(adsbygoogle = window.adsbygoogle || []).push({}); Curvature form with respect to principal connection

Hi all,

I have a question. Let us suppose that [itex]P[/itex] is a principal bundle with [itex]G[/itex] standard group, [itex]\omega[/itex] a principal connection (as a split of tangent space in direct sum of vertical and horizontal vectors, at every point in a differential way) and [itex]\theta \in \Omega^k(P,g)[/itex] a k-form valued in the lie algebra ([itex]g[/itex] is lie algebra of [itex]G[/itex]). I tried to compute curvature of [itex]\theta[/itex] with respect to [itex]\omega[/itex] but I'm already stuck.

By definition, curvature is a operator [tex]D:\Omega^k(P,g) \to \Omega_{Hor}^{k+1}(P,g):D\theta (X_1, X_2, \ldots, X_{k+1}):= d\theta (X_1^{Hor}, X_2^{Hor}, \ldots, X_{k+1}^{Hor})[/tex]

where [itex]d[/itex] is external differential of k-form and [itex]X_j^{Hor}[/itex] is horizontal part of [itex]X_j[/itex] with respect to principal connection.

So, in the simplest case, we take [itex]\theta[/itex] a zero form valued in the lie algebra (i.e. a function). If we compute the curvature we must take the exterior differential: [itex]d\theta(x,e) = \partial_\mu \theta dx^\mu + \partial_a \theta de^a[/itex] where [itex](x^\mu;e^a)[/itex] are local fibered coordinates.

Obviously, in the same fibered coordinate, principal connection is in the form:

[tex]\omega = dx^\mu \otimes (\partial_\mu - \omega^A_\mu (x) \rho_A)[/tex]

and [itex]X_j^{Hor} := \omega(X_j)[/itex]

where [itex]\partial_\mu = \frac{\partial}{\partial x^\mu}[/itex] and [itex]\rho_A[/itex] is a basis of local right invariant vector.

Ok, could you give me a hint to complete the result?

Thanks in advance!

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# Curvature form respect to principal connection

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