# Help with covariant differentiation

• A
I'm having trouble evaluating the following expression (LATEX):

##\nabla_{i}\nabla_{j}T^{k}= \nabla_{i} \frac{\delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{i m} \frac{\delta T^{m}}{\delta z^{i}} + \Gamma^{k}_{i m} \Gamma^{m}_{i l} T^{l}##

What are the next steps to complete the covariant differentiation? (This is not a homework assignment)

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Orodruin
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What are the next steps to complete the covariant differentiation?
Getting the first step correct. Your expression even has different free indices on both sides.

wrobel
If you will
I'm having trouble evaluating the following expression (LATEX):

##\nabla_{i}\nabla_{j}T^{k}= \nabla_{i} \frac{\delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{i m} \frac{\delta T^{m}}{\delta z^{i}} + \Gamma^{k}_{i m} \Gamma^{m}_{i l} T^{l}##

What are the next steps to complete the covariant differentiation? (This is not a homework assignment)

Just use basis and these formulas will be easy to write. For example let ##A=a_i^je^i\otimes e_j## then $$(\nabla_k A)e^k=(\frac{\partial a_i^j}{\partial x^k}e^i\otimes e_j+a_i^j(\nabla_ke^i)\otimes e_j+a_i^je^i\otimes(\nabla_k e_j))\otimes e^k$$ and remember that $$\nabla_ke_i=\Gamma_{ik}^se_s,\quad \nabla_ke^i=-\Gamma_{sk}^ie^s$$

Sorry. Better?

##\nabla_{i}\nabla_{j}T^{k}= \nabla_{i} \frac{\delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{j l} \frac{\delta T^{l}}{\delta z^{i}} + \Gamma^{l}_{i m} \Gamma^{k}_{j l} T^{m}##

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The first covariant derivative I get. I'm having trouble with the second covariant differentiation ##\nabla_{i}## of ##\nabla_{j} T^{k}##, where ##\nabla_{j} T^{k} = \frac{ \delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{j l} T^{l}##.

Orodruin
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The first covariant derivative I get. I'm having trouble with the second covariant differentiation ##\nabla_{i}## of ##\nabla_{j} T^{k}##, where ##\nabla_{j} T^{k} = \frac{ \delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{j l} T^{l}##.

How does the covariant derivative act on a general type (1,1) tensor? How can you apply that to the type (1,1) tensor ##\nabla_j T^k##?

How does the covariant derivative act on a general type (1,1) tensor? How can you apply that to the type (1,1) tensor ##\nabla_j T^k##?

You mean the invariant ##T##? Where ##T=T^{k} z_{k}##?

##\nabla_{j} T = \nabla_{j} T^{k} z_{k} = (\frac{ \delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{j l} T^{l}) z_{k}##?

Orodruin
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You mean the invariant ##T##? Where ##T=T^{k} z_{k}##?

##\nabla_{j} T = \nabla_{j} T^{k} z_{k} = (\frac{ \delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{j l} T^{l}) z_{k}##?
No, I mean a general type (1,1) tensor ##T_i^j##.

##\nabla_{k} T^{i}_{j} = \frac{ \delta T^{i}_{j}}{\delta z^{k}} + \Gamma^{i}_{k m} T^{m}_{j} - \Gamma^{m}_{j k} T^{i}_{m} ##

The tensor ##T^{i}_{j}## can be contracted ## T^{i}_{j} z^{j} = T^{i} ##?

##\nabla_{i}\nabla_{j}T^{k}= \nabla_{i} \frac{\delta T^{k}}{\delta z^{j}} + \Gamma^{k}_{j l} \frac{\delta T^{l}}{\delta z^{i}} + \Gamma^{l}_{i m} \Gamma^{k}_{j l} T^{m}##

##\frac{\delta T }{\delta z^{j}} = \frac{\delta (T^{k} z_{k})}{\delta z^{j}} ## and then solve for ##\frac{\delta T^{k}}{\delta z^{j}}##?

One thought I had was to consider ##T^{k}## a function of ##z_{k}##, i.e., ##T^{k} = T^{k}(z_{k})##, such that ##\frac{\delta T^{k}}{\delta z^{j}} = \frac{\delta T^{k}}{\delta z^{k}} \frac{\delta z_{k}}{\delta z^{j}}## and ##\frac{\delta z_{k}}{\delta z^{j}} = \Gamma^{m}_{k j} z_{m}##

Orodruin
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