THE SIZE OF THE SYMMETRIC GROUP NATHAN WILLIAMS 1. Introduction In our study of the descent algebra and the symmetric group, we haven’t seen a lot of geometry. I’d like to fix that by proving (geometrically!) that the symmetric group has n! elements (interpreting those descent classes geometrically). Let Sn be the symmetric group on [n] := f1; 2; : : : ; ng, defined to be the group of bijections [n] ! [n], where the group operation is composition. You probably know that jSnj = n!—for any element w 2 Sn there are n choices for w(1), leaving n − 1 choices for w(2), and so on. We will take a more geometric point of view—one that generalizes to other (crystallographic) reflection groups. We will interpret Sn+1 as an n-dimensional convex polytope P, so that each of the (n + 1)! permutations corresponds to a certain n-dimensional simplex that is similar to ( n ) X A = (x1; x2; : : : ; xn): xi ≥ 0; xi ≤ 1 : i=1 The union of all of these simplices will be the entire polytope P. 1.1. The simplex A. Let’s look at A a little more closely. 1 Proposition 1. Vol(A) = n! . cn Proof. We show by induction on dimension that Vol(cA) = n! . This is true for n−1 n c R c (c−x) (c−x) the line [0; c]. Then (the slices method) Vol(A) = (n−1)! = − n(n−1)! = x=0 x=0 cn n! : Proof. The number of integer points in cA is the number nonnegative integer so- c+n (c+n)···(c+1) lutions to the equation x0 + x1 + x2 + ··· + xn = c, which is n = n! . 1 By Ehrhart theory, Vol(A) is the leading coefficient of this polynomial, or n! . Proof. For each w 2 Sn, define the region Aw = f(x1; x2; : : : ; xn) : 1 > xw(1) > xw(2) > ··· > xw(n) > 0g: In fancier language, the order polytope of an n-element antichain is a parallelepiped of volume one. It has a triangulation given by the linear extensions of the antichain. There are n! of these. You might be unhappy—why was I allowed to change A to Aw for this last proof? First, I don’t care about the boundaries since those have smaller dimension and therefore don’t change the overall volume. Definition 2. For R a “region” with finite volume and T an n×n invertible matrix, Vol(T R) = jdet(T )jVol(R). If jdet(T )j = 1, then the linear transformation T is volume-preserving. 1 2 N. WILLIAMS Ae1 1 Ee6 Ee7 Aen Ben Ee8 Cen Fe4 Den Ge2 Figure 1. The finite and affine Dynkin diagrams (the affine node is marked in gray). Pi Now A is spanned by ei, while Ae is spanned by j=1 ej (we only need consider Ae because it is similar to all the other Aw). So to map A to Ae, we can just send Pn+1−i ei 7! j=1 ej. The matrix of this transformation is the matrix with ones on and above the diagonal, and zeros below the diagonal. Therefore, its determinant is one and the map that sends A to Ae is volume-preserving. Exercise 3. What is the inverse of this matrix? 1.2. Weyl’s formula. Our proof will cut P by the descent classes of Sn+1, and rearrange them so that they form an (n + 1)-fold cover of the parallelepiped Bn = f(x1; x2; : : : ; xn) : 0 ≤ xi ≤ 1g : This implies that (n + 1)Vol(Bn) = Vol(Pn) = jSn+1j · Vol(A) 1 (n + 1) = jS j · ; n+1 n! so that jSn+1j = (n + 1)!. Example 4. Show Mathematica stuff. Theorem 5. Let W be a Weyl group. Then jW j = f · n! · (a1a2 ··· an); where • f is the index of connection; • n! is the volume of an n-dimensional parallelepiped; where • a1; a2; : : : ; an are the coefficients of the highest root αe when expressed as a sum of simple roots (scaling factors for the parallelepiped). The Weyl groups are classified in Figure 1. The proof I want to show you is due to Hermann Weyl, but we’ll do everything explicitly for the symmetric group. 1.3. Eulerian Numbers and Stanley’s Bijection. A related map was used by Stanley to answer a question of Foata, and it is so nice that I’ll give it right away. We will come back to it at the end. Recall that a descent of a permutation w is a position i such that w(i) > w(i+1). If you subdivide the unit n-dimensional cube Bn = f(x1; x2; : : : ; xn) : 0 ≤ xi ≤ 1g THE SIZE OF THE SYMMETRIC GROUP 3 into level sets n X Rn;k := f(x1; x2; : : : ; xn) 2 Bn : k ≤ xi ≤ k + 1g; i=1 the volume of Rn;k is equal to the number of permutations in Sn with k descents. These numbers are called the Eulerian numbers, denoted Vol(Rn;k) = A(n; k). If Pn−1 k we let An(t) := k=0 A(n; k)t be the Eulerian polynomials, then we have 1 X xn t − 1 (1) A (t) = : n n! t − e(t−1)x n=0 Exercise 6. Use Equation (1) to show that 1 X An(t) mntm = : (1 − t)n+1 m=0 The volumes Vol(Rn;k) had been known since Laplace, but it was Foata who gave this combinatorial interpretation of the Eulerian numbers as permutations. Foata asked for a bijective proof: the third proof of Proposition 1 gave a decomposition of the unit cube into permutations—is there a way to rearrange this decomposition (without changing volume) so that the permutations with k descents are sent to the kth level set Rk? Stanley provided a positive solution. Example 7. Draw cubes and explain bijection. Exercise 8. Check that sending a “generic” point (x1; x2; : : : ; xn) in Aw (a point such that xi 6= xi+1 and xn 6= 1) to the point (y1; y2; : : : ; yn) such that ( xi−1 − xi if xi−1 > xi yi := 1 + xi−1 − xi if xi−1 < xi gives the desired map (where x0 = 1). Pn Answer to Exercise 8. We telescope i=1 yi = jdes(w)j + 1 − x1, so the image does indeed lie in the correct level set Rk. The map sends a point x = (x1; x2; : : : ; xn) 2 Aw to the point 0 1 −1 0 ··· 0 0 1 0 1 B 0 1 −1 ··· 0 0 C B C X B . C y = @ eiA + B . ··· . C x: B C i:i2des(w) @ 0 0 0 ··· 1 −1 A 0 0 0 ··· 0 1 Since the determinant of the linear part of this is absolute value one, it is volume- preserving by Definition 2. Note that this matrix is just the inverse of the matrix that we used to prove A and Ae had the same volume! We now invert the map. Starting with y1 = 1 − x1, we obtain y1. In general, we set ( xi−1 − yi if xi−1 > yi xi = 1 + xi−1 − yi if xi−1 < yi; Pi so that 0 < xi < 1. This is the same as xi = 1 − j=1 yj mod 1 4 N. WILLIAMS 1 1 For example, if we have x = 5 (3; 1; 5; 2; 4), then y = 5 (2; 2; 1; 3; 3). And we compute 5x1 = 3 = 5 − 5y5 = 5 − 2 = 3; 5x2 = 5 − (2 + 2) = 1; 5x3 = 5 − (2 + 2 + 1 mod 5) = 5; 5x4 = 5 − (2 + 2 + 1 + 3 mod 5) = 2; 5x5 = 5 − (2 + 2 + 1 + 3 + 3 mod 5) = 4: 2. The Symmetric Group and the Regular Simplex Proposition 9. Sn is isomorphic to the group of symmetries of a regular simplex in Rn−1. Proof. We embed the regular (n − 1)-dimensional simplex into Rn—with basis n feigi=1—as ( n n ) X X ∆n−1 := ciei : ci ≥ 0; ci = 1 : i=1 i=1 n Then the vertices of ∆n−1 are the feigi=1, any permutation of these vertices gives a symmetry of ∆n−1, and these are the only symmetries of ∆n−1. In other words, w 2 Sn acts on the set of vertices of ∆n−1 by w(ei) = ew(i) and therefore on all of Rn by n ! n X X w xiei = xiew(i): i=1 i=1 Example 10. Draw ∆2. n Exercise 11. Check that this representation of Sn on R is not irreducible by Pn showing that any permutation w fixes the space h i=1 eii. Can you write down Pn n generators for the complement of h i=1 eii in R ? Exercise 12. Show that S4 acting on vector space spanned by the six edges heij : 1 ≤ i < j ≤ 4i of ∆3 has the following three invariant subspaces: he12 + e13 + e14 + e23 + e24 + e34i; he12 − e14 − e23 + e34; e13 − e14 − e23 + e24i; and he12 − e34; e13 − e24; e14 − e23i: Can you decompose Sn acting on the (k − 1)-dimensional faces of ∆n−1 into irre- ducible representations? (Hint: expand the product of two homogenous symmetric functions hk · hn−k in the Schur function basis.) Exercise 13. S4 acts by permuting the four long diagonals of a cube. Decompose the corresponding permutation representations on the faces, vertices, and edges of the cube. The point I’m trying to make is that there are lots of “natural” things to act on, and it isn’t clear why the standard permutation representation of Sn ought to be favored! For example, why shouldn’t we just let Sn act on itself? This is called the regular representation.