^l-g_{il}\Gamma_{kj}^l$$ Since $\partial_k g_{ij}=\Gamma_{jik}+\Gamma_{kji}$, thus: $$\nabla_{k}g_{ij}=\Gamma_{jik}+\Gamma_{kji}-\Gamma_{kji}-\Gamma_{jik}=0$$ Therefore : $$\nabla_{k}g_{ij}=0$$ For $g^{ij}$ we have : $$\nabla_k g^{ij}=\partial_k g^{ij}+g^{lj}\Gamma_{lk}^i+g^{il}\Gamma_{lk}^j$$ Where : $$\partial_k g^{ij}=-g^{li}\Gamma_{lk}^j-g^{lj}\Gamma_{kl}^i$$ Hence : $$\nabla_k g^{ij}=0$$ Now you have your derivatives try to answer your question, note that $\mathrm{D} g_{ij}=\nabla_k \ g_{ij} \ du^k$.

Good luck with that.