## Archive for March, 2010

### 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 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

### Whitney’s extension theorem revisited

March 9, 2010

I (very surprisingly) bumped into Charles Fefferman at Northwestern this afternoon…Hence we talked math for a little bit. Among other things I mentioned that I’ve been trying to extend $C^1$ functions to the disc volume-preservingly. After trying on the board for a while, he laughs out loud when he saw that this may be obtained applying his favorite Whitney’s extension theorem. (I’ll discuss what he did later in the poster)

Mean while, it’s a pity that I’ve never written a post on Whitney’s extension theorem, hence here it is~

Given a compact subset $K$ in $\mathbb{R}^n$ and a function $f: K \rightarrow \mathbb{R}$, when can we extend it to a $C^r$ function on the whole $\mathbb{R}^n$?

First we note that there are obvious cases for which this can’t be done: for example, if we take $E$ to be a segment in $\mathbb{R}^2$ and $f$ a one-variable function of lower regularity than $r$, then of course there are no way to find a $C^r$ extension.

Hence it’s only reasonable to restrict our attention to those $f$ that has ‘candidate derivatives’ of all orders no larger than $r$ at all points in $E$.

i.e. For any $k$-fold subscript $d= (d_1, d_2, \cdots, d_k)$ with $d_1+d_2+ \cdots +d_k \leq r$ (we will denote $d_1+d_2+ \cdots +d_k = |d|$, there is a continuous function $f_d: K \rightarrow \mathbb{R}$ with the following property:

For all $x_o \in K$, $\displaystyle f_d(x) = \sum_{|l| \leq r-|d|} \frac{f_{l+d}(x)}{l!}(x-x_0)^l+R_d(x, x_0)$  where $R_d(x, x_0) \sim o(|x-x_0|^{r-|d|})$ as $x \rightarrow x_0$ and is uniform in $x_0$.

i.e. The functions $f_\alpha$ are compatible as Taylor coefficients of some $C^r$ function on $\mathbb{R}^n$, which is absolutely necessary for a $C^r$ extension to exist.

Whitney’s extension theorem: (classical version)

Suppose a set of functions $f_\alpha$ with all multi-index $| \alpha | \leq r$ satisfying the above Taylor condition at all points in $K$. Then there is a $C^r$ function $\hat{f}: \mathbb{R}^n \rightarrow \mathbb{R}$ s.t. $\hat{f}|_K = f_{\bar{0}}$ and for all $\alpha \leq r$, $(D^\alpha \hat{f})|_K = f_\alpha$. Furthermore, $\hat{f}$ can be taken real analytic on $\mathbb{R}^n \backslash K$.

This is indeed the best one could hope for. i.e. there is a $C^r$ extension whenever possible, furthermore the extension is at worst $C^r$ at the points which it is given to be only $C^r$ and much better (analytic) everywhere else.

However, sometimes we would like to control the $C^r$ norm of the resulting function in terms of the $C^r$ norm of the function on $K$.

Theorem: (Fefferman)

For any $n, \ r$, there exists $C$ such that the extension $||\hat{f}|| \leq C \cdot ||f||$ where the norm is the $C^r$ norm.

### Systolic inequality on the 2-torus

March 1, 2010

Starting last summer with professor Guth, I’ve been interested in the systolic inequality for Riemannian manifolds. As a starting point of a sequence of short posts I plan to write on little observations I had related to the subject, here I’ll talk about the baby case where we find the lower bound of the systole on the $2$-torus in terms of the area of the torus.

Given a Riemannian manifold $(M, g)$ where $g$ is the Riemannian metric.

Definition: The systole of $M$ is the length of smallest homotopically nontrivial loop in $M$.

We are interested in bounding the systole in terms of the $n$-th root of the volume of the manifold ( where $n$ is the dimension of $M$ ).

Note that the systole is only defined when our manifold has non-trivial fundamental group. I wish to remark that for the case of n-torus, having an inequality of the form $(\mbox{Sys}(\mathbb{T}^n))^n \leq C \cdot \mbox{Vol}(\mathbb{T}^n)$ is intuitive as we can see in the case of an embedded $2$-torus in $\mathbb{R}^3$, we may deform the metric (hence the embedding) to make a non-contactable loop as small as we want while keep the volume constant, however when we attempt to make the smallest such loop large when not changing the volume, we can see that we will run into trouble. Hence it’s expected that there is an upper bound for the length of the smallest loop.

Since if only one loop in some homotopy class achieves that minimal length, we should be able to enlarge it and contract some other loops in that class to enlarge the systole and keep the volume constant. Hence it’s tempting to assume that all loops in the same class are of the same length. In the $2$-torus case, such thing is the flat torus. Since any flat torus has systole proportional to $(\mbox{Vol}(\mathbb{T}^2))^{\frac{1}{2}}$, we have reasons to expect the optimal case fall inside this family. i.e.

$(\mbox{Sys}(\mathbb{T}^2))^2 \leq C \cdot \mbox{Vol}(\mathbb{T}^2)$.

This is indeed the case. The result was given in an early unpublished result by Loewner.

Let’s first optimize in the class of flat torus:

My first guess was that $C$ cannot be made less than $1$ i.e. the torus $\mathbb{R}^2/ \mathbb{Z}^2$ is the optimum case. However, this is not true. Let’s be more careful:

$\mathbb{T}^2 = \mathbb{R}^2 / (0,c)\mathbb{Z} \times (a, b)\mathbb{Z}$

Since by scaling does not change ratio between $(\mbox{Vol}(\mathbb{T}^2))$ and $(\mbox{Sys}(\mathbb{T}^2))^2$, we may normalize and let $c=1$

Let $\alpha, \beta$ be generators of the fundamental group of $\mathbb{T}^2$ length of all geodesic loops in class $[\alpha], \ [\beta]$ are the side lengths of the parallelepiped i.e. $1$ and $||(a,b)||$. W.L.O.G we suppose $a, b > 0$. $\alpha \beta^{-1}$ has length $||((1-a),b)||$ and all geodesics in other classes are at least twice as long as one of the above three.

Hence the systole is maximized when those three are equal, we get $a=1/2, b=\sqrt{3}/2$. The systole in this case is $1$ and the volume is $\sqrt{3}/2$. Hence for any flat torus, we have

$(\mbox{Sys}(\mathbb{T}^2))^2 \leq \frac{2}{\sqrt{3}} \cdot \mbox{Vol}(\mathbb{T}^2)$.

Theorem (Loewner): This bound holds for any metric $g$ on $\mathbb{T}^2$.

Proof: We will show this by reducing the case to flat metric.

$g$ induced an almost complex structure on $\mathbb{T}^2$, on surfaces, any almost complex structure is integrable. Hence there exists $f:\mathbb{T}^2 \rightarrow \mathbb{R}^+$ and $g= f \dot g_0$ where $(\mathbb{T}^2, g_0)$ is a Riemann surface.

By uniformization theorem, $(\mathbb{T}^2, g_0)$ is the quotient of $\mathbb{C}$ by a discrete lattice. i.e. $(\mathbb{T}^2, g_0)$ is a flat torus $\mathbb{R}^2 / (0,c)\mathbb{Z} \times (a, b)\mathbb{Z}$.

By scaling of the torus, we may assume the volume of the manifold is $1$ i.e.

$\displaystyle \int_{\mathbb{T}^2} f \ dV_{g_0} = 1 = \mbox{Vol}(\mathbb{T}^2, g_0)$

Any nontrivial homotopy class of loops on $(\mathbb{T}^2, g)$ can be represented by a straight loop on the flat torus. The length of such a loop in $(\mathbb{T}^2, g)$ is merely integration of $f$ along the segment.

Here we have a family of loops in the homotopy class that is straight, by taking a segment of appropriate length orthogonal to the loops, we have the one-parameter family of parallel loops foliate the torus. Hence integrating over the segment of the length of the loops gives us the total volume of the torus. By Fubini, we have at least one loop is longer than volume of the torus over length of the segment we integrated on, which is the length of the straight loop in the flat torus.

Therefore the systole of $(\mathbb{T}^2, g)$ is smaller than the minimum length of straight loops which is smaller than that of the flat torus. While the volume are the same. Hence it suffice to optimize the ratio in the class of flat tori. Establishes the theorem.

Combining the pervious statement, we get

$(\mbox{Sys}(\mathbb{T}^2))^2 \leq \frac{2}{\sqrt{3}} \cdot \mbox{Vol}(\mathbb{T}^2)$

for any metric on the torus.