Condition for section of the frame bundle of $M$ to be the 1-jet of a local diffeomorphism $\phi: M \rightarrow \mathbb{R}^n$?

Question

We're given an n-dimensional Riemannian manifold $M$ and its frame bundle $FM$. The tangent bundle can be locally regarded as an invertible map $\phi:M\rightarrow\mathbb{R}^{n}$. Let's confine $\phi$ to be $C^{\infty}$.

Choosing some arbitrary section $ \sigma:M\rightarrow FM$ of the frame bundle (and likewise for coframes) we have a set of basis and their dual 1-forms $\left(\sigma_{a},\sigma^{a}\right)$. My question is: when is $\sigma$ the 1-jet $j^{1}\phi$ of $\phi$?

I'm pretty sure the above is enough to say that $\phi$ is a local diffeomorphism fixing a point $x\in M$. The question then becomes when is $\sigma$ the 1-jet of a local diffeomorphism on $M$?

Note: In this context our basis can be considered moving frames as in Cartans' method du repere mobile

My thought process is as follows:

Let us choose a torsion free connection on $M$ (i.e. $d\sigma^{a}=\omega^{ab}\sigma_{b}$). Now consider the Ricci (curvature) scalar $R$ on $M$. $R$ should be invariant under the map $\phi $ (there is some subtlety here in how you apply $\phi$ discussed for example here).

Then I would expect the variation of the curvature with respect to the variation of $\sigma_{a}$ to vanish precisely when our basis are the 1-jets $j^1\phi$. In other words, the curvature shouldn't change as we vary a diffeomorphism (or Taylor expansions/jets of it) that is when:

$$\delta R/\delta\sigma_{a}=0$$

at every point in $M$. Does this make sense? If this is off base, what is a sufficient condition?

NOTE: Any form of Ricci curvature would have sufficed, I only chose the Ricci scalar for it's lack of unmatched indices.

Answer 1

Most of what you wrote in the question is either wrong (e.g. the sentence "The tangent bundle can be locally regarded as an invertible map $\phi:\to {\mathbb ℝ}^$" in the opening paragraph) or irrelevant.

Your real question seems to be about finding necessary and sufficient conditions for the existence of a local diffeomorphism $M\to {\mathbb R}^n$ with the given "infinitesimal information." In the question you were trying to formulate the problem in terms of sections of the frame bundle, which is unnatural, the natural setting (as we will see below) is that of the coframe bundle of $M$. Of course, by introducing a (semi)Riemannian metric on $M$, one gets and isomorphism between the two bundles (and, hence, allows one to restate everything in terms of sections of the frame bundle of $M$), but this only obscures the nature of the problem.

Now, suppose we are given an $n$-dimensional manifold $M$ and a $C^\infty$-map $\phi: M\to {\mathbb R}^n$, $\phi=(\phi_1,...,\phi_n)$, where $\phi_i\in C^\infty(M), i=1,...,n$. This yields an $n$-tuple of differential forms on $M$: $$ \omega_i=d \phi_i, i=1,...,n. $$ A map $\phi$ is a local diffeomorphism if and only if $\omega_1\wedge ...\wedge \omega_n$ is a volume form on $M$ (i.e. a nowhere vanishing $n$-form), if and only if the $n$-tuple
$(\omega_1,...,\omega_n)$ is a section of the coframe bundle of $M$.

Conversely, given an $n$-tuple of 1-forms $$(\omega_1,...,\omega_n)$$ on $M$, there exists a smooth map $\phi=(\phi_1,...,\phi_n): M\to {\mathbb R}^n$ with $\omega_i=d\phi_i, i=1,...,n$, if and only if:

(a) Each form $\omega_i$ is closed, $d\omega_i=0, i=1,...,n$. (This condition is sufficient locally.)

(b) Each form $\omega_i$ represents zero de Rham cohomology class in $H^1(M; {\mathbb R})$, in other words, each $\omega_i$ is exact. For instance, this condition is automatically satisfied if $H^1(M; {\mathbb R})=0$.

None of this has anything to do with Cartan's method of moving frames, or curvature(s) of some Riemannian metric on $M$.