# Rotational Killling vectors

1. Sep 2, 2011

### lennyleonard

Hi everyone, i hope you'll help me with this doubt i have, which came up when reading weinberg's "gravitation and cosmology".

We are talking about killing vectors.
No problem for the ordinary traslational killing vectors,the problems begin with the rotational ones and the concept of isotropy.

Now, weinberg says that all of the transformation that build a generic killing vector V are of the form$$x^\mu\rightarrow x^\mu+\varepsilon V^\mu$$and that's fine.

He then says that a rotational killing vector leaves the origin fixed, so that if you take a point X on the manifold and you consider a rotational killing vector at X you have $V^\mu(X)=0$ - that's kind of fine too, because in the relation $$x^\mu\rightarrow x^\mu+\varepsilon V^\mu$$I don't want to change the point; but then what are the components of that killing vector, all zero??? clearly they are not, but why then if i imposed $V^\mu(X)=0$??

He says that you have to let the first derivatives take all possible values, but what does this mean then??

And, more of all, why is regarding those killing vectors as antisymmetric matricies??
he in fact says that rotational killing vectors must verify these proprieties:
\begin{align} &V_\lambda{}^{\mu\nu}(x;X)=-V_\lambda{}^{\nu\mu}(x;X)\\ &V_\lambda{}^{\mu\nu}(X;X)=0\\ &V_\lambda;\rho{}^{\mu\nu}(x;X)=\delta^\mu _\lambda\delta^\nu_\rho-\delta^\mu_\rho\delta^\nu_\lambda \end{align}

and, one more thing, what does the round parenthesis notation $(x;X)$ in $V_\lambda{}^{\mu\nu}(x;X)$ mean?? the capital X is the point in which you consider the vector, but the other x????