# help on connections

In the book on Riemannian Geometry by John Lee ("Riemannian manifolds: an Introduction to Curvature") the author gives an exercise on page 54 involving connections:

Let $\triangledown$ be a linear connection. If $\omega$ us a 1-form and $X$ is a vector field, show that the coordinate expression for $\triangledown_X \,\omega$ is $$\triangledown_X \, \omega = \left(X^i \partial_i \omega_k - X^i w_j \Gamma^j_{ik} \right) \, dx^k$$ where $\{\, \Gamma^k_{ij} \,\}$ are the Christoffel symbols of the given connection $\triangledown$ on TM.

I am a little stuck here, I appologize as I sense this should be really easy to solve. My guess is that I could use the property of the natural pairing $$\langle \partial_j , \ dx^k \, \rangle = \delta_k^j$$ together with the property $$\begin{eqnarray} &0 = \triangledown_{\partial_i} \delta_k^j = \triangledown_{\partial_i} \langle dx^k, \delta_j \, \rangle = \langle \triangledown_{\partial_i} \, dx^k, \partial_j \, \rangle + \langle dx^k, \triangledown_{\partial_i} \partial_j \, \rangle \\ &= \langle \triangledown_{\partial_i} \, dx^k, \partial_j \, \rangle + \langle dx^k, \Gamma^k_{ij} \partial_k \, \rangle = \langle \triangledown_{\partial_i} \, dx^k, \partial_j \, \rangle + \Gamma^k_{ij} \end{eqnarray} \quad \text{(is this correct ?)}$$ but I am not sure how to proceed or where I am having errors in my argument so far.. any help would be great ! sorry again if this is obvious .. thanks !

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In the book (directly above the exercise on the same page) there is formula (4.7) - which for a tensor $F$ says that $$(\nabla_X F)(Y) = X(F(Y)) - F(\nabla_XY)$$ Just set $F=\omega$ to get the formula from Thomas answer $$(\nabla_X \omega)(Y) = X(\omega(Y)) - \omega(\nabla_XY)$$ and plug in $\omega=\omega_i dx^i$ and $X=X^j\partial_j$
I don't know that book, so I don't know how it introduces the nabla operator on forms. But assuming you know the corresponding formula for the covariant derivative of vector fields and the fact that, for each smooth vector field $Y$, $$X(\omega(Y)) = (\nabla_X \omega)(Y) + \omega(\nabla_XY)$$ (the left hand side being the derivative of the function $\omega(Y)$ in direction $X$) you can just solve this for $(\nabla_X \omega)(Y)$ and choose $Y=\partial_i$, say. Then plug in the coordinate representations of the other terms.