INDIRECT, IS Exp Scissors Congruence: An Introduction Gregory Leibon Department of Mathematics Dartmouth College Scissors Congruence: An Introduction – p.1/109 Scissors Congruence For this talk, let Gn 2 fEn; Sn; Hng Scissors Congruence: An Introduction – p.2/109 Scissors Congruence Let P and Q be geodesic polyhedra in Gn (finite sided, compact). We say that P is scissors congruent to Q (P ∼ Q) provided we can chop P up into a finite number of pieces and reassemble these pieces to form Q. Scissors Congruence: An Introduction – p.3/109 Example: P ∼ Mirror(P ) Scissors Congruence: An Introduction – p.4/109 Example: P ∼ Mirror(P ) Scissors Congruence: An Introduction – p.5/109 Example: P ∼ Mirror(P ) Scissors Congruence: An Introduction – p.6/109 Example: P ∼ Mirror(P ) Scissors Congruence: An Introduction – p.7/109 Example: P ∼ Mirror(P ) Scissors Congruence: An Introduction – p.8/109 Example: P ∼ Mirror(P ) Scissors Congruence: An Introduction – p.9/109 Example: P ∼ Mirror(P ) Scissors Congruence: An Introduction – p.10/109 Example: P ∼ Mirror(P ) Scissors Congruence: An Introduction – p.11/109 Example: P ∼ Mirror(P ) Scissors Congruence: An Introduction – p.12/109 Example: P ∼ Mirror(P ) Scissors Congruence: An Introduction – p.13/109 Example: P ∼ Mirror(P ) Scissors Congruence: An Introduction – p.14/109 Example: P ∼ Mirror(P ) Scissors Congruence: An Introduction – p.15/109 Example: P ∼ Mirror(P ) Scissors Congruence: An Introduction – p.16/109 Example: P ∼ Mirror(P ) Scissors Congruence: An Introduction – p.17/109 Example: P ∼ Mirror(P ) Scissors Congruence: An Introduction – p.18/109 Example: P ∼ Mirror(P ) Scissors Congruence: An Introduction – p.19/109 Some Notation Here we see a polygon A. 0 0 1 2 3 1 2 3 Scissors Congruence: An Introduction – p.20/109 Some Notation Here we see a second polygon C. 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.21/109 Some Notation We let P F Q be P S Q provided int(P T Q) = ;: Here we see A F C. 0 0 1 2 3 4 4 1 5 6 7 7 8 2 5 9 3 8 9 6 Scissors Congruence: An Introduction – p.22/109 Some Notation Now we take a new view of A F C. We start with this polygon B, 0 0 4 5 6 4 5 6 Scissors Congruence: An Introduction – p.23/109 Some Notation and here is D. 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.24/109 Some Notation Here we have B F D. 0 0 4 5 6 4 1 1 2 3 7 7 8 2 5 9 3 8 9 6 Scissors Congruence: An Introduction – p.25/109 Some Notation Note A F C ∼ B F D, since they are equal. 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.26/109 Some Notation Further more C ∼ D since they differ by a translation. 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.27/109 Zylev's Theorem n Zylev's Theorem asserts that in G , if A F C ∼ B F D and C ∼ D; then A ∼ B: 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.28/109 Realizing Zylev's Theorem To prove it we divide the region using our correspondences. Here we pair upC and D. 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.29/109 Realizing Zylev's Theorem Here we pair up A F C and B F D. 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.30/109 Realizing Zylev's Theorem Using our pairings, let us "subtract". Note the 0 regions are paired via A F C ∼ B F D. 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.31/109 Realizing Zylev's Theorem Note 1 can be viewed in D via A F C ∼ B F D. 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.32/109 Realizing Zylev's Theorem From C ∼ D, 1 can be identi®ed with 4. 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.33/109 Realizing Zylev's Theorem Now 4 can be viewed in B via A F C ∼ B F D. 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.34/109 Realizing Zylev's Theorem Hence we have identi®ed 1 and 4. 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.35/109 Realizing Zylev's Theorem Similarly we can do the same for 2. 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.36/109 Realizing Zylev's Theorem This process is called "subtraction"... 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.37/109 Realizing Zylev's Theorem and is a way to construct new bijection from old. 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.38/109 Realizing Zylev's Theorem The key is that by moving to the right using A F C ∼ B F D.... 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.39/109 Realizing Zylev's Theorem and to the left using C ∼ D.... 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.40/109 Realizing Zylev's Theorem that we can never get caught in loop. 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.41/109 Realizing Zylev's Theorem The reason is simple namely each node has one entering path from A F C ∼ B F D.... 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.42/109 Realizing Zylev's Theorem and one exiting path from C ∼ D. 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.43/109 Realizing Zylev's Theorem Hence there is no way to enter such a loop staring in A.... 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.44/109 Realizing Zylev's Theorem and... 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.45/109 Realizing Zylev's Theorem and... 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.46/109 Realizing Zylev's Theorem and... Similarly 0 0 0 1 4 2 5 3 6 4 4 1 1 5 2 6 3 7 7 7 8 8 2 5 9 9 3 8 9 6 Scissors Congruence: An Introduction – p.47/109 Realizing Zylev's Theorem we must land in B.