Let $ (M, g) $ be a Riemannian manifold and $ \omega $ a connection on the tangent bundle of $ M $ at a given point.
Then the curvature form is the $\mathfrak{g} $-valued 2-form $ \Omega \in \Omega^2(M) $ where $ \mathfrak{g} $ is the Lie algebra of the structure group in the tangent bundle $ TM $.
The connection form is given by
$ \Omega = d\omega + \frac{1}{2} \left[ \omega \wedge \omega \right] $
From what i understand the curvature form is an equally valid description of curvature as the Riemann tensor.
Is there a way of equaling both?
In the wikipedia page "curvature on Riemannian manifolds" there's this equation
$ R(\boldsymbol{u}, \boldsymbol{v}) \boldsymbol{w} = \Omega(\boldsymbol{u} \wedge \boldsymbol{v}) \boldsymbol{w} $
I would like to use this to get an equation in terms of the components, however i don't know about how to write the curvature form in terms of its components.
I know that
$ \Omega^i_j = d\omega^i_j + \omega^i_k \wedge \omega^k_j $
And also that using the definition of $ \Omega $
$ \Omega (X, Y) = d\omega(X, Y) + \frac{1}{2} \left[\omega(X), \omega(Y) \right] $
However i don't know what to do or how to continue. I tried replacing the u, v and w with basis vectors and got
$ R(\boldsymbol{e}_i, \boldsymbol{e}_j) \boldsymbol{e}_k = \Omega(\boldsymbol{e}_i \wedge \boldsymbol{e}_j) \boldsymbol{e}_k $
$ R^\ell_{ijk} = \Omega(\boldsymbol{e}_i \wedge \boldsymbol{e}_j) \boldsymbol{e}_k $
But i don't know what to do about that Omega there.
I will work with the coordinate vector fields $\partial_i$. Let $(x^1,\dots,x^n)$ be local coordinates on $M$, and $\partial_i = \frac{\partial}{\partial x^i}$ the local basis of coordinate vector fields. Define the Christoffel symbols for the Levi-Civita connection by $$ \nabla_{\partial_i} \partial_j = \Gamma_{ij}^k \partial_k. $$ Then by $[\partial_i, \partial_j]=0$, $\Gamma_{ij}^k = \Gamma_{ji}^k$.
Note that the connection form is $$ \omega_j^k = \Gamma_{ij}^k dx^i. $$
Now we define the curvature tensor by $$ R(\partial_i, \partial_j)\partial_k = \nabla_{\partial_i}\nabla_{\partial_j}\partial_k - \nabla_{\partial_j}\nabla_{\partial_i}\partial_k = R_{ijk}^\ell \partial_\ell. $$ A direct calculation gives $$ R_{ijk}^\ell = \Gamma_{k[j,i]}^\ell + \Gamma_{k[j}^m\Gamma_{i]m}^\ell= \Gamma_{kj,i}^\ell - \Gamma_{ki,j}^\ell + \Gamma_{kj}^m\Gamma_{im}^\ell -\Gamma_{ki}^m\Gamma_{jm}^\ell, $$ where $[\cdot,\cdot]$ is physicists' notation for skew-symmetrization. (Note again $\Gamma_{ij}^k$ is symmetric in $i$ and $j$.)
So in $R_{ijk}^\ell$, $i$ and $j$ are the form indices, and $k$, $\ell$ are the endomorphism indices.
Now the End-valued curvature 2-form is \begin{align*} \Omega_k^\ell &= d\omega_k^\ell + \omega^\ell_m\wedge \omega^m_k\\ &= d(\Gamma_{jk}^\ell dx^j) + \Gamma_{im}^\ell\Gamma_{jk}^mdx^i\wedge dx^j\\ &= \big(\Gamma_{kj,i}^\ell + \Gamma_{kj}^m\Gamma_{im}^\ell\big)dx^i\wedge dx^j\\ &=\frac{1}{2} R_{ijk}^\ell dx^i\wedge dx^j. \end{align*}
At the end, we have $$ R(\partial_i, \partial_j)\partial_k = R_{ijk}^\ell \partial_\ell = \Omega(\partial_i\wedge \partial_j)\partial_k, $$ and by linearity that $$ R({\mathbf u}, {\mathbf v}){\mathbf w}= \Omega({\mathbf u}\wedge {\mathbf v}){\mathbf w}. $$