# General relativity from the general linear group

I am looking at this answer: https://physics.stackexchange.com/a/225417/747. It states:

Let $$f\colon U\to V$$ be any coordinates transformation on charts of a manifold $$U,V\subset\mathcal{M}$$ (i. e. a change of coordinates). Under such transformation fields $$\phi(x)$$ are sent into $$\phi'(f(x)) = S(x)\phi(x)$$.

In order the equations of motion to be satisfied, one must require certain appropriate conditions on the factor $$S(x)$$ (in particular one can see that these could be related to the representations of the underlying group of transformations $$f$$). The set of all allowed operators $$S(x)$$ defines the symmetry group of the theory for the general mapping $$f$$ as defined above. In the case of general relativity $$f$$ are the diffeomorphisms and $$S(x)$$ span the general linear group (up to isomorphisms and cartesian products).

My question is how far the expression $$\phi'(f(x)) = S(x)\phi(x)$$ is from general relativity? Can we derive the equations of motion using said expression as a starting point? What steps and additional assumptions to said expression must be made to reach the appropriate equation of motion?

The question I have linked provides a related answer, but not quite: Ideally, I am looking for a step-by-step process starting from $$\phi'(f(x)) = S(x)\phi(x)$$ as step 1, and the field equations for the final step. If I need additional baggage to get GR, what specifically is the additional baggage, then such baggage should be shown in intermediary steps. Specifically, I work with a system whose solutions is the set of all expressions of the type $$\phi'(f(x)) = S(x)\phi(x)$$, where $$S(x) \in GL(4,R)$$. I suspect it is relatable to GR but I do not know how close to it this is. Is a general expression of this type $$\phi'(f(x)) = S(x)\phi(x)$$ equivalent to GR, enough to derive GR, or merely a solution compatible with GR?

• Does this answer your question? physics.stackexchange.com/q/225413 Dec 27, 2020 at 14:49
• @N.Steinle I think the difference is that I am asking for the next step (in more beginner friendly terms): how to get to GR from $\phi'(f(x)) = S(x)\phi(x)$, whereas the linked question is about the relationship between GL(4,R) and Diff(M). Dec 27, 2020 at 14:55
• Then you've answered your own question: "In the case of general relativity f are the diffeomorphisms and S(x) span the general linear group (up to isomorphisms and cartesian products)." So are you really asking for a justification of this statement? Dec 27, 2020 at 14:59
• @N.Steinle Yes, but more an explanation in beginner friendlier terms. I have a situation where all I have to work with is an expression of the type $\phi'(f(x))= S(x)\phi(x)$. I suspect it is relatable to GR but I do not know how close to it this is. Can one derive the field equations (or the equations of motion) from said expression? If I need additional baggage to get GR, what specifically is the additional baggage. Ideally, I am looking for a step-by-step process starting from $\phi'(f(x))= S(x)\phi(x)$ as step 1, and the field equations for last step. Dec 27, 2020 at 15:07
• If I were you, I would write that exactly in your question. The clearer you describe your question the more likely you'll get an answer and an answer that is helpful. I do not have an answer though :) Hope my upvote helps Dec 27, 2020 at 15:30