Derived inverse-pullback representation

Statement

Let a Lie group $G$ act smoothly from the left on a smooth manifold $M$ . On $C^\infty(M;\mathbb C)$ , define

(\pi(g)f)(p)=f(g^{-1}\cdot p).

For the fundamental field $X_M(p)=\left.\frac d{dt}\right|_0\exp(tX)\cdot p$ , differentiation gives the Lie-algebra action

d\pi(X)f=-X_Mf=-\mathcal L_{X_M}f,

where $\mathcal L$ is the Lie derivative on functions.

Chain-rule calculation

The inverse in the group action produces the minus sign:

\left.\frac d{dt}\right|_0 f(\exp(-tX)\cdot p) =-df_p(X_M(p)).

Inverse pullback is a left representation: $\pi(g)\pi(h)=\pi(gh)$ . The identity $[X_M,Y_M]=-[X,Y]_M$ then gives the usual derived commutator identity.

Hilbert-space realizations

If the action preserves a positive smooth measure on $M$ , inverse pullback also defines a strongly continuous unitary representation on $L^2(M)$ . The same derivative formula holds on $C_c^\infty(M)$ and on other smooth functions when their orbit maps have the required $L^2$ derivatives. A pointwise smooth $L^2$ function need not itself be a smooth vector.

For a flow, the forward Koopman convention $U_t f=f\circ T_t$ instead differentiates to $+X_Mf$ . Specifying forward or inverse pullback prevents a sign mismatch.