Curvature and torsion on principal bundles - Lec 24 - Frederic Schuller

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Recommends Ali · 2020-07 · Mathematics Stack Exchange · on “How can a vector field act on a Lie Algebra element?”

We have the definition of a vector field as a smooth section of the tangent bundle $$X:P\longrightarrow TP,$$ where $(TP,\pi',P)$ is the tangent bundle over the total space of the principal G-bundle $(P,\pi,M)$ . I.e, a vector field is an assignment of a vector ( $\in TP$ ) to every point of a smooth manifold $P$ . I have, however, come across expressions like " $X(A)$ " where $X$ is (supposedly) a vector field and $A$ is a Lie algebra element (for example at 19:31 here [this video] ). How is the vector field defined in this case? Do we simply take for granted that the lie algebra is given a manifold structure, at which point the vector field is defined in the same way, something like $X:T_eG\lon

Recommends Ali · 2020-08 · Mathematics Stack Exchange · on “How is the wedge product in the Bianchi identity for torsion defined?”

The torsion $\Theta:=\mathcal{D}\theta$ of a connection one-form $\omega$ for a principal $G$ -bundle with respect to a solder(ing) form $\theta$ is given by $$\Theta=d\theta+\omega\wedge\theta$$ where $d$ is the exterior derivative and $\mathcal{D}$ is the exterior convariant derivative. The Bianchi identity for torsion is given by $$\mathcal{D}\Theta=\Omega \wedge\theta$$ where $\Omega$ is the curvature of the connection one-form. My question is: in both of these statements, how is $\wedge$ defined? $\omega$ is a Lie-algebra-valued one-form and $\theta$ is a $V$ -valued two-form, where $V$ is the representation space of a linear representation of the Lie group $G$ , so how can we wedge the

Cites user169903 · 2018-05 · Mathematics Stack Exchange · on “How curvature and field strength are exactly the same?”

I am watching these lecture series by Fredric Schuller. Curvature and torsion on principal bundles - Lec 24 - Frederic Schuller [this video] @minute 34:00 In this part he discusses the Lie algebra valued one and two forms on the principal bundle that are pulled back to the base manifold. He shows the relationship between general relativity and electromagnetism in the classical theory. He emphasize that for instance that the curvature of space-time is exactly the same as the non-abelian Yang-Mills field strength. It is not clear to me in what sense they are "exactly" the same. Isn't the physical electromagnetism filed strength tensor a real valued object and abelian?

Cites Wyatt Kuehster · 2025-05 · Mathematics Stack Exchange · on “Can connections on a principal bundle have torsion?”

Yes, there is a notion of torsion on a principal $G$ -bundle! In the formulation I know, it requires extra structure, namely a representation $\rho:G\rightarrow GL(V)$ with $\dim(V)=\dim(M)$ . One then defines a "soldering form" $\theta\in \Omega^1(P;V)$ , which is horizontal, $G$ -equivariant in the appropriate sense, and induces an isomorphism $TM\cong P\times_G V$ . The $\textbf{torsion}$ of the connection is $D\theta\in \Omega^2(P;V)$ . Notice that the dependence on the connection is hiding in the "D" here which is the exterior covariant derivative . The curvature is $D\omega$ where $\omega\in \Omega^1(P;\mathfrak{g})$ is connection form, which is vertical, and ad-equivariant. One can sh