C. Ernst Department of Mathematics and Statistics, Northern Arizona University Catalan Numbers For n ≥ 0, the Catalan numbers are defined by Cn = 1 n+1 2n n = (2n)! (n+1)!n! The first few Catalan numbers are: 1, 1, 2, 5, 14, 132, 429, 1430, . . . The Symmetric Group The symmetric group, Sym(n), is the collection of bijections from {1, 2, . . . , n} to {1, 2, . . . , n} where the operation is function composition (left ← right). Each element rearranges a string of n objects; called a permutation. One example is taking 1 2 3 and rearranging it to 2 1 3 . In this case, the permutation is “swap first and second.” Representations of Permutations Two ways of representing elements from Sym(n) are via permutation diagrams and via cycle notation. Here are some examples from Sym(5). α = = (1 2 3 4 5) β = = (2 4 3) γ = = (1 5) Let’s try composing. Using diagrams: αβ = = βα = = Using cycle notation: αβ = (1 2 3 4 5)(2 4 3) = (1 2 5) βα = (2 4 3)(1 2 3 4 5) = (1 4 5) We see that compositions of permutations do not necessarily commute (order matters). However, sometimes permutations do commute. βγ = (2 4 3)(1 5) = = γβ = (1 5)(2 4 3) = = So, β and γ commute. Theorem Compositions of disjoint cycles commute. Example Consider the following products in S4 : (1 2)(3 4)(2 3)(1 2)(2 3) and (3 4)(2 3)(1 2). It turns out that these are both expressions for the element (1 4 3 2). This is easily verified by just composing each expression. The Adjacent 2-Cycles In Sym(n), the adjacent 2-cycles are defined as follows. s1 = (1 2), s2 = (2 3), s3 = (3 4), . . . , sn−1 = (n − 1 n) We use s1 to mean ”swap position 1 with position 2”. Similarly, we use si to mean ”swap position i with position i + 1.” The adjacent 2-cycles in Sym(5) are s1, s2, s3 and s4 , or (1 2), (2 3), (3 4), and (4 5). Theorem Every permutation in Sym(n) can be written as a composition of the adjacent 2-cycles. That is, the adjacent 2-cycles generate Sym(n). Remark It is important to note that there are potentially many different ways to express a given permutation as a product of adjacent 2-cycles, but we only need a few tools to get from one expression for a permutation to another. Theorem The symmetric group Sn is generated by the adjacent 2-cycles subject only to the following relations. (si )2 = (i i + 1)2 = (1) (2-cycles have order two) sisj = sjsi → (i i + 1)(j j + 1) = (j j + 1)(i i + 1), where |i − j| > 1 (disjoint cycles commute) sisjsi = sjsisj → (i i + 1)(i + 1 i + 2)(i i + 1) = (i + 1 i + 2)(i i + 1)(i + 1 i + 2), where |i − j| = 1 (braid relation) Picture Proofs sisi = = = (1) sisi = = = = sjsi sisjsi = = = = sjsisj Example Let’s play with an example. s1s3s2s1s2 = s1s3s1s2s1 = s1s3s1s2s1 = s3s1s1s2s1 = s3s1s1s2s1 = s3s2s1 Reduced Expressions If we express a permutation as a product of adjacent 2-cycles in the most efficient way possible (i.e., there is not a way to write the product with fewer factors), then we call the expression a reduced expression. Example There are 11 reduced expressions for (1 3 5 4) that split into 2 “classes”: s1s2s1s4s3 s1s2s4s1s3 s1s4s2s1s3 s1s2s4s3s1 s1s4s2s3s1 s4s1s2s3s1 s4s1s2s1s3 s2s1s2s4s3 s2s1s4s2s3 s2s4s1s2s3 s4s2s1s2s3 Heaps We can visually represent each class using heaps. 1 3 2 4 1 1 3 2 2 4 Unique Heaps Let’s look at an example that has only one heap. (13)(254) = s2s4s1s3s2 can be represented by 2 1 3 2 4 This is special because there is a unique heap for the corresponding element. Theorem The nth Catalan number corresponds to the number of permutations in Sym(n) with unique heaps. These are the permutations with no opportunity to apply a braid relation. Definition The longest element, denoted by w0 , in Sym(n) is the (unique) element having maximal “length”. The string diagram for w0 is of the form (n = 5 here): The number of reduced expressions for w0 is known. Open problem How many heaps does the longest element in the Sym(n) have? For a given n, we could work really hard to figure out the answer (the bigger n is, the harder we’d have to work). But what we want is a general solution. A (good) solution would either be a function of n or a recurrence relation. Example In Sym(4), the longest element is (1 4)(2 3). In this case, there are 8 distinct heaps. 1 1 3 3 2 2 1 1 3 3 2 2 1 1 1 2 2 3 3 1 3 2 2 3 1 1 3 2 2 2 3 3 1 2 2 2 1 1 3 2 2 2 3 3 1 2 2 2 Remark According to the On-Line Encyclopedia of Integer Sequences, the number of heaps of the longest element follows this sequence: 1, 1, 2, 8, 62, 908, 24698, 1232944, 112018190, 18410581880, 5449192389984. What is this sequence?!! It’s not known. Email:
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