# Invariance group of Morse charts

Suppose I have a smooth function $\varphi$ that vanishes at $p$ and has a positive definite Hessian at that point (suppose that we are on a smooth manifold of dimension $M$). Then the Morse lemma tells us that we can find a chart $x$ (let us call it Morse chart) such that $$\varphi = (x^1)^2 + \dots + (x^n)^2 = \langle x, x \rangle.$$ What is the transformation group of Morse charts?

To be more precise, I am looking for a group that acts freely and transitively on the set of Morse charts.

Obviously, the group $O(n)$ acting on the set of Morse charts via $(Q, x) \mapsto Q\cdot x$ is a subgroup of this group. But are there more such transformations?

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Aren't you looking for the group of diffeomorphisms $f$ satisfying $\varphi \circ f= \varphi$?
The only difficulty that I can see is if the subset of $\mathbb R^n$ on which the chart is defined can vary. In that case, I do not think you will get a free transitive action that is natural in any way. Instead, there will be a groupoid action, with the morphisms $U \to V$ being diffeomorphisms $f: U \to V$ preserving $\varphi \circ f=\varphi$.
(2.) Yes, we're interested in the diffeomorphisms satisfying $\varphi \circ f = \varphi$. Such a diffeo $f$ has the property that, on each level set $\{x:|x|^2=r\}$, $f$ restricts to a diffeo of the level set. Moreover the possible $f$ are basically characterized by this property. Think of $f$ as a (germ of a) 1-parameter family of diffeomorphisms of the $(n-1)$-sphere, modulo some boundary conditions (tending to the identity near 0) to ensure smoothness at $p$. – macbeth Oct 18 '12 at 3:20
If you take the derivatives at the origin you get three power series that formally satisfy the equation $f^2+g^2+h^2=x^2+y^2+z^2$. There are lots of power series that formally satisfy this equation, as you can see by building it up step by step: degree $1$ terms, then degree $2$, then degree $3$, etc. It seems to me that many of them should have positive radius of convergence but I haven't checked. If some of them do have positive radius of convergence that gets you a lot of germs that aren't $O(n)$. – Will Sawin Oct 19 '12 at 0:47
But it's not clear at all to me what should be the boundary conditions near $0$. – Will Sawin Oct 19 '12 at 0:49