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Lie derivative of a function

  1. Dec 24, 2007 #1
    Suppose we define the Lie derivative on a tensor [itex]T[/itex] at a point p in a manifold by

    [itex]\mathcal{L}_V (T) = \lim_{\epsilon \to 0}\frac{\varphi_{-\epsilon \ast}T(\varphi_\epsilon(p))- T(p)}{\epsilon}[/itex]

    where V is the vector field which generates the family of diffeomorphisms [itex]\varphi_t[/itex].

    If T is just an ordinary function [itex]f:M \to \mathbb{R}[/itex] then it seems like the numerator of the above expression is [itex]f(p) - f(p) = 0[/itex] which is unusual since I thought the lie derivative of a function was the ordinary derivative [itex]\mathcal{L}_V f = V^\mu \partial_\mu f[/itex]. Can anyone reconcile this?

  2. jcsd
  3. Dec 24, 2007 #2


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    You've already given yourself a hint! :smile:

    Why do you think that numerator seems like that?

    Also... have you tried any special cases, to see what happens?
  4. Dec 25, 2007 #3
    I don't understand why the numerator is zero. The numerator should be a small quantity but definitely nonzero. Consider the ordinary derivative of a function [itex]f : \mathbb{R} \to \mathbb{R}[/itex]:

    [itex]\frac{df}{dx} = \lim_{\epsilon \to 0} \frac{f(x + \epsilon) - f(x)}{\epsilon}.[/itex]

    Note that prior to evaluation of the limit the numerator is nonzero, as it should be. But with the Lie derivative the numerator is already zero before we even evaluate the limit!

    [itex]\varphi_{-\epsilon\ast}(f(\varphi_\epsilon(p))) - f(p)= f (\varphi_{-\epsilon}\circ \varphi_\epsilon(p)) -f(p)= 0.[/itex]

    My formula for the Lie derivative must not apply when considering functions.
  5. Dec 25, 2007 #4
    The tangent mapping

    \varphi_{-\epsilon *}:T_{\varphi(\epsilon,p)}M\to T_p M

    gets extended naturally to

    \varphi_{-\epsilon *}:(T_{\varphi(\epsilon,p)}M)^k\to (T_p M)^k,

    but for k=0 you need to stop for a moment to wonder what that mapping is supposed to mean. In fact IMO it doesn't even make fully sense to say that a smooth mapping

    M\to \bigsqcup_{p\in M}(T_p M)^0

    would be a real valued [itex]M\to\mathbb{R}[/itex], but isn't that the convention anyway? And then

    \varphi_{-\epsilon *} (f(\varphi_{\epsilon}(p)})) = f(\varphi_{\epsilon}(p))

    is convention too?
    Last edited: Dec 25, 2007
  6. Dec 25, 2007 #5


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    I see you've noticed that real-valued functions are not scalar fields! Just as (tangent) vector fields are sections of the tangent bundle1, scalar fields are sections of the trivial line bundle1 [itex]M \times \mathbb{R} \to M[/itex]. (note that [itex]M \times \mathbb{R} \cong \coprod_{p \in M} \mathb{R}[/itex])

    Recall that [itex]\varphi_{\epsilon*}[/itex] was initially only defined on the tangent spaces; do you recall how you extended it to the entire tensor algebra? That will explain why [itex]\varphi_{\epsilon*}(p, x) = (\varphi_\epsilon(p), x)[/itex] for any point p and scalar x. (I've used (p, x) to denote the element x of the fiber at p of the trivial line bundle)

    Generally, you want to work with scalar fields, and not functions [itex]M \to R[/itex]; happily, there is a natural conversion between the two concepts, so it's easy to go back and forth between them when it's convenient to do so.

    1: these things are, of course, not unique -- but all such objects are isomorphic.
  7. Jan 27, 2008 #6
    May I ask a couple very basic and simple questions related to this topic? Since I'm currently working on a project requiring advanced tensor math knowlede but I'm from engineering background :(

    I need to obtain a diffeomorphism [itex]\varphi_t[/itex] that maps a tensor field T_1 to another tensor field T_2, assume they are symmetric tensors in [itex]R^3\times R^3[/itex]R^3*R^3. The function is to map each tensor in T_1 to the corresponding one in T_2 by translating and rotating the tensor at location p to p', but doesn't change the shape of the tensor.

    I think the derivative of [itex]T_1(\varphi(p))-T_2(p)[/itex] should be computed to estimate the function [itex]\varphi[/itex]. Is the Lie derivative the right one to use?

    Also during the numerical implimentation, at each iteration, how can I update [itex]\varphi[/itex]? If it's a function from scalar to scalar, then I only need obtain the gradient of some cost function, and update [itex]\varphi[/itex] which only contains translation in R^3.

    Sorry if my questions are confusing or totally nonsense. Any suggestion will be appreciated. I've been struggling with this for several weeks, and don't have any clue. I feel so frustrated:frown:
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