Curvature form respect to principal connection

In summary, the curvature of a k-form \theta with respect to a principal connection \omega is given by the 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}) and X_j^{Hor} := \omega(X_j). This can be written as D \eta = d \eta + \omega \wedge \eta - (-1)^{\mathrm{deg} \; \eta} \, \eta
  • #1
Simon_G
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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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  • #2


Simon_G said:
[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]

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

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:

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

Related to Curvature form respect to principal connection

What is the curvature form with respect to a principal connection?

The curvature form with respect to a principal connection is a mathematical concept that describes the curvature of a connection on a principal bundle. It is a 2-form that represents the curvature of the bundle's connection, and it is used to measure how much the bundle twists or turns when parallel transported along a closed loop on its base space.

How is the curvature form calculated?

The curvature form is calculated using the connection 1-form and the structure group of the principal bundle. The connection 1-form is a matrix of coefficients that describes how the sections of the bundle change when parallel transported along a given direction. The structure group is a group of transformations that act on the fibers of the bundle. The curvature form is then obtained by taking the exterior derivative of the connection 1-form and using the structure group to "project out" the unwanted components of the resulting 2-form.

Why is the curvature form important?

The curvature form is important because it allows us to quantify the amount of curvature present in a principal bundle. This information is crucial in many areas of mathematics and physics, such as differential geometry, general relativity, and gauge theories. It is also a fundamental tool in the study of connections and their associated geometric structures.

What is the relationship between the curvature form and the curvature tensor?

The curvature form is related to the curvature tensor through the Bianchi identity. This identity states that the exterior derivative of the curvature form is zero, which is equivalent to saying that the curvature tensor is skew-symmetric in its first two indices. In other words, the curvature form encodes all the information about the curvature tensor, and it is a more concise and elegant way of describing the curvature of a connection.

How does the curvature form change under a gauge transformation?

The curvature form changes under a gauge transformation in a specific way that is dictated by the structure group of the principal bundle. In particular, the curvature form is invariant under transformations that belong to the center of the structure group. This means that the curvature form behaves like a true geometric object and is not affected by local changes in the coordinates or gauge transformations.

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