Question

Covariant derivative

Proof that covariant derivative of a tensor is a tensor.

Answer 1

What I got is\[
\nabla_k T^{i'}=J^{i'}_i\left(\frac{\partial T^i}{\partial Z^{k'}}+\Gamma^i_{kj}T^j\right)+T^iJ^{i'}_{ik}
\]which does not transform according the rule of tensors.

Answer 2

@Kainui I am dying...

Answer 3

Let's start with a vector that's a contraction of two tensors, the components with the basis vectors at that point.

\[\vec V = V^i \vec Z_i\]
Now we already know the derivative of something with no indices is a tensor, so we can just write it here to show it no problem!
\[\frac{\partial \vec V}{\partial Z^i} = \frac{\partial \vec V}{\partial Z^{i'}} \frac{\partial Z^{i'}}{\partial Z^i}\]
Or written slightly nicer using the definition of Jacobian, here we go, the tensor transformation law, so indeed this is a tensor!

\[\frac{\partial \vec V}{\partial Z^i} = \frac{\partial \vec V}{\partial Z^{i'}} J_i^{i'}\]

Ok but I realize you want the covariant derivative, so we take this further and write the vector in terms of its components and use the product rule now.

\[\frac{\partial V^i \vec Z_i}{\partial Z^j}=\frac{\partial V^i}{\partial Z^j}\vec Z_i + V^i \frac{\partial \vec Z_i}{\partial Z^j}\]
Remember all of that above is a tensor. Alright so now what? Well we have to recognize or define it at this point that the derivative of the covariant basis with respect to the coordinates can be expressed as components in terms of the covariant basis and we call these components the Christoffel symbols:

\[\frac{\partial \vec Z_i}{\partial Z^j} = \Gamma^k_{ij} \vec Z_k\]

We can plug this in and since \(i\) and \(k\) are dummy indices, we can change them and it won't matter just so that we can factor out the covariant basis vectors.

\[\frac{\partial \vec V }{\partial Z^j}=\frac{\partial V^i}{\partial Z^j}\vec Z_i + V^i \Gamma^k_{ij}\vec Z_k\]\[\frac{\partial \vec V }{\partial Z^j}=\frac{\partial V^i}{\partial Z^j}\vec Z_i + V^k \Gamma^i_{kj}\vec Z_i\]\[\frac{\partial \vec V }{\partial Z^j}=\left(\frac{\partial V^i}{\partial Z^j}+ V^k \Gamma^i_{kj}\right)\vec Z_i\]

Now we know that the left side of the equation is a tensor. We know the covariant basis is a tensor. Therefore the thing to the left of it is a tensor. We can perhaps show this a bet more nicely by dotting both sides of this with the contravariant basis but I think it's unnecessary. This shows that thing is a tensor. But that thing is the definition of the covariant derivative!

\[\nabla_j V^i = \frac{\partial V^i}{\partial Z^j}+ V^k \Gamma^i_{kj}\]
There are of course other ways to prove this but this is most insightful in my opinion and you can check each piece individually but at this point we have really just entirely by construction made a tensor and called that tensor to be the covariant derivative.

Answer 4

For the way you're doing it, I can give you some more advice on that but it will ultimately boil down to using a lot of the same things I've mentioned above. You have to transform all the indices, and you'll have to show that the covariant derivative of a Jacobian is 0. You need to use more chain rules here, you should only ever have mixed components in your Jacobians, if you see primed with unprimed indices anywhere else you have a problem.

\[\nabla_i T^j =J_i^{i'}\nabla_{i'} (J_{j'}^j T^{j'})\]
Here's your starting point, if you want to do it this way.

Answer 5

I mistyped. What I got is this:
\[
\nabla_k T^{i'}=J^{i'}_i\left(\frac{\partial T^i}{\partial Z^k}+\Gamma^i_{kj}T^j\right)+T^iJ^{i'}_{ik}=J^{i'}_i\nabla_k T^{i}+T^iJ^{i'}_{ik}
\]
The other term clearly is not 0. I tried multiple times but it just won't work.

Answer 6

What does \(\displaystyle \nabla_kT^{i'}\) equals to? In particular how should the Christoffel symbol change from unprimed \(T^i\) to primed \(T^{i'}\)? Maybe I am missing something in the Christoffel symbol because there is nowhere else I could get a \(J^{i'}_{jk}\) to cancel the second term.

Answer 7

I am trying to see how the covariant derivative transform under coordinate transformation. Somehow I made a big mess out of the stuff. The worst part is I seemingly got half of it right. I got this in addition to the mess above.
\[
\nabla_{k'} T^{i'}=J^{i'}_iJ^k_{k'}\nabla_k T^{i}+\text{nonsense}\]

Answer 8

Does it even make sense to talk about \(\nabla_k T^{i'}\)? Granted it is possible to write \(\displaystyle \nabla_k \left(T^{i'}\left(Z\left(Z'\right)\right)\right)\) but is it a meaningful object?

Answer 9

Yeah it's meaningful it's just that it is not as explicit. for instance, if I want you to calculate

\[\frac{\partial}{\partial x} (r^3 - 2 \theta)\]
you'd have to represent it just like you're saying where you plug in what its dependence is of r and theta on x to actually proceed in calculating it.

Answer 10

Yeah this is what you're wanting to prove: \[\nabla_{k'} T^{i'}=J^{i'}_iJ^k_{k'}\nabla_k T^{i}\]
But I think you're running into some tensor algebra problems along the way so we just gotta sort those out. So what you've typed is: \[\nabla_k T^{i'}=J^{i'}_i\left(\frac{\partial T^i}{\partial Z^k} + \Gamma^i_{kj}T^j \right) + T^iJ^{i'}_{ik} = J^{i'}_i\nabla_k T^{i}+T^iJ^{i'}_{ik}\]

I'll take it a little slower and write out what you should be getting:

\[\nabla_k T^{i'} = \nabla_k (J^{i'}_iT^i) = \nabla_k(J_i^{i'}) T^i + J_i^{i'} \nabla_k( T^i)\]
So I've used the product rule here, I don't know if logically it follows to use it right now or not, but you can derive it as well. You'll have to use it in some way or another.

At this point you need to know how to take the covariant derivative of something with multiple indices and a covariant and contravariant index, since specifically this term here will cause you problems.
\[\nabla_k J_i^{i'} \]

Give this a shot, cause you'll run into more issues with it since the chain rule will be necessary in here.

Answer 11

I think you'll find out pretty quickly why I was avoiding this pathway and went the other way around the mountain. I think you should try to hijack the proof I've already given and turn it into the proof you want it to look like instead of starting from scratch, since fundamentally they are the same.

Answer 12

So what you are saying is that \(\displaystyle
\nabla_{k} T^{i'}\neq\frac{\partial T^{i'}}{\partial Z^k}+\Gamma^{i'}_{kj'}T^{j'}
\)?

Answer 13

I'm not saying that, I'm saying
\[
\nabla_k T^{i'} \ne J^{i'}_i\left(\frac{\partial T^i}{\partial Z^k}+\Gamma^i_{kj}T^j\right)+T^iJ^{i'}_{ik}=J^{i'}_i\nabla_k T^{i}+T^iJ^{i'}_{ik}
\]
because that final term is not right, you've just done a single derivative not a covariant derivative on the Jacobian term.

Answer 14

What you should've done is end up here:
\[\nabla_k T^{i'} = J^{i'}_i\nabla_kT^i+T^i \nabla_k J^{i'}_{i}\]
and then now you have to show: \[\nabla_k J^{i'}_{i}=0\]

Answer 15

So when I do this proof I already know the point of the covariant derivative is to free yourself from local baggage of the metric tensor and the basis and move around freely, so all of these (and other related objects) are all zero.

\[\nabla_k Z_{ij} = 0\]\[\nabla_k J_i^{i'}=0\]\[\nabla_k \delta^i_j=0\]\[\nabla_k \vec Z_i = 0\]\[\nabla_k \varepsilon^{ij} = 0\] just to throw a handful of symbols out there.

The proof knowing these along with the product rule for the covariant derivative works is:

\[\nabla_k T^i = J_k^{k'} \nabla_{k'}(J^i_{i'}T^{i'}) =J_k^{k'} \nabla_{k'}(J^i_{i'})T^{i'}+J_k^{k'} J^i_{i'}\nabla_{k'}(T^{i'})=J_k^{k'} J^i_{i'}\nabla_{k'}T^{i'}\]

So if you can prove all these parts, you have proven it.

Answer 16

I am telling you all this because otherwise you really just seem to be fumbling around in the dark lost in the symbols and indices, hopefully this gives you some structure of how your proof should follow. You don't have to prove the product rule independently, you can prove it during your proof of showing the covariant derivative is a tensor, but it will continue to be unmanageable.

Answer 17

Is this correct?
\[
\nabla_k J^{i'}_i=\frac{\partial J^{i'}_i}{\partial Z^k}+\Gamma^{i'}_{km'}J^{m'}_{i}-\Gamma^{m}_{ki}J^{i'}_{m}
\]

Answer 18

Yup looks good!

Answer 19

Then I did the second term,\[
\begin{align*}
\Gamma^{i'}_{km'}J^{m'}_i
&=\left(\frac{\partial\vec{Z_k}}{\partial Z^{m'}}\cdot\vec{Z^{i'}}\right)J^{m'}_i\\
&=\left(\frac{\partial\vec{Z_k}}{\partial Z^j}\frac{\partial Z^j}{\partial Z^{m'}}\cdot\vec{Z^{i'}}\right)J^{m'}_i\\
&=\frac{\partial\vec{Z_k}}{\partial Z^i}\cdot\vec{Z^{i'}}\\
&=\frac{\partial\vec{Z_k}}{\partial Z^i}\cdot\vec{Z^m}J^{i'}_m\\
&=\Gamma^m_{ki}J^{i'}_m\\
\end{align*}
\]And I am screwed because this term cancels the third term, leaving the first term \(J^{i'}_{ik}\).

Answer 20

In the third step, I contracted the two Jacobians together and contracted the resulting delta with the derivative.

Answer 21

If what I did was correct, I will end up in the same situation as \(\nabla_k T^{i'}=J^{i'}_i\nabla_k T^{i}+T^iJ^{i'}_{ik}\) because \(\nabla_kJ^{i'}_i=J^{i'}_{ik}\).

Answer 22

Yeah I think I just realized what the problem is, I didn't remember that jacobian covariant derivative correctly but turns out all the same since even though it's not zero what you end up with is the extra terms.

So if you're ending up with the same situation but with a minus sign on that extra term cancel each other out.

It ends up being this pair here:
\[J_{ik}^{i'}J_{j'}^i + J_i^{i'}J_{j'k'}^iJ_k^{k'} = 0 \]
Which cancel out cause it just follows from differentiating this:
\[J_i^{i'}J_{j'}^i = \delta_j^i\]
Whoops lol, oh well effectively the same.

Answer 23

By hijacking the proof,
\[
\begin{align*}
\frac{\partial \vec{V}}{\partial Z^k}&=\nabla_kV^i\vec{Z_i}\\
&=\nabla_kV^i\vec{Z_{i'}}J^{i'}_i\\
&=\nabla_kV^iJ^{i'}_i\vec{Z_{i'}}
\end{align*}\]which suggest \(\nabla_kV^{i'}=J^{i'}_i\nabla_kV^i\). Then what is up with the extra \(J^{i'}_{ik}\)?

Answer 24

I dunno, I guess I should just sit down and work out the details from scratch haha, I think I'm making some mistake somewhere in my reasoning.

Answer 25

The problem I have is that \(\displaystyle \nabla_k J^{i'}_i=\frac{\partial J^{i'}_i}{\partial Z^k}+\Gamma^{i'}_{km'}J^{m'}_{i}-\Gamma^{m}_{ki}J^{i'}_{m}=\frac{\partial J^{i'}_i}{\partial Z^k}=J^{i'}_{jk}\) because second and third term cancels out. Needless to say this is BAD!

Answer 26

Well ok as you've written it, it's ambiguous since you haven't used parentheses to indicate what the covariant derivative is acting on so I can't tell ya.

Answer 27

\[
\begin{align*}
\frac{\partial \vec{V}}{\partial Z^k}&=\left(\nabla_kV^i\right)\vec{Z_i}\\
&=\left(\nabla_kV^i\right)\vec{Z_{i'}}J^{i'}_i\\
&=\left(J^{i'}_i\nabla_kV^i\right)\vec{Z_{i'}}
\end{align*}
\]

Answer 28

I think there are probably problems coming in when trying to use crossed indices on the Christoffel symbols, I think that's where the issue is.

Answer 29

Probably. But I don't see what is the problem of the crossed indices Christoffel either...

Answer 30

Algebraically it seemed to work except the answer is wrong...

Answer 31

Well the Christoffel symbol is not a tensor so I guess that might be part of it. Although it seemed pretty good using some of the things, I think the main problem was just throwing the thing on it like that I don't think that's quite right. I have some ideas for what might be right but I gotta work it out first.

Answer 32

If there is something wrong with the Christoffel, then it must be the lower k. Only if the lower k is wrong we can possibly get something like \(-J^{i'}_{ik}\) to cancel the first term out.

Answer 33

Hahaha ok it's simple it's right just not complete.

Answer 34

\[\nabla_k J_i^{i'} = J_{ik}^{i'}+J_i^{m'}\Gamma_{m'k}^{i'}-J_m^{i'}\Gamma_{ik}^m\]
Just let the Jacobians do what they do and contract to change variables and get:
\[\nabla_k J_i^{i'} = J_{ik}^{i'}+\Gamma_{ik}^{i'}-\Gamma_{ik}^{i'}\]\[\nabla_k J_i^{i'} = J_{ik}^{i'}\]
So far so good, same thing as last time. Now what's left? Well, starting from here:
\[J_i^{i'}J_{i'}^j = \delta_i^j\]
differentiate it to get:
\[J_{ik}^{i'}J_{i'}^j + J_i^{i'}J_{i'k}^j = 0\]\[J_{ik}^{i'}J_{i'}^j=- J_i^{i'}J_{i'k}^j \]
Again let the Jacobians do what they do and change coordinates so we get new indices:
\[J_{ik}^j =-J_{ik}^j\]
But this means it's zero!
\[J_{ik}^j=0\]
So you can see where we got this from or just contract back to where it came from by contracting both sides with \(J^{i'}_{j}\)
\[J_{ik}^{i'}=0\]
So plug that in and we have shown afterall:
\[\nabla_k J_i^{i'} =0\]

Answer 35

idk maybe I'm cheating and I can't actually do that with Jacobians, I guess I'm getting rusty on this

Answer 36

But clearly \(J^{i'}_{ik}\) couldn't be 0. Just calculate \(J^{i'}_{ik}\) and it is pretty obvious that it is not 0!

Answer 37

The Jacobians do not cancel out. You get two Jacobians on one side and none on the other.

Answer 38

Yeah true, wishful thinking here on that being 0 lol.

Answer 39

I really don't see what else can go wrong except the Christoffel. Suppose \(\displaystyle \nabla_k J^{i'}_{i}=0\), then it must be the case that \(\displaystyle \Gamma^{i'}_{km'}J^{m'}_i=\Gamma^m_{ki}J^{i'}_m-J^{i'}_{ik}\). The question is how to get there.

Answer 40

\[\vec Z_j = J_j^{j'}\vec Z_{j'}\]\[\frac{\partial \vec Z_j}{\partial Z^k} = \frac{\partial J_j^{j'}}{\partial Z^k} \vec Z_{j'} + J_j^{j'} \frac{\partial \vec Z_{j'}}{\partial Z^k}\]\[\Gamma_{jk}^i \vec Z_i =J_{jk}^{j'} \vec Z_{j'}+J_j^{j'}J_k^{k'} \Gamma_{j'k'}^{i'}J_{i'}^i \]

I am looking at this, probably where our problem lies.

Answer 41

Whoops I think I left out a covariant basis vector there on the last term haha whoops, oh well, I think if we use this to transform rather than whatever we were doing, we'll snag that extra term right there.

Answer 42

\[
\nabla_k J_i^{i'} = J_{ik}^{i'}+J_i^{m'}\Gamma_{m'k}^{i'}-J_m^{i'}\Gamma_{ik}^m
\]Transform which term using that? The third?

Answer 43

You live in America right? I think it is pretty late/early by now. Maybe we should do this tomorrow. Meanwhile I have to get some lunch.

Answer 44

Can we even take the covariant derivative of Jacobian? Googling "covariant derivative jacobian" returned an ominously small number of relevant results...

PS I was not sure whether it is "small amount" or "little amount". It is "small number".