I'm reading the notes on gauge theory by José Figueroa-O'Farrill and got stuck on an exercise. To state it, let me first explain my notation.
Let $G$ be a Lie group, $P\to M$ a principal $G$-bundle, $H\subset TP$ a horizontal distribution, $\omega$ the connection $1$-form. Let $h:TP\to TP$ be the horizontal projection, the tensor which is identity on $H$ and annihilates the vertical tangent vectors. Given a linear representation $\rho$ of $G$ on a vector space $V$, one defines covariant differentiation as follows. For a horizontal $G$-equivariant $k$-form $\alpha$ on $P$ (i.e., $h^*\alpha=\alpha$, $R_g^*\alpha=\rho(g^{-1})\circ\alpha$ for all $g\in G$), define $d^H\alpha:=h^*d\alpha$.
Show that $d^H\alpha=d\alpha+\rho(\omega)\wedge\alpha$.
In the notes, the case $k=0$ is proved, and the reader is asked to prove the general case.
I tried to prove it by modifying the proof for $k=0$, but there are terms involving Lie brackets which I can't get rid of. Here's what I did. Let $u_0,\ldots,u_k$ be vector fields on $P$. For ech $i$, write $u_i=hu_i+u_i^V$ where $u_i^V$ is vertical. Then $d^H\alpha(u_0,\ldots,u_k)=d\alpha(hu_0,\ldots,u_k)=d\alpha(u_0-u_0^V,\ldots,u_k-u_k^V)$. Expanding using the multilinearity gives $d\alpha(u_0,\ldots,u_k)$, which is the first term on the right hand side, and terms of the form $\pm d\alpha(u_i^V,\ldots)$. We have $d\alpha(v_0,\ldots,v_k)=\sum(-1)^iv_i\alpha(v_0,\ldots,\hat{v}_i,\ldots,v_k)+\sum(-1)^{i+j}\alpha([v_i,v_j],v_0,\ldots,\hat{v}_i,\ldots,\hat{v}_j,\ldots,v_k)$. Since $h^*\alpha=\alpha$, $h$ annihilates vertical vectors, and the Lie bracket of vertical fields is still vertical, we see that $d\alpha(v_0,\ldots,v_k)=0$ if at least two $v_i$'s are vertical. Thus only the terms $-d\alpha(u_0,\ldots,u_i^V,\ldots,u_k)$ survive in the above expansion. Expanding these gives $\rho(\omega)\wedge\alpha$ plus terms of the form $\pm\alpha([u_i^V,u_j],u_0,\ldots,\hat{u}_i^V,\ldots,\hat{u}_j,\ldots,u_k)$, and the latter does not seem to cancel out. What's the problem here?
Thanks in advance!
I found the proof myself. It is Theorem 31.19 in Differential Geometry: Connections, Curvature and Characteristic Classes by Loring W. Tu. Basically, the solution is that one may assume that the vector fields $u_i$ are invariant under the action of $G$, in which case the Lie brackets $[u_i^V,u_j]$ all vanish.