# Many Markov components

In my recent preprint, I showed that once $r \ge c / n$ the Metropolis Markov chain on configurations of hard discs is no longer ergodic. In particular, even for reasonably small radius there are stable configurations of discs. In this note the goal is to make a more quantitative statement about the number of Markov components in particular when $r > c_2 / \sqrt{n}$ (which is the usual range of interest in statistical mechanics.

For the main construction, start with the following configuration of $24$ discs in a square. (It is not relevant to the following discussion, but it is fun to point out that this is the densest packing of $24$ equal sized discs in a square.)

This configuration of $24$ discs is easily seen to be stable (meaning that each disc is individually held in place by its neighboring discs and the boundary of the square), and in fact it is still stable if we delete a carefully chosen set of four discs as follows.

By stacking $k^2$ of these together one has stable configurations with $20 k^2$ discs.

Finally one can add discs in any of the “holes” in these kinds of configurations, leading to more and more possibilities for stable configurations. For example, here is one of the ${36 \choose 18} \approx 9.08 \times 10^9$ possibilities for dropping discs in $18$ of the $36$ holes in the previous illustration.

By continuing constructions like this for larger and larger $k$ we observe two facts.

(1) As $n \to \infty$ the number of components can grow quite fast. To make this quantitative we expand on the example. Suppose there are $20 k^2$ discs to start out, then there are $4 k^2$ holes. Suppose we drop $2 k^2$ discs into those holes. Then the number of discs $n = 22 k^2$ in total, and then the number of Markov components is at least ${4 k ^2 \choose 2 k^2 } = { 2n/11 \choose n/11 }$. (This is for unlabeled discs. For labeled discs, multiply this by $n!$.) By Stirling’s formula, this number grows exponentially fast in $n$ (something like $1.134^n$).

In this particular example the discs have total area $\pi r^2 n = \approx 0.67$, but that could obviously adjusted by changing the ratio of extra discs to holes.

(2) For every $\lambda \in (0.60, 0.75)$ there is an $\alpha = \alpha(\lambda) > 1$ and a sequence of configurations $C_1, C_2, \ldots$ with $C_i \in$ Config(i, r(i)), the number of Markov components greater than $\alpha ^ i$ for sufficiently large $i$, and $\pi r^2 n \to \lambda$ as $i \to \infty$.

It turns out that what are described here are only separate components in the sense that a Markov chain that only moves one disc at a time can not mix from one state to the other. (The four discs in the center of a square can rotate about the center of the square!) But a small modification of this construction would seem to give exponentially many components in the sense of $\pi_0$, each with small positive measure.

Is exponentially many components the maximal possible? The Milnor-Thom Theorem gives an upper bound much larger than this, something like $O(e^{c n^2})$ for some $c > 0$. It would also be very interesting to know any bounds on the topological complexity of these path components.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 32, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.949518084526062, "perplexity_flag": "head"}