Let $G$ be a connected Lie group. We need to show that the following are equivalent:
-
$\omega$ is an exact left-invariant 1-form.
-
$\omega$ is an exact right-invariant 1-form.
-
$\omega$ is an exact bi-invariant 1-form.
-
$\omega = df$, where $f: G \rightarrow (\mathbb{R},+)$ is continuous (hence $C^{\infty}(G)$) homomorphism.
I have done some calculations in proving (1) $\implies$ (2).
Let $\omega = df$ and $v \in T_h G$. $L_g$ and $R_g$ denotes left-translation and right translation by $g$ respectively.
$(df)_h(v) = (df)_e (D_h L_{h^{-1}} (v))$ using left-invariance. This says that $\omega$ depends solely on its value at $e$.
$(R_{g}^{\ast} df)_h(v) = (df)_e (D_e (L_{g^{-1}} \circ R_g )(D_h L_{h^{-1}})(v)) $.
I don't know how to proceed further.
Best Answer
Let's assume that $df$ is left-invariant. Thus
$$ L_g^* df = df \Rightarrow d(L_g^* f) = df.$$
Since $G$ is connected, this implies that $L_g^* f - f$ is constant. So
$$ L_g^* f = f + C(g)\Rightarrow f(gh) = f(h) + C(g) , \ \ \ \forall h, g\in G.$$
Now assume that $f(e) = 0$ (this can be done by adding a constant to $f$, which does not change $df$). Then with $h = e$,
$$ f(g) = f(e) + C(g) \Rightarrow f(g) = C(g).$$
Thus $f(gh) = f(h)+ f(g)$: that is, $f$ is a homomorphism. So we have $(1)\Rightarrow (4)$. $(2)\Rightarrow (4)$ is similar.
On the other hand, if $f$ is a smooth homomorphism, then for each fixed $g$,
$$ L_g^* f = f + f(g)\Rightarrow d (L_g^* f) = df\Rightarrow L_g^* df = df. $$
So $df$ is left-invariant. Similarly one can show $(4)\Rightarrow (2)$.