Covariant derivative of a 1-form

[semigroups]

Let $$f:U \to \mathbb{R}^3$$ be a surface with local coordinates $$f_i=\frac{\partial f}{\partial u^i}$$. Let $\omega$ be a one-form. I want to express $$\nabla \omega$$ in terms of local coordinates and Christoffel symboles. Where $$\nabla$$ is the Levi-Civita connection (thus it coincides with the covariant derivative).

Let $$X=X^if_i,Y=Y^if_i$$ be two tangent vector fields on $f$, by definition one has:
$$(\nabla_X \omega)(Y)=X^i \frac{\partial (\omega(Y))}{\partial u^i}-\omega (\nabla_XY)=X^i \frac{\partial (\omega(Y^if_i))}{\partial u^i}-\omega(\nabla_{X^if_i} Y^if_i)$$

I know $$\omega(\nabla_{X^if_i} Y^if_i)=\omega (X^i[Y^{j}(\nabla_{f_i}f_j)+Y^{j}_{,i}f_j])=\omega (X^i[Y^{j}(\Gamma^{k}_{ij}f_{k})+Y^{j}_{,i}f_j])=X^{i}Y^{j}\Gamma^{k}_{ij} \omega (f_k)+X^iY^i_{,i} \omega(f_j)$$

Now to compute $$\frac{\partial (\omega(Y^if_i))}{\partial u^i}=Y^{i}_{i} \omega (f_i)+Y^i \frac{\partial(\omega (f_i)))}{\partial u^i}$$.

Substitute them into the formula for $$\nabla_X \omega (Y)$$ to get a result dependent on the partial derivatives of $Y$ (after cancellations), but this should not be the case, where did I go wrong?

 

[quasar987]

The covariant derivative of a 1-form \omega is a 1-form \nabla_X\omega. And a 1-form (i.e. a field of covectors) eating a vector field Y does not depend on the partial derivatives of the components of Y:

\alpha(Y=\sum Y^if_i)=\sum Y^i\alpha(f_i)

So why do you expect \nabla_X\omega (Y) to behave differently? ;)

 

[semigroups]

Thanks for pointing it out! My answer should be $$(\nabla_X \omega)(Y)=X^iY^j(\omega_{j,i}-\Gamma^{k}_{ij} \omega_k)$$

I'm now trying to differentiate a covariant 2-tensor (may not be a 2-form), namely $$(\nabla_{X}S)(Y,Z)=X^i \frac {\partial S(Y,Z)}{\partial u^i}-S(\nabla_{X}Y,Z)-S(Y,\nabla_{X}Z)$$

If denote S(f_i,f_j) to be S_{ij}, is it still valid to partial differentiate with respect to $$u^k$$: $$\frac{\partial S(f_i,f_j)} {\partial u^k}=S_{ij,k}$$?? If so, then the covariant derivative of a 2-tensor field will depend on the partial derivatives of the vector fields $$X,Y,Z$$ then?

 

[dextercioby]

Again, why would the result depend on derivatives of components of the vectors ? The vectors are merely <helping tools> to make things general.