# Killing vector in Stephani

1. ### Pengwuino

6,942
In Stephani's "Relativity", section 33.3, equation (33.9), he has the Killing equations for cartesian coordinates as

$$\xi_{a,b}+\xi_{b,a}=0$$

From there he says upon differentiation, you can get the following three equations

$$\xi_{a,bc}+\xi_{b,ac}=0$$
$$\xi_{b,ca}+\xi_{c,ba}=0$$
$$\xi_{c,ab}+\xi_{a,cb}=0$$

Now, I'm not use to the ,; notation, but doesn't the first equation mean

$$\partial_b \xi_a + \partial_a \xi_b=0$$?

If so, I don't understand the other 3 equations then. If for example, the first one is suppose to be subsequent differentiation by $$\partial_c$$, then wouldn't it be$$\xi_{a,b,c}+\xi_{b,a,c}=0$$?

2. ### cepheid

5,189
Staff Emeritus
I think that it is supposed to be a second derivative, and the second comma is omitted. So:

$$\xi_{b,ca} = \partial_a(\partial_c\xi_b)$$

EDIT: If you assume that, then does it work?

3. ### Pengwuino

6,942
As far as i can tell, no. He seems to be permuting the indices but I don't know what about the killing vector allows one to do that.

4. ### dextercioby

12,327
If you're working in a flat space without a torsion, then the partial derivatives commute when applied to any covector, be it Killing or not.

So from the Killing equation $\xi_{(a,b)} = 0$, differentiating it by $x^c$, one obtains succesively

$$\xi_{(a,b)c} = \xi_{(a,bc)} = 0 {}$$ ,

thing which allows you, Stephani and everybody else to permute the indices in every of the 6 possible cases, without changing anything.