# Curvature form respect to principal connection

1. Oct 13, 2011

### 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?

Last edited: Oct 13, 2011
2. Oct 13, 2011

### Ben Niehoff

Re: Curvature form with respect to principal connection

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:

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