Calculation of Lie derivative - follow up

[cianfa72]

TL;DR Summary

About the calculation of Lie derivative - follow up

Hi, a doubt related to the calculation done in this old thread.

$$\left(L_{\mathbf{X}} \dfrac{\partial}{\partial x^i} \right)^j = -\dfrac{\partial X^j}{\partial x^i}$$
$$L_{\mathbf{X}} {T^a}_b = {(L_{\mathbf{X}} \mathbf{T})^a}_b + {T^{i}}_b \langle L_{\mathbf{X}} \mathbf{e}^a, \mathbf{e}_i \rangle + {T^{a}}_j \langle \mathbf{e}^j, L_{\mathbf{X}} \mathbf{e}_b \rangle$$
The second term on RHS should actually be ${T^{i}}_b \left(L_{\mathbf{X}} \dfrac{\partial x_i}{\partial x^a} \right)^j= -{T^{i}}_b\dfrac{\partial X^i}{\partial x^a}$

I'am confused about it.

 

[ergospherical]

I'm pretty sure I'm right (but I would say that, wouldn't I...). Taking the coordinate basis $\mathbf{e}^a = dx^a$ you have obviously $(\mathbf{e}^a)_i = \delta^a_i$, and since by this point we would have already worked out the expression for the Lie derivative of a covector (having first derived it for functions, then vectors, using normal coordinates or otherwise), you have
\begin{align*}
(L_X \mathbf{e}^a)_i = \frac{\partial \delta^a_i}{\partial x^j} X^j + \frac{\partial X^j}{\partial x^i}\delta^a_j = 0 + \frac{\partial X^a}{\partial x^i}
\end{align*}
with no minus.

 

[cianfa72]

Taking the coordinate basis $\mathbf{e}^a = dx^a$ you have obviously $(\mathbf{e}^a)_i = \delta^a_i$

Ah ok, so for you $\mathbf{e}^a$ is an element of covector basis. I was confused by that.

The point was that an upper index to me resulted in a vector element (see for instance abstract index notation that uses Latin letters to assign names to slots).

To me mixing up different notation for tensor calculus is really a nightmare

 

[cianfa72]

As far as I can tell, the contraction for instance of the second term on RHS in post#1 is done contracting the first 'slot' of tensor $\mathbf T$ with the covector $L_{\mathbf X}\mathbf e^a$ and the second 'slot' of $\mathbf T$ with the vector $\mathbf e_b$. Right?

 

[cianfa72]

Another doubt about the calculation of covariant derivative of metric tensor evaluated on a pair of vector fields.

Take $\nabla$ as a metric compatible covariant derivative operator (i.e. the covariant derivative operator associated to the Levi-Civita connection). Then is abstract index notation:
$$\nabla_X \, g(X,Y)= X^c\nabla_cg_{ab}X^aY^b = X^aY^bX^c \nabla_c g_{ab} + g_{ab}[Y^bX^c\nabla_cX^a + X^aX^c\nabla_cY^b]$$ The first term on RHS vanishes since the covariant derivative operator is metric compatible by assumption. The other two terms instead correspond to the appropriate contractions of the metric tensor. So coming back to the first notation we get:
$$\nabla_X \, g(X,Y) = g(\nabla_XX,Y) + g(X,\nabla_XY)$$ Is the above correct ?

 

[cianfa72]

Nobody? Thanks.

 

[ergospherical]

There's no question to answer, you've just been stating things.

 

[cianfa72]

There's no question to answer, you've just been stating things.

Yes, I would kindly ask if my calculation in post#5 is correct, thanks.

 

[PeterDonis]

Yes, I would kindly ask if my calculation in post#5 is correct, thanks.

Does contraction (or, equivalently, taking an inner product) commute with covariant differentiation?

 

[cianfa72]

Does contraction (or, equivalently, taking an inner product) commute with covariant differentiation?

Yes, it does. However I'm not sure if contraction is actually needed in the calculation.

 

[PeterDonis]

I'm not sure if contraction is actually needed to do the calculation.

An expression like $g(X, Y)$ is contraction (or, as I said, taking an inner product; they're the same thing when you have two vectors).

 

[PeterDonis]

Yes, it does.

Ok, good. Next question: does covariant differentiation obey the chain rule?

 

[cianfa72]

does covariant differentiation obey the chain rule?

Yes, it does. However I believe the Leibniz rule is actually involved here.

$\nabla_X \, g(X,Y)= X^c\nabla_cg_{ab}X^aY^b$

 

[PeterDonis]

Yes, it does.

Ok, good.

However I believe the Leibniz rule is actually involved here.

$\nabla_X \, g(X,Y)= X^c\nabla_cg_{ab}X^aY^b$

The Leibniz rule involves differentiating under an integral sign. There are no integrals here.

The chain rule involves differentiating a product. That's what you are doing here--differentiating an inner product. And since you have agreed that covariant differentiation commutes with the inner product, and obeys the chain rule, you now have the answer to the question you posed at the end of post #5. Do you see why?

 

[cianfa72]

The Leibniz rule involves differentiating under an integral sign.

Sean Carrol and Wald talk of Leibniz rule for covariant derivative of products (including tensor products). The chain rule, instead, applies to composition of maps.

 

[Orodruin]

The Leibniz rule involves differentiating under an integral sign. There are no integrals here.

Not that Leibniz rule. The other Leibniz rule.

 

[PeterDonis]

Not that Leibniz rule. The other Leibniz rule.

Ah, ok, otherwise known as the "product rule".

 

[PeterDonis]

The chain rule, instead, applies to composition of maps.

The inner product is a map (from a pair of vectors to a scalar). But I agree that "product rule" is more appropriate for this case.

 

[Orodruin]

Ah, ok, otherwise known as the "product rule".

You'd have thought Leibniz would be fine with one rule, but noooo ... Had to go and make more contributions to science ...

 

[cianfa72]

So an expression like $X^c\nabla_cg_{ab}X^aY^b$ is actually the covariant derivative $X^c\nabla_c$ applied to the tensor product $g_{ab}X^dY^e$ followed by the contraction according abstract indexes $d=a$, $e=b$.

 

[PeterDonis]

So an expression like $X^c\nabla_cg_{ab}X^aY^b$ is actually the covariant/directional derivative $X^c\nabla_c$ applied to the tensor product $g_{ab}X^dY^e$ followed by the contraction according abstract indexes $d=a$, $e=b$.

As the expression is written, the contraction occurs first, then the derivative (since we normally apply operators right to left). But since contraction commutes with covariant differentiation, it doesn't matter which occurs first. That's why it's fine to pull the covariant derivatives inside the inner product when you apply the product rule, as you did in post #5.

 

[cianfa72]

I was following the abstract index notation from Wald pag 31. He uses a bracket () to mean contraction "inside" the covariant derivative operator. Therefore a writing as in post #20 should mean: take the covariant derivative along $X$ of the tensor product $g \otimes X \otimes Y$ and then contract.

 

[cianfa72]

Another point related to this. In coordinate-free notation the contraction of a covariant "slot" with a controvariant "slot" of a tensor should mean fill in the slots accordingly using any vector basis and its associated dual bases. Just pick any basis for the vector space and the associated dual basis.

 

[PeterDonis]

In coordinate-free notation the contraction of a covariant "slot" with a controvariant "slot" of a tensor should mean fill in the slots accordingly using any vector basis and its associated dual bases.

Yes, that's fine, but, as I pointed out in my previous post just now, neither $A$ nor $B$ are covectors, so neither one has a covariant slot. Both have only contravariant slots. That's why you need the metric to contract them (which means, to take their inner product).

 

[cianfa72]

Yes, that's fine, but, as I pointed out in my previous post just now, neither $A$ nor $B$ are covectors, so neither one has a covariant slot. Both have only contravariant slots. That's why you need the metric to contract them (which means, to take their inner product).

Yes of course. Nevertheless in last posts I was talking of cases in which the contraction actually makes sense (e.g. $A$ is a vector and $B$ is dual vector). Coming back to post#22 I was simply talking about the notation used by Wald p.31 number 3) that is general and applies to all kind of tensor (not just to a tensor product involving the metric tensor $g$).

 

[PeterDonis]

in last posts I was talking of cases in which the contraction actually makes sense (e.g. $A$ is a vector and $B$ is dual vector).

Where have you said from your post #5 on (since that's the subthread in which we are having this discussion) that anything is a dual vector? Look at your post #5: you gave $X$ and $Y$ upper indexes. Where does anything appear in any of your posts from then on that has a lower index other than the metric and the nabla operator?

 

[PeterDonis]

Coming back to post#22 I was simply talking about the notation used by Wald p.31 number 3) that is general

His notation is general, but you were saying you were using it in a specific post of yours, in which everything except the metric and the nabla operator has an upper index. Not only that, your post #5 involves inner products of vectors, which, as I have already pointed out, requires a metric.

If you want to ask about an expression that involves contracting a vector and a covector, with no metric involved, you can of course do that. But so far in this thread you have not written a single such expression.

 

[cianfa72]

His notation is general, but you were saying you were using it in a specific post of yours, in which everything except the metric and the nabla operator has an upper index. Not only that, your post #5 involves inner products of vectors, which, as I have already pointed out, requires a metric.

Maybe I caused confusion, sorry. W.r.t. post#5 I was just saying that following the Wald's notation $X^c\nabla_cg_{ab}X^aY^b$ actually means take the covariant derivative of tensor product $g \otimes X \otimes Y$ along $X$ first and then take the contraction.

 

[PeterDonis]

Maybe I caused confusion

Not to me. I understood your post #5 perfectly. The only thing I was questioning was this from your post #25:

in last posts I was talking of cases in which the contraction actually makes sense

I was simply pointing out that so far in this thread you have not described any such cases. Every specific case you have described involves two vectors, not a vector and a covector.