Is there a quartic polynomial in two variables that have multiple local minima and no other critical points?

Question

Question. Can a degree 4 polynomial $p:\mathbb R^2 \to \mathbb R$ have $N\geq 2$ local minima and no other critical points?

I got this question when trying to answer: How many strict local minima a quartic polynomial in two variables might have?

A closely related question asks: If a two variable smooth function has two global minima, will it necessarily have a third critical point? Among the answers @RiverLi gives an example of the function $$ f(x,y)=(x^2-1)^2+(x^2y-x-1)^2, \tag{*}\label{RL} $$ which is a polynomial, but of degree 6.

Notice that polynomials like \eqref{RL} can be written as $p(x,y)=u(x,y)^2 + v(x,y)^2$. Since we need $p$ to be degree 4 polynomial, $u$ and $v$ must be quadratic polynomials. We need there to be a set $P\subset \Bbb R^2, |P|=N$, of points such that:

1. $(x,y)\in P \ \ \Longleftrightarrow \ \ u(x,y)=v(x,y)=0$;
2. $(x,y)\in P \ \ \Longleftrightarrow \ \ u_x u + v_x v = u_y u + v_y v =0$ when evaluated at $(x,y)$.

For what values of $N\geq 2$ is there a set $P\subset \Bbb R^2, |P|=N$ and quadratic polynomials $u,v:\Bbb R^2 \to \Bbb R$ satisfying conditions 1 and 2?

My guess is that for $N=2$ at least one of the polynomials $u,v$ has to be of the third order as in \eqref{RL}. However, it is possible that there would be desirable $u,v$ for $N=3$ or $N=4$. Note that by Bézout's theorem condition 1 can not be satisfied at more than 4 points given that $u,v$ are quadratic.

Alternatively, is there another form than $p=u^2+v^2$ that could represent a quartic polynomial $p$ with $N\geq 2$ local minima and no other critical points?

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Disclaimer. I asked about the special case of $N=2$ on https://math.codidact.com.

Answer 1

The answer appears to be no.

This paper: "Counting Critical Points of Real Polynomials in Two Variables":

http://www.jstor.org/stable/2324459

gives a partial answer for the case $N=4$, which they prove is not possible. (For the full answer see edit further below.)

The paper defines the global index of the gradient of a polynomial as the "turning number" of this vector field, see:

https://en.wikipedia.org/wiki/Vector_field#:~:text=The%20index%20of%20a%20vector,a%20source%20or%20sink%20singularity

and

https://www.researchgate.net/figure/A-critical-point-of-a-normalized-vector-field-is-a-point-where-its-direction-is-undefined_fig5_358687586.

They give the formula for the global index: $$i=m+n-s$$ where $i$ is the index, $m$ is the number of local maxima, $n$ is the number of local minima, and $s$ is the number of saddle points.

It further states that the local index of an isolated non-degenerate maximum or minimum can be at most $1$, so isolated non-degenerate maximum and minimum doesn't have any index multiplicity for a polynomial.

In the paper, proposition 2.5 states that the index $i$ of a polynomial $f(x,y)$ of degree $d$ about a circle $C$ in the plane is bounded according to $$|i| \leq d-1$$

If we take the circle large enough it will enclose all critical points and give us the global index.

If we insert $d=4$ for a degree $4$ polymomial we get that the index is bounded as:

$$|i| \leq 3$$

Moreover, if we only have local minima ($m=0$, $s=0$) then we have:

$$i=n$$

Now, for $n=N=4$ we get $i=4>3$, so that $N=4$ is not possible.

Later on, the authors comment that they don't believe $i=d-1$ is possible which would also exclude the case $N=3$ for $d=4$, however this is only a conjecture. It would be very nice if someone could find a counter example to this. Very little seems to be written on the cases $N=3$ and $N=2$.

Edit

This paper: "The Index of grad f (x, y)", Alan H. Durfee, Topology Vol. 37, No. 6, pp. 1339Ð1361, 1998 gives the full answer. There it is proved that $$i \leq max(1,d-3)$$ so that for a quartic polynomial the index can be at most 1. This means that for a polynomial with only isolated minima and no other critical points you can only have one minimum. If you have more than one minima then you also have saddle points (and potentially some maxima).

Side Note

For those interested: Bezout's theorem tells us that for a real quartic polynomial in two variables you can have at most $(d-1)^2=9$ critical points. This is one such polynomial: $$(y^2 - x^2)(y^2 -xy + x^2)+x^2-y^2$$ and a plot below where the red curves are the loci of the zeros of each gradient component. Where they intersect you have a critical point. $$(y^2 - x^2)(y^2 -xy + x^2)+x^2-y^2$$" />