# Killing Tensor of Friedman-Robertson-Walker Metric

I would like help showing that the tensor, $$K_{\mu\nu}=a^2(g_{\mu\nu}+u_\mu u_\nu),$$ where $$u^\mu =(1,0,0,0)$$, is a Killing tensor of the spatially flat FRW metric,

$$ds^2=-dt^2+a(t)^2\left(dr^2+d\Omega^2\right).$$

Specifically it must satisfy $$\nabla_{(a}K_{\mu\nu)}=0.$$

I can see that the tensor is basically $$a^2\times(\text{spatial projection matrix})$$, but not sure if there is a trick or symmetry argument to show it is Killing?

Only source I can find is Carroll pg 344, claiming it is easy to check.

• @QuirkyTurtle98 Are you only interested in a clever argument rather than an explicit calculation? – G. Smith Feb 19 '20 at 2:30
• The $d\Omega^2$ in the metric should be multiplied by $r^2$. – G. Smith Feb 19 '20 at 4:06

I don't know a clever way to do this. But it is "easy to check" by just verifying that

$$\nabla_{(\alpha}K_{\mu\nu)}=0$$

is satisfied. It took me about half an hour on paper. No computer algebra is necessary!

The spatially flat FRW metric metric is actually

$$ds^2=-dt^2+a(t)^2(dr^2+r^2d\Omega^2)$$

which is equivalent to

$$ds^2=-dt^2+a(t)^2(dx^2+dy^2+dz^2).$$

The calculation is particularly straightforward in $$t,x,y,z$$ coordinates. The spatial coordinates are all equivalent, so we can just consider indices to be either $$0$$ (temporal) or $$i$$ (spatial).

We have

$$g_{00}=-1;\quad g_{0j}=0;\quad g_{ij}=a^2\delta_{ij}$$

and

$$g^{00}=-1;\quad g^{0j}=0;\quad g^{ij}=a^{-2}\delta^{ij}$$

from which we find that the only nonzero Christoffel symbols are

$$\Gamma^0_{ij}=-\frac{\dot a}{a^3}\delta_{ij}$$

and

$$\Gamma^i_{0j}=\frac{\dot a}{a}\delta^i_j.$$

(There are six cases to consider and each calculation is a line or two.)

Next using $$u_\mu=(-1,0,0,0)$$, we find that

$$K_{00}=0;\quad K_{0j}=0;\quad K_{ij}=a^4\delta_{ij}.$$

Using the usual formula for the covariant derivative of a tensor with two covariant indices, we can proceed to calculate that the only nonzero covariant derivatives of $$K_{\mu\nu}$$ are

$$\nabla_0K_{ij}=2a^3\dot{a}\,\delta_{ij}$$

and

$$\nabla_i K_{0j}=-a^3\dot{a}\,\delta_{ij}.$$

(Again, there are six cases to consider. Each takes no more than a few lines. Note the conceptually interesting second result, where the covariant derivative of a zero component is nonzero, due to nonzero Christoffel symbols multiplying nonzero other components.)

Finally, the Killing tensor condition has to be checked for four cases. Recall that parenthesized indices are to be symmetrized by summing over permutations. Since $$K_{\mu\nu}$$ is symmetric, we need consider only three of the six permutations; we'll just "rotate" the indices.

When the three indices are all temporal, it reduces to one term which we have found to vanish:

$$\nabla_0 K_{00}=0.$$

When two indices are temporal and one is spatial, it is trivially zero because all the terms are zero:

$$\nabla_0 K_{0i}+\nabla_0 K_{i0}+\nabla_i K_{00}=0.$$

When one index is temporal and two are spatial, it is nontrivially zero because the three terms — mirabile dictu! — cancel:

$$\nabla_0 K_{ij}+\nabla_i K_{j0}+\nabla_j K_{0i}=2a^3\dot{a}\,\delta_{ij}-a^3\dot{a}\,\delta_{ij}-a^3\dot{a}\,\delta_{ij}=0.$$

When all three indices are spatial, it is trivially zero again:

$$\nabla_i K_{jk}+\nabla_j K_{ki}+\nabla_k K_{ij}=0.$$

So

$$\nabla_{(\alpha}K_{\mu\nu)}=0$$

is satisfied for all possible values of $$\alpha$$, $$\mu$$, and $$\nu$$.