MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

In the real analytic case the problem reduces to an algebraic one. Denote by $\newcommand{\eA}{\mathscr{A}}$ $\eA$ the ring of germs at $\newcommand{\bR}{\mathbb{R}}$ $0\in\bR^n$ of real analytic functions defined in some neighborhood of the origin. To such a function $f$ we associate its Jacobian ideal $J_f\subset\eA$ generated by the germs at $0$ $\newcommand{\pa}{\partial}$ of the first order partial derivatives of $f$, $\pa_{x_1} f,\dotsc, \pa_{x_n} f$. Then

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

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.