TESSELLATION OF A TRIANGLE BY REPEATED BARYCENTRIC SUBDIVISION

Under iterated barycentric subdivision of a triangle, most triangles become ﬂat in the sense that the largest angle tends to π . By analyzing a random walk on SL 2 ( R ) we give asymptotics with explicit constants for the number of ﬂat triangles and the degree of ﬂatness at a given stage of subdivision. In particular, we prove analytical bounds for the upper Lyapunov constant of the walk.


Introduction
The iterated barycentric subdivision of a triangle ∆ is defined as follows. In the first stage the three medians are drawn resulting in six smaller triangles ∆ 1 through ∆ 6 . At stage 2 the process is repeated in each ∆ i , 1 ≤ i ≤ 6, producing in 36 triangles ∆ i j , 1 ≤ i, j ≤ 6. In general, on the nth stage the medians are drawn in each triangle produced in the n − 1st stage, so that 6 n triangles ∆ i 1 ...i n ,1 ≤ i 1 , ..., i n ≤ 6, result. We ask: as n grows do the triangles in stage n become "flat", in the sense that they have largest angle approaching π? It is not hard to show that not all triangles that result from barycentric subdivision become flat: if ∆ = ABC has m(A) ≤ m(B) ≤ m(C), with centroid O and X the midpoint of BC, then it is a nice exercise in Euclidean geometry to check that the least angle in COX is at least min(m(A), 30 • ); in particular, min(m(A), 30 • ) is a lower bound on the maximum least angle appearing among triangles in the nth stage of subdivision, for any n. Nonetheless, Stakhovskii has proposed that as n → ∞ nearly one hundred percent of all triangles at stage n have largest angle near π and in 1996 Bárány, Beardon and Carne [1] gave an elegant proof of this conjecture. They identify iterated barycentric subdivision with a random walk X n on S L 2 ( ), the triangles at the nth stage of subdivision being similar to the images of ∆ under X n . Their key observation is that the generators of the random walk generate a dense subgroup of S L 2 ( ), and so a theorem of Furstenberg [6] implies that there exists γ > 0 such that for any v ∈ 2 , lim n→∞  [4] with some generalizations) gives a very satisfactory characterization of barycentric subdivision as n → ∞, it has the disadvantage that the Ergodic Theoretic approach does not readily yield information for "finite" n. Recently Diaconis and Miclo [5] have given another formulation using iterated random function Markov chains that allows them to deduce more information about the limiting shape of triangles after many subdivisions, but this approach also does not translate easily to giving bounds for the number of triangles at a given stage having largest angle of a certain size. The strongest approach in this finite direction is due to David Blackwell [2], who has proved (at least computationally) that barycentric subdivision decreases a certain "pseudo-fatness" metric on average. In the following I give a hybrid of the methods of [2] and [1] together with a relatively simple geometric construction, which yields an explicit lower bound tending to 1 for the proportion of triangles in stage n having largest angle in a certain range. This range converges to π exponentially quickly with respect to n; in particular, I obtain analytic upper and lower bounds for the constant γ, which, while not matching, can be made to give γ to arbitary precision by working with later stages of subdivision.
Acknowledgements. I would like to thank Persi Diaconis for introducing me to this problem, for several references, and for his advice in writing this paper. I am also grateful to Persi and his collaborators Curt McMullen and Laurent Miclo for sharing manuscripts with me that are still in progress. I owe my concluding argument with Azuma's inequality to the first referee. Finally, I thank Fai Chandee, Leo Goldmakher and Xiannan Li for helpful comments.

Formalism
Embed ∆ ⊂ 2 with the standard basis, and let ∆ 0 be the equilateral triangle with vertices Let T : 2 → 2 be the affine linear map defined by that carries ∆ 0 onto the uppermost right triangle formed by drawing the medians of ∆ 0 . For w ∈ 2 , Tw = 1 given by reflection in the x-axis and rotation by 2π 3 respectively, then writing 6 , a collection of affine linear maps carrying ∆ 0 onto ∆ 1 0 , . . . , ∆ 6 0 is and A D 6 is the corresponding collection of normalized linear components of these maps. For affine linear map Sw = λBw + v, det B = 1 carrying ∆ 0 onto ∆, linearity of barycentric subdivision implies that the six triangles ∆ i are the images under S of the six triangles ∆ i 0 . In particular, if we let µ be the uniform distribution on A D 6 and define the usual n-fold convolution then, up to dilation and translation, the distribution of triangles appearing in ∆ 0 after n stages of barycentric subdivision is given by Y ∆ 0 , where Y is chosen from S L 2 ( ) according to µ (n) , and the distribution of triangles in ∆ is given by BY ∆ 0 .
The following observations allow us to reformulate statements about barycentric subdivision in terms of operators in S L 2 ( ). For a proof of these calculations see the notes at the end. Observation 1 shows that the difference between π and the largest angle in ∆ where Y is the corresponding operator chosen from µ (n) and s 1 = (0, 1), are unit vectors in the direction of the sides of ∆ 0 . Meanwhile, observation 1 and is sufficiently flat, the distances from π of the largest angles in ∆ i 1 ...i n 0 and (B∆ 0 ) i 1 ...i n are within a factor of π 2 B 2 , so that henceforth we consider only barcentric subdivision of ∆ 0 and restrict our attention to describing the distribution of Y v for v a unit vector and Y chosen from µ (n) . Theorem 1. With µ defined as above, v an arbitrary unit vector in 2 and for any λ > 0 Here the numerical constants 0.05857 and 0.09461 come from solving a certain algebraic equation pertaining to the dilatation of a circle under the operator A, above, while 0.8874 bounds the step size in a random walk. Specializing our result to λ = 2 log n we have

Proof of Theorem
For v an arbitary unit vector in 2 , the walk Y n induces a natural random walk Y n v on 2 . The random difference We show that independent of the direction v n−1 , C ≥ E[Z n ] ≥ c > 0 and deduce the listed bounds for Y n v by the method of bounded differences.
Let w ∈ 2 \ {0} be arbitrary. Given X chosen from A D 6 Dw Dr w Dr 2 w .
It remains to describe the bounded differences argument proving almost certain exponential growth. Recall that Z n = log Y n v − log Y n−1 v is the log-length increment of each step of our walk, and put W n = Z n − E[Z n |v n−1 ]. The sum is a Martingale with increments bounded by and hence by Azuma's inequality is itself a random quantity, but in view of Lemma 2, We recover and in view of the bounds on δ − , δ + , and log A + δ + < 0.8874 this proves the theorem.

Remarks and related problems
Our constants δ − ≈ 0.05857 and δ + ≈ 0.09461 give lower and upper bounds for the upper Lyapunov constant γ := lim n→∞ We could improve our constants by averaging over the linear operators resulting from later stages of subdivision, but, as it stands, a result of type Y n v = exp(γn − o(n)) with probability 1 − o(1) is beyond the limitations of our method. Based upon computer calculations, γ ≈ 0.071. For a nice exposition of growth of random walks in matrix groups and Lyapunov constants see [3]. Barycentric subdivision also produces flat simplices in higher dimension (this is done in [4]). In dimension k the corresponding random walk is on S L k ( ) and the analogue of the dihedral group D 6 is the symmetric group S k+1 . Furstenberg's theorem applies as before, giving lim n 1 n log Y n v = γ > 0. One might again expect to obtain upper and lower bounds for γ by allowing a (k + 1)cycle σ to play the role of the rotation r in Lemma 2, although the explicit calculations become increasingly difficult. Returning to 2 , there remain interesting questions. Put θ n for the maximum of the least angles among triangles at stage n; in our introduction we sketched an argument that θ n ≥ min(θ 0 , 30 • ). Pictures suggest that more is true, e.g. in stage 3 of subdivision of an equilateral triangle we already have sub-triangles that are almost equilateral. One conjectures that θ n → 60 • as n → ∞ and exponentially quickly, although the best argument that I know applies only to subsequences of stages and is polynomial in n. Finally, our theorem shows that after many subdivisions most of the resulting triangles are relatively flat, so that they take on the direction of their longest side. One could ask for the distribution of these sides, as n → ∞, which is closely related to stationary distribution of the random walk on P 1 defined by choosing v 0 ∈ P 1 according to some distribution and putting v n = X n v n−1 , where the X n are chosen independently uniformly from A D 6 . Such a limiting distribution exists independent of the initial distribution by the application of Furstenberg's work [6] in [1] and [4]. Proving anything else about this distribution, e.g. an explicit description, existence of a density, a rate of convergence, however, seems to be a nice and challenging open problem. A Monte Carlo