4

Let $P$ be a principal bundle with structure group $G$, equipped with a principal connection, whose connection $1$-form we denote $\omega$. Suppose we also have a linear representation $\rho:G\to GL(V)$ (which then gives rise to a right-action $(g,v)\mapsto\rho(g^{-1})(v)$ and by taking the tangent at the identity gives us $\rho_*:\mathfrak{g}\to\text{End}(V)$, which can be thought of as a bilinear map $\rho’:\mathfrak{g}\times V\to V$). Finally, let $\alpha$ be a $V$-valued $k$-form on $P$.

My question is if there is any “nice” formula for the exterior covariant derivative $D\alpha$ in general? I know some special cases:

  • if $\alpha=\omega$ is the connection 1-form itself, then we get Cartan’s structure equation for the curvature.
  • if $k\geq 2$ and $\alpha$ is vertical then $D\alpha=0$ (a useful remark when proving Bianchi’s identity $D\Omega=0$).
  • if $\alpha$ is horizontal and $G$-equivariant (for the naturally induced right action of $G$ on the $k^{th}$ exterior power of $TR$, and the right action of $G$ on $V$ described above), i.e a “tensorial form on $P$ of type $(V,\rho)$”, then we have $D\alpha=d\alpha+\omega\wedge_{\rho’}\alpha$.

Is there a more general formula which addresses other cases as well? My guess is not because the definition of $D\alpha$ involves the various horizontal projections, so if we make no further assumptions on $\alpha$, then it’s going to be hard to simplify (and annihilate many of the terms when unwinding definitions). In fact, even for $k=1$ and assuming $\alpha$ is vertical I run into trouble trying to obtain a “nice” formula, because unlike the connection 1-form, we don’t have the nice property that $\omega(X_{\xi})=\xi$ (where $X_{\xi}$ is the “fundamental” vertical vector field on $P$ induced by $\xi\in\mathfrak{g}$) so the proof of Cartan’s structure equation doesn’t really go through here (unless I’m missing something).

Note I’m also fine with accepting there isn’t really a general formula (though an accompanying (heuristic even) explanation would be nice), but just curious if there exists one.

peek-a-boo
  • 65,833
  • What is the exterior covariant derivative on $P$? I am aware of such a construction only for associated vector bundles on which we have a covariant derivative induced by the connection one form. – Chris Jul 12 '23 at 06:58
  • @Chris it’s defined by composing the exterior derivative $d\alpha$ with the horizontal projection corresponding to the direct sum decomposition of the tangent bundle into a horizontal and vertical part (hence taking care of the “covariant” part). So, $(D\alpha)(s_1,\dots, s_{k+1}):=(d\alpha)(h(s_1),\dots, h(s_{k+1}))$. – peek-a-boo Jul 12 '23 at 12:32

0 Answers0