Curvature form respect to principal connection

[Simon_G]

Curvature form with respect to principal connection

Hi all,

I have a question. Let us suppose that P is a principal bundle with G standard group, \omega 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 \theta \in \Omega^k(P,g) a k-form valued in the lie algebra (g is lie algebra of G). I tried to compute curvature of \theta with respect to \omega but I'm already stuck.

By definition, curvature is a operator 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})

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

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

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

\omega = dx^\mu \otimes (\partial_\mu - \omega^A_\mu (x) \rho_A)

and X_j^{Hor} := \omega(X_j)

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

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

Thanks in advance!

 

[Ben Niehoff]

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})

and X_j^{Hor} := \omega(X_j)

It seems that these two facts are sufficient to give you the answer. Did you want to write the answer another way?

I'm afraid I'm not very familiar with the "horizontal vs. vertical" way of thinking of principal connections, so I can only tell you what the result is:

D \eta = d \eta + \omega \wedge \eta - (-1)^{\mathrm{deg} \; \eta} \, \eta \wedge \omega