I am an analyst struggling through some geometry used in physics.

Some background: For some Lie group $G$, let $P$ be a principal $G$-bundle over the smooth manifold $M$. Let $\omega$ be a connection 1-form on $P$. This is a Lie algebra-valued 1-form. Given an open covering $U_i$ of $M$ together with local sections $\sigma_i$ on $U_i$ of $P$, we can pullback the connection one-form to get connection one-forms $A_i$ on each $U_i$ which 'transform in a certain way' on the overlaps $U_i\cap U_j$. If the bundle is trivial, which for principal bundles is the same as the existence of a global section, then you get a genuine global one-form on $M$. In gauge theory this (local) one-form is called the gauge potential.

I find the usual definition of the curvature two-form I keep seeing unsatisfactory. Either you see definitions with no explanation, e.g. "The curvature is given by $\Omega = d\omega + \omega\wedge\omega$". Or one defines something called the exterior covariant derivative $D$: This is the usual exterior derivative on forms but applied in a straightforward way (see wiki) to 'vector valued' forms, including our Lie algebra-valued connection one-form. The curvature is simply the exterior covariant derivative of the one form. Either way you then pullback this two-form as in the previous paragraph to get a Lie algebra-valued curvature two-form on $M$ called the gauge field strength and you can write it in terms of the gauge potential as one would hope.

Question: I am a bit puzzled about the definition of the curvature two-form on a principal bundle using 'exterior covariant differentiation' though because it feels like a cheat: It feels like you are doing covariant differentiation without a connection. You have basically differentiated a $\mathfrak{g}$-valued one-form covariantly, no? And the exterior covariant derivative $D$ satisfies $D^2\phi = \Omega\wedge\phi$... So you have some $D$ for which $D^2$ is the curvature... but without going via a connection...

Is there a connection on some kind of $\mathfrak{g}$-bundle on $P$ lurking in the background somewhere... for which this curvature two-form is the curvature? Or is it at least possible to interpret things this way?

I am an analyst struggling through some geometry used in physics.

Some background: For some Lie group $G$, let $P$ be a principal $G$-bundle over the smooth manifold $M$. Let $\omega$ be a connection 1-form on $P$ (a "principal connection"). This is a Lie algebra-valued 1-form.

As for the curvature two-form, either you see definitions with no explanation at all, e.g. "The curvature is given by $\Omega = d\omega + \frac{1}{2}[\omega,\omega]$". This is obviously less than ideal for improving one's intuitive appreciation. Or one defines something called the exterior covariant derivative $D$ (see wiki) and then the curvature is simply the exterior covariant derivative of the connection one-form.

The issue: I can't get round the following observation though: From the point of view of the manifold $P$, $\omega$ is just a one-form with values in some vector space which happens to be $\mathfrak{g}$. Usually when you need to covariantly differentiate such an object, you would need a connection in a bundle $E \to P$ with fibre $\mathfrak{g}$, no? Then $\omega$ would be an $E$-valued one-form on $P$, i.e. in $\Gamma(E) \otimes\Omega^1(P)$, and you can differentiate covariantly in the normal way using the connection. Why is this scenario different?

The exterior covariant derivative $D$ satisfies $D^2\phi = \Omega\wedge\phi$... So you have some sort of covariant differentiation $D$ which differentiates forms $\eta$ taking values in $\mathfrak{g}$ and for which $D^2$ is some sort of curvature... but of $P \to M$ and not of the bundle in which $\eta$ is taking values. Isn't this strange? Or is this indeed just how things are? This prompts my more precise question:

Is the Lie algebra-valued curvature two-form on a principal bundle P the curvature of some vector bundle over P with fibre $\mathfrak{g}$?

I am an analyst struggling through some geometry used in physics.

Some background: For some Lie group $G$, let $P$ be a principal $G$-bundle over the smooth manifold $M$. Let $\omega$ be a connection 1-form on $P$. This is a Lie algebra-valued 1-form. Given an open covering $U_i$ of $M$ together with local sections $\sigma_i$ on $U_i$ of $P$, we can pullback the connection one-form to get curvature one-forms $A_i$ on each $U_i$ which 'transform in a certain way' on the overlaps $U_i\cap U_j$. If the bundle is trivial, which for principal bundles is the same as the existence of a global section, then you get a genuine global one-form on $M$. In gauge theory this (local) one-form is called the gauge potential.

I find the usual definition of the curvature two-form I keep seeing unsatisfactory. Either you see definitions with no explanation, e.g. "The curvature is given by $\Omega = d\omega + \omega\wedge\omega$". Or one defines something called the exterior covariant derivative $D$: This is the usual exterior derivative on forms but applied in a straightforward way (see wiki) to 'vector valued' forms, including our Lie algebra-valued connection one-form. The curvature is simply the exterior covariant derivative of the one form. Either way you then pullback this two-form as in the previous paragraph to get a Lie algebra-valued curvature two-form on $M$ called the gauge field strength and you can write it in terms of the gauge potential as one would hope.

Question: I am a bit puzzled about the definition of the curvature two-form on a principal bundle using 'exterior covariant differentiation' though because it feels like a cheat: It feels like you are doing covariant differentiation without a connection. You have basically differentiated a $\mathfrak{g}$-valued one-form covariantly, no? And the exterior covariant derivative $D$ satisfies $D^2\phi = \Omega\wedge\phi$... So you have some $D$ for which $D^2$ is the curvature... but without going via a connection...

Is there a connection on some kind of $\mathfrak{g}$-bundle on $P$ lurking in the background somewhere... for which this curvature two-form is the curvature? Or is it at least possible to interpret things this way?

I am an analyst struggling through some geometry used in physics.

Some background: For some Lie group $G$, let $P$ be a principal $G$-bundle over the smooth manifold $M$. Let $\omega$ be a connection 1-form on $P$. This is a Lie algebra-valued 1-form. Given an open covering $U_i$ of $M$ together with local sections $\sigma_i$ on $U_i$ of $P$, we can pullback the connection one-form to get connection one-forms $A_i$ on each $U_i$ which 'transform in a certain way' on the overlaps $U_i\cap U_j$. If the bundle is trivial, which for principal bundles is the same as the existence of a global section, then you get a genuine global one-form on $M$. In gauge theory this (local) one-form is called the gauge potential.

I find the usual definition of the curvature two-form I keep seeing unsatisfactory. Either you see definitions with no explanation, e.g. "The curvature is given by $\Omega = d\omega + \omega\wedge\omega$". Or one defines something called the exterior covariant derivative $D$: This is the usual exterior derivative on forms but applied in a straightforward way (see wiki) to 'vector valued' forms, including our Lie algebra-valued connection one-form. The curvature is simply the exterior covariant derivative of the one form. Either way you then pullback this two-form as in the previous paragraph to get a Lie algebra-valued curvature two-form on $M$ called the gauge field strength and you can write it in terms of the gauge potential as one would hope.

Question: I am a bit puzzled about the definition of the curvature two-form on a principal bundle using 'exterior covariant differentiation' though because it feels like a cheat: It feels like you are doing covariant differentiation without a connection. You have basically differentiated a $\mathfrak{g}$-valued one-form covariantly, no? And the exterior covariant derivative $D$ satisfies $D^2\phi = \Omega\wedge\phi$... So you have some $D$ for which $D^2$ is the curvature... but without going via a connection...

Is there a connection on some kind of $\mathfrak{g}$-bundle on $P$ lurking in the background somewhere... for which this curvature two-form is the curvature? Or is it at least possible to interpret things this way?