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Let $(E,\pi,B)$ be a principal bundle with structure group $G$. The connection $1$-form can be thought of as a projection on the vertical part. It allows us to characterize the horizontal subspaces as $H_p E = \ker \omega_p$ then.

Apart from that, there's the curvature $2$-form. This object is defined as follows: let $\operatorname{hor}$ mean the projection operator taking a vector field to its horizontal part. Then, if $\eta$ is a $k$-form on $E$ its exterior covariant derivative is the $k+1$ form $D\eta$ defined by

$$D\eta(X_1,\dots,X_{k+1}) = d\eta(\operatorname{hor}X_1,\dots,\operatorname{hor}X_{k+1}).$$

We call the curvature $2$-form then the differential form $\Omega = D\omega$ where $\omega$ is the connection $1$-form.

Although the definition is perfectly clear I can't understand what this object really represents. I mean, when we define curvature for curves on space, the curvature is meant to represent how much the curve deviates from a straight line. On the other hand, when reading books about General Relativity some time ago, I read that the curvature of the Levi-Civita connection is intended to encode the information of the difference between a paralel transported vector around a loop and the starting vector. Those two ideas are more geometric. This curvature $2$-form, on the other hand, I can't understand what it really represents.

So, what the curvature $2$-form really represents? How it relates to those other ideas of curvature? And how this intuition is captured by the rigorous definition?

Gold
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    $\Omega$ defines a curvature tensor $R$. Take E be the tangent bundle and $\omega$ be the Levi-Civita connection form this is the same curvature tensor of a Riemann manifold. – Xipan Xiao Apr 23 '15 at 03:30
  • Would you say in which book you've read about the geometric interpretation of the Levi-Civita connection? It would be helpful, thank you. – lou Jun 15 '23 at 03:51
  • It is discussed in most GR books. See e.g. Carroll's Spacetime and Geometry and Wald's General Relativity – Gold Jun 15 '23 at 11:05

1 Answers1

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Remember we are working with principal bundles. Thus, contrary to what other people seem to think, a curvature $2$-form is not the same thing as a Riemann curvature tensor. However, we can get a Riemann curvature tensor from it, in specific cases. We shall discuss that construction momentarily.

Let $P$ be a principal $G$-bundle with projection $\pi:P\to B$. Fixing a connection form $\omega$ on $P$ and a projection $h:TP\to HP$, we define the curvature $2$-form $\Omega$ by $\Omega(X\wedge Y)=\mathrm{d}\omega(hX\wedge hY)$. Expanding this a little further, we get $$\Omega(X\wedge Y)=hX(\omega(hY))-hY(\omega(hX))-\omega([hX,hY])=-\omega([hX,hY]).$$

How might we interpret that? Well, I like to think of it as a measure of how vertical the bracket of the horizontal projections is.

Now, if we consider the frame bundle $F_{GL}M$ of a smooth manifold $M$, then we can get a connection on $F_{GL}M$ and pass that information down to $M$. However, in order to do this, we need to talk in terms of specific frames to get specific projections, leading to la méthode du repère mobile or, roughly translated, "the method of moving frames." Using this, we can essentially interpret $R$ as the "passed-down version of $\Omega$ on $M$".

Robin Goodfellow
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