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Is there a cubic polynomial $c(x,y)$ with exactly 3 saddle point critical points?

In other words, can a cubic polynomial in two variables have three critical points, all of which are saddle points? Or could it have the maximum 4 critical points (as per Bézout's theorem), with only one of them being a local extremum?

A broader formulation of the question could be:

What number of saddle points can a cubic polynomial in two variables have without having any local maximum?

Observations:

  • By Bézout's Theorem a cubic polynomial (in two variables) has at most 4 critical points.
  • It is easy to see that a cubic polynomial has at most one maximum / minimum. Thus every polynomial $c(x,y)$ with 4 critical points has at least 2 saddle points. For example, the polynomial $c(x,y) = x^3-3x + y^3-3y$ has four critical points: Local maximum at $(-1,-1)$; local minimum at $(1,1)$ and saddle points at $(1,-1)$ and $(-1,1)$.
  • A quartic polynomial can have all 9 critical points without having any local maximum: Can a real quartic polynomial in two variables have more than 4 isolated local minima?
  • As argued in the answer to question https://math.stackexchange.com/q/4620663/1134951, there is an upper bound on the number of $M-s$, where $M$ is the number local extrema (local minima and maxima) and $s$ is the number of saddle points. Here we are interested in the lower bound.
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  • $\begingroup$ There is a somehow related question on Math.SE on whether a cubic polynomial in two variables can have exactly three critical points (and not four). The example I gave there has a degenerate critical point at $(0,0)$. $\endgroup$
    – cs89
    May 16 at 11:17

1 Answer 1

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The cubic $x^3 - xy^2 - 2x^2 + x$ has critical points in $(1,0)$, $(0,-1)$, $(0,1)$ and $(1/3, 0)$. The determinant of the Hessian matrix is $-4(3x^2 + y^2 - 2x)$. It assumes the values $-4$, $-4$, $-4$ and $4/3$ in these four critical points. Thus the first three of them are saddle points.

Added later: A simpler example is the cubic $xy(x+y-1)$ with saddle points in $(0,0)$, $(1,0)$, $(0,1)$ (arising from setting $\alpha=\beta=0$, $\gamma=1$ in the comments below).

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  • $\begingroup$ Thanks a lot for this nice example. Could you share some insights on how you found this example? I'd really appreciate some intuition. $\endgroup$
    – Pavel Kocourek
    May 15 at 21:10
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    $\begingroup$ @PavelKocourek The three saddle points we are looking for are either on a line (not considered here), or by an affine transformation we may and do assume that they are $(0,0)$, $(1,0)$, $(0,1)$. Furthermore, we may assume that $(0,0)$ is on the curve. All this gives linear equations for the unknown coefficients. Now the Hessian determinants in terms of a parametrization of the linear solution space are simple enough as to see how to choose the parameters to make them negative. $\endgroup$
    – Peter Mueller
    May 15 at 21:29
  • $\begingroup$ The edited answer now has a different choice of saddle points yielding a shorter cubic. $\endgroup$
    – Peter Mueller
    May 15 at 22:06
  • $\begingroup$ Thanks for the explanations. Following the procedure you mentioned by hand I found that the general formula for a cubic polynomial $c(x,y)$ with $c(0,0)=0$ and critical points $(0,0),(0,1),$ and $(1,0)$ is $$P(x, y) = \alpha \left(x^3 - \frac{3}{2}x^2\right) + \beta \left(y^3 - \frac{3}{2}y^2\right) + \gamma \left(x^2y + xy^2 - xy\right)$$, so one only need to guess the coefficients $\alpha,\beta,\gamma$ to make the Hessian determinants negative at the three critical points. Did you use Sage for finding the right parameters? $\endgroup$
    – Pavel Kocourek
    May 16 at 8:54
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    $\begingroup$ @PavelKocourek Yes, I used Sage for the mere calculation. In your notation, the Hessian determinants in the three critical points are $9\alpha\beta - \gamma^2$, $-9\alpha\beta + 6\alpha\gamma - \gamma^2$ and $-9\alpha\beta + 6\beta\gamma - \gamma^2$. And by just looking at them one sees that one can even have $\alpha=\beta$ and all of them negative. $\endgroup$
    – Peter Mueller
    May 16 at 11:46

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