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Coordinate independence of Lie derivative

  1. May 27, 2010 #1
    Hello Forum,

    since my GR tutor can't help me with some issues arising I thought it is time to register here.

    I am very confused about the phrase "coordinate independence". Especially regarding the Lie Derivative and the Commutator of two vector fields.

    The Lie Derivative is said to be coordinate independant, right?
    Let's have a look at [tex]\mathcal{S}^2[/tex]. I know that I need at least 2 charts for it, but lets stick to usual polar coordinates
    [tex]x = \cos \phi \sin \theta[/tex]
    [tex]y = \sin \phi \sin \theta[/tex]

    for identifying points on the sphere with points in [tex]\mathbb{R}^2 \supset [0,2 \pi]\times[0, \pi][/tex]

    Then [tex]g_{ij} = diag(1, \sin^2 \theta)[/tex], taking the 1-coordiante to be [tex]\theta[/tex]

    Now, we know that [tex]\mathscr{L}_{\partial_\phi} g = 0[/tex] but [tex]\mathscr{L}_{\partial_\theta} g \neq 0[/tex]

    I am pretty sure (I didn't check it though) that if I was doing some coordinate change now, and I was changing the components of the metric tensor accordingly to [tex]g'_{ij}[/tex] and changing the vectors too, I would still get [tex]\mathscr{L}_{\parital'_\phi} g' = 0.[/tex]
    This looks good then and one is tempted to say that we checked that the Lie derivative is independent of coordinates for an example.

    BUT what confuses me now is that somehow the coordinates have been important here. Right from the beginning. We have chosen some polar axis to define our coordinate chart. We broke the symmetry by doing that, I guess, but I don't know what this actually means.
    In my mind I imagine a sphere, perfectly round, no axis, no broken symmetry. Now I choose an axis. I now know that the corresponding "perpendicular" vectors [tex]\partial_\phi[/tex] are killing vectors.
    Now I remember these killing vectors but begin with another axis. In these coordinates the vector looks different but the metric tensor does not. Hence the Lie derivative won't be still 0. It changed although we didn't change the manifold (the nice sphere in the head) and the vector (also fixed in the head). In this sense it is not coordinate independent.
    What has happened here?

    So I know that the coordinates we have chosen are "ill" because we just need one single map, but I can't imagine this alone gives rise to my confusion. For example one has to choose an axis for a stereographic projection as well.
    Still, a bonus question would be, what problems _are_ arising due to this one-chart-are-no-atlas-problem.

    Let's say Lie Derivative and thus the commutator of 2 vector fields are coordinate independent.
    I take an arbitrary point out of my manifold. I want to know the commutator of two given fields. I can choose any coordinates. I choose the coordinates in such a way that the 2 vectors of the 2 fields in this point of the manifold point in the same direction as the coordinate axes.
    So the vectors are just partial derivatives. But this means the commutator vanishes. But this means every commutator vanishes everywhere if I do this for every point (assume the fields to be smooth).

    Am I right, that the flaw here is, that the vector_fields_ are not just partial derivatives everywhere in the chart but just in one point. And I have to take the derivative first and then put in the point. So basically my argumentation would also mean that the ordinary derivative of every function from R to R is zero in every point because in a point the function is just a constant number.
    The thing is, that the upper argumentation was used in some proof in the lecture to make a commutator vanish. But probably the vector fields have been special.
    Is there a common case where one might use a similar argumentation? Or is there not, meaning the professor made a mistake?

    Thank you very much for clearing up the confusions. Struggling with them for 2 weeks now :)
  2. jcsd
  3. May 27, 2010 #2
    Once you know that something is coordinate independent, you are free to use whatever coordinates you want to actually carry out the computation. On the other hand, there are invariant formulas that give you the Lie derivative (like the standard dynamical definition or that it is a derivation with respect to the tensor product).

    You are correct. Indeed, it is a theorem that given any two independent vector fields X and Y, you can find coordinates (x^i say) such that [tex]X = \partial_1, Y = \partial_2[/tex] if and only if [X,Y] = 0.

    Is there a way you can post the proof given in class?
    Last edited: May 27, 2010
  4. Jun 9, 2010 #3
    Thank you for your reply, eok20. I am sorry it took me so long to get the photographs.

    Ah, now that you say it, I can remember having heard that.
    Do you know any source where I can find the proof?

    http://j.imagehost.org/view/0383/IMG_3862s [Broken]
    http://h.imagehost.org/view/0843/IMG_3863s [Broken]
    The interesting part is the upper part of the second page.
    Would be interesting to know if the Prof made a mistake here. Any other time I thought so first he's been right at the end ;)

    To the other, first, question:

    Yes. But here the answer to the question if a vector is a Killing vector, which should be independent of coordinates because it is something like Lie derivative = 0, _is_ basis dependent.

    Not, if we do a calculation, change coordinates (changes vector and metric tensor) and calculate again (case A). But when we choose coordinates and calculate and now start over again, choose coordinates differently, and give the metric tensor the same functional dependence on the new coordinates as the old metric tensor had on the old coordinates (what I mean is, choose a different axis and write down g_ij exactly as before but with theta' instead of theta. but what has been vector partial_phi now is partial_theta for example) (case B).
    Somehow I already know part of the answer: I just have to compare with the case A where I learn that the _same_ metric tensor looks different in the new coordinates. As a reason the metric tensor of B (with the same functional dependence) has to be a different object. So the Lie derivative has changed. Still, some clarification would be useful. Why does the metric tensor of the sphere depends on the axis chosen? Why do I get a new different tensor object if I choose a different axis? Has it anything to do with the "illness" (one chart only for the sphere) of the coordinates? What general information can be extracted out of Killing vector fields then, if they depend on the initial coordinate choice if the metric tensor depends on that? (Or formulated in another way: When I have a Killing vector field, how can I be sure I have not just discovered a coordinate effect (like thinking: ahhh i can rotate the sphere around one axis only (might be 3 or so, I think there are 2 more Killing vectors)).
    Last edited by a moderator: May 4, 2017
  5. Jun 11, 2010 #4
    I have a tough time imagining what he means since the use of terminology is very sloppy, but I believe he's referring to the fact that the covariant derivative is torsion-free.
  6. Jun 12, 2010 #5
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