Let $M$ be a manifold and let $g$ be a tensor on it, say for example a metric $g\in\Gamma(T^{\ast}M\otimes T^{\ast}M)$. I know how to perform any computation on $g$. For instance, taking its derivative respect to the a connection $\nabla$, evaluating it at a point, taking its Lie derivative, obtaining the curvature of the Levi-Civita connection etc.

However, there is a dual formulation on the frame bundle $F(M)$ of $M$, but I never knew how to do the same calculations on the frame bundle, and as I understand it is sometimes simpler to work on the frame bundle. I would like to know how a tensor on $M$ is represented from the point of view of the frame bundle, and how are the typical operations (curvature, Lie derivative etc) implemented. A tensor in $M$ is a section of the corresponding tensor vector bundle. How is this mapped to the frame bundle?

For example, given an open set $U$ of the atlas of $M$ I can write $g$ in coordinates as follows

$g = g_{ab}\,dx^{a}\otimes dx^{b}$

What would be the analog local expression from the point of view of the frame bundle?

Finally, I would like to know a reference where these things are explained in detail.


  • $\begingroup$ Hi! I need some thought into this. But a book you can check is S.S.Chern's "Lectures on Differential Geometry". It has a section on the frame bundle. $\endgroup$ – Bombyx mori Feb 16 '15 at 3:03

A caveat: I'm a physicist, and only crudely trained in the language of differential geometry as it is commonly taught to mathematicians, but I'll do the best I can.

For simplicity, I'll deal with only orthonormal frames. This involves choosing some frame--a fiber of the frame bundle, as it were?--in which set of vector fields in the frame is an orthonormal set. Hence, if you have a frame $f = \{E_{(1)}, E_{(2)}, \ldots, E_{(n)} \}$, then the metric must obey

$$g(E_{(i)}, E_{(j)}) = \delta_{(i)(j)}$$

Here, the indices in parentheses are used to keep these indices obviously distinct from coordinate indices.

When you have such a frame, there is a natural coframe of dual vector fields $E^{(i)}$ such that $g_{(i)(j)} = \delta_{(i)(j)} E^{(i)} \otimes E^{(j)}$.

However, one of the main objects used when working with frames isn't the metric (since it's just the Kronecker delta). Rather, consider a linear map $h:TM \mapsto TM$ such that $h(\partial_i) = E_{(i)}$. One name for this map is the vielbein ("many legs"). The vielbein helps us switch back and forth between the coordinate basis and the orthonormal basis. In particular, see that

$$g[h^{-1}(E_{(i)}), h^{-1}(E_{(j)})] = g(\partial_i, \partial_j) = g_{ij}$$

When we have $h$ and its inverse available to us, we no longer need the coordinate components of the metric; the frame components of the metric and the vielbein have all that information--and more.

Given a tensor, one need only change basis from the bases of coordinate vectors and covectors to the frame vectors and covectors. We already established the change of basis $h$ between coordinate and frame vectors. The corresponding law for covectors is just $h^\dagger(E^{(i)}) = dx^i$, where $h^\dagger$ means the adjoint of $h$. So for instance, if we have a $(1,1)$ tensor ${T_i}^j = T(\partial_i, dx^j)$, then we can find the frame components instead by ${T_{(i)}}^{(j)} = T(E_{(i)}, E^{(j)})$. Alternatively, one could evaluate the coordinate components while still working with the frame by using the vielbein:

$${T_i}^j = T[h^{-1}(E_{(i)}), h^\dagger(E^{(j)})]$$

This could be considered to define a new tensor $\bar T$ such that ${\bar T_{(i)}}^{(j)} = {T_i}^j$. Each of these ways of looking at the tensor $T$ could be useful for different problems.

One of the most interesting differences between using coordinates versus a frame is the covariant derivative. Let $v = v^{(i)} E_{(i)}$, and the covarint derivative of another vector field $A$ takes the form

$$\nabla_v A = v^{(i)} {{h^\dagger}_{(i)}}^{(j)} \frac{\partial A^{(k)}} {\partial x^j} E_{(k)} + \omega_{(k)(j)(i)} v^{(i)} A^{(j)} E_{(k)}$$

Instead of the usual Christoffel symbols, we get the "Ricci rotation coefficients" $\omega$ instead. They're antisymmetric on the first two frame indices. If you have a little exterior algebra knowledge, it can be interpreted as a bivector-(field)-valued linear function of a vector (field). Being antisymmetric on its first two indices, there are fewer Ricci rotation coefficients than there are Christoffel symbols. Moreover, they're called rotation coefficients for a reason: they're intimately tied to smooth choices for rotating your frame basis across the manifold.

A similar approach can be used for converting Lie derivatives to frame form, but so far we've done nothing really special: we've simply used the formulas and equations for Riemannian geometry that you know, except with a different basis. The only real wrinkle to keep in mind was the different connection coefficients.

What the frame formalism opens up to you, though, is the power of differential forms applied to Riemannian geometry. This is the heart of Cartan's moving frames approach. In the moving frames approach, we define the following connection 1-form ${\omega_{(i)}}^{(j)}$ and curvature 2-form ${\Omega_{(i)}}^{(j)}$:

$$\begin{align*} {\omega_{(i)}}^{(j)}(X) &= g[\nabla_X E_{(i)},E_{(j)}] \\ {\Omega_{(i)}}^{(j)}(X, Y) &= \frac{1}{2} g[R(X,Y) E_{(i)}, E_{(j)}] \end{align*}$$

These are pretty different from conventional one-forms and two-forms, in the sense that they're really functions of three or four vectors that we just reinterpret by feeding in some of the frame basis vectors accordingly. But this is pretty standard, as it leads to the Cartan structure equations:

$$\begin{align*} dE^{(i)} &= E^{(i)} \wedge {\omega_{(i)}}^{(j)} \\ d{\omega_{(i)}}^{(j)} &= {\Omega_{(i)}}^{(j)} + {\omega_{(i)}}^{(k)} \wedge {\omega_{(k)}}^{(j)}\end{align*}$$

This approach offers an alternative to computing the connection and Riemann tensor through the usual index manipulations.

I can't speak to solid references for these concepts outside of the usual suspects on the internet, but hopefully this gives you a beginner's guide to the practical use of frames.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.