Hi all,

We all know that the lie derivative of the metric tensor along a Killing Vector vanishes, by definition. I am trying to show that the Lie derivative of the Ricci tensor along a Killing vector also vanishes, and I am hoping to interpret it physically.

What might be a good direction to proceed? Thanks!


Recall that the definition of the Lie derivative of a tensor field $T$ with respect to a vector field $X$ is given by "dragging" $T$ with respect to the one-parameter (quasi) group $\phi_t$ generated by $X$, i.e., computing $\phi_t^*(T)$, and differentiating wrt $t$ at $t = 0$. But to say that $X$ is a Killing field means that the $\phi_t$ are (partial) isometries, and so not only preserve the metric tensor but also the Riemann curvature tensor and its contraction the Ricci tensor or any other tensor field that is defined canonically from the metric tensor and so preserved by isometries. Thus any such tensor field is preserved by dragging, i.e., $\phi_t^*(T)$ is constant in $t$ and so has a zero derivative.

Regarding the physical interpretation, let me try to answer a slightly different question. Recall that the Ricci tensor comes up as the Euler-Lagrange expression for the Einstein-Hilbert functional, and that the latter is invariant under the group of ALL diffeomorphisms. So it is natural to ask what the Noether Theorem (connecting one-parameter groups that preserve a Lagrangian to constants of the motion of the corresponding Euler-Lagrange equations) leads to in this case. The answer is that it gives the contracted Bianchi identity for the Ricci tensor. Perhaps this is what your question about physical significance was aiming at.

| cite | improve this answer | |
  • 2
    $\begingroup$ Dick and Andrei's answers are very good, but I confess to being somewhat surprised that there is no obvious infinitesimal tensor calculation to demonstrate this fact. At least I am unable to find one. $\endgroup$ – Deane Yang Nov 25 '10 at 17:44
  • $\begingroup$ I was intrigued by Deane's remark, so I worked out a purely tensorial proof. I will edit my answer correspondingly. $\endgroup$ – Andrei Moroianu Nov 25 '10 at 19:47
  • $\begingroup$ @Andrei: Just a small question or quibble---why do you put the minus sign in the definition of the Lie derivative? That would make the Lie derivative of a function $f(x)$ on the line wrt the vector field $\partial/\partial x$ equal to $-f^'(x)$ rather than $f^'(x)$ which seems a bit strange. $\endgroup$ – Dick Palais Nov 25 '10 at 20:23
  • $\begingroup$ @Dick: Oh, that's because the definition of the "push forward" $f_*$, which is defined by $(f_*K)_x=df(K_{f^{-1}(x)})$. Here $df$ is the natural extension to tensors, of course. In particular, $f_*(g)$ is just $g\circ f^{-1}$, so the minus sign disappears when you differentiate, so you get the usual formula $L_\xi g=\xi.g$, with the right sign... $\endgroup$ – Andrei Moroianu Nov 25 '10 at 20:39

The Lie derivative of any tensor field $K$ with respect to a vector field $\xi$ is by definition $$L_\xi(K)=-\frac d{dt}|_{t=0}(\phi_t)_*K,$$ where $\phi_t$ is the local flow of $\xi$. Now, if $M$ has a Riemannian metric $g$ and $\xi$ is Killing with respect to $g$, each $\phi_t$ is a local isometry of $(M,g)$. From the uniqueness of the Levi-Civita connection, it follows that every isometry $\phi$ is affine, i.e. $$\phi_*(\nabla_X Y)=\nabla_{\phi_*X}\phi_*Y.$$ From here you get immediately $\phi_*R=R$ for the Riemannian curvature, and since the Ricci tensor of $R$ is obtained by a trace: $$Ric(X,Y)=trace(V\mapsto R_{V,X}Y),$$

one gets $\phi_*Ric=Ric$ for every isometry $\phi$. The first formula thus shows that $L_\xi Ric=0$ for every Killing vector field $\xi$.

Edit: Here is another, purely tensorial, proof of the same statement. Let $\xi$ be Killing, in the sense that $g(\nabla_X\xi,Y)+g(X,\nabla_Y\xi)=0$ for all vector fields $X,Y$. After taking the covariant derivative wrt some vector field $Z$, and doing some standard manipulations, one gets the usual Kostant formula: $$\nabla^2_{X,Y}\xi=R_{\xi,X}Y,\qquad\forall X,Y\in C^\infty(TM).$$ This is just a rewriting of $$L_\xi(\nabla_XY)=\nabla_{L_\xi X}Y+\nabla_X(L_\xi Y),$$ i.e. some sort of Leibniz formula. Applying this formula several times eventually yields the corresponding Leibniz formula for $R$: $$L_\xi(R_{X,Y}Z)=R_{L_\xi X,Y}Z+R_{X,L_\xi Y}Z+R_{X,Y}(L_\xi Z),$$ i.e. $L_\xi R=0$, and finally $L_\xi Ric=0$ after taking the trace.

Of course, this is just the infinitesimal version of the first proof...

| cite | improve this answer | |
  • $\begingroup$ Andrei, thanks for the infinitesimal version, but what I really meant was a proof that follows directly from the properties of the Ricci tensor itself, rather than using its definition in terms of covariant derivatives of vector fields. But your proof is very nice. $\endgroup$ – Deane Yang Nov 26 '10 at 4:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.