Let $(M,g)$ be an $n$-dimensioal riemannian manifold. A vector field $X$ on $M$ is said to be a Killing vector if the flow it generates is an isometry; that is, it preserves the metric $g$. There are many ways of writing this. The one which is relevant for this question is the following. If we let $\nabla$ denote the Levi-Civita connection, then $X$ is Killing if and only if the endomorphism $A_X : TM \to TM$ defined by $$A_X(Y) = - \nabla_Y X$$ is skewsymmetric, so that for all vector fields $Y,Z$ on $M$ one has that $$ g(A_X(Y),Z) = - g(Y,A_X(Z)).$$

In summary, if we let $\mathfrak{so}(TM)$ denote the bundle of skewsymmetric endomorphisms of $TM$, then $X$ is Killing if and only if $A_X$ defines a section of $\mathfrak{so}(TM)$.

Let $\mathrm{SO}(TM)$ denote the bundle of oriented orthonormal frames of $TM$. It is a principal $\mathrm{SO}(n)$ bundle over $M$. In the case I'm mostly interested in, $M$ is a spin manifold, so that there is a principal $\mathrm{Spin}(n)$ bundle $\mathrm{Spin}(TM)$ and a bundle surjection $\mathrm{Spin}(TM) \to \mathrm{SO}(TM)$ which restricts fibrewise to the covering homomorphism $\mathrm{Spin}(n) \to \mathrm{SO}(n)$.

If $\rho : \mathrm{Spin}(n) \to \mathrm{GL}(V)$ is a representation, then we can form the associated vector bundle $$E := \mathrm{Spin}(TM) \times_\rho V.$$ Attached to every Killing vector $X$ on $M$ we have a Lie derivative $\mathcal{L}_X$ on sections of $E$. Explicitly, this Lie derivative takes the form $$ \mathcal{L}_X \sigma = \nabla_X \sigma + \rho(A_X) \sigma,$$ where I am using $\rho : \mathfrak{so}(n) \to \mathfrak{gl}(V)$ also to denote the derivative map of the the representation. (I am also identifying $\mathfrak{so}(TM)$ with $\mathfrak{so}(n)$ via a choice of local frame.)

For example, in the case of the tangent bundle itself viewed as an associated bundle where $\rho$ is the defining representation of $\mathfrak{so}(TM)$, then as expected, we find $$ \mathcal{L}_X Y = \nabla_X Y + A_X(Y) = \nabla_X Y - \nabla_Y X = [X,Y] .$$


Although I quite often use the formula for the Lie derivative $\mathcal{L}_X$ along a Killing vector, I do not feel I have a good conceptual understanding of it.

Could someone enlighten me?

  • 2
    $\begingroup$ Just some pointless speculation: if you look at an arbitrary diffeomorphism of $M$ to itself, there's no obvious way of lifting it to your associated bundle $E$. But if you have an isometry on $(M,g)$, this induces a canonical map from $SO(TM)$ to itself, and can be lifted to a map from $Spin(TM)$, and hence $E$ to itself. So when you have a Killing vector field $X$ you can actually define the Lie derivative on $E$ using the standard formula $\lim_{t\to 0} ({}^X\phi^*_t - Id)/t$ which probably works out to be your formula. But I am not quite sure what kind of an answer you are looking for. $\endgroup$ – Willie Wong May 13 '10 at 14:39
  • $\begingroup$ Thanks. I think that you are right and that this ought to be simply a result of the calculation you mention. My question, which perhaps is a little vague, is whether that's all there is to it or whether there is an alternative explanation for the formula. $\endgroup$ – José Figueroa-O'Farrill May 13 '10 at 15:24

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