Curvature Matrix on a Smooth Vector Bundle

Question

Suppose $\pi:E\rightarrow M$ is a smooth vector bundle with a metric connection $\nabla$, and let $(\sigma_i)$ be a local frame over some open $U\in M$. Why is the curvature \begin{equation} R(X,Y)\sigma_i=\nabla_X\nabla_Y\sigma_i-\nabla_Y\nabla_X\sigma_i-\nabla_{[X,Y]}\sigma_i \end{equation} a $\text{Hom}(E,E)$-valued 2-form? Further, how does this translate into a matrix form \begin{equation} \Omega(\sigma_i)=\Omega_{ij}\sigma_j, \end{equation} where $\Omega_{ij}$ is an antisymmetric matrix? I had thought I understood how to construct $\Omega_{ij}$, but I can't come up with any reason for \begin{equation} \Omega_{ij}=-\Omega_{ji} \end{equation} to be true, unless there is some way of indicial symmetry which I don't see for the Christoffel symbols, defined here by \begin{equation} \nabla_X\sigma_i=\Gamma_{ik}^jX^k\sigma_j, \end{equation} where $X$ is a smooth vector field on $M$.

Answer 1

Why is the curvature \begin{equation} R(X,Y)\sigma_i=\nabla_X\nabla_Y\sigma_i-\nabla_Y\nabla_X\sigma_i-\nabla_{[X,Y]}\sigma_i \end{equation} a $\text{Hom}(E,E)$-valued 2-form?

The point is that the map $(X,Y,\sigma) \mapsto R(X,Y)\sigma$ is multilinear over $C^\infty(M)$, which you can check by inserting an expression like $f_1 X_1 + f_2 X_2$ in place of $X$ and checking that all the terms involving derivatives of $f_1,f_2$ cancel out. Because it depends antisymmetrically on $X,Y$, and the result for any fixed $X,Y$ is a map from $\Gamma(E)$ to itself that's linear over $C^\infty(M)$, this shows that it's actually a smooth $\text{End}(E)$-valued $2$-form. The basic ideas are spelled out in the proof of Lemma 10.29 in my Introduction to Smooth Manifolds, 2nd edition.

Further, how does this translate into a matrix form \begin{equation} \Omega(\sigma_i)=\Omega_{ij}\sigma_j, \end{equation} where $\Omega_{ij}$ is an antisymmetric matrix?

The formula you wrote is really a shorthand for $$ R(X,Y)\sigma_i = \Omega_{ij}(X,Y)\sigma_j. $$ This defines $\Omega_{ij}$ for each $i$ and $j$ as a $2$-form on $U$. It will be antisymmetric in $i,j$ provided the frame $\boldsymbol{\sigma_j}$ is orthonormal. I don't have time to write out a complete proof now, but it follows from the fact that with respect to an orthonormal basis, the connection $1$-forms take values in the Lie algebra $\mathfrak{so}(n)$. This in turn follows from the fact that the connection is metric-compatible.