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As introduced in most standard GR textbooks, the exterior derivative of a p-form $X$ is defined in a coordinate basis as $(dX)_{\mu_1....\mu_{{p+1}}}:= (p+1) \partial_{[\mu_1}X_{\mu_2....\mu_{{p+1}]}}$ (1).

One can show that this transforms as a tensor (or more specifically a p+1 form) when changing from one coordinate basis to another coordinate basis. However, in GR, the definition of a tensor means that it must also transform appropriately when we go to an anholonomic basis (non-coordinate basis).

Since in GR we assume that the connection is torsion free then we can write the exterior derivative in the form $(dX)_{a_1....a_{{p+1}}}= (p+1) \nabla_{[a_1}X_{a_2....a_{{p+1}]}}$ (2) which is a tensor in the full sense (ie in coordinate and anholonomic basis).

When we define the exterior derivative on manifolds with torsion-full connections, how do we do this? Do we just use the definition (2)? If so, it seems odd that we should start the definiton of (1) if (2) is more general. Also I thought the exterior derivative was useful because it is not meant to be related to the connection but (2) is related to the connection.

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In differential geometry, checking that $(1)$ does not depend on the coordinate chart chosen is enough to show that it defines a tensor $\rm{d}X$ which is independent of the chart used. This is the "right" definition for the exterior derivative. However, it will not keep the same form in a frame which is not a coordinate basis (which is to be expected since you are using $\partial_\mu$).

When you have a torsion-free connection, you can then derive $(2)$ which does hold for general frames, but is actually less general than $(1)$ since you need additional structure on the manifold (the connection).

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  • $\begingroup$ But when you don't have a torsion free connection you can't make the connection between (1) and (2) so how would the exterior derivative be defined in this case? (and defined I mean defined in all frames not just coordinate frames) $\endgroup$ – Dan May 3 at 14:57
  • $\begingroup$ @Dan If you look at the common definitions of the exterior derivative on Wiki, only the second relies on choosing coordinates, the first and third are perfectly coordinate-free and hence automatically "for all frames". $\endgroup$ – ACuriousMind May 3 at 15:43

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