Identity for coefficient functions of Riemann Curvature Tensor.

Question

In the solution to one exercise, we have the identification $$R_{\partial_i,\partial_j}\partial_k\Big|_p = R^m_{ijk}(p) \partial_{m}\Big|_p$$ for the riemann-curvature tensor with respect to a point $p \in M$, M SR manifold, and coordinate-vector-fields $\partial_i,\partial_j,\partial_k \in \mathfrak{X}(M)$. Now, if I try to show this explicitly, we get $$R^m_{ijk} = dx^m(R_{\partial_i,\partial_j}\partial_k) = \ldots = \partial_{i}\Gamma^m_{jk}+\Gamma^p_{jk}\Gamma^{m}_{ip}-\partial_{j}\Gamma^m_{ik}-\Gamma^{p}_{ik}\Gamma^{m}_{jp}$$

Multiplying by $\partial_{m}$ we get $$R^{m}_{ijk} \partial_{m} = \partial_{i}\Gamma^m_{jk}\partial_{m}+\Gamma^p_{jk}\Gamma^{m}_{ip} \partial_{m}-\partial_{j}\Gamma^m_{ik} \partial_{m}-\Gamma^{p}_{ik}\Gamma^{m}_{jp} \partial_{m} $$

Okay, so far so good. Now, if we compute $R_{\partial_i,\partial_j}\partial_{k}$ we get $$R_{\partial_i,\partial_j}\partial_{k} = \nabla_{\partial_i}\nabla_{\partial_j}\partial_k-\nabla_{\partial_j}\nabla_{\partial_i}\partial_k \quad (\text{since} \quad [\partial_i,\partial_j] = 0)$$ $$=\nabla_{\partial_i}(\Gamma^p_{jk} \partial_p)-\nabla_{\partial_j}(\Gamma^p_{ik} \partial_p)$$ $$=\partial_i(\Gamma^p_{jk})\partial_p+\Gamma^{p}_{jk}\nabla_{\partial_i}\partial_p-\partial_{j}(\Gamma^{p}_{ik})\partial_p-\Gamma^{p}_{ik}\nabla_{\partial_j}\partial_{p}$$$$= \partial_i(\Gamma^{p}_{jk})\partial_p+\Gamma^{p}_{jk}\Gamma^{c}_{ip}\partial_c-\partial_j(\Gamma^{p}_{ik})\partial_p-\Gamma^{p}_{ik}\Gamma^{d}_{jp} \partial_d$$

Now, in the last step, I presume I am allowed to change the first term $$\partial_i(\Gamma^{p}_{jk})\partial_p$$ to $$\partial_i(\Gamma^{m}_{jk})\partial_{m}$$ without changing the indices from $p$ to $m$ in the second term $$\Gamma^{p}_{jk}\Gamma^{c}_{ip}\partial_c?$$ Because then, I can proceed, setting $c = m$ in the second term, $p=m$ in the third term (without changing $p$ for the other terms) and finally $d=m$ to find the identity I want to show (se prelude). So my main question is, am I allowed to change $p$ individually? And is there a simpler way to show this identity?

Answer 1

The short answer to your second question is that there is nothing to show. The definition of the components of a tensor is $$T_{ijk...}^{lmn...}:= T(e_i,e_j,e_k,...,\epsilon^l,\epsilon^m,\epsilon^n,...)$$ where the $e$'s are the basis vectors and the $\epsilon$'s are the basis covectors. So when you write

$$R_{\partial_i,\partial_j}\partial_k\Big|_p = R^m_{ijk}(p) \partial_{m}\Big|_p$$

This is already just a definition.
Now, presumably you first defined the Riemann tensor by its components, so this ends up being an "identity", but, really, it is backwards.

As for the actual calculation of the components, yours is as simple as it gets

Answer 2

Is there a simpler way to show this identity?

Yes: In your proof you used the fact that \begin{equation}\tag{1} R{}^m_{ijk} = dx{}^m(R_{\partial_i,\partial_j}\partial_k). \end{equation} Let $U$ be the domain of $x$, then \begin{equation}\tag{2} \forall X\in\mathfrak{X}(U)=dx{}^mX\partial_m. \end{equation} You are simply considering the case $X=R_{\partial_i,\partial_j}\partial_k$: \begin{equation} R_{\partial_i,\partial_j}\partial_k\overset{(2)}{=}(dx{}^mR_{\partial_i,\partial_j}\partial_k)\partial_m\overset{(1)}{=}R^m_{ijk}\partial _m \end{equation}