Cartan’s connection 1-forms

Difficulty: ★★★★☆ undergraduate / graduate

The connection 1-forms $\boldsymbol\omega{_\mu^{\hphantom\mu\nu}}$ are one way to express a connection $\nabla$ on a manifold. The connection coefficients $\Gamma_{\sigma\mu}^{\hphantom{\sigma\mu}\nu}$ are more familiar and achieve the same purpose, but package the information differently. Connection forms are part of Cartan’s efficient and elegant “moving frames” approach to derivatives and curvature.

[I am only just learning this material, so this article is for my own notes, consolidation of understanding, and checking of conventions. It is a work in progress. There is not much actual derivation in what follows, so don’t be intimidated by its length, and most formulae here are just notation.]

Write $\mathbf e_\mu$ for a vector basis at each point, and $\mathbf e^\nu$ for its dual basis. For now, we do not assume these frames are orthonormal (in fact, we don’t even need a metric, for now). The connection forms for this basis are: $\boldsymbol\omega_\mu^{\hphantom\mu\nu}(\mathbf X) := \langle\nabla_{\mathbf X}\mathbf e_\mu,\mathbf e^\nu\rangle$ , where $\mathbf X$ is any input vector. (I will sometimes write $\langle\cdot,\cdot\rangle$ for the contraction between a vector and covector, because the unified notation with the metric scalar product is convenient, although it is sometimes worth reminding oneself that no metric is needed in this case.) To find the components, substitute basis vectors $\mathbf X \rightarrow \mathbf e_\sigma$ :

$\omega_{\mu\hphantom\nu\sigma}^{\hphantom\mu\nu} := \boldsymbol\omega_\mu^{\hphantom\mu\nu}(\mathbf e_\sigma) = \langle\nabla_\sigma\mathbf e_\mu,\mathbf e^\nu\rangle =: \Gamma_{\sigma\mu}^{\hphantom{\sigma\mu}\nu},$

where $\nabla_\sigma := \nabla_{\mathbf e_\sigma}$ as usual. Hence with our conventions, the $\mu$ -index specifies which basis vector field is being differentiated, $\sigma$ specifies the direction it is being differentiated in, and $\nu$ specifies the component of the resulting vector. (Lee 2018 book Problem 4-14 uses the same convention. MTW book §14.5, Frankel 2012 book §9.3b, and Tu 2017 book §11.1 would write $\boldsymbol\omega^\nu_{\hphantom\nu\mu}$ for our expression — which swaps the index order.)

We could define other connection 1-forms $\boldsymbol\omega^\nu_{\hphantom\nu\mu}$ for the dual basis. Note the different index placement. These are:

$\omega\indices^\nu_{\hphantom\nu\mu\sigma} := \boldsymbol\omega^\nu_{\hphantom\nu\mu}(\mathbf e_\sigma) := \langle\nabla_\sigma\mathbf e^\nu,\mathbf e_\mu\rangle = -\langle\nabla_\sigma\mathbf e_\mu,\mathbf e^\nu\rangle = -\omega_{\mu\hphantom\nu\sigma}^{\hphantom\mu\nu}.$

Hence the two sets of connection forms are related:

$\boldsymbol\omega_\mu^{\hphantom\mu\nu} = -\boldsymbol\omega^\nu_{\hphantom\nu\mu}.$

Caution: This is not the same antisymmetric relation which most texts refer to. These texts refer to a single set of connection forms for an orthonormal basis. In that case, we have additionally $\boldsymbol\omega_\mu^{\hphantom\mu\nu} = -\boldsymbol\omega_\nu^{\hphantom\nu\mu}$ , and similarly for our alternate connection forms.

This used:

$0 = \nabla_\sigma\langle\mathbf e_\mu,\mathbf e^\nu\rangle = \langle\nabla_\sigma\mathbf e_\mu,\mathbf e^\nu\rangle + \langle\mathbf e_\mu,\nabla_\sigma\mathbf e^\nu\rangle.$

For the first equality, $\langle\mathbf e_\mu,\mathbf e^\nu\rangle = \delta_\mu^{\hphantom\mu\nu}$ is constant, so its gradient vanishes. The second equality follows from the defining properties of the covariant derivative, i.e. the extension of the connection to covectors and other tensors (e.g. Lee 2018 book Prop. 4.15).

[Regarding index placement, and their raising and lowering, I was formerly confused by this issue in the context of vector bases, for a previous blog article. Specifically, to express an arbitrary frame in terms of a coordinate basis, some references write the components as $\mathbf e_a^{\hphantom a\mu}$ . The Latin index is raised and lowered using the metric components in the arbitrary frame, whereas the Greek index uses the metric components in the coordinate frame. However textbooks were not clear on what was definition vs. what was derived. I eventually concluded the various indices and their placements are best treated as a definition of components, with any formulae for swapping/raising/lowering being obtained from that.]

It is interesting to express various covariant derivatives in terms of the connection forms. But firstly:

$\boldsymbol\omega_\mu^{\hphantom\mu\nu} = \omega_{\mu\hphantom\nu\tau}^{\hphantom\mu\nu}\mathbf e^\tau, \qquad\qquad \boldsymbol\omega^\nu_{\hphantom\nu\mu} = \omega^\nu_{\hphantom\nu\mu\tau}\mathbf e^\tau,$

simply from the definition of covector components. But to check anyway, contract both sides of either equality with $\mathbf e_\sigma$ , to recover the defining formulae. It follows:

$\nabla_\sigma\mathbf e_\mu = \langle\nabla_\sigma\mathbf e_\mu,\mathbf e^\tau\rangle\,\mathbf e_\tau = \omega_{\mu\hphantom\tau\sigma}^{\hphantom\mu\tau}\mathbf e_\tau.$

To check: the first equality is just components of the vector ` $\nabla_\sigma\mathbf e_\mu$ ‘. But can check it holds by contracting both sides with $\mathbf e^\nu$ . Similarly for the covector gradients,

$\nabla_\sigma\mathbf e^\nu = \langle\nabla_\sigma\mathbf e^\nu,\mathbf e_\tau\rangle\,\mathbf e^\tau = \omega^\nu_{\hphantom\nu\tau\sigma}\mathbf e^\tau = -\omega_{\tau\hphantom\nu\sigma}^{\hphantom\tau\nu}\mathbf e^\tau.$

Now, $\nabla = \mathbf e^\sigma\otimes\nabla_\sigma$ . Because: substitute $\mathbf X$ into the left slot (in our convention) of both sides. The RHS becomes: $\langle\mathbf e^\sigma,\mathbf X\rangle\nabla_\sigma = X^\sigma\nabla_\sigma = \nabla_{\mathbf X}$ , by linearity of this slot. Now, apply this identity to $\mathbf e_\mu$ :

$\nabla\mathbf e_\mu = \mathbf e^\sigma\otimes\nabla_\sigma\mathbf e_\mu = \omega_{\mu\hphantom\tau\sigma}^{\hphantom\mu\tau}\mathbf e^\sigma\otimes\mathbf e_\tau = \boldsymbol\omega_\mu^{\hphantom\mu\tau}\otimes\mathbf e_\tau.$

And:

$\begin{split} \nabla\mathbf e^\nu &= e^\sigma\otimes\nabla_\sigma\mathbf e^\nu = \omega^\nu_{\hphantom\nu\tau\sigma}\mathbf e^\sigma\otimes\mathbf e^\tau = -\omega_{\tau\hphantom\nu\sigma}^{\hphantom\tau\nu}\mathbf e^\sigma\otimes\mathbf e^\tau \\ &= \boldsymbol\omega^\nu_{\hphantom\nu\tau}\otimes\mathbf e^\tau = -\boldsymbol\omega_\tau^{\hphantom\tau\nu}\otimes\mathbf e^\tau. \end{split}$

The antisymmetric part of the covariant derivative is (one half times) the exterior derivative:

$\begin{split} d(\mathbf e^\nu) &= \boldsymbol\omega^\nu_{\hphantom\nu\sigma}\wedge\mathbf e^\sigma = \omega^\nu_{\hphantom\nu\sigma\tau}\mathbf e^\tau\wedge\mathbf e^\sigma = -\omega^\nu_{\hphantom\nu\sigma\tau}\mathbf e^\sigma\wedge\mathbf e^\tau \\ &= \omega_{\sigma\hphantom\nu\tau}^{\hphantom\sigma\nu}\mathbf e^\sigma\wedge\mathbf e^\tau. \end{split}$

This is Cartan’s first structural equation! We have assumed a connection with no torsion.

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