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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 $\varphi \circ f=\varphi$.

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Hi Will (and Kofi). (1.) It seems to me that one can work with the set of germs of charts near p, and the group of germs of diffeomorphisms fixing p. Then the action is well-defined and free and transitive. –  macbeth Oct 18 '12 at 3:19
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(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
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