# Isolated critical points

Is the following statement true or false?

Let $f:U\subset{\bf R}^n\to{\bf R}$ be a $C^2$-function (or $C^k$, with $k>2$; or real analytic) defined in a neighborhood $U$ of $0$. Assume that $0$ is the unique critical point of $f$ in $U$, and that it is totally degenerate (i.e., $\mathrm d^2f(0)=0$). Let $V\subset{\bf R}^n$ be a subspace, denote by $\pi_V:{\bf R}^n\to V$ the orthogonal projection. Then, there exists a positive definite quadratic form $Q$ on $V$ such that the function $\widetilde f(x)=f(x)+Q\big(\pi_V(x)\big)$ has $0$ as an isolated critical point.

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$$\mbox{\dim \eA/J_f <\infty }\;\;\Rightarrow 0 \;\;\mbox{is an isolated critical point of f} .$$

The dimension of the $\bR$-algebra $\eA/J_f$ is called the Milnor number of $f$ at the critical point $0$ and it is denoted by $\mu(f,0)$. (A similar result holds if $\eA$ is replaced with the ring of germs at $0$ of smooth functions.) The natural condition to impose is the finite multiplicity of the critical point, $\mu(f,0)<\infty$.

Thus, in the real analytic case, one can ask the closely related question: given that $\mu(f,0)<\infty$, is it true that for any homogeneous quadratic polynomial $q:\bR^n\to \bR$ we have $\mu(f+q,0)<\infty$.

Here is a general perturbation result: if $p:\bR^n\to\bR$ is a polynomial function then there exists $\newcommand{\ve}{\varepsilon}$ $\ve_0>0$ such that, for any $|s|<\ve_0$ we have

$$\mu(f+s p,0)<\infty .$$

Much more refined results are available. I refer you two two sources I found useful.

T. de Jong, G. Pfister: Local Analytic Geometry, Vieweg, 2000

Arnold, Gusein-Zade, Varchenko: Singularities of Differentiable Maps. Vol.1

Theorem 6.4.5. in deJong-Pfister is particularly relevant.

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Thank you so much, Liviu. In the last centered formula, you mean: \mu(f+sp)<\infty, correct? –  Paolo Piccione Nov 24 '13 at 20:21