How weird can the level set of a polynomial be?

Question

Let $p_n(\cdot)$ be a $d-$variate polynomial of degree $n$. How many connected components does the set

$$\{x:\ p_n(x)>0\}$$

have? Are these components also convex?

Note that, in case $d=1$ there are at most $n$ connected components, all convex, as it follows from the fundamental theorem of algebra. In fact, as $p_n$ has at most $n$ many zeros $x_1,\dots x_n$, the set $\{x:\ p_n(x)>0\}$ cannot be made by more than $n$ open intervals.

EDIT:

No more convexity, the question is now only on the connected components (I suspect they are of order $O(n^d)$). Also, I am satisfied with the order of magnitude in $n,d$, without the precise upper bound.

Answer 1

Let me explain an upper bound on the number of components, which is asymptotically consistent with your conjecture.

I will be using your notation where the dimension is $d$ and the degree is $n$ even though it is inconsistent with the reference below.

The following is Theorem 3 in

Milnor, John W., On the Betti numbers of real varieties, Proc. Am. Math. Soc. 15, 275-280 (1964). ZBL0123.38302.

Theorem: If $X\subset \mathbb R^d$ is defined by a polynomial inequality of the form $f\ge 0$ of degree $n$, then the sum of Betti numbers of $X$ is at most $$ B_{n,d}:=\frac{1}{2}(2+n)(1+n)^{d-1} $$
which is asymptotically (as $n, d\to\infty$) $\frac{1}{2}n^d$.

In particular, the number of path-connected components of $X$ is $\le B_{n,d}$, since this number is the zeroth Betti number of $X$.

Milnor in his paper proves a more general result in the case of several polynomial inequalities. Milnor's work was motivated by an earlier work of Rene Thom who obtained an upper bound on the total Betti numbers of algebraic subsets of $\mathbb R^d$.

Corollary. In the above setting, the number of connected components of the subset $X:=\{x\in \mathbb R^d: f(x)>0\}$ is at most $B_{n,d}$.

Proof. Consider the family of polynomials $f_t(x):= f(x) - t$, $t>0$. For every polynomial $f_t$ define the subset $X_t:=\{x\in \mathbb R^d: f_t(x)\ge 0\}$. Then, clearly, $$ X= \bigcup_{t>0} X_t. $$ According to Milnor's theorem, each $X_t$ has at most $B_{n,d}$ components.

Lemma. Suppose that $X$ is a topological space represented as an increasing union of subsets $X_t$, $t>0$, where each $X_t$ has at most $N$ path-components. Then $X$ also has at most $N$ path-components.

Proof. We need to show that if $x_0,...,x_N$ are points in $X$, then at least two of these points belong to the same path-component. There exists $t$ such that $x_0,...,x_N\in X_t$. Since $X_t$ has at most $N$ components, there is a path in $X_t$ connecting two of these points, say, $x_0, x_1$. But then $x_0, x_1$ belong to the same path-component of $X$ as well. qed.

Applying this lemma, we obtain the corollary. qed

Answer 2

Consider the following univariate polynomial:

$$p_n(x)=(x-1)(x-2)...(x-n)$$

Then $p_n(x)>0$ has $\lceil \frac{n+1}{2}\rceil$ connected components and it's pretty obvious what they are. Then consider the following d-variate polynomial:

$$P_{n,d}(x_1, x_2,..., x_d)=p_n(x_1)p_n(x_2)...p_n(x_d)$$

Then $P_{n,d}(x_1, x_2,..., x_d)>0$ should have ${\lceil\frac{(n+1)^d}{2}\rceil}$ connected components which are again pretty obvious. This is thus at least a minimum. I suspect, like you did, that it's also a maximum but I don't have a rigorous proof for that part.

EDIT:

I'll add some motivation for why I think this is the maximum. If you look at any line, it can intersect with at most $n$ separate component edges. In case of a bivariate polynomial and thus 2D plane, you can consider such line sweeping the plane. How many times can the components intersecting with the sweeping line change. It would seem intuititive that it's as many times as an orthogonal line can intersect with the component edges which is again $n$.

Then for e.g. a bivariate polynomial you can consider $n$ such lines in the direction of one coordinate axis and $n$ lines in the direction of the other coordinate axis. These intersect at $n^2$ points and you can get $n^2$ components arranged into 2D grid. Then it goes similarly for $d$ dimensions. It would seem intuitive that you can't have more but I don't know how to make this argument more rigorous.

Answer 3

Not a complete answer, but a lower bound is $$\Big\lceil \frac{1}{2}\sum_{i=0}^d {n \choose i}\Big\rceil.$$ To see this, let $$p(x) = q_1(x) q_2(x) \cdots q_n(x)$$ where each $q_i(x) = q_i(x_1, \ldots, x_d) = a_i + \sum_{j=1}^d c_{ij} x_i$ where the $c_{ij}$ and $a_i$ are random real numbers according to some continuous distribution, e.g. uniform on $[0,1]$. Then $\{x: p(x) = 0\} = \bigcup_{i=1}^n \{x | p_i(x) = 0\}$ is a union of hyperplanes in general position, and the number of components of $\{x: p(x) \neq 0\}$ is $$c = \sum_{i=0}^d {n \choose i}$$ by the argument here. Then, changing the sign of $p(x)$ if necessary, $\lceil c/2 \rceil$ of these components will have $p(x) > 0$.

Answer 4

If you're not restricting the degree of the polynomial, it can be pretty weird.

The Stone-Weierstrass theorem is true in multiple variables, so you can take any continuous function you want (on a bounded subset of your domain) and approximate it with a polynomial to an arbitrary degree of accuracy. In particular, you can have any number of connected components, and they do not need to be convex.