Combinatorial Species Combinatorial species S is any sort of labelled structure built from a finite set A which does not depend on the names or properties of the elements of A. Let’s see some examples. 2 / 49 Example: Trees Let A = f1; 2; 3; 4g. Members of the species Trees built from A: 3 / 49 Example: Linear Orders Let A = fa; b; c; d; eg. Structures of type Linear Orders put on A: 4 / 49 Example: Partitions Let A = f1; 2; 3; 4g. Members of Partitions constructed from A: 5 / 49 Example: Permutations Let A = f1; 2; 3; 4g. Structures of type Permutations built from A: 6 / 49 Combinatorial Species, again Combinatorial species S is any sort of labelled structure built from a finite set A which does not depend on the names or properties of the elements of A. More concretely, S is a function which sends a finite set A to S(A) the set of all structures of S built from A. For the technocrats: a combinatorial species is a functor S : Fin0 ! Fin0 from the groupoid of finite sets with bijections to itself. 7 / 49 Example: Linear Orders L is the species of Linear Orders. 8 / 49 Example: Permutations P is the species of Permutations. 9 / 49 What’s in a name? Members of a species S built from A “can’t interpret” the names of the elements of A. For example, if A = f1; 2; 3; 4; 5g and B = fa; b; c; d; eg, then Trees “knows” how to use i to transform structures built from A into structures built from B. 10 / 49 To Specify a Species Let [n] = f1; 2; 3;:::; ng. We only need to know S[n] the members of S built from [n] for each n ≥ 0. Say species S and T are equivalent and write S ≈ T if there is a “natural” way to get a correspondence between members of S and T. Technocrats: S and T are equivalent if S and T are naturally isomorphic as functors. 11 / 49 Example The species of Red and Blue Colorings of a Set. The species of Functions to [2]. The species of Subsets. 12 / 49 Counting Members of a Species Write jS[n]j for the number S structures on [n]. If S ≈ T, then jS[n]j = jT[n]j for all n ≥ 0. 13 / 49 Counting Members of a Species Write jS[n]j for the number S structures on [n]. If S ≈ T, then jS[n]j = jT[n]j for all n ≥ 0. Caution: jS[n]j = jT[n]j for each n ≥ 0 does not imply S ≈ T! 14 / 49 Non-Example Linear Orders and Permutations have the same number of members jL[n]j = jP[n]j = n!, but they are not equivalent! Question: How would you show two species were inequivalent? Hint: consider the symmetries of members of the two species. 15 / 49 Generating Function of a Species To any species S we associate the generating function X xn S(x) = jS[n]j : n! n≥0 “S(x) is an algebraic manifestation of the entire species S.” I hope to explain this idea through examples. 16 / 49 Example: Permutations Let P be the species of Permutations. Since jP[n]j = n! we have X xn X xn X 1 P(x) = jP[n]j = n! = xn = n! n! 1 − x n≥0 n≥0 n≥0 1 The same calculation shows L(x) = 1−x when L is the species of Linear Orders. 17 / 49 Example: Cyclic Permutations Let C be the species of Cyclic Permutations. Note that jC[n]j = (n − 1)!. X xn X xn X xn 1 C(x) = jC[n]j = (n − 1)! = = log : n! n! n 1 − x n≥0 n≥0 n≥0 18 / 49 Example: Finite Sets Let Exp be the species of Finite Sets. Then jExp[n]j = 1 for every n. X xn X xn Exp(x) = jExp[n]j = = ex: n! n! n≥0 n≥0 You could view this as a “reason” why ex shows up so often. It is also related to why some people say things like jFinite Setsj = e: 19 / 49 Examples: m Element Sets Let Em be the species of m Element Sets. Then ( 1 m = n jEm[n]j = 0 m 6= n: Hence X xn xm E (x) = jE [n]j = : m m n! m! n≥0 We write X for the species E1 of singletons. 20 / 49 Operations on Species: Sum Let S and T be species. Sum: Structures of S + T on A are either structures of S or of T. (S + T)(A) = S(A) t T(A): 21 / 49 Operations on Species: Product Let S and T be species. Product: S · T is trickier to define. A structure of (S · T)(A) is formed by partitioning A = B t C, then putting an S structure on B and a T structure on C. G (S · T)(A) = S(B) × T(C): A=BtC 22 / 49 Example: Cyclic Permutations · Linear Orders 23 / 49 Ops on Species = Ops on Generating Functions There’s some justice in this world: (S + T)(x) = S(x) + T(x) (S · T)(x) = S(x)T(x): 24 / 49 Example Consider the identity 1 1 − x + x 1 = = 1 + x · : 1 − x 1 − x 1 − x Its fun to try “lifting” algebraic identities to the level of species. ? L ≈ 1 + X · L: 25 / 49 Example Consider the identity 1 1 − x + x 1 = = 1 + x · : 1 − x 1 − x 1 − x Its fun to try “lifting” algebraic identities to the level of species. L ≈ 1 + X · L: “A linear order is either empty or it may be written as a first element followed by a linear order.” 26 / 49 Composition Let S and T be species and suppose T has no structures on the empty set (jT[0]j = 0.) Composition: S ◦ T takes some explaining. To construct a member of (S ◦ T)(A): 1. Partition A = B1 t B2 t ::: t Bk into non-empty sets. 2. Put a T structure on each Bi. 3. Put an S structure on fB1; B2;:::; Bkg . G Y (S ◦ T)(A) = S[k] × T(Bi): A=ti≤kBi i≤k 27 / 49 Example: Cyclic Perms ◦ Trees C and T are the species of Cyclic Permutations and Trees. Let’s build a member of (C ◦ T)[10]: 28 / 49 Example: Cyclic Perms ◦ Trees 29 / 49 Exponentiation Useful to compose with Exp the species of Finite Sets. Members of (Exp ◦ S)(A) have a simpler description: 1. Partition A = B1 t B2 t ::: t Bk into non-empty sets. 2. Put an S structure on each Bi. 30 / 49 Comp of Species = Comp of Generating Functions (S ◦ T)(x) = S(T(x)) (Exp ◦ S)(x) = eS(x): 31 / 49 Example: Exp ◦ C C is the species of Cyclic Permutations. Here is a member of the species (Exp ◦ C)[10]. ) P ≈ Exp ◦ C, Permutations = Exp(Cyclic Permutations) 1 1 = exp log 1 − x 1 − x 32 / 49 Simple graphs A simple graph is a graph with no edges from a point to itself and at most one edge between a pair of points. Question: Can you express Simple Graphs as the composition of two species? Hint: Try counting the number of simple graphs built from [n] and then work backwards. 33 / 49 Bernoulli Polynomials Bernoulli polynomials Bn(q) arise in the study of the Riemann Zeta function, p-adic integration, classic summation formulas, etc. Their coefficients are the mysterious Bernoulli numbers. 1 1 1 1 1 5 691 7 3617 43867 1; − 2 ; 6 ; 0; − 30 ; 0; 42 ; 0; − 30 ; 0; 66 ; 0; − 2730 ; 0; 6 ; 0; − 510 ; 0; 798 ;::: Bn(q) is defined by the following identity: X xn xeqx = (ex − 1) B (q) : n n! n≥0 Question: Can you lift this to a species identity? 34 / 49 Derangements A permutation with no fixed points is called a derangement. Counting derangements of [n] is a classic problem. Let’s count with species! 35 / 49 Derangements Permutations = Exp(Cyclic Permutations) Derangements = Exp(Cyclic Permutations of Size ≥ 2) 1 C (x) = C(x) − x = log − x: ≥2 1 − x log( 1 )−x 1 −x D(x) = e 1−x = e : 1 − x 36 / 49 1 How to multiply a power series by 1−x: P n Say f(x) = n≥0 anx is a power series. 37 / 49 Derangements 1 1 X (−1)n D(x) = e−x = xn 1 − x 1 − x n! n≥0 0 1 0 1 m m n X X (−1) n X X (−1) x = @ A x = @n! A m! m! n! n≥0 m≤n n≥0 m≤n 38 / 49 Derangements 1 1 X (−1)n D(x) = e−x = xn 1 − x 1 − x n! n≥0 0 1 0 1 m m n X X (−1) n X X (−1) x = @ A x = @n! A m! m! n! n≥0 m≤n n≥0 m≤n Therefore, 39 / 49 Differentiation The derivative of S is the species: DS(A) = S(A t {∗}) S structures on A together with a distinguished point ∗. Here are members of DT[4] where T is the species of Trees. 40 / 49 Example: Derivative of Linear Orders Let L be the species of Linear Orders. Example member of DL[5]: 1 < 5 < 2 < ∗ < 4 < 3 Notice there’s a natural correspondence between members of DL and members of L2 = L · L: 1 < 5 < 2 < ∗ < 4 < 3 ! (1 < 5 < 2; 4 < 3) Thus DL ≈ L2.