What is a $G$-invariant map and why the Lie derivative is zero?

Question

I was reading the lecture notes on symplectic geometry by Anna Cannas. In the Noether Theorem below, I can't find what exactly is a G-invariant map. Is it a map $f$ such that $f(g \cdot x)=f(x)$? Then why $\mathcal{L}_{X^\#}f=0$?

Answer 1

$G$-invariance of a function $f:M\rightarrow\mathbb{R}$ indeed means that $f(g\cdot p)=f(p)$ for $g\in G, p\in M$. In other words, the function is constant on $G$-orbits.

Fix a point $p\in M$, and let's check that $(\mathcal{L}_{X^{\sharp}}f)(p)=0$. Consider the curve $ \gamma(t):=\exp(-tX)\cdot p, $ which satisfies \begin{cases} \gamma(0)=p,\\ \left.\frac{d}{dt}\right|_{t=0}\gamma(t)=X^{\sharp}(p). \end{cases} Now, since $f$ is $G$-invariant, we have $f(\gamma(t))=f(p)$. Differentiating this relation at time $t=0$ gives \begin{align*} 0&=\left.\frac{d}{dt}\right|_{t=0}f(\gamma(t))\\ &=(df)_{\gamma(0)}\left(\left.\frac{d}{dt}\right|_{t=0}\gamma(t)\right)\\ &=(df)_{p}\left(X^{\sharp}(p)\right)\\ &=(\mathcal{L}_{X^{\sharp}}f)(p). \end{align*}