Pentagramma Mirificum old wine into new wineskins

Transcription

1 Pentagramma Mirificum old wine into new wineskins R. Schwartz and S. Tabachnikov (arxiv: ; Math. Intelligencer, to appear) Lyon, November 2009, Claude s B-Day 1

2 Three classical theorems, Pappus ( 300 A.D.), Pascal (1639) and Briançon (1806): 2

3 Concerning the subtitle, a New Testament parable, in Luke s version, reads: And no one puts new wine into old wineskins; otherwise the new wine will burst the skins and will be spilled, and the skins will be destroyed. But new wine must be put into fresh wineskins. And no one after drinking old wine desires new wine, but says, The old is good. We discovered eight (new?) configuration theorems of projective geometry ( old wine ) by computer experimentation ( new wineskins ); most of the proofs are also by a computer algebra system. And we believe, this old (wine) is good. 3

4 A polygon P RP 2 is a cyclically ordered collection {p 1,..., p n } of points, its vertices. The sides {l 1,..., l n } are the lines l i = p i p i+1 in RP 2. The dual polygon P (RP 2 ) has vertices {l 1,..., l n }; the sides of the dual polygon are {p 1,..., p n } (considered as lines in (RP 2 ). One has: (P ) = P. Let P n and P n be the sets of n-gons in RP 2 and (RP 2 ). Define the k-diagonal map T k : P n P n: for P = {p 1,..., p n }, let T k (P ) = {p 1 p k+1, p 2 p k+2,..., p n p k+n }. Then T k is an involution. The map T 1 is the projective duality. Extend the notation: T ab = T a T b, T abc = T a T b T c, and so on. 4

5 Theorem 1: (i) If P is an inscribed 6-gon, then P T 2 (P ). (ii) If P is an inscribed 7-gon, then P T 212 (P ). (iii) If P is an inscribed 8-gon, then P T (P ). (and it doesn t continue!) 5

6 6

7 Theorem 2: If P is a circumscribed 9-gon, then P T 313 (P ). Theorem 3: If P is an inscribed 12-gon, then P T (P ). 7

8 Theorem 4: (i) If P is an inscribed 8-gon, then T 3 (P ) is circumscribed. (ii) If P is an inscribed 10-gon, then T 313 (P ) is circumscribed. (iii)* If P is an inscribed 12-gon, then T (P ) is circumscribed. (and it doesn t continue either!) 8

9 9

10 Odds and Ends (i) The Pentagram Map: T 12. Studied by R. Schwartz in The pentagram map, Experim. Math. 1 (1992), 71 81; The pentagram map is recurrent, Experim. Math. 10 (2001), ; Discrete monodromy, pentagrams, and the method of condensation, J. Fixed Point Th. Appl. 3 (2008), ; and by V. Ovsienko, R. Schwartz and S. Tabachnikov in Quasiperiodic motion for the pentagram map, E.R.A. 16, 2009, 1 8; The pentagram map: a discrete integrable system, arxiv:

11 (ii) Pentagons: 1) Every pentagon is inscribed and circumscribed. 2) Every pentagon is projectively equivalent to its dual. 3) The pentagram map is the identity: T 12 (P ) = P. Thus 5-gons can be added to Theorem 1. C.-F. Gauss studied self-polar pentagons in: 11

12 (iii) Palindroms: in all the theorems, the words w are palindromic and the maps T w are involutions. (iv) Self-dual polygons: are classified in D. Fuchs, S. Tabachnikov. Self-dual polygons and self-dual curves, Funct. Anal. & Other Math., 2 (2009), For odd n, the space of projective equivalence classes is n 3- dimensional, just like the moduli space of inscribed (or circumscribed) n-gons. Theorem 1 (ii) can be rephrased: P is an inscribed heptagon iff T 2 (P ) is projectively self-dual. Theorem 2 can be also rephrased: P is an inscribed nonagon iff T 3 (P ) is projectively self-dual. 12

13 (v) Relabeling: Relabel the vertices of a dodecagon as follows: i 5i. Then Theorem 4 (iii), which reads: If P is an inscribed dodecagon then T (P ) is also inscribed, transforms to: If P is an inscribed dodecagon then T (P ) is also inscribed. And so on. (vi) A similar theorem: R. Schwartz has a theorem in Discrete monodromy, pentagrams, and the method of condensation that can be rephrased as follows: If P is a 4n-gon inscribed into a degenerate conic then (T 1 T 2 T 1 T 2... T 1 )(P ) (4n 3 terms) is also inscribed into a degenerate conic. 13

14 The Pentagram Map Two spaces: P n, space of closed n-gons; dim = 2n 8; T n, space of twisted n-gons; dim = 2n: φ : Z RP 2 s.t. φ(k + n) = M φ(k); k. M is the monodromy. Twisted polygons φ 1 and φ 2 are equivalent if there is Ψ P GL(3, R) such that φ 2 = Ψ φ 1. Then M 2 = ΨM 1 Ψ 1. Main Theorem: The Pentagram Map on T n is completely integrable: 1). There are 2[n/2] + 2 algebraically independent integrals. 2). There is an invariant Poisson structure of corank 2 if n is odd, and corank 4 if n is even, such that the integrals Poisson commute. 14

15 Corner coordinates: left and right cross-ratios x 1, y 1,..., x n, y n. v i 2 v i+2 v i 1 v i v i+1 Cross-ratio: [t 1, t 2, t 3, t 4 ] = (t 1 t 2 ) (t 3 t 4 ) (t 1 t 3 ) (t 2 t 4 ). 15

16 The map T 12 (x, y) = (x, y ) is given by: x i = x i 1 x i 1 y i 1 1 x i+1 y i+1, y i = y i+1 1 x i+2 y i+2 1 x i y i. Two consequences: 1). Hidden scaling symmetry (x 1, y 1,..., x n, y n ) (tx 1, t 1 y 1,..., tx n, t 1 y n ) commutes with the map. 16

17 2). Easy invariants: O n = and, for even n, O n/2 = i even x i + n i=1 x i, E n = n i=1 y i, x i, E n/2 = y i + i odd i even i odd y i. These are the Casimir functions of the Poisson bracket. 17

18 Monodromy invariants: O 2 ne n (Tr M) 3 det M = [n/2] k=1 are polynomials in (x i, y i ), decomposed into homogeneous components; likewise, for E k with M 1 replacing M (with negative weights). O k They are algebraically independent. Poisson bracket: {x i, x i+1 } = x i x i+1, {y i, y i+1 } = y i y i+1, and the rest = 0. The Jacobi identity is automatic. 18

19 Inscribed polygons Computer experiments suggest: if a (twisted) polygon is inscribed into a conic then E k = O k for all k (and the same for circumscribed polygons). Why?? Theorem: the space of inscribed n-gons is a coisotropic subspace of the Poisson manifold T n. This leads to a Poisson mapping of the moduli space of twisted n-gons in RP 1 which takes the left corner coordinates x i to the right ones y i (this is also given by rational functions). Conjecture: this map is completely integrable for each n. 19

20 Continuous limit Object of study: the space P of non-degenerate twisted parameterized curves in RP 2 modulo projective equivalence: Lift so that Then γ(x + 1) = M(γ(x)). Γ(x) Γ (x) Γ (x) = 1. Γ (x) + u(x) Γ (x) + v(x) Γ(x) = 0. Thus P identifies with the space of linear differential operators on R: ( ) d 3 A = + u(x) d dx dx + v(x), where u and v are smooth 1-periodic functions. 20

21 Rewrite as A = ( d dx ) ( u(x) d dx + d ) dx u(x) + w(x) where w(x) = v(x) u (x) 2 (sum of a skew-symmetric and zeroorder symmetric operators). The functions u and w are projective-differential invariants of the curve γ ( projective curvature and projective length element ). Want to study P n P as n. 21

22 Construction:!(x)!(x+")!(x-")! (x) " It turns out that u ε = u + ε 2 ũ + (ε 3 ), w ε = w + ε 2 w + (ε 3 ), giving the flow: u = ũ, ẇ = w. A computation reveals : u = w, ẇ = u u 3 u 12, 22

23 or ü + ( u 2 ) 6 + u(iv ) 12 = 0, the Boussinesq equation! The continuous limit of the scaling is: u(x) u(x), w(x) w(x) + t. 23

24 Happy Birthday, Claude! 24