# Factor 1/2 in the Curvature Two-form of a Connection Principal Bundle

1. Aug 6, 2014

### center o bass

In the formulation of connections on principal bundles, one derives an
expression for the covariant exterior derivative of lie-algebra valued forms which is given by
$$D\alpha = d \alpha + \rho(\omega) \wedge \alpha,$$
where $\rho: \mathfrak g \to \mathfrak{gl}(\mathfrak g)$ is a representation on the Lie algebra. Now, one often encounters the following formula for the curvature of the connection $\omega$:
$$\Omega = D\omega = d\omega + \frac{1}2 [\omega, \omega].$$

However, if we use the representation $\text{ad}:\mathfrak g \to \mathfrak{gl}(\mathfrak g)$, then the covariant exterior derivative of $\omega$ gives
$$D\alpha = d \alpha + \rho(\omega) \wedge \alpha = d \alpha + [\xi_k, \xi_l] \omega^k \wedge \omega^l = d\alpha + [\omega, \omega]$$.

But where have the factor $1/2$ gone?

I suspect my error might lie in one of the following:
(1): my definition of $[\omega, \omega]$ as $[\xi_k, \xi_l] \omega^k \wedge \omega^l$.
(2): or, that what is meant by the expression $\rho(\omega) \wedge \omega$ is perhaps not $[\xi_k, \xi_l] \omega^k \wedge \omega^l$.

But which one is it? And why?

2. Aug 6, 2014

### Ben Niehoff

Your error is in assuming that you can take a covariant derivative of the connection $\omega$. But $\omega$ is not gauge-covariant!

To be clear, this line is in error:

It is true that

$$\Omega = d\omega + \frac12 [\omega,\omega] = d \omega + \omega \wedge \omega.$$
However, you cannot write this as "$D\omega$", because there is no such object as "$D\omega$". And as you have shown, attempting to write out "$D\omega$" from the standard formula gives you the wrong answer.

You should be able to show, however, for a gauge-covariant form $\alpha$, that

$$DD\alpha = \Omega \wedge \alpha.$$

Last edited: Aug 6, 2014
3. Aug 7, 2014

### center o bass

When you say $\omega$ is not gauge covariant, what do you mean? $\omega$ transforms as $\omega \mapsto Ad_{g^{-1}} \omega$ under pullback of the right action on $P$.

Furthermore, in http://empg.maths.ed.ac.uk/Activities/GT/Lect2.pdf, he defines the curvature as
$\omega = d\omega \circ \text{hor}$ where $\hor$ picks out the horizontal component of any vector field according to $\omega$. This is also how he defined the exterior covariant derivative on the same page.

Is it the fact that $\omega$ is not a horizontal 1-form that is the source of my error? Indeed, http://empg.maths.ed.ac.uk/Activities/GT/Lect2.pdf, defines the exterior covariant derivative for horizontal and covariant forms.

4. Aug 7, 2014

### Ben Niehoff

Sorry, I'm not as fluent in "horizontal/vertical" language. In the notation I'm familiar with, the gauge connection $\omega$ transforms as

$$\omega \to g^{-1} \omega g + g^{-1}dg,$$
which fails to be covariant (because it involves derivatives of the transformation). Writing "$D\omega$" is analagous to taking the covariant derivative of a Christoffel symbol; it doesn't make sense.

5. Aug 7, 2014

### center o bass

Ah, okay. On the principal it does transform as $\omega \mapsto g\omega g^{-1}$. It is the pullback of $\omega$ with a local section that transform according to
$$\omega \mapsto g^{-1} \omega g + g^{-1}dg.$$

6. Sep 23, 2014

### lavinia

The curvature 2 form of a connection on a principal bundle may be defined as the exterior covariant derivative of the connection 1 form. (By definition the exterior covariant derivative is the exterior derivative composed with horizontal projection.) One can verify that your formula is the same thing by checking three cases: the two vectors are both horizontal,both vertical,one horizontal one vertical.

$$\Omega = D\omega = d\omega + \frac{1}2 [\omega, \omega].$$

is the correct equation in the principal bundle. If you check the three cases, the factor of 1/2 will be clear.

Notice that $$d\omega(X,Y) = X\omega(Y) - Y\omega(X) - \omega[X,Y]$$

And $$\omega[X,Y] = [\omega(X),\omega(Y)]$$

The trivial case is when both vectors are horizontal. Then the equation is just

$$\Omega = D\omega = d\omega.$$

This because $$\Omega(X,Y) = d\omega(hX,hY) + 0= d\omega(X,Y).$$

I have always seen the exterior covariant derivative denoted with a large D.

Last edited: Sep 26, 2014