Moving frames method

Question

I want to grasp the moving frames method but I find some obstacles. I don't know if this question is suitable for MO, if it is not the case please let me know and I will move it.
I am aware there are other related questions here like this one or this one, but they don't answer my doubts.

Given a Lie group $G$ and a homogeneous space $X\equiv G/H$, the goal of the moving frames method is to study submanifolds $M$ of $X$. In particular we want to know if two given submanifolds $M$ and $\tilde{M}$ are "congruent", in the sense that there is a "movement" $g\in G$ such that $g(M)=\tilde{M}$.

I know I have a $\mathfrak{g}$-valued differential form on $G$ which is left invariant, the Maurer-Cartan form $\theta$. The left-invariance of $\theta$ allows us to show (see griffiths1974cartan, lemma (1.3)) that given two maps $f,\tilde{f}$ from, let's say, $\mathbb{R}^n$ to $G$ then $\tilde{f}(x)=g\cdot f(x)$ if and only if $\tilde{f}^*(\theta)=f^*(\theta)$. This way $f^*(\theta)$ provide a set of invariants to characterize submanifolds of $G$ (not of $X$!).

Suppose our submanifolds $M$ and $\tilde{M}$ are parametrized respectively by maps $\alpha:\mathbb{R}^n \to X$ and $\tilde{\alpha}:\mathbb{R}^n \to X$. If we are in a particular case, for example $G=E(2)$, $X=\mathbb{E}^2$ and $M,\tilde{M}$ curves, we have a "canonical" way to lift $\alpha$ and $\tilde{\alpha}$ to $f_{\alpha}, f_{\tilde{\alpha}}:\mathbb{R} \to G$ (the unitary tangent vector and its orthogonal, together with the curve point itself). This way, if $$ f_{\alpha}^*(\theta)={f}_{\tilde{\alpha}}^*(\theta) $$ we conclude that the "curves of frames" $f_{\alpha}$ and $f_{\tilde{\alpha}}$ are congruent, and therefore $\alpha$ and $\tilde{\alpha}$ are congruent.

The key fact here is, I think, that the assignment $$ \alpha \mapsto f_{\alpha} $$ is $G$-invariant, in the sense that $\tilde{\alpha}=g\alpha$ if and only if $f_{\tilde{\alpha}}=g f_{\alpha}$. Otherwise we could have congruent curves in $E(2)$ which couldn't be detected by the invariants (because of the "bad assignment" of frames to the curves).

Question 1
Back to the general case: can we always find such a "canonical lift"? Is there a method to find it? Or is the moving frames method restricted to a bunch of particular cases?
$\blacksquare$

Question 2
Can you provide at least a brief list of examples of these assignments? For example:

• Curves in $\mathbb{R}^3$ with Euclidean movements: the Frenet frame.
• Surfaces in $\mathbb{R}^3$ with Euclidean movements: a frame made with the point, the normal vector and two ortogonal vectors aligned with the principal directions of the surface (or is not this necessary?).
• ...

$\blacksquare$

I have read the article of Griffiths, the corresponding chapter of Cartan for begginers and From Frenet to Cartan, but I am still blocked with these doubts.

Answer 1

I think you might want to read a couple of articles on the moving frame that carefully discuss this issue (and show that it is more subtle than most people realize).

The first is a paper by Mark Green, The moving frame, differential invariants and rigidity theorems for curves in homogeneous spaces (Duke Math. J. 45 (1978), no. 4, 735–779). Mark discusses the 1-dimensional case in great detail and shows, by example, that, in fact, there are cases in which a `canonical' lifting by Cartan's method does not exist, and one must extend the method by introducing a family of liftings in order to find all of the invariants.

The second is a set of lecture notes by Gary Jensen, Higher order contact of submanifolds of homogeneous spaces, (Lecture Notes in Mathematics, Vol. 610. Springer-Verlag, Berlin-New York, 1977.) He discusses the foundations of the subject in great detail and gives a number of illuminating examples.

Of course, there are lots of more recent papers and articles, but I think that these two illustrate a lot of the issues and point out some of the pitfalls that aren't apparent at first sight.

Answer 2

I am going to answer my own question with my conclusion after some reading of Bryant references and other papers I have found.

Question 1

The answer to question 1 is no, there is not an standard way to assign a lift which is equivariant.

Indeed, this lack has given rise to lots of works on this topic. For example, together with the works cited by Robert Bryant, in the paper of Griffiths it is said that for some cases, "by going to a sufficiently high order jet or contact element of a submanifold $M$ of a homogeneous space, there will naturally appear a good frame over $M$ in a similar manner to the appearance of the Frenet frames of a curve in Euclidean space".

In two more recent works, Moving Coframes I and II, by Fels and Olver, it is said that "the group theoretical basis for the method has hindered the theoretical foundations from covering all the situations to which the practical algorithm could be applied". I understand that what they are doing in these works is to give a different approach to the method, in such a way that they solve this problem of the "standard assignment". They even say that "our formulation of the framework goes beyond what Griffiths envisioned, and successfully realizes Cartan’s original vision".

I am looking forward to studying these works!

Question 2

I will try to keep updated the list of examples as I understand new cases:

• Curves $\alpha$ in equi-affine space $\mathbb{A}^3\equiv \mathbb{R}^3 \rtimes SL(3)/SL(3)$: an unimodular frame built from $\alpha'(t)$, $\alpha''(t)$ and $\alpha'''(t)$. I find it a particularly iluminating example. Is very well explained in Clelland's book, From Frenet to Cartan, page 172.

• Curves in $T^2=\mathbb{R}^2 \rtimes\{id\}/\{id\}$: a plane where the only allowed transformations are the translations. A "standard assignment of frames" would consist simply of the point of the curve itself. The obtained invariants are the components of the tangent vector of the curve (a curve in $\mathbb{R}^3$ can be carried into another if they both share the same tangent vectors. How do we compare this tangent vectors? For example, by carrying them to the tangent space at the identity, i.e., applying the MC form!). This example illustrates very well why the problem is solved when we can lift the manifold $M$ to $G$ in a standard way.