# Connection 1-forms?

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

$$$$\nabla_XE_i=\omega_i^j(X)E_j$$$$ for all $$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 $$X$$?

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 $$\nabla_XE_i(x)=\sum_j\omega_i^j(x)E_i(x)$$.
• 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? – y9QQ Oct 7 '19 at 23:05
• 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. – Tsemo Aristide Oct 7 '19 at 23:10