Deriving the Lie Derivative of a Covector: The Leibniz Rule

[latentcorpse]

Use the Leibniz rule to derive the formula for the Lie derivative of a covector $\omega$ valid in any coordinate basis:
$(L_X \omega)_\mu = X^\nu \partial_\nu \omega_\mu + \omega_\nu \partial_\mu X^\nu$
(Hint: consider $(L_X \omega)(Y)$ for a vector fi eld $Y$).

Well I have the formula $L_X(Y) = [X,Y]$ but how do i deal with it when there is that $\omega$ thrown in there as well?

 

[fzero]

Do you know how to express the product of a covector with a vector, $$\omega Y$$? What degree does the resulting object have?

 

[latentcorpse]

Do you know how to express the product of a covector with a vector, $$\omega Y$$? What degree does the resulting object have?

I'm sure the product bit is in my notes but I can't find it. As for the degree bit, I'm fairly sure we haven't covered that.

 

[fzero]

$$\omega Y = \omega_\mu Y^\mu$$ is degree (or rank) zero, so it's just a function. What you want to do is write

$$L_X (\omega Y) = (L_X \omega) Y + \omega (L_X Y).$$

You should already have learned how to compute the Lie derivative of a function and vector field, so you can solve for $$(L_X \omega) Y$$ and hence $$L_X \omega$$.

 

[latentcorpse]

$$\omega Y = \omega_\mu Y^\mu$$ is degree (or rank) zero, so it's just a function. What you want to do is write

$$L_X (\omega Y) = (L_X \omega) Y + \omega (L_X Y).$$

You should already have learned how to compute the Lie derivative of a function and vector field, so you can solve for $$(L_X \omega) Y$$ and hence $$L_X \omega$$.

well $L_X f = X(f) \Rightarrow L_X ( \omega Y) = X( \omega Y)$
and
$L_X Y = [X,Y]$

Therefore, $(L_X \omega)(Y) = L_X ( \omega Y) - \omega ( L_X Y ) = X ( \omega Y) - \omega [X,Y]$

Is this correct? If so, how do I get to the formula I wrote down in post 1?

 

[dextercioby]

Well, use the formulas from here http://en.wikipedia.org/wiki/Lie_derivative

 

[latentcorpse]

Well, use the formulas from here http://en.wikipedia.org/wiki/Lie_derivative

How is that going to help though? Then I'm going back a step again am I not? As I'll have a $L_X Y$term again, no?

 

[dextercioby]

What's

$$\mathcal{L}_{X}\omega(Y)$$ equal to ?

What about

$$\omega\left(\mathcal{L}_{X}Y\right)$$ ?

 

[fzero]

well $L_X f = X(f) \Rightarrow L_X ( \omega Y) = X( \omega Y)$
and
$L_X Y = [X,Y]$

Therefore, $(L_X \omega)(Y) = L_X ( \omega Y) - \omega ( L_X Y ) = X ( \omega Y) - \omega [X,Y]$

Is this correct? If so, how do I get to the formula I wrote down in post 1?

You'll need to express the equations in index notation. For instance

$$X f = X^\nu \partial_\nu f.$$

 

[latentcorpse]

You'll need to express the equations in index notation. For instance

$$X f = X^\nu \partial_\nu f.$$

so $(L_X \omega) (Y) = X^\nu \partial_\nu \omega_\mu Y^\mu - \omega_\nu [X,Y]^\nu$

can this be written as

$(L_X \omega) (Y) = X^\nu \partial_\nu \omega_\mu Y^\mu - \omega_\nu (X^\mu \partial_\mu Y^\nu - Y^\mu \partial_\mu X^\nu)$?

Thanks.

 

[dextercioby]

There's one term missing from your formula. The derivative in the "+" term of the RHS is acting on the scalar, not on the 1-form's components.

 

[latentcorpse]

i think that's waht i was meaning. how's this:

$(L_X \omega)(Y) = X^\nu \partial_\nu ( \omega_\mu Y^\mu) - \omega_\nu ( X^\mu \partial_\mu Y^\nu - Y^\mu \partial_\mu X^\nu )$
$= X^\nu \partial_\nu \omega^\mu Y^\mu + X^\nu \omega_\mu \partial_\nu Y^\mu - \omega_\nu ( X^\mu \partial_\mu Y^\nu - Y^\mu \partial_\mu X^\nu )$?

is that ok?
why do we expand the commutator in this way? i mean i find this part of the course a bit confusing for example,

$(XY)^\mu = X^\nu \partial_\nu Y^\mu$ but why do i not exapnd the Y term with a partial derivative as well?

Thanks

 

[latentcorpse]

Do you know how to express the product of a covector with a vector, $$\omega Y$$? What degree does the resulting object have?

hey. thanks for your help. do you have any advice on my above post?

thanks.

 

[fzero]

i think that's waht i was meaning. how's this:

$(L_X \omega)(Y) = X^\nu \partial_\nu ( \omega_\mu Y^\mu) - \omega_\nu ( X^\mu \partial_\mu Y^\nu - Y^\mu \partial_\mu X^\nu )$
$= X^\nu \partial_\nu \omega^\mu Y^\mu + X^\nu \omega_\mu \partial_\nu Y^\mu - \omega_\nu ( X^\mu \partial_\mu Y^\nu - Y^\mu \partial_\mu X^\nu )$?

Why don't you try simplifying that last line?

is that ok?
why do we expand the commutator in this way? i mean i find this part of the course a bit confusing for example,

$(XY)^\mu = X^\nu \partial_\nu Y^\mu$ but why do i not exapnd the Y term with a partial derivative as well?

Thanks

The partial derivative you're talking about corresponds to the free index $$\mu$$. You could write

$(XY) = (XY)^\mu \partial_\mu$

to get the form that you are thinking about.

 

[latentcorpse]

Why don't you try simplifying that last line?



The partial derivative you're talking about corresponds to the free index $$\mu$$. You could write

$(XY) = (XY)^\mu \partial_\mu$

to get the form that you are thinking about.

ok so can i cancel the 2nd and 3rd terms in my expression since in the 3rd term the omega and the X have different indices and therefore can be swapped giving the 3rd term equal to the 2nd term but with a minus sign.

therefore i get $(L_X \omega)(Y)=X^\nu \partial_\nu \omega_\mu Y^\mu + \omega_\nu Y^\mu \partial_\mu X^\nu$

now i assume it's ok to move the $Y^\mu$ to the end of the second term giving
$(L_X \omega)(Y)=X^\nu \partial_\nu \omega_\mu Y^\mu + \omega_\nu \partial_\mu X^\nu Y^\mu$
$(L_X \omega)(Y)=(L_X \omega)_\mu Y^\mu$
$\Rightarrow (L_X \omega)_\mu = X^\nu \partial_\nu \omega_\mu + \omega_\nu \partial_\mu X^\nu$ as required.

Is this all ok? Am I right in my reasoning for being able to move these terms about?

And one other point about the $(XY)=(XY)^\mu \partial_\mu$ thing.
Can you run through the algebra on how to get from $(XY)^\mu \partial_mu$ to $X^\nu \partial_\nu Y^\mu$

Thanks a lot for your time.

 

[fzero]

Is this all ok? Am I right in my reasoning for being able to move these terms about?

That's fine.

And one other point about the $(XY)=(XY)^\mu \partial_\mu$ thing.
Can you run through the algebra on how to get from $(XY)^\mu \partial_mu$ to $X^\nu \partial_\nu Y^\mu$

Yeah actually there's a big problem with your formula for $$(XY)^\mu$$ that I missed earlier. The question is really what kind of object is $$XY$$ supposed to be? Explicitly

$$XY = (X^\nu \partial_\nu) ( (Y^\mu \partial_\mu) = X^\nu (\partial_\nu Y^\mu ) \partial_\mu + X^\nu Y^\mu \partial_\nu\partial_\mu,$$

since $$X$$ is a derivation. Now the first term is a vector field, but the second term is not. This expression doesn't make a whole lot of sense, but if we define

$$[X,Y] = XY - YX = X^\nu (\partial_\nu Y^\mu ) \partial_\mu - Y^\mu (\partial_\mu X^\nu ) \partial_\nu = \left( X^\nu (\partial_\nu Y^\mu -Y^\nu (\partial_\nu X^\mu ) \right) \partial_\mu$$

the annoying terms cancel and we obtain a new vector field. That's the initial reason that we introduce the Lie bracket.

The algebra that you were asking about was based on a wrong formula, but the algebra above in the bracket should be easy enough to follow.

 

[latentcorpse]

That's fine.



Yeah actually there's a big problem with your formula for $$(XY)^\mu$$ that I missed earlier. The question is really what kind of object is $$XY$$ supposed to be? Explicitly

$$XY = (X^\nu \partial_\nu) ( (Y^\mu \partial_\mu) = X^\nu (\partial_\nu Y^\mu ) \partial_\mu + X^\nu Y^\mu \partial_\nu\partial_\mu,$$

since $$X$$ is a derivation. Now the first term is a vector field, but the second term is not. This expression doesn't make a whole lot of sense, but if we define

$$[X,Y] = XY - YX = X^\nu (\partial_\nu Y^\mu ) \partial_\mu - Y^\mu (\partial_\mu X^\nu ) \partial_\nu = \left( X^\nu (\partial_\nu Y^\mu -Y^\nu (\partial_\nu X^\mu ) \right) \partial_\mu$$

the annoying terms cancel and we obtain a new vector field. That's the initial reason that we introduce the Lie bracket.

The algebra that you were asking about was based on a wrong formula, but the algebra above in the bracket should be easy enough to follow.

Great. That worked out fine and helped clear a bunch of stuff up!

Next I have to use normal coordinates to argue that one can replace partial derivatives with co-variant derivatives to obtain the basis-independent result (where $\nabla$ is the Levi-Civita connection)
$(L_X \omega)_a = X^b \nabla_b \omega_a + \omega_b \nabla_a X^b$

Now normal coords are another thing that confuse me a bit. My understanding is that when we are in normal coords all partial derivatives become covariant derivatives but I don't understand why this is justified. The other thing I was hoping you could shed some light on is the whole thing about changing indices from $\mu,\nu$ etc to $a,b$ etc when something is true in any basis (is that correct?)

Thanks again.

 

[fzero]

Great. That worked out fine and helped clear a bunch of stuff up!

Next I have to use normal coordinates to argue that one can replace partial derivatives with co-variant derivatives to obtain the basis-independent result (where $\nabla$ is the Levi-Civita connection)
$(L_X \omega)_a = X^b \nabla_b \omega_a + \omega_b \nabla_a X^b$

Now normal coords are another thing that confuse me a bit. My understanding is that when we are in normal coords all partial derivatives become covariant derivatives but I don't understand why this is justified. The other thing I was hoping you could shed some light on is the whole thing about changing indices from $\mu,\nu$ etc to $a,b$ etc when something is true in any basis (is that correct?)

It's not a question of the formulas being true in any basis. The problem is that the tangent space at two different points is not really the same vector space. They are isomorphic, but a vector at a point p undergoes parallel transport when you move to a point q. The decomposition of a vector field at point p into basis elements will in general be different from the decomposition at point q.

Are you using a vierbein to define local coordinates? As in

$$g_{ab} = {e_a}^\mu {e_b}^\nu \delta_{\mu\nu}.$$

If so, I think that you can probably show that

$$\partial_\mu \omega_\nu = {e^a}_\mu \partial_a ( {e^b}_\nu \omega_b ) = {e^a}_\mu{e^b}_\nu \nabla_a \omega_b,$$

but the calculation seems a bit tedious.

In your case, the calculation is much simpler, since you can just expand

$$X^b \nabla_b \omega_a + \omega_b \nabla_a X^b$$

using the Christoffel symbols to obtain the result.

 

[latentcorpse]

It's not a question of the formulas being true in any basis. The problem is that the tangent space at two different points is not really the same vector space. They are isomorphic, but a vector at a point p undergoes parallel transport when you move to a point q. The decomposition of a vector field at point p into basis elements will in general be different from the decomposition at point q.

Are you using a vierbein to define local coordinates? As in

$$g_{ab} = {e_a}^\mu {e_b}^\nu \delta_{\mu\nu}.$$

If so, I think that you can probably show that

$$\partial_\mu \omega_\nu = {e^a}_\mu \partial_a ( {e^b}_\nu \omega_b ) = {e^a}_\mu{e^b}_\nu \nabla_a \omega_b,$$

but the calculation seems a bit tedious.

In your case, the calculation is much simpler, since you can just expand

$$X^b \nabla_b \omega_a + \omega_b \nabla_a X^b$$

using the Christoffel symbols to obtain the result.

ok. we definitely don't have anything on "vierbein" in our notes.

So what is the reasoning for being able to replace partial by covariants in normal coordinates?

And when you say expand that in terms of Christoffel symbols, do I make use of
$(\nabla_X Y)^\mu = X^\nu e_\nu ( Y^\mu) + \Gamma^\mu{}_{\rho \nu} Y^\rho X^\nu$?

 

[fzero]

ok. we definitely don't have anything on "vierbein" in our notes.

So what is the reasoning for being able to replace partial by covariants in normal coordinates?

And when you say expand that in terms of Christoffel symbols, do I make use of
$(\nabla_X Y)^\mu = X^\nu e_\nu ( Y^\mu) + \Gamma^\mu{}_{\rho \nu} Y^\rho X^\nu$?

You want to use the covariant derivative in the normal coordinates. On a vector field

$$\nabla_a X^b = \partial_a X^b + {\Gamma^b}_{ac} X^c .$$

 

[latentcorpse]

You want to use the covariant derivative in the normal coordinates. On a vector field

$$\nabla_a X^b = \partial_a X^b + {\Gamma^b}_{ac} X^c .$$

ok and is defined
$\nabla_a \omega_b = \partial_a \omega_b - \Gamma^c{}_{ab} \omega_c$ for a covector field.

Then
$X^b \nabla_b \omega_a + \omega_b \nabla_a X^b \\ =X^b ( \partial_b \omega_a - \Gamma^c{}_{ab} \omega_c ) + \omega_b ( \partial_a X^b + \Gamma^b{}_{ac} X^c ) \\ = X^b \partial_b \omega_a + \omega_b \partial_a X^b$ be cancelling the $\Gamma$ terms.

Now the bit that I don't really understand (because I don't understand the difference between a,b,... and mu,nu,... labels) - how do I justify saying that this is equal to
$X^\nu \partial_\nu \omega_\mu + \omega_\nu \partial_\mu X^\nu$

I also have to work out $(L_X g)_{\mu \nu}$

So the formula for the Lie derivative along X of a type (r,s) tensor field is defined, in coordinate notation, as

$L_X T^{a_1 \dots \_r}{}_{b_1 \dots b_s} = X^c ( \nabla_c T^{a_1 \dots a_r}{}_{b_1 \dots b_s} ) - ( \nabla_c X^{a_1} ) T^{c a_2 \dots a_r}{}_{b_1 \dots b_s} - \dots - ( \nabla_c X^{a_r} ) T^{a_1 \dots a_{r-1} c}{}_{b_1 \dots b_s} + ( \nabla_{b_1} X^c ) T^{a_1 \dots a_r}{}_{c b_2 \dots b_s} + \dots + ( \nabla_{b_s} X^c ) T^{a_1 \dots a_r}{}_{b_1 \dots b_{s-1} c}$

so applying thsi to the metric tensor I get

$L_x g_{\mu} = X^\rho ( \nabla_\rho g_{\mu \nu} ) + ( \nabla_\mu X^\rho ) g_{\rho \nu} + ( \nabla_\nu X^\rho ) g_{\mu \rho}$

now the answer I'm looking for involves partial derivatives instead of covariant ones. Do you know where I have messed up?Thanks a lot.

 

[latentcorpse]

bump.