Birth-death : Always more than 1 bifurcation?

Question

Say I have a (smooth) function $f : \mathbb{R}^n \to \mathbb{R}$, and a critical point $x$ (ie, $f'(x) = 0$). I call this point degenerate if $\det \text{Hess}_x f = 0$ (so, equivalently, if the kernel of the Hessian at $x$ is non-trivial). An example is $x = 0$ for $f : \mathbb{R} \to \mathbb{R} : x \mapsto x^3$.

Then, if I perturb my function $f$ generically, I should observe such a phenomenon:

Called a "birth-death bifurcation": my degenerate critical point will either bifurcate into multiple non-degenerate critical points (birth), or die. (In the picture, I drew the bifurcations $f(x) \pm \varepsilon x$ for $\varepsilon > 0$. We can easily show that, for $f(x) = x^3$, $f - \varepsilon x$ has two critical points near $0$, while $f + \varepsilon x$ has none.)

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My question is then the following: is it standard knowledge that my degenerate critical point will either die, or bifurcate into strictly more than one critical point? (For a generic bifurcation). If so, where can I find a proof of this statement?

It seems to be the picture that everyone always draws when speaking of birth-death bifurcations, but I haven't been able to find a proof, or an intuition behind this.

PS: provided that the degenerate critical point does indeed bifurcate into multiple critical points, I already know that these will be (generically) non-degenerate. So I only need the statement that:

"If the degeneracy doesn't die along the perturbation, then it will bifurcate into $N > 1$ critical points."

Any ideas, or references would be greatly appreciated :). Have a nice day!

Answer 1

Eric's answer is correct.

The $x^3$ can not be perturbed into function with one non-degenerate singularity essentially because then that point is either a global max or a global min, which does not match the behavior of $x^3$ far away from $0$ ("at infinity"). For the function $x^4$ we don't have such an obstruction.

In general, any function whose critical points lie in a compact region can be perturbed into one with non-degenerate critical points (this is studied in Morse theory, and the resulting function is called Morse). Moreover, if the original function had isolated singularities, then one can associate each such point a Hopf index (for non-degenerate points it's $(-1)^k$, where k is the Morse index, the number of negative eigenvalues of the Hessian), and if the perturbation is through functions fixed at infinity (or more generally, transverse to the boundary of a region enclosing all singularities at all times), the Poincare–Hopf theorem says the sum of all indexes stays constant. In your case, the $x^3$ has index $0$ singularity at $0$, so we again see that it can not be perturbed into a Morse function with exactly one non-degenerate critical point (as that would have Hopf index of $\pm 1$); whereas $x^4$ has an index $1$ singularity at $0$, and can indeed be perturbed into a function with unique non-degenerate minimum (as in Eric's answer).

In general, in higher dimension, for isolated singularities the Hopf index gives some restrictions in the same vein. I believe Conley index should give some obstructions in non-isolated case.

Answer 2

Your statement is false. Consider $x^4$. The perturbation $x^4+ax^3+bx^2+c$ has a derivative which is a cubic and so can anywhere from 1 to 3 real roots in the neighborhood of $0$ for small $a,b,c$. For example, $x^4+.1x^2$ has one critical point, $x^4-.1 x^3$ has 2 (double root at 0), and $x^4-.1x^2$ has 3.

In more dimensions, you’re looking at zeroes of the determinant of a hessian, so you’re considering zeroes in the neighborhood of a zero of a real function on $\mathbb R^n$. You can get anywhere from zero critical points to infinitely many locally to the perturbation of a degenerate critical point.