Olver "Application of Lie Groups to Differential equations" Proposition 2.18. I have trouble with proving (and even intuitively understanding) the following proposition in Olver's book "Application of Lie Groups to Differential equations".
Proposition 2.18. Let $G$ act semi-regularly on $M$ and let $\xi^1(x), ..., \xi^{m-s}(x)$ be a complete set of functionally independent invariants defined on an open subset $W \subset M$. If a subvariety $S_F = \{x: F(x) = 0\}$ is $G$-invariant, then for each solution $x_0 \in S_F$ there is a neighbourhood $U \subset W$ of $x_0$, and an equivalent $G$-invariant function $\hat{F}(x) = \hat{F}(\xi^1(x), ..., \xi^{m-s}(x))$ whose solution set coincides with that of $F$ in $U$:
$$ S_F \cap U = S_{\hat{F}}\cap U = \{ x \in U : \hat{F}(\xi^1(x), ..., \xi^{m-s}(x)) = 0 \}.$$
The solution goes (approximately) as follows. We consider a complete set of invariants near $x_0$. Then, under certain relabeling of coordinates, we can change coordinates from $(x^1, ..., x^m)$ to $(\xi^1(x), ..., \xi^{m-s}(x), x^{m-s+1}, ..., x^m)$. It has the form $\psi(x) = (\xi(x), \hat{x})$, where $\xi(x) = (\xi^1(x), ..., \xi^{m-s}(x))$, $\hat{x} = (x^{m-s+1}, ..., x^m)$. In the new coordinates, function that defines subvariety becomes $F^*(\xi(x), \hat{x}) = F(x) $. Consider arbitrary $x_0 \in S_F$ and define function (on some suitable subset around $x_0$) as $\hat{F}(\xi(x)) = F^*(\xi(x), \hat{x}_0)$, where $\hat{x}_0$ is set of last $s$ values in new coordinates of a point $x_0$.
So far, I feel comfortable and understand how $\hat{F}(\xi(x))$ is well-defined. I have problem in the following short argument where we try to show that claim is true for such choice of $\hat{F}$.
Olver's approach: since $S_F$ is $G$-invariant, and the orbits of $G$ in these coordinates are the common level sets of the invariants $(\{ \xi(x) = c \})$, we find $F^* (\xi(x), \hat{x}) = 0$ if and only if $F^* (\xi(x), \hat{x}_0) = 0$ since both points lie on the same slice.
My confusion: I think I understand comment about orbits, as I am able to prove that if $O$ is some orbit then $(O \cap U) \subset \{ \xi(x) = c \}$ for some constant $c$. Now, consider any point $x$ such that $F^*(\xi(x), \hat{x}) = 0$. This means (by definition of $F^*$) that $F(x) = 0$ and so $x \in S_F$.
We would like to argue that $F^*(\xi(x), \hat{x}_0) = 0$. If somehow one could show that $x$ and $x_0$ are on the same orbit then $x_0 = g \cdot x$ for some $g \in G$ then $\xi(x_0) = \xi(g \cdot x) = \xi(x)$ because these are invariant functions and so because $x_0 \in S_F$, $0 = F(x_0) = F^*(\xi(x_0), \hat{x}_0) = F^*(\xi(x), \hat{x}_0)$.
Question 1: Is it possible to show that $x_0$ and $x$ are on the same orbit? Intuitively, I think that is not true as my function could be $F(x) = 0$ for all $x$. Then, $S_F = M$ would be the invariant surface. But then, if I consider any two points, they do not have be on the same orbit. Am I making a mistake? If not, what alternative reasoning is used here? Should we define $x_0$ for each orbit? But then, I guess, $\hat{F}$ is not well-defined as it is not enough to know invariant function values as we also need to specify on which orbit we are.
Now, consider the other direction. Assume that  $F^*(\xi(x), \hat{x}_0) = 0$. Again, if I knew that $x$ and $x_0$ are on the same orbit, then proof is clear. But as I mentioned, I do not see why that should be true.
Question 2: How to prove this direction? What happens if $x_0$ and $x$ are not on the same orbit? Is there some conditions (maybe implied by semi-regularity) that complete set of invariants uniquely determines orbit? I think this is true only for a regular case.
 A: In general the points $x_0$ and $x$ do not lie on the same $G$-orbit.
The point here is that the 'hatted' coordinates enter in an inessential way in determining the zero set of $F$.
An example going through the elements in the above construction:  $M={\Bbb R}^3$ and $G=\{g\in \Bbb R\}$ acting by  $g.(x,y,z)=(x,y,z+g)$, translation in the $z$-direction. $\xi^1(x,y,z)=x$, $\xi^2(x,y,z)=y$ is a complete pair of invariants for $G$.
Let $S$ be the zero-set of $F(x,y,z)=x(1+y^2+z^2)$. (A somewhat silly way to describe the $G$-invariant plane: $x=0$).
Along the lines you describe, we may rewrite $F$ using the invariants plus  one remaining coordinate as $F^*(\xi^1,\xi^2,\hat{z})= \xi^1(1+(\xi^2)^2+\hat{z}^2)$. Pick a point $(0,y_0,z_0)\in S$. Then
$$\hat{F}(x,y,z)=\hat{F}(\xi^1,\xi^2) = F^*(\xi^1,\xi^2,z_0)= \xi^1(1+  (\xi^2)^2+z_0^2)=x(1+y^2+z_0^2)$$
is now $G$-invariant and will have the same zero set as $F$. But points need not be on the same $G$-orbit (because of $y$).
Regarding details in the proof of Olver: Consider $x$ in a small neighborhood of $x_0$. When $F^*(\xi(x),\hat{x})=0$ then $F^*(\xi(g.x),\widehat{g.x})=F^*(\xi(x),\widehat{g.x})=0$ for $g$ in a neighborhood of the identity. By semi-regularity of the group action (assuming that this is in the differential sense) and since $\xi$ is invariant, $g\mapsto \widehat{g.x} \in {\Bbb R}^s$ must give rise to a local diffeo from a neighborhood of ${\rm Id}\in G$ to a neighborhood of $\hat{x_0}\in {\Bbb R}^s$. So in fact, $F^*(\xi(x),\hat{y})=0$ for all $\hat{y}$ in a neighborhood of $\hat{x_0}$. In particular, $F^*(\xi(x),\hat{x})=0$ iff $F^*(\xi(x),\hat{x_0})=0$ for $x$ in a neighborhood of $x_0$. This is the key point in the proof. Note again, that $x$ and $x_0$ need not be on the same $G$-orbit.
