# Physical intuition for curvature on higher order frame bundles?

$$\DeclareMathOperator\SO{SO}$$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:

$$d\theta^{a}=\omega_{b}^{a}\theta^{b}$$

(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.

• I'm not sure that a question about "physics based intuition regarding higher order frame bundles" is really well-posed here. Jan 15 at 22:20
• @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 Jan 15 at 23:48