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I'm working through Lee's "Riemannian Manifolds: an Introduction to Curvature" by myself, and I'm a bit stuck on problem 4-5:

Let $\nabla$ be a connection on $M$, let $\{E_i\}$ be a local frame on some open subset $U\subset M$, and let $\{\phi^i\}$ be the dual coframe. Show that there is a uniquely determined matrix of $1$-forms $\omega_i^j$ on $U$, called the connection $1$-forms for this frame, such that

\begin{equation} \nabla_XE_i=\omega_i^j(X)E_j \end{equation} for all $X\in TM$.

I know my understanding of differential forms is lacking, but I don't even understand where they come into this problem... Are we just saying that, since the covariant derivative on a section produces a new section, there must be a linear transformation between them, and this linear transformation depends on the direction $X$?

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It is exactly what you think, since $E_i$ is a local frame $\{E_i(x)\}$ is a basis of $T_xM$ for every $x\in U$; $\nabla_XE_i$ is an element of $TM_{\mid U}$, there exits $\omega_i(x)$ such that $\nabla_XE_i(x)=\sum_j\omega_i^j(x)E_i(x)$.

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  • $\begingroup$ I'm trying to think of this in terms of arbitrary smooth vector bundles, so that generally these sections are in $\Gamma(E)\neq\Gamma(TM)$. I see no problems with this, since all of these objects can be extended from a local trivialization of $E$ via partitions of unity, so long as $M$ is paracompact. In this case, what are the differential forms in the connection matrix? I can't imagine that they are the same as the forms from the cotangent bundle... Further, how does asymmetry in $i$ and $j$ come from this definition? $\endgroup$ – y9QQ Oct 7 '19 at 23:05
  • $\begingroup$ The same idea can be applied to any vector bundle. You can show that the form $\omega_i$ is a linear of $X$ (use the definition of the covariant derivative) and differentiable. $\endgroup$ – Tsemo Aristide Oct 7 '19 at 23:10
  • $\begingroup$ Where does the asymmetry come from? $\endgroup$ – y9QQ Oct 8 '19 at 0:06

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