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.

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    $\begingroup$ It's quite possible to use approximate methods in a completely rigorous proof. I think to make this rigorous you need something like the fact (given in Wikipedia) that "A Killing field is determined uniquely by a vector at some point and its gradient (i.e. all covariant derivatives of the field at the point)." I have no idea if Weinberg used something similar in his argument, so I can't say whether it is completely rigorous. $\endgroup$ – Peter Shor Jul 13 '16 at 19:02
  • $\begingroup$ Yes, you are right in that the initial conditions are just the values of the Killing vector field components and its covariant derivatives at X. He does establish this quite early. Only, working with infinitesimal seems to be the trick that does the job but spoils the fun(the conditions become loose as first order correction terms are the only ones retained in the expansion). $\endgroup$ – Subho Jul 13 '16 at 19:18
  • $\begingroup$ It's a general principle that the structure of a Lie group is mostly contained in its Lie algebra, which is the first order transformations. ("Infinitesimal rotations" and such in physics speak.) The reason for this is that the flow of the vector fields preserve the group structure. Thus if the first order approximation is good enough in a radius $r$, you can do a new first order expansion around $r-\epsilon$ and get almost twice as far. In the physical example of Killing vectors we can be more concrete. The Killing vector generates a symmetry, so the spacetime looks the same at the point the $\endgroup$ – Robin Ekman Jul 13 '16 at 19:25
  • $\begingroup$ first order flow takes you to. Thus the first order approximation is good there, too! $\endgroup$ – Robin Ekman Jul 13 '16 at 19:28
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    $\begingroup$ If what we were talking about were an arbitrary function, your doubts would be justified. But it's not; we're talking about a manifold with a group (that generates the Killing field) acting on it. That adds much more structure, and lets the value at one point determine the value everywhere. Offhand, I don't know any easy references for this. Maybe somebody else does. $\endgroup$ – Peter Shor Jul 14 '16 at 11:36

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.

  • $\begingroup$ $f(x)=e^{-1/x}$... $\endgroup$ – MannyC May 7 at 4:10
  • $\begingroup$ You assume analyticity with this statement, which is untenable in reality. $\endgroup$ – Bence Racskó May 7 at 9:36

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