# Moving frame method with non-matrix Lie group

I am trying to understand the modern formulation of the moving frame method for Lie group acting on a manifold. I know the following theorem

Let be $$M$$ a manifold, $$G$$ a Lie group and $$\omega$$ the Maurer-Cartan form of $$G$$. If $$f_1, f_2: M \to G$$ are two functions, than $$f_1^*\omega = f_2^*\omega$$ if and only if $$f_1 = gf_2$$ for a fixed $$g \in G$$.

I have always seen this theorem proved for matrix Lie group. In this case that's easy because $$G$$ can play directly with his Lie algebra $$\frak{g}$$ (for example we have $$\omega_g = g^{-1}dg$$). My question is:

Is this theorem true for a general Lie group? How is this proved?

There are various proofs in textbooks. My favorite proof is to use E. Cartan's graph trick and the Frobenius Theorem.

As you stated it, things aren't quite right. You need $$M$$ connected.

If $$\omega^1,\dots,\omega^n$$ are $$n$$ basis left-invariant forms (so we pick a basis for $$\mathfrak g^*$$ and pull back by $$L_g$$), consider the differential ideal generated by the $$1$$-forms $$\eta^i = f_1^*\omega^i-f_2^*\omega^i.$$ Conceptually, we're looking at the map $$F=(f_1,f_2)\colon M\to G\times G$$ and pulling back the forms $$\phi^i=\pi_1^*\omega^i - \pi_2^*\omega^i$$ by the product map. The differential system $$\phi^1=\dots=\phi^n=0$$ is completely integrable, since $$d\phi = \pi_1^*[\omega,\omega] - \pi_2^*[\omega,\omega] = [\phi,\pi_1^*\omega] + [\pi_2^*\omega,\phi] \equiv 0 \pmod\phi.$$ Indeed, integral manifolds of $$\phi^i=0$$ give left cosets of the diagonal subgroup $$\Delta\subset G\times G$$. Since $$F^*\phi^i = 0$$ by hypothesis, and since $$M$$ is connected the image of $$F$$ must be contained in one of those integral manifolds, which says that $$f_1=gf_2$$ for some $$g\in G$$.

• Thanks, as always, for your answer! Is necessary to use Frobenius theorem? I have always seen it used for the existence part. And which textbooks do you have in mind? – Marco All-in Nervo Dec 3 '18 at 23:18
• Well, in this case you can get the integral manifolds by inspection. I was thinking of the more interesting parallel result that you construct $f\colon M\to G$ if you have a system of $1$-forms on $M$ satisfying the Maurer-Cartan equations of $G$. ... This stuff is in Warner and Spivak (volume 1). There's also a beautiful paper of Griffiths on Lie Groups and Moving Frames (Duke, 41, no. 4, pp. 775-814). Also, see Chern/Chen/Lam, pp. 198 ff. – Ted Shifrin Dec 3 '18 at 23:47
• Yes, I got your answer and your proof is really beautiful. I was looking for a simpler proof (as in Spivak, thank you!) with instruments that I can manage with my little experience. The paper of Griffiths seems really ...wow. Thank you again, next time I will start my question with "Dear Shifrin, here my new doubts about moving frames..." – Marco All-in Nervo Dec 4 '18 at 20:20

Here Spivak's proof from A Comprehensive Introduction to Differential Geometry, Vol. 1  