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On a vector bundle, I can define the Riemannian curvature tensor

$$\nabla_X \nabla_Y - \nabla_Y \nabla_X - \nabla_{[X, Y]}$$

In the frame bundle of this vector bundle, I can define the curvature form

$$\textrm{d}\omega(X^H, Y^H) = \textrm{d}\omega(X, Y) + \frac{1}{2}[\omega\wedge\omega](X, Y)$$

How to directly see (a geometrical interpretation) that the two is equivalent?

Alex
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  • What do you mean by equivalent? Are you aware of the isomorphism of horizontal, $W$-valued, type-$\rho$ $k$-forms on the frame bundle $P$ onto $k$-forms on the base manifold which take values in the associated bundle $P\times_{\rho}W$, where $\rho$ is some representation on a space $W$? – nicrot000 Dec 07 '22 at 15:13
  • Yes I am aware you can construct the associated vector bundle $E$ for a Principal bundle $P$. If $P$ is the frame bundle of a vector bundle $E'$, I think then $E$ is $E'$ (isomorphic). By equivalent I mean they describe the same physical thing. Or is there some kind of 1-to-1 correspondence between each term of these two expressions? – Alex Dec 08 '22 at 00:21
  • They are connected under the homomorphism described in the answer below here (https://math.stackexchange.com/questions/4583981/natural-identification-between-the-difference-of-two-principal-connections-and-t/4601844#4601844), which for the identity representation is an isomorphism. – nicrot000 Dec 22 '22 at 17:07
  • I'm not aware of a geometric interpretation of the single terms, but note that (generally) neither are $\nabla_X\nabla_Y-\nabla_Y\nabla_X$ and $\nabla_{[X,Y]}$ differential forms with values in $\text{End}(E)$, nor are $\text{d}\omega$ and $[\omega\wedge\omega]$ themselves horizontal, $G$-invariant forms. – nicrot000 Dec 22 '22 at 17:12
  • @nicrot000 I couldn't figure out how the link answers my question. On your second comment, are you saying because individual terms are not tensorial they don't have a particular geometric interpretation? – Alex Dec 27 '22 at 12:06
  • Quick comment: The curvature of a connection on a vector bundle is usually called simply a connection. The adjective Riemann is usually reserved for the curvature tensor of a Riemannian metric. – Deane Jan 02 '23 at 17:02
  • Yeah, $Riemannian$ is reserved, but the curvature of a connection is not called connection again, is it? – nicrot000 Jan 03 '23 at 13:52

1 Answers1

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I write this as an answer since it is too long for the comment section.

The curvature form of a connection form $\omega\in\Omega^1(P,\mathfrak{g})$, that is $$F(X,Y):=\textrm{d}\omega(X^H, Y^H) = \textrm{d}\omega(X, Y) + \frac{1}{2}[\omega\wedge\omega](X, Y)$$ is also a differential form $F\in\Omega^2(P,\mathfrak{g})$ with values in the Lie algebra $\mathfrak{g}$ of the group $G$, but contrary to the connection form it is horizontal. Moreover, along the fibers of $P$ it transforms under the the adjoint representation of $G$, hence by the isomorphism in the first two comments it is associated with a differential form on the base manifold $M$ which takes values in the adjoint bundle $$\textrm{Ad}(P)=P\times_\text{ad}\mathfrak{g}.$$ In the link in the comments you can now find how some other representation $\rho$ of $G$ on some space $W$ gives rise to a bundle homomorphism from $\textrm{Ad}(P)$ to the endomorphism bundle of $E=P\times_\rho W$. This bundle homomorphism extends to the section spaces and tothe spaces of differential forms with values in these bundles.

The curvature form of $\omega$, as the $\textrm{Ad}(P)$-valued form on $M$, and the curvature form $\nabla^\omega_X \nabla^\omega_Y - \nabla^\omega_Y \nabla^\omega_X - \nabla^\omega_{[X, Y]}$ of the covariant derivative $\nabla^\omega$ on $E$ induced by $\omega$, as the $\mathrm{End}(E)$-valued form on $M$, are related by the the above homomorphism. If I remember correctly I'm missing some details, but it's essentially that.

And no, I don't say that they don't have a geometrical interpretation, but the single terms cannot have a rigorous correspondence term-wise, since the natural identifications and bundle homomorphisms from above only identify forms with the respective behaviors (e.g. being horizontal and transforming under $\rho$ for forms on $P$).

However, I guess to work out whether there is some correspondence you'd have to dive into the dirty work and prove the above correspondence of the different notions of curvature (and work out the details that I have skipped). I wouldn't expect so, but maybe there is somebody with a deeper view on that.

nicrot000
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  • Quick comment: To answer the original question, it is unnecessary to introduce the group $G$, its Lie algebra $\mathcal{g}$, and a representation of $G$. It suffices to consider all frames (i.e., let $G = GL(n)$ with its standard representation). – Deane Jan 02 '23 at 17:21