Killing vectors and Ricci Tensor - MathOverflow

Killing vectors and Ricci Tensor

2010-11-25T15:21:39Z

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!

Answer by Dick Palais for Killing vectors and Ricci Tensor

2010-11-25T16:30:36Z

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.

Answer by Andrei Moroianu for Killing vectors and Ricci Tensor

2010-11-25T16:33:29Z

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...