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

  1. Aug 6, 2014 #1
    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. jcsd
  3. Aug 6, 2014 #2

    Ben Niehoff

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    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
  4. Aug 7, 2014 #3
    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.
     
  5. Aug 7, 2014 #4

    Ben Niehoff

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    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.
     
  6. Aug 7, 2014 #5
    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
    $$\omega \mapsto g^{-1} \omega g + g^{-1}dg.$$
     
  7. Sep 23, 2014 #6

    lavinia

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    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
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