I have been studying semi-riemannian geometry from the Barrett O'Neill's book and general relativity from different sources, but I haven't been able to understand what physicist mean by diffeomorphism invariance.
Let $M$ be a lorentzian manifold with metric tensor $g$. If $\nabla$ is the Levi-Civita connection and $R_{XY}(Z)$ the curvature associated with $\nabla$, then the Ricci tensor is given by
$$\text{Ric}(X,Y)=\sum_{k}\epsilon_{k}g(R_{XE_k}(Y),E_k),$$
where $\{E_k\}$ is a frame field, $\epsilon_k:=g(E_k,E_k)$, and $X, Y\in\mathfrak{X}(M)$ are two arbitrary differentiable vectors fields in $M$. In a similar way, the scalar curvature is given by:
$$S:=\sum_{k}\epsilon_{k}\text{Ric}(E_k, E_k),$$
and the Einstein's field equations in vacuum are $Ric(X,Y)-\frac{1}{2}Sg(X,Y)=0.$
In physics, the Einstein's equations are used to determine the metric tensor $g$, so we can write them as a function of g:
$$Ric(g)(X,Y)-\frac{1}{2}S(g)g(X,Y)=0.$$
Now, the physicist say that if $f:M\rightarrow M$ is a diffeomorphism and $g$ is a solution of the Einstein's equations, then $\widetilde{g}:=f^{*}(g)$ (the pullback of g under f) is also a solution. But this part is very confusing to me because we can try to evaluate the Ricci tensor and the scalar curvature associated with $\widetilde{g}$, or we can apply $f$ directly to Ric (because is a tensor) and to $S$ in the form:
$$\widetilde{\text{Ric}}:=f^{*}(\text{Ric}), \widetilde{S}:=f^{*}(S)$$
It is clear that $Ric(\widetilde{g})$ (the Ricci tensor associated with $\tilde{g}$), $\widetilde{\text{Ric}}(g)$ (the Ricci tensor associated with $g$ but pulled back by $f$) and $\widetilde{\text{Ric}}(\widetilde{g})$ (the Ricci tensor associated with $\tilde{g}$ but pulled back by $f$) are different. I haven't been able to find a relation between them. Moreover, I have made some calculations and I think I can say that:
$$\text{Ric}(\widetilde{g})(X,Y)-\dfrac{1}{2}S(\widetilde{g})\widetilde{g}(X,Y)\neq 0$$
I have tried many different combinations using $\widetilde{\text{Ric}}(\widetilde{g})$, $Ric(\widetilde{g})$ and $\widetilde{\text{Ric}}(g)$ (and the correspondings for the scalar curvature $S$), but I haven't been able to find the field equation satisfied by $\widetilde{g}$.
What is the field equation satisfied by $\widetilde{g}$?, where is my mistake?, how can I see explicitly the diffeomorphism invariance of general relativity using the language used by mathematicians (like O'Neill or Spivak)?, what is the relation between $Ric(\widetilde{g})$ , $\widetilde{\text{Ric}}(g)$ and $\widetilde{\text{Ric}}(\widetilde{g})$?. Maybe the answer to the second question is something easy for a mathematician, however I cannot see it explicitly making calculations using the definitions.
I'd like to be able to understand all this directly from the field equations and not from the Einstein-Hilbert action 😀 .
Best Answer
This may not be the answer you are asking for, but regardless: if you pull back the metric with $f$, then, by definition $$\tilde g(f(p))(f_{*,p} X, f_{*,p}Y)=f^* g(p)(X,Y)$$ which means, (again) by definition, that $f$ is an isometry between $(M, g)$ and $(M,\tilde{g})$. This implies that you cannot geometrically distinguish quantities which are defined through the metric in these manifolds, which then implies (without further computation!) that your curvature related equations are fulfilled in $(M,g)$ if and only if that's true for $(M, \tilde{g})$.
This is assuming you are using the Levi Civita connection, which is what you have claimed. Of course you have to use the Levi Civita connection with respect to $\tilde{g}$ in $(M, \tilde{g})$ and the one with respect to $g$ in $(M,g)$.
The less formal reasoning is that the mathematical description of a physical process must not depend on the mathematical representation, which is, by the way, also a paradigm in (differential) geometry. The way the earth is moving around the sun does not (must not) change just because you switch from, say, cartesian to cylindrical coordinates.
(I can easily believe that you have trouble to verify the computations, especially in the coordinate free formalism, as these calculations offer many opportunities for errors. I don't know, though, where your mistake is...).