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In General Relativity is there a TE symmetry similar to CPT symmetry in the Standard Model ? It's pretty easy to understand that by flipping charge and parity you merely get a time reversed equivalent of your system, so flipping time as well would lead to an equivalent description. Similarly, since metric perturbations in GR are sourced by energy density, it seems to me that GR is invariant if we operate the transformation $(t,\rho) \rightarrow (-t,-\rho)$. Is this correct or am I missing something here ? Do the other $T_{\mu \nu}$ terms come into play in ways which I haven't considered here ? Is the symmetry actually $(g_{\mu \nu},T_{\mu \nu}) \rightarrow (-g_{\mu \nu},-T_{\mu \nu})$ ?

The reason I think this might work is if you make the energy density negative in the Friedmann equations expansion turns into contraction, then you time flip it and it turns back into expansion. Is this a ubiquitous behavior ? If this is indeed generally true can we say there is indeed an ET symmetry ?

ticster
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The structure of general relativity does not allow the theory itself to be classified in terms of global, discrete symmetries such as time-reversal. The Einstein field equations don't refer to a time coordinate; they're expressed tensorially, which means that they are completely independent of what coordinates you choose. Since there is no guarantee that you have a preferred time coordinate, there is no guarantee in general that you can define a global time-reversal operator. You can, for example, have spacetimes that aren't time-orientable.

This is similar to an issue that came up in your other question, which was about Lorentz symmetry. Operators like Lorentz boosts and time reversal are local, not global.

The Friedmann equations are not the field equations of GR, and they don't tell you anything about the symmetry properties of GR.

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