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Lie derivative of vector field = commutator

  1. Nov 19, 2008 #1

    Fredrik

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    Can somone remind me how to see that the Lie derivative of a vector field, defined as

    [tex](L_XY)_p=\lim_{t\rightarrow 0}\frac{\phi_{-t}_*Y_{\phi_t(p)}-Y_p}{t}[/tex]

    is actually equal to [itex][X,Y]_p[/itex]?
     
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  3. Nov 19, 2008 #2

    Fredrik

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    Nevermind, I got it. Nakahara's book had the clue I needed to figure out the rest.
     
  4. Oct 7, 2009 #3

    Fredrik

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    I tried to prove this again the other day, and I couldn't remember how. Then I remembered that I had started a thread about it some time ago. Unfortunately it turned out to be useless. I'm not sure what clue in Nakahara's book I was referring to in #2. I suspect it was the result that the Lie derivative can also be expressed as

    [tex](\mathcal L_XY)_pf=\lim_{t\rightarrow 0}\frac{Y_p-\phi_t_*Y_{\phi_{-t}(p)}}{t}[/tex]

    but I don't see how to use it (or prove it).

    I would still like to prove the result I mentioned in #1, but I need some help filling in some missing details. A trick that seems promising is to add and subtract a term in the numerator:

    [tex](\mathcal L_XY)_pf=\lim_{t\rightarrow 0}\frac{\phi_{-t*}Y_{\phi_t(p)}f-Y_pf}{t}=\lim_{t\rightarrow 0}\frac{Y_{\phi_t(p)}(f\circ\phi_{-t})-Y_{\phi_t(p)}f+Y_{\phi_t(p)}f-Y_pf}{t}[/tex]

    Now the last two terms give me one of the terms of the commutator:

    [tex]X_p(Yf)=\phi(p)_*D_0 (Yf)=D_0(Yf\circ\phi(p))=\frac{d}{dt}\bigg|_0 Y_{\phi_t(p)}f[/tex]

    Here [itex]\phi[/itex] is the flow of X, and [itex]\phi(p)[/itex] is the map [itex]t\mapsto\phi_t(p)[/itex], i.e. the integral curve of X through p. [itex]D_0[/itex] is the operator that takes a real-valued function of one real variable to its derivative at 0.

    So far so good, but I'm having difficulties with the rest. I think I have proved that

    [tex]Y_p(Xf)=-\frac{d}{dt}\bigg|_0 Y_p(f\circ\phi_{-t})[/tex]

    Edit: I said something dumb here, so I deleted it.
     
    Last edited: Oct 7, 2009
  5. Oct 7, 2009 #4

    dx

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    There's a simple proof using a special coordinate system. If you introduce a coordinate sytem {t, qa} such that X = ∂t, then the coordinates of the pushforward of Yp to φ(t) are the same as those of Yp, so the difference between Y at φ(t) at the pushforward of Yp is simply the ordinary derivative of Y with respect to t (since X is equal to ∂t) multiplied by t, so

    [tex] \mathcal{L}_{X}{Y} = \frac{\partial Y^t}{\partial t}\partial_t + \sum_{a}\frac{\partial Y^a}{\partial t}\partial_a [/tex]

    Evaluating the Lie bracket [∂t, Ytt + ∑Yaa], we get the same result:

    [tex] [X, Y] = \frac{\partial Y^t}{\partial t}\partial_t + \sum_{a}\frac{\partial Y^a}{\partial t}\partial_a [/tex]

    So, LXY = [X, Y] in this coordinate system, and this must be true in all coordinate systems because both sides are coordinate independent.
     
    Last edited: Oct 7, 2009
  6. Oct 8, 2009 #5

    mma

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  7. Oct 13, 2009 #6

    Fredrik

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    Thanks for the replies. I haven't been able to access the internet for a few days, so I haven't been able to reply. The Spivak proof was just what I needed. When I finally understood it (including the mistake he made in the definition of g), I was able to use the same idea to finish the proof I started in #3.

    I was having some difficulties with the expression

    [tex]\lim_{t\rightarrow 0}\frac{Y_{\phi_t(p)}(f\circ\phi_{-t})-Y_{\phi_t(p)}f}{t}[/tex]

    To deal with this, we note that [itex]f\circ\phi_{-t}[/itex] is the function that takes [itex]q[/itex] to

    [tex]f(\phi_{-t}(q))=f(q)+t\frac{d}{dt}\bigg|_0 f(\phi_{-t}(q))+\mathcal O(t^2)=f(q)-tX_qf+\mathcal O(t^2)[/tex]

    so we have

    [tex]f\circ\phi_{-t}=f-tXf+\mathcal O(t^2)[/tex]

    and the rest is easy.
     
  8. Oct 13, 2009 #7

    Fredrik

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    It's funny how dumb I can be sometimes. This identity seemed difficult to prove, but I eventually realized that you get this result immediately if you just set t'=-t in the definition in #1.
     
  9. Oct 14, 2009 #8

    haushofer

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    Yes, that was something which also took some time for me when I first read Nakahara :D It's nice to see that other people have similar issues ;)
     
  10. Oct 14, 2009 #9

    haushofer

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    I have a small question on this, which is about Lie derivatives of scalar fields (for simplicity). I'll simply write down my reasoning, and if someone finds a mistake I'll be happy to hear!

    We can define a Lie derivative of a scalar field with respect to a vector field xi as

    [tex]
    \delta_{\xi}\phi(x) = \phi'(x) - \phi(x)
    [/tex]

    Here xi induces the coordinate transformation. I see this as the following: We first induce a coordinate transformation via xi,

    [tex]
    \phi(x) \rightarrow \phi'(x') , \ \ \ \ \ \ via \ \ \ \ \ \xi = x'-x
    [/tex]

    which means that we are switching from a coordinate system F to a coordinate system F'. From my point of view, {x} in the coordinate system F and {x'} in the coordinate system F' refer to the same point P on the manifold M. After that, we change our coordinate {x'} in F back to its original value x,

    [tex]
    x' \rightarrow x
    [/tex]
    Obviously, in the coordinate system F' the coordinate {x'} refers to the point p on M, so the coordinate {x} in F' refers to a new point q on the manifold M, right?

    So, am I right if I say that in looking at the variation

    [tex]
    \delta\phi(x) = \phi'(x) - \phi(x)
    [/tex]

    we are comparing the fields at two different points p and q on M, and that this variation is basically a Lie derivative? Something goes wrong here, because if I define a Lie derivative via a diffeomorphism I see that I evaluate everything at one and the same point: you have a point p, and a point q=f(p) where f is a diffeomorphism on the manifold M, and you evaluate for a tensor field T the object T(q)=T(f(p)); after that, you use the differential map f* induced by f to pull this whole thing back to the original point p to compare f*T(f(p)) with T(p).

    What's wrong with my reasoning?
     
  11. Oct 14, 2009 #10

    dx

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    What's the contradiction? By pulling back structures by the infinitesimal diffeomorphism induced by the vector field, you are using information not present at point p.
     
  12. Oct 14, 2009 #11

    haushofer

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    -edit

    The contradiction I'm seeing is that in my definition in the right hand side

    [tex]
    \delta\phi(x) = \phi'(x) - \phi(x)
    [/tex]

    I'm really evaluating the two fields at different points on the manifold, while in the definition with diffeomorphisms I'm not; I compare two tensor fields at one and the same point on the manifold. How can this be?
     
    Last edited: Oct 14, 2009
  13. Oct 14, 2009 #12

    dx

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    T(f(p) is the value of T at the point q = f(p), as you said. q and p are not the same point.
     
  14. Oct 15, 2009 #13

    haushofer

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    No, but I pull the field T(f(p))=T(q) back to the original point p with the differential map, and f*(T(f(p)) IS evaluated in the point p. And this field you compare with T(p). And that's what I can't match up with my definition in terms of coordinates.

    I'm missing something very silly here, I suppose.
     
  15. Oct 15, 2009 #14

    dx

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    But f(p) = q! Just because you write T(q) as T(f(p)) doesn't mean you're evaluating the field at p. The field is T, and it is evaluated at q. To compare with the field at the original point, it must be pulled-pulled back to that point, or you can push-forward T at the original point to q and compare it there.

    This is what Lie differentiation means: A vector field is a family infinitesimal transformation of the manifold (call them φε). Think of vectors as equivalence classes of parametrized curves. Such a set at one point is mapped by φ to another point, and gives a vector at the new point. You can imagine this being done to each vector of the tangent space at a point p. Thus we have a map between tangent spaces at two points, P and φε(P); call this map f. Given another vector field v, the Lie derivative is lim ε → 0 of the vector (1/ε)(v(φε(P)) - f(v(P)). Similarly, you can define the Lie derivative of 1-forms, which are equivalence classes of functions. Once you have the Lie derivative of vectors anf 1-forms, you can Lie differentiate any tensor field.
     
  16. Oct 15, 2009 #15

    haushofer

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    Yeah, I see that now. But now the question is: how do you rephrase this in our "coordinate" definition of the Lie derivative which you often see,

    [tex]
    \delta_{\xi}T = T'(x) - T(x)
    [/tex]

    or T'(x') - T(x'), which shouldn't make a difference I suppose, because in the end you take the limit of xi = x'-x goes to 0.
     
  17. Oct 16, 2009 #16

    haushofer

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    So obviously this is a nasty question. Let me compare the passive and the active way of defining Lie derivative. That is, the coordinate way and the diffeomorphism way.

    In

    [tex]
    \delta\phi(x) = \phi'(x) -\phi(x)
    [/tex]

    I first do a coordinate transformation [itex]\phi(x) \rightarrow \phi'(x') [/itex] and after that I set back the value of the coordinate x' back to x. This I compare with [itex]\phi(x)[/itex].

    In the diffeomorphism expression I calculate the field at the point [itex]f(p)[/itex] and after that I use the differential map [itex]f_*[/itex] to pull this expression back to the point p via [itex]f_*\phi(f(p))[/itex]. This I compare with [itex]\phi(x)[/itex].

    How can i rephrase these two situations into eachother? There seems to be some sort of duality, but I still can't see it, and that was a problem I already had quite some time ago.
     
  18. Oct 16, 2009 #17

    dx

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    Can you explain more clearly what the 'coordinate definition' is? What is the 'coordinate definition' of the lie derivative of a vector field v with respect to a vector field w?
     
  19. Oct 17, 2009 #18

    haushofer

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    The "coordinate definition" (by which I mean a definition in terms of the coordinates, without mentioning diffeomorphisms on the manifold) can be found in books like d'Inverno, and is defined as
    [tex]
    \delta_{\xi}T = T'(x) - T(x)
    [/tex]

    or
    [tex]
    \delta_{\xi}T = T'(x') - T(x')
    [/tex]

    with xi infinitesimally.
     
  20. Oct 17, 2009 #19

    dx

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    Well, that's just notation. You haven't explained what x' and x are and how they are related. Where does the vector field with respect to which the differentiation is being performed come in? What is T'? I don't have d'Inverno and I have found no pages online which describes this. I also don't see any limits, which is weird because the Lie derivative is defined as a limit.
     
    Last edited: Oct 17, 2009
  21. Oct 17, 2009 #20

    Fredrik

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    I'll be using the same notation as I did earlier. The Lie derivative of a function f in the direction of a vector field X is defined as

    [tex]\mathcal L_Xf=\lim_{t\rightarrow 0}\frac{\phi_t^*f-f}{t}[/tex]

    So if x is a coordinate system, we have

    [tex]\mathcal L_Xf(p)=\lim_{t\rightarrow 0}\frac{f\circ\phi_t(p)-f(p)}{t}=\lim_{t\rightarrow 0}\frac{f\circ\phi_t\circ x^{-1}\circ x(p)-f\circ x^{-1}\circ x(p)}{t}[/tex]

    [tex]=\lim_{t\rightarrow 0}\frac{F_t(x(p))-F_0(x(p))}{t}[/tex]

    where I have defined [itex]F_t=f\circ\phi_t\circ x^{-1}[/itex]. The above takes the form

    [tex]=\lim_{t\rightarrow 0}\frac{F'(x)-F(x)}{t}[/tex]

    if we agree to write x instead of x(p), F instead of F0, and F' instead of Ft.

    This is probably what you're looking for, at least for scalar fields. I'm sure we can do something similar for arbitrary tensor fields, but I don't see why we would want to. I consider d'Inverno's notation and terminology obsolete.
     
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