- #1

semigroups

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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 in to 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?

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 in to 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?

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