## Lie derivatives and Diffeomorphism

Over the last days I was mulling over the invariance of general relativity under the whole diffeomorphism group. Now, after looking up some textbooks and after some conversations i had with Sandro and Rainer Häußling i come to the conclusion that the subject is not simple at all. Before I bring up my thoughts I should convince myself that its not only my incompetence holding me back from understanding. Here´s a link to a discussion where some much brighter guys struggle with the same problem ; )

The whole thing started with a task about the Lie derivative given on the actual GR problem sheet. So first I will make some comments on the task:

One can derive the isometry group of a given space from the Lie derivative . If the Lie derivative of the metric tensor along some vector field vanishes, that vector field represents a generator of the isometry group. To understand this consider the definition of the Lie derivative :

$\mathcal{L}_X T = \displaystyle\lim_{\lambda \to 0} \frac{1}{\lambda} \left( \Phi_{\lambda }^* T - T \right)$

Where $\Phi_{\lambda }^*$ indicates the flow of $X$ and acts as a pullback on T. The Lie derivative therefore compares the tensor components evaluated at some point $p$ to the tensor components at another point $p\prime$, where $p$ and $p\prime$ are in general related by a diffeomorphism.

Back to the metric tensor. It is easy to accept now that the condition $\mathcal{L}_Xg=0$ corresponds to the statement that a shift of the metric tensor along the flow of $X$ doesn´t change it. The next task is to sort out the diffeomorphism which fullfill this condition. First we write down the local coordinate expression of the lie derivative:

$\mathcal{L}_XT^{\mu_1 \dots \mu_r}_{\nu_1 \dots \nu_s} = X^{\lambda}\partial_{\lambda}T^{\mu_1 \dots \mu_r}_{\nu_1 \dots \nu_s}\newline \hspace*{18mm} -T^{\lambda\,\mu_2 \dots \mu_r}_{\nu_1 \dots \nu_s}\partial_{\lambda}X^{\mu_1}-\text{all upper indices}\newline \hspace*{18mm} +T^{\mu_1 \dots \mu_r}_{\lambda\, \dots \nu_s}\partial_{\nu_1}X^{\lambda}+\text{all lower indices}$

(The derivation of the above expression can be found in Straumanns textbook).

In case of a fixed solution of Einsteins equations (i. e. Minkowski space) the condition reduces to the famous killing equations

$\partial^{\mu} X^{\nu}+\partial^{\nu} X^{\mu}= 0$

In both equations the partial derivative can be replaced by the covariant one (the christoffel terms chancel). And because the Levy-Civita connection is metric the $(Xg)$ term always vanishes.

The left hand side of the killing equation is a symmetric 2-tensor $A_{\mu\nu}$ , which leaves 10 possible solutions. Because the Poincare group is the isometry group of the Minkowski spacetime, there are actually 10 vector fields. Minkowski space time is a maximal symmetric space. There are solutions of the Einstein equation with less or even no killing vector fields at all. Beside the generator of transitions, its not obvious how the residual killing vectors look like. The construction can be done in the vielbein formalism. I will tell you where an instruction can be found and present an easy example instead:

${\Bbb R}^2 :\quad ds^2= dx^2+dy^2$

A symmetric tensor $A_{i j}$ in 2 dimensions has 3 components and we expects therefore 3 killing vector fields. Two of them are obvious again and the third one also if we take a look at the metric in polar coordinates:

$X_1=(1,0)^T\, ,\, X_2=(0,1)^T\, ,\,X_3=(-x , y)^T$

So much for the resume. What I want to know is if there is any way to show the diffeomorphism invariance of GR in this context?
My question is: why won´t the equation $\mathcal{L}_Xg(x)=0$ , not inserting some solution, but the general dynamical metric tensor field in GR spit out the whole diffeomorphism group?

I talked to Florian Scheck today about the problem and he told me the problem is badly posed. I ask for the vector fields which leave every solution of Einsteins equation invariant, which should turn out as no fields at all instead of all vector fields.

Any further input is heavily appreciated.