Covariant derivative of a (co)vector field

  • #1
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Homework Statement:
Calculate ##(\nabla_X Y)^i## and ##(\nabla_X \omega)_i##, where ##X## is a vector field and ##\omega## is a covector field.
Relevant Equations:
Definition of the covariant derivative:
A covariant differentiation on a manifold ##\mathcal{M}## is a mapping ##\nabla## which assigns to every pair ##X, Y## of ##C^\infty## vector fields on ##\mathcal{M}## another ##C^\infty## vector field ##\nabla_X Y## with the following properties:

1) ##\nabla_{X} Y \text { is bilinear in } X \text { and } Y##,
2) ##\text { if } f \in \mathcal{F}(\mathcal{M}), \text { then }##
$$\begin{array}{l}{\nabla_{f X} Y=f \nabla_{X} Y} \\ {\nabla_{X}(f Y)=f \nabla_{X} Y+X(f) Y}\end{array}.$$
My attempt so far:
$$\begin{align*}
(\nabla_X Y)^i &= (\nabla_{X^l \partial_l}(Y^k\partial_k))^i=(X^l \nabla_{\partial_l}(Y^k\partial_k))^i\\
&\overset{2)}{=} (X^l (Y^k\nabla_{\partial_l}(\partial_k) + (\partial_l Y^k)\partial_k))^i = (X^lY^k\Gamma^n_{lk}\partial_n + X^lY^k{}_{,l}\partial_k)^i\\
&= (X^lY^k\Gamma^n_{lk}\partial_n + X^lY^k{}_{,l}\partial_k) x^i\\
&= X^lY^k\Gamma^i_{lk}+ X^lY^i{}_{,l}
\end{align*}$$
This seems reasonable to me, at least all the dummy indices disappear. But when I try to do the same for the covector field I quickly run into problems...
$$\begin{align*}
(\nabla_X \omega)_i &= (\nabla_{X^l \partial_l}(\omega_kdx^k))_i = (X^l\nabla_{\partial_l}(\omega_kdx^k))_i \\
&= (X^l(\omega_k \nabla_{\partial_l}(dx^k) +( \partial_l \omega_k) dx^k))_i \overset{(*)}{=}(X^l\omega_k \nabla_{\partial_l}(dx^k) +X^l \omega_{k,l} dx^k)\partial_i \\
&= X^l\omega_k \nabla_{\partial_l}(dx^k) (\partial_i) +X^l \omega_{k,l} dx^k(\partial_i) = X^l\omega_k \nabla_{\partial_l}(dx^k) (\partial_i) +X^l \omega_{i,l}
\end{align*}$$
After ##(*)## I started guessing, but now I have to evaluate ##\nabla_{\partial_l}(dx^k) ## and I'm not really sure what this is... If I had to guess I'd probably say
$$\nabla_{\partial_l}(dx^k) = U_{ln}^{k}dx^n,$$
where ##U_{ln}^{k}## is somehow related to ##\Gamma^n_{lk}##. Can someone maybe help me here? How exactly is ##\nabla_{\partial_l}(dx^k) ## defined?
 

Answers and Replies

  • #2
Orodruin
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What is ##\nabla_X(dx^a(\partial_a))##?
 
  • #3
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I think it's ##\nabla_X (dx^a(\partial_a)) = \nabla_X \delta^a_a = \nabla_X##, or isn't it?
 
  • #4
Orodruin
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Sorry, I meant to write ##\nabla_X(dx^a(\partial_b))##. Either way, you cannot just remove the delta as you have done in the last step.
 
  • #5
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I'm sorry, I can't follow you... I mean we then would have ##\nabla _X (dx^a (\partial_b))= \nabla_X \delta^a_b = X^i\nabla_{\partial_i}\delta^a_b##. But what now? Also, why exactly are we considering this? I mean, there is a difference between ##\nabla _X (dx^a (\partial_b)) ## and ##(\nabla _X dx^a) (\partial_b)##, or isn't there?
 
  • #6
Orodruin
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We are considering this because there are two ways of computing it. The delta is a constant. What is the covariant derivative of a constant? Can you think of a second way of computing the original expression? (Yes, the two expressions you mentioned are inequivalent, and that is a further hint.)
 
  • #7
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The delta is a constant. What is the covariant derivative of a constant?
I don't know. Usually I'd say the "derivative" of a constant is zero, but I'm really not sure if this is true when talking about the covariant derivative... The only things I know about this derivative are written above, and we defined ##\nabla_{\partial_i}\partial_j := \Gamma^k_{ij}\partial_k##.

Can you think of a second way of computing the original expression?
With this you mean ##\nabla_{\partial_i}dx^a##? I'm sorry, but I don't really have any other ideas..
 
  • #8
Orodruin
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One of the conditions on the covariant derivative is that it reduces to a directional derivative on scalar fields.

Another is that ##\nabla_X(\omega(V)) = \omega(\nabla_X V) + (\nabla_X \omega)(V)##, where ##\omega## is a one-form and V and X vectors.
 
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  • #9
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Alright, so if I don't misunderstand we have
$$\begin{align*}
(\nabla_{\partial_l}dx^k)(\partial_i) &= \nabla_{\partial_l}(dx^k(\partial_i)) - dx^k(\nabla_{\partial_l}\partial_i)\\
&= \nabla_{\partial_l}\delta^k_i - dx^k(\Gamma^u_{li}\partial_u)\\
&= \nabla_{\partial_l}\delta^k_i - \Gamma^k_{li} = -\Gamma^k_{li},
\end{align*}$$
which then leads directly to
$$(\nabla_X \omega)_i= X^k\omega_{i,k} - \Gamma^j_{ki}X^k\omega_j.$$
Thank you very much for the help!
 
  • #10
Orodruin
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Yes, that is correct.
 

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