In the previous article I introduced Cartan’s connection 1-forms. It is interesting to express various covariant derivatives in terms of them. But firstly:
![\[ \boldsymbol\omega_\mu^{\hphantom\mu\nu} = \omega_{\mu\hphantom\nu\tau}^{\hphantom\mu\nu}\mathbf e^\tau, \qquad\qquad \boldsymbol\omega^\nu_{\hphantom\nu\mu} = \omega^\nu_{\hphantom\nu\mu\tau}\mathbf e^\tau, \]](http://cmaclaurin.com/cosmos/wp-content/ql-cache/quicklatex.com-f397e1fbf79d43687701700aa42670d2_l3.png "Rendered by QuickLaTeX.com")
simply from the definition of covector components. But to check anyway, contract both sides of either equality with 
, to recover the defining formulae. It follows:
![\[ \nabla_\sigma\mathbf e_\mu = \langle\nabla_\sigma\mathbf e_\mu,\mathbf e^\tau\rangle\,\mathbf e_\tau = \omega_{\mu\hphantom\tau\sigma}^{\hphantom\mu\tau}\mathbf e_\tau. \]](http://cmaclaurin.com/cosmos/wp-content/ql-cache/quicklatex.com-2d5e1b1e197351d1ed7f9e679381e304_l3.png "Rendered by QuickLaTeX.com")
To check: the first equality is just components of the vector `
‘. But can check it holds by contracting both sides with 
. Similarly for the covector gradients,
![\[ \nabla_\sigma\mathbf e^\nu = \langle\nabla_\sigma\mathbf e^\nu,\mathbf e_\tau\rangle\,\mathbf e^\tau = \omega^\nu_{\hphantom\nu\tau\sigma}\mathbf e^\tau = -\omega_{\tau\hphantom\nu\sigma}^{\hphantom\tau\nu}\mathbf e^\tau. \]](http://cmaclaurin.com/cosmos/wp-content/ql-cache/quicklatex.com-21127b0f5cf22f54c65756c7ef1ccf83_l3.png "Rendered by QuickLaTeX.com")
Now, 
. Because: substitute 
into the left slot (in our convention) of both sides. The RHS becomes: 
, by linearity of this slot. Now, apply this identity to 
:
![\[ \nabla\mathbf e_\mu = \mathbf e^\sigma\otimes\nabla_\sigma\mathbf e_\mu = \omega_{\mu\hphantom\tau\sigma}^{\hphantom\mu\tau}\mathbf e^\sigma\otimes\mathbf e_\tau = \boldsymbol\omega_\mu^{\hphantom\mu\tau}\otimes\mathbf e_\tau. \]](http://cmaclaurin.com/cosmos/wp-content/ql-cache/quicklatex.com-8f2df818e811d8acf1ea5b7f97bb442a_l3.png "Rendered by QuickLaTeX.com")
And:
![\[\begin{split} \nabla\mathbf e^\nu &= e^\sigma\otimes\nabla_\sigma\mathbf e^\nu = \omega^\nu_{\hphantom\nu\tau\sigma}\mathbf e^\sigma\otimes\mathbf e^\tau = -\omega_{\tau\hphantom\nu\sigma}^{\hphantom\tau\nu}\mathbf e^\sigma\otimes\mathbf e^\tau \\ &= \boldsymbol\omega^\nu_{\hphantom\nu\tau}\otimes\mathbf e^\tau = -\boldsymbol\omega_\tau^{\hphantom\tau\nu}\otimes\mathbf e^\tau. \end{split}\]](http://cmaclaurin.com/cosmos/wp-content/ql-cache/quicklatex.com-9faf56c751b9f2cc0c5b5ad6fee9bd9e_l3.png "Rendered by QuickLaTeX.com")
The antisymmetric part of the covariant derivative is (one half times) the exterior derivative:
![\[\begin{split} d(\mathbf e^\nu) &= \boldsymbol\omega^\nu_{\hphantom\nu\sigma}\wedge\mathbf e^\sigma = \omega^\nu_{\hphantom\nu\sigma\tau}\mathbf e^\tau\wedge\mathbf e^\sigma = -\omega^\nu_{\hphantom\nu\sigma\tau}\mathbf e^\sigma\wedge\mathbf e^\tau \\ &= \omega_{\sigma\hphantom\nu\tau}^{\hphantom\sigma\nu}\mathbf e^\sigma\wedge\mathbf e^\tau. \end{split}\]](http://cmaclaurin.com/cosmos/wp-content/ql-cache/quicklatex.com-b93a68da7af62887c0b6ec1cf1a7a72c_l3.png "Rendered by QuickLaTeX.com")
This is Cartan’s first structural equation! We have assumed a connection with no torsion.