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

A(P)\coloneqq TP/G \;\longrightarrow\; M,

whose fiber over $x\in M$ consists of $G$ -orbits of tangent vectors $v_p\in T_pP$ with $p\in P_x$ .

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

a:A(P)\to TM.

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

[\![\sigma,\tau]\!]\;\coloneqq\;\text{the class of }[X,Y],

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 $P$ .

Right-action convention

For the right-action convention, invariant vector fields along a group fiber are right-invariant, whose bracket is the opposite of the usual Lie-algebra bracket. With the standard bracket on $\operatorname{ad}(P)$ , the Atiyah sequence is embedded by $[p,\xi]\mapsto[-\xi^\#_p]$ ; the minus sign compensates for the right-action convention.

Examples

Bundle over a point. If $M=\{\ast\}$ and $P=G$ , then $TP/G$ identifies with $\mathfrak{g}$ via right translation; right-invariant vector fields have the opposite of the usual Lie-algebra bracket. Composing this identification with $\xi\mapsto-\xi$ gives the standard Lie-algebra bracket used for the adjoint-bundle inclusion.

Trivial bundle. For $P=M\times G$ , there is a vector bundle isomorphism $TP/G \cong TM \oplus (M\times \mathfrak{g}),$ and, using the splitting in which $(X,\phi)$ represents the invariant field with vertical part $-\phi$ , the bracket on sections $(X,\phi)$ , $(Y,\psi)$ is $[\![(X,\phi),(Y,\psi)]\!] = \bigl([X,Y],\, X(\psi)-Y(\phi)+[\phi,\psi]\bigr).$

Principal $U(1)$ -bundle. Since the adjoint action of $U(1)$ on $\mathfrak{u}(1)\cong i\mathbb{R}$ is trivial, the adjoint bundle is the trivial line bundle $M\times\mathfrak{u}(1)$ . Thus the Atiyah sequence is an extension of $TM$ by this bundle; after choosing a connection, its curvature appears in the resulting bracket formula.