$\DeclareMathOperator\SO{SO}$A priori: I apologize if this isn't up to Mathoverflow standards, I've had very little luck getting questions on this subject answered elsewhere.

I'm looking for a physics based intuition regarding higher order frame bundles. For example: given some Riemannian manifold $M$ it is standard in physics to work on the bundle of 1-frames $FM$. This is a principal $G$-bundle over $M$.

In general relativity (forgetting the pseudo-Riemannianicity for a moment) we can write the fundamental field variables as sections $\theta$ of the bundle of (co-) frames. One also chooses a (usually) linear, metric compatible $g$-valued connection $\omega$ defined by:


(wedge product implied) which defines a splitting at each point of the tangent space

$$T(FM)=TM\oplus TG$$

For example, if we're on the oriented orthonormal frame bundle $F_{o}M$ in dimension 4 we have:

$$T(F_{o}M)=TM\oplus T(\SO(4))$$

So that the (co)frames can be considered as basis of the (co)tangent space $TM$ or $T^{*}M$. Then the $\omega_{b}^{a}$ may be thought of as a set of (co)frames/basis on $TG$ and $T^{*}G$. Physically we're taught that the connection acts on some vector/tensor on $M$ as it is parallel transported around. I think of a 1-frame then as tied to some point following a world-line through space-time. Then a connection says how it is transported on $M$. Notions of curvature and the Einstein–Hilbert action follow from this.

What I don't understand is how to interpret general n-frames. Suppose we have the same situation as above but now we're on the bundle of (anholonomic) 2-frames $F^{2}M$. This is a principal $G_{2}$ bundle over $M$ where $G\subset G_{2}$. Choosing a linear connection on this bundle (together with the connection on $FM$) then determines a splitting:

$$T(F^{2}M)=TM\oplus TG_{2}\rightarrow TM\oplus TG\oplus T\left(\frac{G_{2}}{G}\right)$$

For example, If we're again choosing (anholonomic) oriented orthonormal frames:

$$T(F_{o}^{2}M)=T\left(F_{o}M\right)\oplus \SO(10)\rightarrow T\left(F_{o}M\right)\oplus \SO(4)\oplus \SO(6)$$

Where the last deformation retraction is due to the natural splitting of frames here into 1-frame and 2-frames. Physically what are two frames and what would be the meaning of the curvature of a connection on such a bundle $F^{2}M$ compared to $FM$? I'm having trouble understanding what the curvature represents?

Are the $n$-coframes also like solder forms as in the 1-coframe case. I think this is related to an $n$-contact structure on $M$.

Don't be afraid to delve deep into it, I have Kobayashi & Nomizu as well as many other texts (including the former author's earlier book on transformations)

I should note here that end game-wise, I'm interested in a possible relation between this and the Chern index on the original frame bundle.

  • $\begingroup$ I'm not sure that a question about "physics based intuition regarding higher order frame bundles" is really well-posed here. $\endgroup$ Jan 15 at 22:20
  • 1
    $\begingroup$ @IgorKhavkine Thank you, I'll edit it appropriately after work. More along the lines of what geometrically do higher order frames represent, and curvature of connections on such bundles $\endgroup$
    – R. Rankin
    Jan 15 at 23:48


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.