# Linearization of a gradient field

Setup: Suppose we are given a smooth function $\phi$ that has a nondegenerate minimum at $x=0$. Then we can choose a coordinate system $x$ such that the gradient is given by $$X = \mathrm{grad} \phi = \sum_i a_i x^i \frac{\partial}{\partial x^i} + O(|x|^2)$$ where the $a_i>0$ are the eigenvalues of the Hessian of $\phi$ at $x=0$.

Now we look for smooth functions $a$ and numbers $\lambda$ such that $$X^i \frac{\partial a}{\partial x^i}= \lambda a.$$

If we can somehow (smoothly) linearize $X$, then it is straight forward that the only solutions are $\lambda = k^1 a_1 + \dots + k^n a_n$, together with $a = x^k$, and we also have the expected multiplicities of eigenvalues, meaning, to each multiindex $k$ corresponds exactly one eigenvalue.

Question: The question is now if a linearization is possible. There is a theorem by Sternberg and Poincaré that gives a positive answer under the additional assumption that the condition $$a_i \neq m^1 a_1 + \dots + m^{i-1} a_{i-1} + m^{i+1} a_{i+1} +\dots + m^n a_n$$ with all $m^i\geq 0$ is satisfied, but does not use that $X$ is a gradient field. So I am hoping that one can maybe drop the condition in this case?

Motivation: The question arose when following the WKB method outlined in the literature. Helffer (Semi-Classical Analysis for the Schrödinger Operator and Applications) does cite the Sternberg theorem but doesn't even mention the condition above. Dimassi and Sjöstrand (Spectral Asymptotics in the Semiclassical Limit) use other methods than linearization but require instead that all numbers of the form $$m^1 a_i + \dots + m^n a_n$$ are different, which is even a stronger condition. The whole situation seems a bit curious here.

P.S.: Even if a Linearization is not possible, I would like to know if it is true that the eigenvalues are exactly $k^1 a_1 + \dots k^n a_n$ counted with multiplicity in the general setup or if one actually needs some of these diophantine-type conditions above. Easy linear examples show that the statement can still be true if Sternberg's condition is violated.

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