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In a paper that I was reading, I stumbled across the following theorem:

Let $X$ be a vector field with $$X= > a^ix^i\partial_{x^i} + > \mathcal{O}(|x|^2),$$ where $x$ is some chart and $a^i>0$. Then there exist a chart $y$ such that $X$ is linear with respect to $y$, meaning $$X =a^iy^i\partial_{y^i}.$$

It was referenced as "Sternbergs linearization theorem" and it sounded like common knowledge in the paper, but till now I couldn't find a proof anywhere.

Does anyone know a proof or a reference to one?

Also though my intuition is that this theorem does hold, I don't really have an understanding how it is important that the $a^i$ are greater $0$. Why does this make a difference?

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google flow box theorem – Orbicular Dec 7 '11 at 21:57
@Oribicular: The Flow Box Theorem won't apply near a singular point, which is the issue here. What you really want to google is the Hartman–Grobman Theorem. By the way, the original paper of Hartman (Proc. Amer. Math. Soc. 11 (1960), 610-620) is quite readable. – Robert Bryant Dec 8 '11 at 0:22
up vote 12 down vote accepted

Unfortunately, you need to be careful, as this is false as stated, unless you allow the chart to be only Lipschitz. Consider, for example, my answer to this question, where a smooth example in dimension $2$ with $a_1=1$ and $a_2=2$ is given that cannot be linearized by a smooth (or even $C^2$) change of coordinates.

For a proof of the linearization result, you can look in any good book on dynamical systems, or consult the original paper of Hartman (Proc. Amer. Math. Soc. 11 (1960), 610-620).

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If I see this correctly, the sign of the coefficients doesn't make a difference then? – Matthias Ludewig Dec 8 '11 at 16:24
@Kofi: By the way, I should also have said that it's worth taking a look at Sterberg's original paper, too: S. Sternberg, On contractions and a theorem of Poincare, Amer. J. Math. 79 (1957) 809–824. He does prove a linearization theorem with greater regularity than Lipschitz, in fact, in fact, he proves a smooth linearization theorem, but the hypotheses are more restrictive than just that the $a_i$ are all positive. (They have to be, as the example I quoted shows.) – Robert Bryant Dec 8 '11 at 17:05
@Kofi: That is more or less correct. The Hartman-Grobman theorem says that you can perform this linearization near $p$ as long as the linear part of the vector field has no purely imaginary (in particular, no zero) eigenvalues. You can see why these hypotheses are necessary, since, for example, you can't linearize the vector field $X = x^2\ \partial_x$ at $x=0$, and, similarly, you can't linearize the vector field $X = (y + x^3 + xy^2)\ \partial_x - (x - x^2y - y^3)\ \partial_y$ in the plane at $x=y=0$. – Robert Bryant Dec 8 '11 at 17:09

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