Basic forms theorem

Let $\pi:P\to M$ be a principal $G$-bundle with right action $R_g:P\to P$. A differential form $\alpha \in \Omega^k(P)$ is called basic if it satisfies:

1. $G$-invariance: $R_g^*\alpha = \alpha \quad \text{for all } g\in G.$
2. Horizontality: for every $X\in \mathfrak{g}$, letting $X^\#$ denote the corresponding fundamental vertical vector field on $P$, $\iota_{X^\#}\alpha = 0.$

Theorem (basic forms theorem)

A form $\alpha \in \Omega^k(P)$ is basic if and only if there exists a unique form $\beta \in \Omega^k(M)$ such that

In other words, basic forms are exactly those that “descend” from $P$ to the base $M$.

Why the conditions are exactly right (proof idea)

• If $\alpha=\pi^*\beta$, then $\alpha$ is automatically $G$-invariant because $\pi\circ R_g=\pi$, and $\alpha$ is horizontal because $\pi_*(X^\#)=0$ for vertical vectors.
• Conversely, if $\alpha$ is horizontal and invariant, choose a local section $s:U\to P$ and define $\beta_U := s^*\alpha$. On overlaps $U\cap V$, two sections differ by a gauge transformation $s_V = s_U \cdot g$ for a map $g:U\cap V\to G$. Invariance and horizontality imply $s_U^*\alpha = s_V^*\alpha$, so the $\beta_U$ glue to a global $\beta$ with $\pi^*\beta=\alpha$.

Example: characteristic forms

If $\omega$ is a connection on $P$ with curvature $\Omega$, then applying an $\mathrm{Ad}$-invariant polynomial $P$ on $\mathfrak{g}$ produces a real-valued form $P(\Omega)$ that is horizontal and $G$-invariant, hence basic, so it descends to $M$. Its closedness is proved using the Bianchi identity .