 Solution

In coordinates, write ${\nabla }_{{\partial }_i}{\partial }_{\ell }={\Gamma }_{i\ell }^k{\partial }_k$. The Riemannian curvature two-form is defined by the matriceswith repeated indices summed. It assigns to $X,Y$ the fiberwise endomorphism $R(X,Y)$ whose entries are $R^k{}_{\ell ij}{X}^i{Y}^j$. We can prove both coordinate independence and the required intrinsic identity as follows.

Define an operator on vector fields by $C(X,Y)=[{\nabla }_X,{\nabla }_Y]-{\nabla }_{[X,Y]}$. It is alternating in $X,Y$. For a smooth function $f$, the identities $[fX,Y]=f[X,Y]-Y(f)X$ and ${\nabla }_Y(fZ)=Y(f)Z+f{\nabla }_{Y}Z$ givethe two terms involving $Y(f){\nabla }_{X}Z$ cancel. Alternation gives linearity over smooth functions in $Y$ as well. Expanding in the final argument yieldsThus $C$ is tensorial in all three arguments, defining a global endomorphism-valued two-form.

For coordinate fields ${\partial }_i,{\partial }_j$, their bracket is zero. Expand $[{\nabla }_{{\partial }_i},{\nabla }_{{\partial }_j}]Z$ using $Z=Z^{\ell }{\partial }_{\ell }$. The second derivatives of $Z^{\ell }$ cancel by equality of mixed partials. The terms containing first derivatives of $Z^{\ell }$ cancel in pairs, and the remaining coefficient of $Z^{\ell }{\partial }_k$ is exactly $R^k{}_{\ell ij}$ above. Hence the coordinate definition equals the global tensor $C$, establishing coordinate independence and provingThis convention for the Riemann curvature tensor is valid for any affine connection; in particular it applies to the Levi-Civita connection.