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arxiv: 2605.14030 · v1 · submitted 2026-05-13 · 🧮 math.DS · math.GT

Recognition: 2 theorem links

· Lean Theorem

Complexity of Billiards in Polygons Associated to Hyperbolic (p,q)-Tilings

Sunrose T. Shrestha , Jane Wang
Authors on Pith no claims yet

Pith reviewed 2026-05-15 04:46 UTC · model grok-4.3

classification 🧮 math.DS math.GT
keywords billiard dynamicshyperbolic polygonssymbolic dynamicstopological entropygrowth ratesgrammar rulestiling paths
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The pith

Billiard languages in hyperbolic (p,q)-polygons have explicit exponential growth rates for even q and complete grammar rules for realizable words.

A machine-rendered reading of the paper's core claim, the machinery that carries it, and where it could break.

This paper examines the symbolic dynamics of billiard trajectories in regular p-sided polygons with internal angle 2π/q embedded in the hyperbolic plane. The complexity of the language formed by sequences of side hits grows exponentially, and this growth rate equals the topological entropy of the billiard flow. The authors derive closed-form expressions for the growth rate when q is even and interval bounds when q is odd. For the even-q case they also supply exact rules that decide which finite, infinite, or bi-infinite words over an alphabet of p letters arise from actual billiard paths.

Core claim

The billiard language of these hyperbolic polygons grows at an exponential rate that can be computed explicitly when q is even; the same rate equals the topological entropy of the system. When q is even, a finite set of grammar rules completely characterizes the words realized by billiard trajectories, proved via minimal tiling paths in the associated hyperbolic tiling.

What carries the argument

Minimal tiling paths, which select shortest representatives in the hyperbolic tiling and thereby determine exactly which sequences of polygon sides can be hit by a billiard trajectory.

If this is right

  • Topological entropy of the billiard flow equals the explicitly computed growth rate for every even q.
  • The full symbolic dynamics of the billiard is given by a regular language when q is even.
  • Growth-rate bounds for odd q restrict the possible entropy values without giving exact formulas.
  • Bi-infinite words correspond to complete geodesics in the tiling that stay inside the polygon unfolding.

Where Pith is reading between the lines

These are editorial extensions of the paper, not claims the author makes directly.

  • The grammar rules may extend to give a presentation of the fundamental group action on the unfolding, connecting billiard complexity to hyperbolic group theory.
  • Closed-form entropy values for even q allow direct comparison with entropies of other hyperbolic flows or geodesic flows on surfaces.
  • The distinction between even and odd q suggests a parity-dependent dichotomy in the recurrence properties of the billiard map.

Load-bearing premise

The assumption that every word permitted by the grammar rules is realized by some billiard trajectory and that no other words occur.

What would settle it

Exhibiting a single word that obeys the stated grammar yet cannot be realized by any billiard path in the polygon, or a realizable path whose word violates the grammar.

Figures

Figures reproduced from arXiv: 2605.14030 by Jane Wang, Sunrose T. Shrestha.

Figure 1
Figure 1. Figure 1: On the left, a generalized diagonal with combinatorial length 4 and billiard word bcd. On the right, the unfolding of this generalized diagonal. Definition 1.2 (Generalized diagonals and combinatorial length). A generalized diagonal in a polygonal billiard table is a billiard trajectory that begins and ends at vertices, without hitting any vertices in between. Edges of the polygon do not count as generaliz… view at source ↗
Figure 2
Figure 2. Figure 2: The vertex v is the unique minimal vertex for the shaded face. Since q ≥ 4, there are at least two new outgoing edges at this stage which form the start of a new open face that v belongs to. For this open face, v is the unique minimal vertex. The case when p = 3 is similar since q ≥ 7, and so at stage k + 1 each vertex v forms at least q − 4 ≥ 3 new edges. □ We are now ready to prove Theorem 1.13. Theorem … view at source ↗
Figure 3
Figure 3. Figure 3: A schematic showing types of vertices in the dual graphs of certain (p, q)-tilings illustrating the case of q being even q odd separately. faces generated when we create the next level of vertices of one greater distance, we see that there are two new type 2 vertices and p − 3 new type 1 vertices. The reasoning is similar for the other types. The one case that is a little bit different is when we are consi… view at source ↗
Figure 4
Figure 4. Figure 4: Left: a portion of a geodesic segment γ and nudged segment γ ′ . Right: A geo￾desic segment γ that is a portion of an edge geodesic, and a nudge γ ′ , and the corresponding tiling path Aγ = (A0, A1, . . . , A6). Definition 3.6 (Combinatorial and tiling lengths of a geodesic segment). Let γ and let γ ′ be a nudge of γ. The tiling length tl(γ) is then defined as the length of the associated tiling path Aγ. T… view at source ↗
Figure 5
Figure 5. Figure 5: A way to turn a tiling path in a (p, q)-tiling into a shorter tiling path if it does not minimally intersect an edge geodesic g. Here, dots represent tiles. The original tiling path is A = (A0, . . . , An) and the shortened tiling path is (A0, . . . , Ai−2, A′ i , . . . , A′ j , Aj+2, . . . , An). Suppose that A crossed an edge geodesic g at two distinct places: from Ai−1 to Ai and Aj to Aj+1, as shown in … view at source ↗
Figure 6
Figure 6. Figure 6: Left: A schematic diagram of part of a zigzag path of edges (in bold) in a (p, q)- tiling with q = 5. Right: Two zigzags ζ1 and ζ2 that intersect at an edge e. However, besides at that edge, such zigzags do not intersect as we state below: Proposition 3.12 (Behavior of intersecting zigzags). Let ζ1 and ζ2 be two zigzags that share an edge e, in a hyperbolic (p, q)-tiling with q odd. Then, ζ2 is the reflect… view at source ↗
Figure 7
Figure 7. Figure 7: In the setting of Case 1 where the edge geodesic ge2 bisects tile A0, there are two ways in which the tiling path A crosses ge2 [PITH_FULL_IMAGE:figures/full_fig_p016_7.png] view at source ↗
Figure 8
Figure 8. Figure 8: In Case 2, the edge geodesic ge2 does not go through tile A0 so we consider the zigzag ζ ′ . By Proposition 3.12, ζ and ζ ′ do not intersect anywhere else other other than e1. In particular, all the edges e2, . . . , em are on the same side of ζ ′ as A0. As em belongs to An+1 we see that A0 and An+1 are on the same side of ζ ′ . Now consider the path A. It crosses ζ ′ once at edge e1. As A1 and An+1 are on… view at source ↗
Figure 9
Figure 9. Figure 9: A schematic for a situation where the geodesic γpA,pB crosses a zigzag ζ separat￾ing the tiles A and B multiple times. Since ζ separates A and B, the excess intersections are even in number (in this illustration, that number is 4). We then pair the excess intersections edges (in this case (e1, e2) and (e3, e4) are paired) so that for each pair, we associate q−3 2 in-between edges (in this illustration q = … view at source ↗
Figure 10
Figure 10. Figure 10: In a (8, 4) tiling, the word 12124141 does not violate rules E1 and E2 of Theorem 1.14, but cannot be realized by a billiard word because the corresponding tiling path double crosses the edge geodesic g. The word is word equivalent to 12121414, which violates rule E2. The goal of this section is to provide an alternate and self-contained proof of a precisely stated billiard path realization theorem for bi… view at source ↗
Figure 11
Figure 11. Figure 11: A schematic of two unfolded billiard trajectories passing near a vertex in a tiling where q = 8. We notice that these trajectories cannot inter￾sect more than 8 2 = 4 edges around the vertex in a row. The two grammar rules are natural. Rule E1 states that a geodesic in the tiling cannot hit the same side twice in a row. When q is even, recall that edge geodesics (see Definition 3.3) are bi-infinite geodes… view at source ↗
Figure 12
Figure 12. Figure 12: A schematic illustrating Lemma 4.17. In both cases, the edge geodesic g does not intersect the geodesic extension of edge e. Lemma 4.19 (Necessarily fellow traveling). Let A = (A0, . . . , An) be a tiling path so that there exists edges e ∈ A0 and f ∈ An that have the same geodesic extension g. If A0 and An are on the same halfspace of g, then the tiling path A is minimal if and only if it fellow travels … view at source ↗
Figure 13
Figure 13. Figure 13: Furthermore, by definition of ge, Aj is in the opposite halfspace of ge as the geodesic g and all tiles that share a vertex with g. In particular, Aj−1 and An are in the opposite halfspace as Aj . Thus A is minimal but double-crosses ge, which contradicts Proposition 3.8. Therefore, the minimal path A must necessarily fellow travel g. Conversely, since a minimal tiling path between A0 and An must exist, i… view at source ↗
Figure 13
Figure 13. Figure 13: If the tiling path A does not fellow travel the geodesic g, then it must double cross the edge geodesic ge [PITH_FULL_IMAGE:figures/full_fig_p029_13.png] view at source ↗
Figure 14
Figure 14. Figure 14: Two minimal tiling paths A and B with the same starting and ending tile. and therefore An must be on the opposite side of geB than A0. Let m ≤ n be such that Am is the first tile of A that is on the opposite side of geB from A0. By Lemma 4.19, A must fellow travel geB before crossing the geodesic at (Am−1, Am) [PITH_FULL_IMAGE:figures/full_fig_p029_14.png] view at source ↗
Figure 15
Figure 15. Figure 15: A schematic depiction of how the path A that fellow travels an edge geodesic geB and then crosses it is vertex sequence equivalent to A′ that first crosses geB and then fellow travels it, via vertex sequence equivalence moves across vertices vk, vk−1, . . . , v1 along geB . By performing repeated vertex sequence equivalence moves across the vertices of geB as depicted in [PITH_FULL_IMAGE:figures/full_fig… view at source ↗
Figure 16
Figure 16. Figure 16: If A crosses g and then fellow travels before crossing g again, then A is word equivalent to a tiling path A′ that backtracks and violates condition E1 for admissibility. By a similar process as in the proof of Proposition 4.15, (A1, A2, . . . , Am), which fellow travels and then crosses g is word equivalent to a path (A1, A′ 2 , . . . , A′ m−1 , Am) that first crosses and then fellow travels g, via a seq… view at source ↗
Figure 17
Figure 17. Figure 17: Two minimal tiling paths A and B with the same starting tile and endpoint in ∂H2 . We first show this map is well defined. Let α, β ∈ [α] be geodesic rays starting in A0 with the same endpoint η ∈ ∂H2 . Since α, β are geodesics rays, the corresponding tiling paths Aα and Aβ are minimal tiling paths with the same base tile and endpoint, and so wα and wβ are admissible by Proposition 4.22. Furthermore, by P… view at source ↗
Figure 18
Figure 18. Figure 18: Tiling paths Aα and Aβ have different endpoints η1 and η2 respectively that are separated by edge geodesic g. Let Bn be the first tile of Aβ that is on the η2 side of g. If the tiling paths Aα and Aβ were word equivalent, then there would exist another minimal tiling path A′ α that matches with Aβ for at least the first n+ 1 tiles, but would differ from Aα by finitely many vertex sequence equivalence move… view at source ↗
Figure 19
Figure 19. Figure 19: A schematic for how a tiling path A that begins and ends in sufficiently far apart sectors relative to a vertex v is vertex sequence equivalent to a tiling path that includes a vertex traversal of v. Proof of Theorem 1.17. Given a centered, bi-infinite, admissible word class, choose a centered element w of the word class, which corresponds to a tiling path (. . . , A−1, A0, A1, . . .) centered at the tile… view at source ↗
Figure 20
Figure 20. Figure 20: Depictions of three of the cases in the proof of Theorem 1.17: Case 2b (left), Case 2c (middle), and Case 3 (right) We will first show that any geodesic connecting the backward and forward limit points η−∞ and η∞ must pass through Ak or A˜ k. To see this, let g1 be an edge geodesic through the vertex ˜v on A˜ k that is the reflection of v through g, such that g1 makes an angle of at least 2π q with each e… view at source ↗
Figure 21
Figure 21. Figure 21: The first two figures show examples of billiard trajectories in a regular hyper￾bolic polygon with 4 sides and 2π/7 internal angle unfolded onto the hyperbolic plane. The third figure shows a word that is not forbidden by Theorem 5.1 but one that cannot be realized as a billiard trajectory. suffix/prefix is not in F. Directed walks in G then represent allowed words in the language. It is known that the gr… view at source ↗
Figure 22
Figure 22. Figure 22: A geodesic γ and how it intersects edges adjacent to a vertex v. Here, γ double￾intersects the zigzag containing edge e1 and e(q+1)/2. A.1. Facts about (p, q)-tilings when q is odd. We first note that when a hyperbolic geodesic γ intersects edges adjacent to a vertex v of the tiling, it intersects edges e1, e2, . . . , en cyclically around v with n ≤ q+1 2 (since any more intersections would cause γ to do… view at source ↗
Figure 23
Figure 23. Figure 23: By the rotational symmetry of a zigzag around a midpoint m2, two segments connecting successive midpoints of zigzag edges must meet at an angle of π. Thus, there is a geodesic connecting all midpoints of a zigzag. In light of Proposition 3.11, we have the following definition. Definition A.1 (Midpoint geodesic of a zigzag). Given a zigzag ζ in a hyperbolic (p, q)-tiling with q odd, let gζ be the geodesic … view at source ↗
Figure 24
Figure 24. Figure 24: Two zigzags that intersect at an edge e and a separate vertex v must be symmetric across a line of symmetry gsym through v and perpendicular to ge. ζ1 and ζ2 are on opposite sides of ge, it follows that the midpoint geodesics gζi , and therefore the zigzags ζi and ζ2, each cross ge at the edge e and therefore cannot share any edge except e with ge. Furthermore, if ζi intersected ge at another point not on… view at source ↗
Figure 25
Figure 25. Figure 25: When n is even, the internal angles of P at v0 . . . , vn−1 sum up to nπ. (a) The case when n is odd and the internal angle of P at v0 is π + π/q. The sum of the internal angles of P is nπ + π/q. (b) The case when n is odd and the internal angle of P at v0 is π − π/q. In this case we make use of a different polygon, Q, contained within P. The sum of the internal angles of Q exceed ( n−1 2 )π [PITH_FULL_I… view at source ↗
Figure 26
Figure 26. Figure 26: When n is odd, we use two polygons, P or Q, depending on the angle at v0 to obtain contradictions to hyperbolicity. Finally, assume angle ∠p0v0v1 is π − π/q < π. See [PITH_FULL_IMAGE:figures/full_fig_p042_26.png] view at source ↗
Figure 27
Figure 27. Figure 27: A diagram of consecutive zigzag intersections Proof. Following [PITH_FULL_IMAGE:figures/full_fig_p043_27.png] view at source ↗
Figure 28
Figure 28. Figure 28: Two cases of non-intersecting edge geodesics illustrated by Lemma 4.17. In both cases, assuming that g intersects the geodesic extension of e gives a contradiction to the hyperbolicity of the polygon P [PITH_FULL_IMAGE:figures/full_fig_p045_28.png] view at source ↗
Figure 29
Figure 29. Figure 29: When p ≥ 4 and q is even, two edge geodesics g1 and g3 that intersect an edge geodesic g2 at distinct vertices v and w will not intersect each other. Lemma 4.27. Consider a hyperbolic (p, q)-tiling with p = 3 and q even. Let g1 and g2 be edge geodesic rays emanating from a vertex v with angle at v that is at least 6π/q. Let g3 be any edge geodesic that intersects g2 at exactly one vertex w ̸= v. Then g3 d… view at source ↗
Figure 30
Figure 30. Figure 30: When p = 3 and q even, schematic depicting two edge geodesic rays g1, g2 emanating from vertex v making angle at least 6π/q. Any geodesic g3 ̸= g2 that intersects g2 at a vertex w other than v cannot intersect g1. Let γ be the edge geodesic corresponding to the edge of A not containing v, passing through v1. We first claim that γ cannot intersect g1. Assume towards contradiction that it does intersect g1 … view at source ↗
read the original abstract

The complexity of the billiard language of regular polygons in the hyperbolic plane with $p$ sides and $2\pi/q$ internal angles is known to grow exponentially and the exponential growth rate is known to equal the topological entropy of the billiard system. In this paper we compute these exponential growth rates explicitly when $q$ is even and give bounds when $q$ is odd. Additionally, for the $q$ even case, we give complete grammar rules that establish when a word (finite, infinite or bi-infinite) in $p$ letters is realized by a billiard path. This latter result is roughly stated and not rigorously proved in a paper of Giannoni and Ullmo (1995). In this paper, we provide a precise statement and a complete proof using new methods relating to minimal tiling paths.

Editorial analysis

A structured set of objections, weighed in public.

Desk editor's note, referee report, simulated authors' rebuttal, and a circularity audit. Tearing a paper down is the easy half of reading it; the pith above is the substance, this is the friction.

Referee Report

2 major / 2 minor

Summary. The manuscript computes the exponential growth rates of the billiard language for regular hyperbolic (p,q)-polygons, giving explicit closed-form expressions when q is even and bounds when q is odd. For even q it also supplies complete grammar rules characterizing which finite, infinite, and bi-infinite words in p letters arise from billiard trajectories, proved via a new combinatorial construction based on minimal tiling paths; this supplies a rigorous version of a sketch appearing in Giannoni-Ullmo (1995).

Significance. If the central claims hold, the work supplies the first explicit growth-rate formulas and a fully rigorous symbolic description for this family of hyperbolic billiards, strengthening the connection between unfolding techniques and subshift complexity. The minimal-tiling-path method is a concrete new tool that could be reused for entropy calculations or periodic-orbit enumeration in other tiling-based dynamical systems.

major comments (2)
  1. [§3] §3 (Minimal tiling paths): the central claim that every billiard trajectory (including infinite and bi-infinite) corresponds to a unique minimal tiling path, and conversely, is load-bearing for both the grammar rules and the growth-rate formulas. The manuscript must explicitly prove that the construction neither omits trajectories whose unfoldings are non-minimal nor includes sequences forbidden by the reflection law; without this bijective verification the grammar is incomplete and the stated growth rates cannot be guaranteed to equal the topological entropy.
  2. [§5] §5 (Growth-rate computation): the explicit formulas for even q are derived from the grammar; any gap in the bijectivity argument of §3 would propagate directly into these formulas. A short independent verification that the number of admissible words of length n generated by the grammar satisfies the claimed recurrence (or closed form) should be added.
minor comments (2)
  1. [Introduction] The introduction should state the precise range of p and q for which the polygons are hyperbolic (i.e., the inequality 1/p + 1/q < 1/2) to avoid any ambiguity for readers.
  2. [§2] Notation for the unfolding and the labeling of the p letters should be fixed once and used consistently in all figures and statements of the grammar rules.

Simulated Author's Rebuttal

2 responses · 0 unresolved

We thank the referee for the detailed and constructive report. We agree that the bijectivity between billiard trajectories and minimal tiling paths in §3 requires a more explicit verification to fully support the grammar rules and growth-rate claims. We will revise the manuscript to address both major comments by strengthening the proof and adding an independent check.

read point-by-point responses

  1. Referee: [§3] §3 (Minimal tiling paths): the central claim that every billiard trajectory (including infinite and bi-infinite) corresponds to a unique minimal tiling path, and conversely, is load-bearing for both the grammar rules and the growth-rate formulas. The manuscript must explicitly prove that the construction neither omits trajectories whose unfoldings are non-minimal nor includes sequences forbidden by the reflection law; without this bijective verification the grammar is incomplete and the stated growth rates cannot be guaranteed to equal the topological entropy.

    Authors: We acknowledge that while the minimal tiling path construction in §3 is intended to establish a bijection (via the unfolding process and the reflection law encoded in the tiling), the current write-up would benefit from a more direct and self-contained argument. In the revised version we will add an explicit lemma proving that (i) every billiard trajectory unfolds to a unique minimal tiling path and (ii) every minimal tiling path projects to a valid billiard trajectory satisfying the reflection law. This will close the potential gap and rigorously justify both the grammar and the entropy formulas. revision: yes

  2. Referee: [§5] §5 (Growth-rate computation): the explicit formulas for even q are derived from the grammar; any gap in the bijectivity argument of §3 would propagate directly into these formulas. A short independent verification that the number of admissible words of length n generated by the grammar satisfies the claimed recurrence (or closed form) should be added.

    Authors: We agree that an independent combinatorial check strengthens the presentation. In the revision we will insert a short subsection (or appendix paragraph) that directly counts the number of admissible words of length n generated by the grammar rules for small n and verifies that these counts satisfy the recurrence relation used to derive the closed-form growth rate. This verification will be independent of the full bijectivity proof and will confirm the formulas for even q. revision: yes

Circularity Check

0 steps flagged

No circularity: growth rates and grammar rules derived via new minimal tiling path methods with external citation support

full rationale

The paper states that exponential growth rates equal topological entropy (a known fact) and computes them explicitly for even q (with bounds for odd q) while supplying complete grammar rules for realizable words. These rules are proved using new methods based on minimal tiling paths, extending but rigorously completing the 1995 Giannoni-Ullmo observation with an independent proof. No self-citations, fitted parameters renamed as predictions, self-definitional constructions, or ansatz smuggling appear in the derivation chain. The central claims rest on combinatorial enumeration of billiard unfoldings rather than reducing to the inputs by construction.

Axiom & Free-Parameter Ledger

0 free parameters · 2 axioms · 0 invented entities

The central claims rest on standard facts about hyperbolic geometry, billiard flows, and topological entropy together with a new combinatorial correspondence between billiard words and minimal tiling paths.

axioms (2)
  • domain assumption The exponential growth rate of the billiard language equals the topological entropy of the billiard flow (taken from prior literature).
    Invoked in the first sentence of the abstract as already known.
  • domain assumption Minimal paths in the (p,q)-tiling correspond to realizable billiard trajectories.
    Used to establish the grammar rules for even q.

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