2

Is it possible to obtain general relativity as a gauge theory from the general linear group?

The starting point is:

$$ M'=GM $$

where $M',M,G$ are elements of $GL(4,\mathbb{R})$.

I believe the second step is to construct a 'pass-throught' derivative (gauge covariant derivative) such that:

$$ D (GM) = GdM \tag{1} $$

According to https://en.wikipedia.org/wiki/Gauge_covariant_derivative#General_relativity, the covariant derivative of general relativity is :

$$ \nabla_j v^I:=\partial_jv^I +\sum_k \Gamma^i_{jk}v^k \tag{2} $$

How can I obtain (2) starting from (1)?

Finally, if you look at this answer To which extent is general relativity a gauge theory? it seems to suggest that you can, but then the comment section of this answer seems to show the author backtracking on the claim.

Anon21
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  • I don't believe that your 'pass-through' derivative is well defined, since given $G,M, A\in\mathrm{GL_n}$ we have $D(GM)=D(GMAA^{-1})=GdM=GMd(AA^{-1})=GMAd(A^{-1})$. I think you have to get a little more into the weeds and consider the action of the Lorentz group on the spacetime manifold since this action preserves the metric structure. – J.V.Gaiter Feb 08 '21 at 19:34

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