A general linear transformation is given by

\begin{align} \psi'(x) \to g \psi(x) g^{-1}, \end{align}

The gauge-covariant derivative associated with this transformation is

\begin{align} D_\mu \psi=\partial_\mu \psi -[iqA_\mu, \psi]. \end{align}

Finally, the field is given as

\begin{align} R_{\mu\nu}= [D_\mu,D_\nu], \end{align}

where, $R_{\mu\nu}$ is the Riemann tensor.

From here, what steps lead me to the Einstein field equation?

  • $\begingroup$ You seem confused about what we mean when we say GR is a gauge theory ($R_{\mu\nu}$ is not the Riemann tensor!). See this answer of mine for the differences between GR and Yang-Mills theories. $\endgroup$
    – ACuriousMind
    Jul 8, 2022 at 12:59

1 Answer 1


None. Nothing of this has anything to do with gravity. Your equation for "R" defines the gauge field curvature $F_{\mu\nu}$, not the Riemann curvature $R_{\alpha \beta\mu\nu}$.

  • $\begingroup$ I am using the general linear group for the gauge. $\endgroup$
    – Anon21
    Jul 5, 2022 at 17:32
  • $\begingroup$ You are still not going to get gravity as GR is not a gauge theory (fibre bundle) in the usual sense. There is a formulation of (non super) suopergravity that gauges the Poincare group, but each fibre conatins the whole of space, so again it's not a conventional gauge theory. $\endgroup$
    – mike stone
    Jul 5, 2022 at 17:52

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