Killing vector in Stephani

by Pengwuino
Tags: killing, stephani, vector
 PF Gold P: 7,120 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$$\xi_{a,b,c}+\xi_{b,a,c}=0$$?
 Emeritus Sci Advisor PF Gold P: 5,196 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?
 Sci Advisor HW Helper P: 11,955 Killing vector in Stephani 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.