Consider a fractal $\mathcal{S}\subset\mathbb{R}^2$ (e.g., a Koch snowflake) and a 1-dim line $\ell$. Marstrand's slicing theorem states that
$$dim_H(\mathcal{S}\cap\ell)\leqslant dim_H(\mathcal{S})-1$$
where $dim_H(\mathcal{S})\in (1,2)$ is the Hausdorff dimension of the fractal. (Let's not consider the case $dim_H(\mathcal{S})<1$, and the cases $dim_H(\mathcal{S})>2$ have their generalization's of the Marstrand's theorem which might, or might not, lead to different/more complicated answers to the below questions.)
E.g., for the Koch snowflake, $dim_H(\mathcal{S}) = \frac{\ln 4}{\ln 3}\approx 1.26$. So if we cut it with a straight line $\ell$ (that isn't the symmetry line) the slice will be Cantor-like with a $dim_H (\mathcal{S}\cap\ell)\lesssim 0.26$.
- Is there any lower bound on $dim_H (\mathcal{S}\cap\ell)$? How small can it get? I have a feeling that the deviation from $dim_H(\mathcal{S})-1$ is relatively small.
- When does the equality occur, i.e., $dim_H(\mathcal{S}\cap\ell) = dim_H(\mathcal{S})-1$?
Best Answer
There are many open questions related to the sharpness of Marstrand's slicing thorem, and some of them are actively studied by the community. Looking at the article linked by D.R., and on articles citing that article, is a good place to start. See also Chapter 6 in Mattila's 2015 book "Fourier Analysis and Hausdorff Dimension".
Regarding your specific questions, it is important to get the statement of Marstrand's slicing theorem right: fixing direction $\theta$ and taking $\ell_x=span(\theta)+x$ for $x\in\theta^\perp$, the estimate $$dim_H(S\cap\ell_x)\le dim_H(S)-1\quad\quad(*)$$ holds for Lebesgue almost every $x\in\theta^\perp$. The theorem is false if we replace "Lebesgue almost every" with "every".
In particular, if you consider the affine Grassmannian $A(2,1)$, which is a 2-dimensional manifold, then the estimate (*) is true for a.e. line $\ell\in A(2,1)$ (where a.e. refers to the natural Lebesgue measure on $A(2,1)$.
Regarding Question 1. If you ask for a lower bound on $dim_H(S\cap\ell)$ that would hold for a.e. $\ell$ in $A(2,1)$, then only the trivial bound $dim_H(S\cap\ell)\ge 0$ holds. One silly example: if $S\cap\ell=\varnothing$, which is true for the vast majority of lines if $S$ is compact, then $dim_H(S\cap\ell)= 0$. Assuming $S\cap\ell\neq\varnothing$ does not help much: if $S$ is a union of a line and a square, and $\ell$ intersects only the line, then we still get $dim_H(S\cap\ell)=0 < 1=dim_H(S)-1$. You can easily arrange for this to happen for a positive proportion of lines in a positive proportion of directions - so for a positive measure subset of $A(2,1)$.
On the other hand, as pointed out by D.R., Marstrand also proved that for a.e. direction $\theta$ there is a positive measure set of $x$ such that $dim_H(S\cap\ell_x)=dim_H(S)-1$, where again $\ell_x=span(\theta)+x$. So you know that the converse to (*) holds for a positive proportion of lines in $A(2,1)$ - but not for a.e. line. This leads us to question 2.
Regarding Question 2. A quite satisfactory answer can be found in the aforementioned book "Fourier Analysis and Hausdorff Dimension", Theorem 6.9., which in your setting says that if $0<H^s(S)<\infty$ for $1<s\le 2$, then there exists an exceptional set of directions $E\subset \mathbb{S}^1$ with $dim_H(E)\le 2-s$ such that for $H^s$-a.e. $x\in S$ and every $\theta\notin E$ we have $$ dim_H(E\cap \ell_{x,\theta})= dim_H(E) - 1,$$ where $\ell_{x,\theta}=x+span(\theta)$.
It is worth stressing, that this result tells you that the converse to (*) holds for "a.e." line, but the meaning of "a.e." is different than we discussed in the answer to Question 1 - it does not refer to the Lebesgue measure on $A(2,1)$, but rather to "a.e." line passing through $E$, and "a.e." refers to the Hausdorff measure on $E$.