# Help with Diffeomorphisms

1. ### latentcorpse

So I'm trying to prove eqn (223) in the notes attached in this thread:

I took the equation $( \phi^* ( \eta) ) ( X) = \eta ( \phi_* (X))$
and expanded in a coordinate basis as follows

$( \phi^* ( \eta) )_\mu dx^\mu X = \eta_\alpha dy^\alpha \phi_*(X)$

So to get the result it seems like it should be a simple case of cross multiplying but unfortunately the $X$ doesn't cancel the $\phi_*(X)$

Thanks for any help.

2. ### fzero

2,601
Re: Diffeomorphisms

The notation $( \phi^* ( \eta) ) ( X) = \eta ( \phi_* (X))$ always means that indices are contracted in the coordinate basis, so

$$( \phi^* ( \eta) ) ( X) = ( \phi^* ( \eta) )_\mu X^\mu,$$

$$\eta ( \phi_* (X)) = \eta_\alpha ( \phi_* (X))^\alpha.$$

Just think of the usual expression for a 1-form acting on a vector: $$\omega(X) = \omega_\mu X^\mu$$.

3. ### latentcorpse

Re: Diffeomorphisms

Ok. So then we get the equality:
$( \phi^*( \eta ) )_\mu X^\mu = \eta_\alpha ( \phi_* ( X ))^\alpha$
I don't see how we can get this into the form of (223)?

Additionally, I am trying to prove (187). Given the hint, I am able to multiply by those vectors and get

$\nabla_X \nabla_Y Z^a- \nabla_Y \nabla_X Z^a = R^a{}_{bcd}Z^bX^cY^d = (R(X,Y)Z)^a$. But according to teh defn of the Riemann tensor (see (178)), I am missing a $\nabla_{[X,Y]}$ piece. Any ideas?

4. ### fzero

2,601
Re: Diffeomorphisms

You can use (221).

You have to use the product rule correctly on the terms $$\nabla_b\nabla_c(Z^a X^bY^c)$$ and (135).

5. ### latentcorpse

Re: Diffeomorphisms

6. ### fzero

2,601
Re: Diffeomorphisms

7. ### latentcorpse

Re: Diffeomorphisms

I get
$( \phi_*(X))^\alpha (\frac{\partial}{\partial y^\alpha})_{\phi(p)} (f) = X^\nu (\frac{\partial}{\partial x^\nu})_p \phi^*(f)$
Again, I'm having trouble getting rid of the $\phi^*$ on the right

Ok. You appear to have multiplied from the left by $Y^dX^c$ on the first term but $X^cY^d$ on the second term. Now we are using abstract index notation so these both represent vectors rather than the components of vectors. How do you know those vectors commute?

Thanks.

8. ### fzero

2,601
Re: Diffeomorphisms

You need to use that $$\phi$$ is the map $$x^\mu\rightarrow y^\alpha$$ and the chain rule.

They represent the components of vectors and commute. Look at the derivation of (130), the 2nd line follows from the first by commuting the components of $$X$$ and $$Y$$.

9. ### latentcorpse

Re: Diffeomorphisms

well $\phi$ has components $y^\alpha(x^\mu)$
I can see we want to introduce something like dy/dx d/dy on the RHS but I just dont see how?

Right. But there he is using basis indices (see chapter 7).
Although he does say in Chapter 7 that we are free to change abstract indices to basis indices to get an equation that is true in an arbitrary basis so I guess everything is ok providing we work this out in greek indices?

Anyway I tried working out the following using the product rule:
$\nabla_Y (X^\mu \nabla_\mu Z^\nu) = Y^\rho \nabla_\rho (X^\mu \nabla_\mu Z^\nu) = Y^\rho X^\mu{}_{; \rho} \nabla_\mu Z^\nu + Y^\rho X^\mu \nabla_\rho \nabla_\mu Z^\nu$
in the hope that one of those terms we get would be useful but I don't see it?

10. ### fzero

2,601
Re: Diffeomorphisms

Use $$\phi^*(f) = f\circ \phi$$ applied to the coordinates $$x^\mu$$.

I'm not sure why you chose that expression. We had

$$Y^d \nabla_X \nabla_d Z^a - X^c \nabla_Y \nabla_c Z^a,$$

so you want to re-express

$$Y^d \nabla_X \nabla_d Z^a = \nabla_X(Y^d \nabla_d Z^a) - ( \nabla_X Y^d )(\nabla_d Z^a)$$

and similarly the other term.

11. ### latentcorpse

Re: Diffeomorphisms

Sorry I still don't get. The $\phi$'s are charts so they have components $x^\mu$ so how would this work?

12. ### latentcorpse

Re: Diffeomorphisms

Sorry, I have been working on some other stuff for the last few days and didn't have time to look at it again until now. I still do not get how either of these work? The second one I keep getting bogged down with what to do with algebra. The first one I don't really follow what you want me to do.

Thanks again.

13. ### fzero

2,601
Re: Diffeomorphisms

There really isn't any algebra to get bogged down with,

$$\nabla_X(Y^d \nabla_d Z^a) = \nabla_X \nabla_Y Z^a,$$

while $$( \nabla_X Y^d )$$ is one term in the expression (130) for $$[X,Y]^d$$.

As for the other part, if $$\phi(x^\mu) = y^\alpha$$, then what is $$(f\circ \phi)(x^\mu)$$?

14. ### latentcorpse

Re: Diffeomorphisms

Ok. I get us down to

$\nabla_X \nabla_Y Z^a - \nabla_Y \nabla_X Z^a - ( \nabla_X Y^d)(\nabla_d Z^a) + ( \nabla_Y X^c)( \nabla_c Z^a)$

Now from our Riemann tensor we want those last terms to give us a $-\nabla_{[X,Y]}Z^a$

So I tried working backwards:

$\nabla_{[X,Y]}Z^a=\nabla_{XY-YX}Z^a = \nabla_{XY}Z^a-\nabla_{YX}Z^a$

Now I don't know how to get to the next line there? Could you explain please?

As for the other bit I would find that $(f \circ \phi)(x^\mu) = f(y^\alpha)$

Thanks.

15. ### fzero

2,601
Re: Diffeomorphisms

Just write

$$- ( \nabla_X Y^d)(\nabla_d Z^a) + ( \nabla_Y X^c)( \nabla_c Z^a) = - ( \nabla_X Y^d-\nabla_Y X^d)(\nabla_d Z^a)$$

and use (130).

The last part makes no sense at all, since $$XY$$ and $$YX$$ are not tangent vectors.

So you need to go back a few posts and figure out what that's good for.

16. ### latentcorpse

Re: Diffeomorphisms

Ok. So I have got the RIcci identity to work out.

As for the transformation, I assume I am meant to go back to

$( \phi_*(X))^\alpha (\frac{\partial}{\partial y^\alpha})_{\phi(p)} (f) = X^\nu (\frac{\partial}{\partial x^\nu})_p \phi^*(f)$

and get rid of the $\phi^*(f)$ on the RHS.
Should I act both sides on the coordinates $x^\mu$? That's the only way I can see of being able to use the $f(x^\mu)=y^\alpha$ property.

Also, on p90, he says that we should think of the singularity as a time rather than a place inside the black hole because r plays the role of the time coordinate. How does this work?

Thanks.

Last edited: Apr 13, 2011
17. ### fzero

2,601
Re: Diffeomorphisms

Then try doing that. It's much better to try to work it out before asking the question.

Hopefully you mean $\phi(x^\mu)=y^\alpha$

You should be able to use the metric to show that the radial vector is timelike for $$r<2M$$.

18. ### latentcorpse

Re: Diffeomorphisms

Ok. I tried taking $f=y^\alpha$ but then on the LHS, this cancelled the d/dy i had and similarly on the RHS it gets pulled back to an $x^\mu$ which cancels the d/dx we have?

This may be a stupid question but what is the form of the radial vector? Would it just be $X=\frac{\partial}{\partial r} \Rightarrow X^\mu(0,1,0,0)$ in Schwarzschild coordinates?
If so, when I calculate $g_{rr}X^rX^r$ I find it to be spacelike so I'm guessing that's wrong.
Moreover, how does showing a vector is timelike tell us it behaves like a time coordinate/that t doesn't behave like a time coordinate?

19. ### fzero

2,601
Re: Diffeomorphisms

Leave $$f$$ arbitrary, but note that you have $$(\partial/\partial y^\alpha )f(y^\alpha)$$ on the LHS and $$(\partial/\partial x^\mu) f(y^\alpha)$$ on the RHS. You want to use the chain rule on the RHS and equate coefficients.

Yes.

You might want to show your work. You should find that it's spacelike for $$r>2M$$ and timelike for $$r<2M$$.

How else would you decide that something is or isn't a time-coordinate?

20. ### latentcorpse

Re: Diffeomorphisms

Leave $$f$$ arbitrary, but note that you have $$(\partial/\partial y^\alpha )f(y^\alpha)$$ on the LHS and $$(\partial/\partial x^\mu) f(y^\alpha)$$ on the RHS. You want to use the chain rule on the RHS and equate coefficients.
[/QUOTE]
What happened to your $\phi^*(f)$ on the RHS - you have just made it $f$, no?

Well in the notes it says to work with the Schwarzschild coordinates so I used the Schwarzschild metric. Now I know that the metric that covers the r<2M is the one defined using ingoing Eddington Finkelstein coordinates but, we know that if we're in the region r<2M and we change back from EF to Schwarz coordinates we should get the Schwarz metric back again (but now applied to the r<2M region)> Isn't this guaranteed by Birkhoff's theorem which tells us that Schwarzschild is unique? So I guess my question is, why does it not work if you use the Schwarzschild metric?

Anyway, let's say I try with the metric in EF coords....

we have $g_{\mu \nu} X^\mu X^\nu$ with $X^\mu=(0,1,0,0)$
so we would have $g_{\mu \nu} X^\mu X^\nu=g_{rr}X^rX^r=g_{rr}=0$?

I'm sure I've done this before. I can't understand what I am doing wrong!

[/QUOTE]
I assume that checking it's timelike is the only way? Presumably then t would be spacelike in this region? How would we prove this since there is no t coordinate in the EF coord metric?

Thanks again.