Mathematical characterisation of diffeomorphisms in General Relativity

Question

Considering the diffeomorphism covariance/invariance of General Relativity, is it possible to characterise mathematically the various kinds of possible transformations $x'^{\mu} = f^{\mu}(x)$? All of them can be regarded passively as coordinate transformations, a subgroup of them can be seen actively as change of observer; some of these could relate any group of physical observers, some others just free falling observers. Elements that enter into play should presumably be the preservation of the nature of worldlines (timelike for physical observers) and geodesic equation. Is it possible to classify these various diffeomorphisms in group structures according to their possible meaning and their mathematical form? What exactly are their mathematical characterisation?

Answer 1

GR is diffeomorphism invariant. If the spacetime manifold $M$ has a metric $g$, then it is possible that its diffeomorphism group $\mathrm{Diff}(M)$ has a subgroup which preserves its metric. This is known as the isometry group, which preserves distance between any two points. The vector fields that generate the isometry group (if exists) are known as the Killing vector fields.

For example, a Minkowki spacetime, has the following Killing vector fields: $$\frac{\partial}{\partial t},\quad\frac{\partial}{\partial x^{i}},\quad t\partial_{i}-x^{i}\partial_{t},\quad\mathrm{and}\quad x^{i}\partial_{j}-x^{j}\partial_{i},$$

which generates energy, momentum, Lorentzian boost, and angular momentum, respectively.

The isometry group in field theory is viewed as a global symmetry. On the other hand, in general, a generic diffeomorphism in GR is viewed as a gauge redundancy.

The Einstein-Hilbert action $$S[g]=\int d^{4}x\sqrt{|\det g|}R$$

should be diffeomorphism invariant because $R$ is a scalar, and $d^{4}x\sqrt{|\det g|}$ is an invariant volume form.

Its diffeomorphism invariance leads to the conservation of stress-energy tensor. i.e $$\nabla^{\mu}T_{\mu\nu}=0,$$

because under a generic diffeomorphism, the action transforms as $$\delta S[g]=\int d^{4}x \frac{\delta S[g]}{ \delta g_{\mu\nu}}\delta g_{\mu\nu}= \int d^{4}x \sqrt{|\det g|} T^{\mu\nu}\nabla_{\mu}\xi_{\nu},$$

where in the last step $$\delta_{\xi}g_{\mu\nu}=\mathcal{L}_{\xi}g_{\mu\nu}=2\nabla_{(\mu}\xi_{\nu)}$$

has been used.

In one-dimensional case, the action of geodesics $$S[g(\tau)]=\int_{a}^{b}d\tau\sqrt{g_{\mu\nu}(q(\tau))\frac{dq^{\mu}}{d\tau}\frac{dq^{\nu}}{d\tau}}$$

is reparameterization invariant. i.e $$S[q(\tau^{\prime})]=S[q(\Phi(\tau)],$$

where $\tau^{\prime}=\Phi(\tau)$ is an arbitrary diffeomorphism on $\mathbb{R}$. This is because the one-form $d\tau$ behaves like a worldline einebein, and the Lagrangian behaves like a worldline vector.

The reparameterization invariance of the geodesic action means that the canonical energy of a free particle is zero. i.e $$H_{c}=\frac{\partial L}{\partial\dot{q}}\dot{q}-L\equiv0,$$

as a result of Euler's theorem for homogeneous functions.

Notice that this does not mean that the geodesic is invariant under the action of $\mathrm{Diff}(M)$, as is pointed out by @Eletie.