How Do Killing Vectors Represent Spacetime Symmetries in Stephani's Relativity?

[Pengwuino]

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?

 

[cepheid]

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?

 

[Pengwuino]

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.

 

[dextercioby]

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.