8
$\begingroup$

By Bézout's theorem a quartic polynomial $p(x,y)$ can have at most 9 isolated critical points. Can all of them be saddle points?

In case of a cubic polynomial there is a mechanical way to answer this type of questions: One can find a general form of such polynomials with critical points at three given points, e.g. at $(0,0),(0,1),$ and $(1,0)$, and then play with the remaining free parameters until a polynomial with the desired properties if found, as there is only one remaining critical point whose position was not normalized by affine transformation [1], [2].

However, the above approach won't work in case of a quartic polynomial, as then affine transformation normalizes position of only 3 out of the 9 potential critical points.

Note that a quartic polynomial in two real variables can have at most 5 minima out of its 9 potential critical points [3], [4]. Can it reversely have no extreme points, so that all 9 of the potential critical points would be saddle points?

$\endgroup$

1 Answer 1

14
$\begingroup$

By (3.1) of Counting Critical Points of Real Polynomials in Two Variables by Alan Durfee, Nathan Kronefeld, Heidi Munson, Jeff Roy, Ina Westby a degree $d$ polynomial with only nondegenerate critical points can contain at most $d(d-1)/2$ saddle points. For $d=4$ this gives a max of $6$ saddle points.

This bound is attained by a product of $d$ linear polynomials, as long as each pair of lines intersects at a different point, as then the $d(d-1)/2$ intersection points are all saddle points.

(This paper was cited in a paper cited in one of the answers you cite.)

$\endgroup$
1
  • 1
    $\begingroup$ Thanks a lot for the answer, I finally read the paper, it is very nicely written with a lot of intuition. $\endgroup$ Commented May 17, 2023 at 17:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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