Atiyah algebroid of a principal bundle

Let $\pi:P\to M$ be a principal G-bundle . The right action of $G$ on $P$ lifts to an action on the tangent bundle $TP$ by differentials $(R_g)_*:TP\to TP$. Form the quotient vector bundle

whose fiber over $x\in M$ is $(T_pP)/G$ for any $p\in P_x$.

The map $d\pi:TP\to TM$ is $G$-equivariant and descends to a bundle map (the anchor)

A section of $A(P)$ can be identified with a $G$-invariant vector field on $P$: explicitly, $\Gamma(A(P)) \cong \mathfrak{X}(P)^G$. Using this, define a bracket on $\Gamma(A(P))$ by

where $X,Y$ are $G$-invariant vector fields representing $\sigma,\tau$ and $[X,Y]$ is their Lie bracket . This is well-defined and makes $A(P)$ into a Lie algebroid over $M$, called the Atiyah algebroid of $P$.

Examples

1. Bundle over a point. If $M=\{\ast\}$ and $P=G$, then $TP/G$ identifies with the Lie algebra $\mathfrak{g}$ (via left translation), and the induced bracket is the usual Lie bracket on $\mathfrak{g}$.

2. Trivial bundle. For $P=M\times G$, there is a vector bundle isomorphism

   $$TP/G \cong TM \oplus (M\times \mathfrak{g}),$$

   and the bracket on sections $(X,\phi)$, $(Y,\psi)$ is

   $$[\![(X,\phi),(Y,\psi)]\!] = \bigl([X,Y],\, X(\psi)-Y(\phi)+[\phi,\psi]\bigr).$$

3. Principal $U(1)$-bundle. Since $\mathfrak{u}(1)\cong i\mathbb{R}$ is abelian, the adjoint part is central. Thus $TP/U(1)$ is (as a Lie algebroid) a central extension of $TM$ by a trivial line bundle, with curvature of a connection measuring the extension class.