I'm trying to solve Problem $4.12$ in John Lee's book Introduction to Smooth Manifolds. It states the following:
Suppose $M$ is a smooth manifold with or without boundary, $\nabla$ is a connection in $TM$, and $A$ is a smooth vector or tensor field on $M$. Then $A$ is parallel if and only if $\nabla A\equiv 0$.
I'm trying to prove this for the vector field case, but I'm having some trouble. Using Lee's notation on Page 98 of the book, for $A\in \mathfrak{X}(M)$ we can write the $(1,1)$-tensor field $\nabla A$ as $$\nabla A= A^i_{;j}E_i\otimes \epsilon^j=\left(E_jA^i+A^k\Gamma^i_{jk}\right)E_i\otimes \epsilon^j$$ by writing $A=A^iE_i$ in some local frame $(E_i)$ and corresponding dual frame $(\epsilon^j)$. Hence if $A$ is parallel, then the covariant derivative $D_tA(t)\equiv 0$ for every smooth curve $\gamma$ on $M$.
Question: From my understanding, if $D_tA(t)\equiv 0$ this means that $$\dot{A}^k(t)+ \dot{\gamma}^i(t)A^j(t)\Gamma^k_{ij}(\gamma(t))=0$$ where $k=1,\ldots, n$. Hence to solve the problem I'd ideally want to have $$\left(\dot{A}^k(t)+ \dot{\gamma}^i(t)A^j(t)\Gamma^k_{ij}(\gamma(t))\right)E_k=\left(E_jA^i+A^k\Gamma^i_{jk}\right)E_i\otimes \epsilon^j$$ somehow, but the righthand side is written in terms of a local frame and coframe while the lefthand side isn't, so I'm not sure how things cancel out here to yield an equality or bring something illuminating to light.
Any help on this problem is appreciated, sorry for the trouble! I appreciate the help.
I know @TedShifrin is trying to guide you through the process, but let me write it out more explicitly.
So your $\text{LHS}\cdot E_k = D_t A(t),$ and your RHS is $\nabla A$.
If you use $(\nabla A)(\dot \gamma) = D_t A(t)$, and $\dot \gamma = \dot \gamma^\ell E_\ell$, you at least will match these two expressions.