From normal coordinate to verify tensor property of Riammnian curvature using different method

Question

From normal coordinate to verify tensor property of Riammnian curvature using different method from existing methods in textbooks and published book or journal article.

Let $(M,g)$ be a riemannian manifold. In normal coordinate $(x^i)$ at $p$, we know the riemann curvature tensor

$$\tag{1} R_{ijkl}(p) = \dfrac{1}{2}(\partial_i\partial_lg_{jk} + \partial_j\partial_kg_{il} - \partial_i\partial_kg_{jl} - \partial_j\partial_lg_{ik})(p). $$ It is well known that the riemann curvature tensor is a tensor, for example, steven weinberg's book gravation and cosmology, page 132,chapter6,(6.1.4). Now I want to using different method to prove it can transform back to its original formula.

I have done three computations.

1. The first method

For example, the first term, $ \partial_i\partial_lg_jk=\partial_i(\Gamma_{lj}^pg_{pk}+\Gamma_{lk}^pg_{pj}) $, then we can swith i,j,k,l, to get the formula $ Rijkl(p) $

2. the second method

   We using two kinds coordinantes, In normal coordinate $ (x^i) $ at p and general coordinates ($e_\alpha$), we should prove from formula (1) to prove it is tensor. for example; I have done like this:

   $\partial_j= A_j^\alpha e_\alpha $, then we can do ad hoc computation: $$ \partial_i\partial_lg_jk=\partial_i\bigg( A_j^\alpha e_\alpha( A_k^\theta A_l^\tau g_{\tau\theta} )\bigg) $$

   then, we can continue to do using the chain rule to expand all the derivateive, at last, we can get 48 terms. Now the trouble is how to combine to get the following formula from left to right

   $$\dfrac{1 }{2}(\partial_i\partial_lg_jk + \partial_j\partial_kg_il - \partial_i\partial_kg_jl - \partial_j\partial_lg_ik)(p)=A_1^\alpha A_j^\beta A_k^\gamma A_l^\theta R_{\alpha \beta\gamma\theta} $$

   Although this follows from immediately fro, steven weinberg's book gravation and cosmology, page 141,chapter6,(6.6.2) ,where, left, one can take coordinate, so the four term can become the four term, on the right, weinberg using the natural coordinate, total four term.

   This kind of compution is terrible, I have tried it , very hard.

   We also know that nomal coordinate is very usefule for verifying the tensor equation and famous formula ,for example, Bochner formula ,since many formula is independent of the choice of frame.

   My question is that if we don't know a priori the correct formula, how we can use the normal coordinate to find the correct formula? since ,during we compute in coordinate, we dicard the connection coefficient and the derivative the metric component.

3. The third method

I have used the relationship,the partial derivate and covariant derivative to compute:

Change $$\partial_i\partial_lg_jk$$ into $$ \nabla_{i}\nabla_{l}g_{jk}+\Gamma g +\Gamma g+ --- $$, Then we get some connection coefficient, at last, I have get the formula

$$\dfrac{1 }{2}(\partial_i\partial_lg_jk + \partial_j\partial_kg_il - \partial_i\partial_kg_jl - \partial_j\partial_lg_ik)(p). =\dfrac{1 }{2}(\dfrac{\partial \Gamma_jk^p}{\partial x_i}g_{pl}-\dfrac{\partial \Gamma_ik^p}{\partial x_j}g_{pl}) $$ however, using the following computation ,we cannot get the same formula,

For example, the first term, $ \partial_i\partial_lg_jk=\partial_i(\Gamma_{lj}^pg_{pk}+\Gamma_{lk}^pg_{pj}) $, then we can swith i,j,k,l, to get the formula $ Rijkl(p) $,but, at last, there are four extra terms compared with the above formula.

My question, the above method is right? Have I made any mistakes?

Any answer is appreciated.