Hausdorff dimension of projections

March 30, 2010

A few days ago, professor Wilkinson asked me the following question on google talk (while I was in Toronto):

Say that a set in \mathbb{R}^n is a k-zero set for some integer k<n if for every k-dimensional subspace P, saturating the set X by planes parallel to P yields a set of n-dimensional Lebesgue measure zero. How big can a k-zero set be?”

On the spot my guess was that the Hausdorff dimension of the set is at most n-k. In deed this is the case:

First let’s note that n-dimensional Lebesgue measure of the P-saturated set is 0 iff the n-k dimensional Lebesgue measure of the projection of our set to the n-k subspace orthogonal to P is 0.

Hence the question can be reformulated as: If a set E \subseteq \mathbb{R}^n has all n-k dimensional projection being n-k zero sets, how big can the set be?

Looking this up in the book ‘The Geometry of Fractal Sets’ by Falconer, indeed it’s a theorem:

Theorem: Let E \subseteq \mathbb{R}^n compact, \dim(E) = s (Hausdorff dimension), let G_{n,k} be the Garssmann manifold consisting of all k-dimensional subspaces of \mathbb{R}^n, then
a) If s \leq k, \dim(\mbox{Proj}_\Pi E) = s for almost all \Pi \in G_{n,k}

b) If s > k, \mbox{Proj}_\Pi E has positive k-dimensional Lebesgue measure for almost all \Pi \in G_{n,k}.

In our case, we have some set with all n-k-dimensional projection having measure 0, hence the set definitely does not satisfy b), i.e. it has dimensional at most n-k. Furthermore, a) also gives that if we have a uniform bound on the dimension of almost all projections, this is also a bound on the dimension of our original set.

This is strict as we can easily find sets that’s n-k dimensional and have all such projections measure 0. For example, take an n-k subspace and take a full-dimension measure 0 Cantor set on the subspace, the set will have all projections having measure 0.

Also, since the Hausdorff dimension of any projection can’t exceed the Hausdorff dimension of the original set, a set with one projection having positive n-k measure implies the dimension of the original set is \geq n-k.

Question 2: If one saturate a k-zero set by any smooth foliations with k-dimensional leaves, do we still get a set of Lebesgue measure 0?

We answer the question in the affective.

Given foliation \mathcal{F} of \mathbb{R}^n and k-zero set E. For any point p \in E, there exists a small neighborhood in which the foliation is diffeomorphic to the subspace foliation of the Euclidean space. i.e. there exists f from a neighborhood U of p to (-\epsilon, \epsilon)^n where the leaves of \mathcal{F} are sent to \{\bar{q}\} \times (-\epsilon, \epsilon)^k, \bar{q} \in (-\epsilon, \epsilon)^{n-k}.

By restricting f to a small neighborhood (for example, by taking \epsilon to be half of the origional \epsilon), we may assume that f is bi-Lipschitz. Hence the measure of the \mathcal{F}-saturated set inside U of U \cap E is the same as f(U \cap E) saturated by parallel k-subspaces inside (-\epsilon, \epsilon)^n. Dimension of f(E) is the same as dimension of E which is \leq n-k, if the inequality is strict, then all projections of f(E) onto n-k dimensional subspaces has measure 0 i.e. the saturated set by k-planes has n dimensional measure 0.

…to be continued

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