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Introduction

Let $K$ be a nice closed domain in $\mathbb{R}^2$, for example the closed unit ball. Recall that the Hausdorff distance on the family $C(K)$ of nonempty compact subsets of $K$ is defined as

$$ D(A,B) = \inf\{ \delta>0: A\subset B_\delta \text{ and } B\subset A_\delta \}, $$

where $X_\delta=\{y:\text{dist}(y,X)<\delta \}$ is the $\delta$ neighborhood of $X$.

Now consider the restriction of $D$ to the family $A_n$ of nonempty intersections of real algebraic curves of degree at most $n$ with $K$. Intuitively, it seems clear to me that the metric space $(A_n,D)$ should be finite dimensional, in any reasonable sense; for example, it should have finite upper box dimension, and even be doubling (meaning that a ball of radius $2\delta$ is covered by a uniformly bounded number of balls of radius $\delta$). However, so far I've failed to find a proof.

Question:

Is the space $(A_n,D)$ doubling/of finite box counting dimension? (the latter means that $A_n$ can be covered by $\delta^{-C}$ $D$-balls of radius $\delta$ for some $C>0$). If so, what is its box counting dimension? (in other words, what is the best $C$?).

Some remarks:

  • An obvious approach is to estimate the Hausdorff distance in terms of the coefficients, but this runs into several problems; for example, there are points of discontinuity: take $P_{\varepsilon}(x)$ with a single real root at $x=0.5$ and such that $P(\varepsilon)=\varepsilon$, and let $Q_\epsilon(x,y)=P_\epsilon(x)$. In the limit $\epsilon=0$, a large new component for $Q_\epsilon=0$ is created.

  • Another natural approach is to fix a large number $N$ (at least $n^2+1$) and consider the map that sends $N$-tuples of points which are at pairwise distances at least $\delta>0$ and which lie on some algebraic curve, to the algebraic curve passing through all of them. If this function was Lipschitz, or even Hölder (from the standard metric on $\mathbb{R}^{2N}$ to $(A_n,d)$), it would follow that the latter is doubling.

  • This is related to effective versions of Łojasiewicz inequality. It is known that for each $n$ there exists $\alpha$, such that for any nonempty real algebraic curve $\gamma$ of degree $n$, if $|P(x)|<\varepsilon$, then $\text{dist}(x,\gamma)\le C \varepsilon^\alpha$, where the constant $C$ depends on the curve. See for example this nice article by Johnson and Kollár.

    If $C$ was a continuous functions of the coefficients, this would easily imply that $(A_n,D)$ is doubling, but $C$ is not continuous in general, as can be seen from the same example as in the first remark.

  • We can prove (with an elementary argument) the following weaker bound: $A_n$ can be covered by $\delta^{O(\log\delta)}$ balls of radius $\delta$ (in the Hausdorff metric). This is enough for our application.

  • Of course, one may ask this question for more general real algebraic varieties, and also for complex varieties (the comple case should be easier).

Motivation

If $\mathcal{F}$ is a class of curves in $\mathbb{R}^2$, a tube of width $\delta$ is the $\delta$-neighborhood of a curve in $\mathcal{F}$. A set $E\subset\mathbb{R}^2$ is then called tube-null (with respect to $\mathcal{F}$) if for every $\delta>0$ it can be covered by countably many tubes so that the sum of their widths is at most $\delta$.

Together with Ville Suomala we prove that if $\mathcal{F}$ is the class of all lines, then there are sets of Hausdorff dimension 1 which are not tube-null (this answers a question of Carbery, Soria and Vargas). The question arose in trying to generalize this result to the family of algebraic curves of degree at most $n$.

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So, for example, an element of $A_1$ could be the union of a finite number of straight lines through the origin? So it would be enough to show that the set of these is infinite-dimensional, when $X$ is the unit disk? –  Gerald Edgar Mar 30 '12 at 14:38
    
Gerald - no, elements of $A_1$ are (single) straight lines ($A_n$ is the family of zero sets $P=0$ where $P\in\mathbb{R}[x,y]$ has degree at most $n$). For $n=1,2$ the answer is easily be seen to be yes (and the actual dimension can be computed) because linear/quadratic algebraic curves admit a simple explicit classification. –  Pablo Shmerkin Mar 30 '12 at 15:02
    
Why are there discontinuities on reducible curves? My intuition says that there are not. For instance, $xy+ax+by+c=(x+b)(y+a)+c-ab$, so the asymptotic and the distance from them both vary continuously. I agree that there is non-differentiability but that is much less bad. –  Will Sawin Mar 30 '12 at 21:15
    
Will - I think you're correct, I've edited the question accordingly. –  Pablo Shmerkin Mar 31 '12 at 0:55
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1 Answer

I think this should follow from some standard (although remarkable) result about semialgebraic sets applied to the universal family (i.e. if P=parameter space for all algebraic curves of degree <=n, look at Q={ (a,x): a is in P, x lies in K and on the curve corresponding to a}.

For example, one can partition P into finitely many semialgebraic pieces such that the map Q->P is semialgebraically trivial over each one -- this is not quite what you want, but getting close. This statement is in Chapter 4 of M. Coste's notes "An introduction to semialgebraic geometry"; the triangulizability statements in Chapter 3 might also be helpful.

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