Burnsides Orbit Counting Lemma Drew Johnson November 17, 2013 Drew - PowerPoint PPT Presentation

Burnside’s Orbit Counting Lemma Drew Johnson November 17, 2013 Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 1 / 23

Motivating Example How many ways is there to fill a tic-tac-toe board with 5 “X”s and 4 “O”s? For example: X O X X O O X X O Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 2 / 23

Motivating Example How many ways is there to fill a tic-tac-toe board with 5 “X”s and 4 “O”s? For example: X O X X O O X X O � 9 = 9 · 8 · 7 · 6 � Answer: = 126 4 4! Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 2 / 23

Motivating Example But we may want to compute the answer up to symmetry, i.e. we wish to consider X O X X O O X X O to be the same as X O X O O X O X X Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 3 / 23

Group actions Definition A group of symmetries acting on a set S is a collection G of bijections from S to itself satisfying 1 id : S → S is in G 2 If g ∈ G then g − 1 ∈ G. 3 If g , h ∈ G, then g ◦ h ∈ G. Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 4 / 23

Group actions Definition A group of symmetries acting on a set S is a collection G of bijections from S to itself satisfying 1 id : S → S is in G 2 If g ∈ G then g − 1 ∈ G. 3 If g , h ∈ G, then g ◦ h ∈ G. Definition The fixed points of a group element g ∈ G are Fix( g ) = { s ∈ S : g ( s ) = s } Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 4 / 23

Group actions Definition A group of symmetries acting on a set S is a collection G of bijections from S to itself satisfying 1 id : S → S is in G 2 If g ∈ G then g − 1 ∈ G. 3 If g , h ∈ G, then g ◦ h ∈ G. Definition The fixed points of a group element g ∈ G are Fix( g ) = { s ∈ S : g ( s ) = s } Definition The orbit of an element s ∈ S is Gs := { g ( s ) : g ∈ G } ⊂ S Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 4 / 23

Back to our example Let S be the set of all 126 possible filled tic-tac-toe boards. Let G be the set with 8 elements: Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 5 / 23

Back to our example Let S be the set of all 126 possible filled tic-tac-toe boards. Let G be the set with 8 elements: The identity Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 5 / 23

Back to our example Let S be the set of all 126 possible filled tic-tac-toe boards. Let G be the set with 8 elements: The identity A quarter turn. Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 5 / 23

Back to our example Let S be the set of all 126 possible filled tic-tac-toe boards. Let G be the set with 8 elements: The identity A quarter turn. A half turn. Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 5 / 23

Back to our example Let S be the set of all 126 possible filled tic-tac-toe boards. Let G be the set with 8 elements: The identity A quarter turn. A half turn. A three quarters turn. Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 5 / 23

Back to our example Let S be the set of all 126 possible filled tic-tac-toe boards. Let G be the set with 8 elements: The identity A quarter turn. A half turn. A three quarters turn. A horizontal reflection. Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 5 / 23

Back to our example Let S be the set of all 126 possible filled tic-tac-toe boards. Let G be the set with 8 elements: The identity A quarter turn. A half turn. A three quarters turn. A horizontal reflection. A vertical reflection. Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 5 / 23

Back to our example Let S be the set of all 126 possible filled tic-tac-toe boards. Let G be the set with 8 elements: The identity A quarter turn. A half turn. A three quarters turn. A horizontal reflection. A vertical reflection. Two diagonal reflections. Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 5 / 23

Back to our example Let S be the set of all 126 possible filled tic-tac-toe boards. Let G be the set with 8 elements: The identity A quarter turn. A half turn. A three quarters turn. A horizontal reflection. A vertical reflection. Two diagonal reflections. One can check that this is a group action. Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 5 / 23

Example of an orbit X O X The boards O O X O X X O O X X O O X X X X X O X O O X O X X X X O O X X O O X O X O O X O X X X O X X O O Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 6 / 23 X X O

But not all orbits are the same size: X O X O X O X O X Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 7 / 23

But not all orbits are the same size: X O X O X O This orbit has only one element. X O X Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 7 / 23

The Main Result Our question can be rephrased as “How many orbits are there?” Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 8 / 23

The Main Result Our question can be rephrased as “How many orbits are there?” The answer is given by Theorem (Burnside’s Lemma) The number of orbits is equal to the average number of fixed points of elements of G, i.e. 1 � # of orbits = | Fix( g ) | | G | g ∈ G Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 8 / 23

Let’s count fixed points Everything is fixed by the identity: 126. Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 9 / 23

Let’s count fixed points Everything is fixed by the identity: 126. For a quarter rotation, (and a three quarter rotation), we must have A B A B C B A B A Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 9 / 23

Let’s count fixed points Everything is fixed by the identity: 126. For a quarter rotation, (and a three quarter rotation), we must have A B A B C B A B A AAAABBBBC: since there are an odd number of Xs, we must have C=X. Then we have two choices: either A = X and B = O or vise versa. Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 9 / 23

Let’s count fixed points Everything is fixed by the identity: 126. For a quarter rotation, (and a three quarter rotation), we must have A B A B C B A B A AAAABBBBC: since there are an odd number of Xs, we must have C=X. Then we have two choices: either A = X and B = O or vise versa. X O X O X O O X O X X X X O X O X O Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 9 / 23

Half Turn A B C D E D C B A AABBCCDDE, so we must pick E=X and two of A, B, C, or D to be X, so � 4 � = 6 choices. 2 Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 10 / 23

Vertical (or horizontal) reflections A D A B E B C F C AABBCCDEF, so we can either: Pick X to be one of D,E,F and two of A,B,C: 3 × 3 = 9 choices Pick D,E,F to all be X, and then one of A,B,C to be X: 3 choices. So we have 12 fixed points. Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 11 / 23

Finally, diagonal reflections For the diagonal reflections, we have A B D C E B F C A AABBCCDEF, so it is again 12. Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 12 / 23

Finally, diagonal reflections For the diagonal reflections, we have A B D C E B F C A AABBCCDEF, so it is again 12. Now we can compute 1 8(126 + 2 · 2 + 6 + 4 · 12) = 23 Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 12 / 23

Coins Problem Suppose you have two indistinguishable coins with indistinguishable sides. You have k colors of paint and can paint each side of a coin a single color. How many different things can you do? (Answer will be a function of k.) Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 13 / 23

Coins Problem Suppose you have two indistinguishable coins with indistinguishable sides. You have k colors of paint and can paint each side of a coin a single color. How many different things can you do? (Answer will be a function of k.) S is the set with k 4 elements consisting of all pairs of painted coins if you remember left/right/top/bottom. There are 2 3 = 8 symmetries— you can swap the coins, and flip each one. Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 13 / 23

Coins id – fixes k 4 . 2 ways to flip one – fixes k 3 (flipped coin must be the same color on both sides). swap the left and right– fixes k 2 (tops and the bottoms must be the same color). swap and then flip both – fixes k 2 (tops and the bottoms must be the same color). flip both – fixes k 2 (each coin must be all one color). 2 ways to swap and then flip one – fixes k (all sides must be the same color). Drew Johnson Burnside’s Orbit Counting Lemma November 17, 2013 14 / 23