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Covariant derivative of a 1-form

  1. Sep 25, 2012 #1
    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?
     
    Last edited: Sep 25, 2012
  2. jcsd
  3. Sep 25, 2012 #2

    quasar987

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    The covariant derivative of a 1-form [itex]\omega[/itex] is a 1-form [itex]\nabla_X\omega[/itex]. 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:

    [tex]\alpha(Y=\sum Y^if_i)=\sum Y^i\alpha(f_i)[/tex]

    So why do you expect [itex]\nabla_X\omega (Y)[/itex] to behave differently? ;)
     
  4. Sep 25, 2012 #3
    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?
     
  5. Sep 25, 2012 #4

    dextercioby

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