In symmetric spaces(for spacetimes of Einsteinian General Relativity) we would like to find the vector space of Killing vectors($\xi^{(n)}_\mu(x)$) for the given metric tensor($g_{\mu\nu})$ at some fixed point $X$.
Now, for infinitesimal coordinate transformations(of the form $x'^\mu=x^\mu +\epsilon\xi^\mu(x) $) which is really a very special case, we determine the corresponding isometries in terms of the associated Killing vector fields.
This special case yields an interesting result about the form of the Killing vector field, especially its possible degrees of freedom in terms of initial values of the field and its first order covariant derivative.
For an infinitesimal patch about a fixed point $X$, the approximate functional form of the $n$-th Killing vector field looks like:
$$ \xi^{(n)}_\rho(x)= A^\lambda_\rho(x;X)\xi^{(n)}_\lambda(X)+B^{\lambda\nu}_\rho(x;X)\xi^{(n)}_{\lambda;\nu}(X)$$
Here $A^\lambda_\rho$ and $B^{\lambda\nu}_\rho$ are functions that depend on $g_{\mu\nu}$ and X and are essentially constants for the set of all Killing vector fields about a point $X$.
So, now the argument goes as follows. :
$$\rm No.\,of\, independent\, killing \,vectors=No.\,of\,independent \,parameters\,uniquely\, identifying \,a\,Killing\,vector\,field\\ =N+\dbinom{N}{2}=\dbinom{N+1}{2} $$
Here, $N$=No. of independent initial values $\xi^{(n)}(X)$ and $\dbinom{N}{2}$=No. of independent initial values of covariant derivatives $\xi^{(n)}_{\lambda;\nu}(X)$(because of antisymmetry condition $\xi_{\sigma;\rho}=-\xi_{\rho;\sigma}$)
So, the question is: Is this approximate calculation(involving Taylor series expansion of Killing vector field components about some fixed point $X$)
correct?
It's disturbing to see that approximate methods are being used to yield very concrete answers like maximal number of independent Killing vectors in a vector space.
These arguments and line of reasoning are mostly drawn from Weinberg's book on Gravitation and Cosmology.
Best Answer
It is not any approximate answer . If you know the value of all derivatives at some point , you can write the function in terms of them . This is not any approximation.