Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 115 / 3

A local flow of $v$ is a smooth family ${\Phi }^t$ defined near $\{0\}\times X$ such that ${\Phi }^0=\mathrm{id}$, ${\partial }_t{\Phi }^t(x)=v({\Phi }^t(x))$, and ${\Phi }^{s+t}={\Phi }^s\circ {\Phi }^t$ whenever defined. The Lie derivative of a differential form isCartan's magic formula is

For $\xi$ in the Lie algebra $\mathfrak{g}$, set ${\Phi }^t=R_{\exp (t\xi )}$. The one-parameter subgroup law gives the flow law, andso this is the global flow of the left-invariant vector field $l_{\xi }$.

For $f(g)=R_g^{\ast }\alpha$, every tangent vector at $g$ is the initial velocity of $g\exp (t\xi )$ for some $\xi$. Since $R_{g\exp (t\xi )}=R_{\exp (t\xi )}\circ R_g$,Thus all ${\mathcal{L}}_{l_{\xi }}\alpha$ vanish exactly when every derivative of $f$ vanishes. This is equivalent to $f$ being a locally constant function.

Suppose first that the left-invariant 1-form $\alpha$ is closed. The scalar function $\alpha (l_{\xi })$ is constant for every $\xi$, because both the form and vector field are left-invariant. Cartan's magic formula therefore givesPart b says that $g\mapsto R_g^{\ast }\alpha$ is locally constant. Since $G$ is connected, it is constant, and its value at the identity is $\alpha$. Hence $R_g^{\ast }\alpha =\alpha$ for every $g$, so $\alpha$ is bi-invariant.

Conversely, if $\alpha$ is bi-invariant, then all these Lie derivatives vanish. Cartan's formula and the constancy of $\alpha (l_{\xi })$ give ${\iota }_{l_{\xi }}d\alpha =0$. The left-invariant vector fields span every tangent space, so $d\alpha =0$.

Connectedness cannot be omitted. The orthogonal group $O(2)$ has an Abelian Lie algebra, so every left-invariant 1-form is closed by the Maurer-Cartan equation in a Lie-algebra basis. Conjugation by a reflection acts as $-1$ on its one-dimensional Lie algebra, so a nonzero left-invariant 1-form is not right-invariant. This is the standard obstruction recorded by the closed left-invariant 1-form criterion.