level sets of multivariate polynomials - MathOverflow

level sets of multivariate polynomials

2009-11-08T20:16:21Z

Let $p:\mathbb R^n \rightarrow \mathbb R$ be a polynomial of degree at most $d$. Restrict $P$ to the unit cube $Q=[0,1]^n\subset\mathbb R^n$. We assume that $p$ has mean value zero on the unit cube $Q$:

$\int_Q f(x) dx = 0.$

For $\alpha>0$ consider the sublevel sets of $P$,

There are several known estimates for the Lebesgue measure of this set which in some sense or another are uniform over some classes of polynomials. For example, we have that

$|E _ \alpha| \lesssim min (pd,n) \frac{ \alpha ^ \frac{1}{d} }{ \|p\| _ { L^p(Q) } ^ \frac{1}{d} } $.

This particular estimate is due to Carbery and Wright and can be found here.

I'm interested in studying the (induced Lebesgue) measure of the boundary of this set

Consider first the easy case of dimension $n=1$. Then the set $E_\alpha$ is a finite union of closed intervals and the question is trivial. It is obvious that in this case there are at most $O(d)$ intervals so the $0$-dimensional measure of the boundary is $O(d)$.

Now in many variables things will be much more complicated. For example can we say that the set $E_\alpha$ has $O(d)$ connected components? Is there an estimate for the measure of the boundary $\partial E _ \alpha $ in terms of $\alpha$, $d$ and $n$, assuming (say) that $\|p\| _ {L^1(Q)}=1$ ?

This question comes up naturally if one tries to study an oscillatory integral with phase $p$ and apply integration by parts (i.e Gauss theorem) imitating the one dimensional method of proving the van der Corput lemma (for example).

Answer by Reid Barton for level sets of multivariate polynomials

2009-11-08T21:22:10Z

I don't know how to estimate the measure of the level sets, but I can answer the question about the number of connected components to an extent. Let $P_f(x) = (x-1)(x-2)\cdots(x-f)$, and define $p(x_1, \ldots, x_n) = P_f(x_1)^2 + \cdots + P_f(x_n)^2$. Then $p$ has degree $2f$, and the level sets $p = \varepsilon$ for small $\varepsilon$ have $f^n = (d/2)^n$ connected components (each $x_i$ must be close to one of the roots of $P_f$).

Theorem 11.5.3 of Bochnak, Coste, & Roy, Real Algebraic Geometry, says the sum of the Betti numbers, hence the number of connected components, of an algebraic set in $R^n$ defined by degree ≤d polynomials is at most $d(2d-1)^{n-1}$. So for fixed n, the maximum number of connected components of the level sets of a degree d polynomial in n variables is $\Theta(d^n)$.

Answer by fedja for level sets of multivariate polynomials

2009-11-08T23:50:24Z

There is a trivial estimate for the measure of the boundary based on the observation that $|p|^2$ is still a polynomial, so the corresponding surface intersects any line at most $2d$ times. Averaging over directions, we get $O(d\sqrt[] n)$ regardless of $\alpha$. Now, the question is what exactly you want: a dimension-independent bound, a bound that is small for large $\alpha$, or anything else. It may really help to try it from the other end: figure out what exactly you need and we'll try to figure out whether it is true or not. Otherwise, you may get plenty of answers with trivial and not so trivial estimates for everything, none of which will fit your real needs.

Answer by fedja for level sets of multivariate polynomials

2009-12-15T14:35:03Z

An encouraging update: Daniel Kane posted his proof for the Gaussian case on arXiv yesterday. The proof is both very simple and brilliant. I won't be surprised if his technique can be modified to cover the cube case.