Is curvature the exterior covariant derivative of the connection?

Question

Let $P\to M$ be a $G$-principal bundle, $G$ a topological group, $\omega$ the connection and $V$ a vector space. We define $d_\omega: \Omega^k_G(P, V)\to\Omega^{k+1}_G(P, V)$ the exterior covariant derivative such that for a vector-valued differential form $\alpha\in\Omega^k_G(P, V)$, we get $d_\omega\alpha=d\alpha+\omega\wedge_\rho\alpha$, where $\rho:\mathfrak g \to \operatorname{End}(V)$ is a representation of the algebra of $G$.

Question: defining the curvature as $\Omega=d\omega+\frac{1}{2}[\omega,\omega]$, is there any representation in which $\Omega=d_\omega\omega$?

My guess: no.

The point is that I have found several forums (as well as in Wikipedia's entry) in which the curvature was expressed as the covariant derivative of the connection, but in the adjoint representation we do not get the $\frac{1}{2}$ in front. Moreover if we notice, we have for matrix groups, $\frac{1}{2}[\omega,\omega]=\omega\wedge\omega$ and thus one might argue that we could get the curvature by using the fundamental representation.

What do you think?

Answer 1

The definition of the exterior covariant derivative isn't $d_\omega\alpha=d\alpha+\omega\wedge_\rho\alpha$.

It is $$ d_\omega\alpha(X...)=d\alpha(\mathrm hX...), $$ where $\mathrm h$ is the horizontal projection.

We then state several things:

• A $k$-form with values in $V$ $\alpha\in\Omega^k(P,V)$ is a pseudotensorial $k$-form of type $\rho$, if it is right-equivariant in the sense that $r_g^\ast\alpha=\rho(g^{-1})\alpha$.
• A $k$-form $\alpha$ with values in $V$ is a tensorial $k$-form of type $\rho$, if it is a pseudotensorial $k$-form of type $\rho$ and is horizontal in the sense that $\alpha(X_1,...,X_k)=0$ whenever all elements are vertical.
• The covariant exterior derivative of a pseudotensorial form is a tensorial form of one higher degree
• If $\omega$ is the connection form, then it is a pseudotensorial 1-form of type $\text{Ad}$, and we have $d_\omega\omega=\Omega$, so the curvature is the covariant exterior derivative.
• If $\alpha$ is a tensorial $k$-form of type $\rho$, then we have the "generalized Cartan structure equation" $$ d_\omega\alpha=d\alpha+\omega\wedge_\rho\alpha, $$ however this is valid only if $\alpha$ is a tensorial form, not when it is a pseudotensorial one.
• In particular, if $\alpha\in\Omega^k(P,\mathfrak g)$ is a tensorial $k$-form of type $\text{Ad}$, then we have $$ d_\omega\alpha=d\alpha+\omega\wedge_{\text{Ad}}\alpha\equiv d\alpha+[\omega\wedge\alpha], $$ where $[\cdot\wedge\cdot]$ is the Lie-exterior product. One may expand (without using the generalized structure equation) that for the curvature we have $$ \Omega=d\omega+\frac{1}{2}[\omega\wedge\omega], $$ where the anomalous factor of 1/2 comes from the fact that $\omega$ is a pseudotensorial form and thus the generalized structure equation is not valid.

Answer 2

Yes, this is the case in synthetic differential geometry if we view the connection 1-form $\omega$ on a principal $G$-bundle as a $G$-valued simplicial 1-form. Then the exterior (Darboux) derivative of $\omega$ is indeed the curvature. Theorem 6.7.1 in Anders Kock's Synthetic Geometry of Manifolds explains how this relates to the usual curvature formula for classical differential forms.