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Chapter IV. Determinants Notes ⃝c F.P. Greenleaf 2014 LAI-f14-det.tex version 8/11/2014 Chapter IV. Determinants. IV.1 The Permutation Group Sn. The permutation group Sn consists of all bijections σ :[1,n] [1,n]where[1,n]= 1,...,n ,withcompositionofoperators → { } σ σ (k)=σ (σ (k)) for 1 k n 1 ◦ 2 1 2 ≤ ≤ as the group operation. The identity element e is the identity map id[1,n] such that e(k)=k,forallk [1,n]. We recall that a group is any set G equipped with a binary operation ( )satisfyingthefollowingaxioms:∈ ∗ 1. Associativity: x (y z)=(x y) z; ∗ ∗ ∗ ∗ 2. Identity element: There is an e G such that e x = x = x e,forallx G; ∈ ∗ ∗ ∈ 3. Inverses: Every x G has a “two-sided inverse,” an element x−1 G such that x−1 x = x x−1 =∈e. ∈ ∗ ∗ We do not assume that the system (G, )iscommutative,withx y = y x;agroupwith this extra property is a commutative group∗ ,alsoreferredtoasan∗ abelian∗ group. Here are some examples of familiar groups. 1. The integers (Z, +) become a commutative group when equipped with (+) as the group operation; multiplication ( ) does not make Z agroup.(Why?) · 2. Any vector space equipped with its (+) operation is a commutatve group, for instance (Kn, +); 3. The set (C×, )=C 0 of nonzero complex numbers equipped with complex multiplication· ( )isacommutativegroup;soisthesubset∼{} S1 = z C : z =1 (unit circle in the· complex plane) because z , w =1 zw ={ z∈ w =1and| | } 1/z =1/ z =1. | | | | ⇒| | | |·| | | | | | 4. General Linear Group. The set GL(n, K)= A M(n, K):det(A) =0 of invertible n n matrices is a group when equipped{ with∈ matrix multiply̸ as} the group operation.× It is noncommutative when n 2. Validity of the group axioms for (GL, )followsbecause ≥ · 1 det(AB)=det(A) det(B)det(I)=1 det(A−1)= , · det(A) and a matrix A has a two-sided inverse det(A) =0. Special Linear Group. These properties⇔ of the̸ determinant imply that the subset SL(n, K)= A M(n, K):det(A)=1 equipped with matrix multiply is also a (noncommutative){ ∈ group; } 5. The set of permutations (Per(X), ), the bijections on a set X of n distinct objects, is also a group when equipped with◦ composition ( )asitsproductoperation.No matter what the nature of the objects being permuted,◦ we can restrict attention to permutations of the set of integers [1,n]bylabelingtheoriginalobjects,andthen we have the group Sn. 68 Permuations. The simplest permutations are the k-cycles. 1.1. Definition. An ordered list (i ,....,ik) of k distinct indices in [1,n]= 1,...,n 1 { } determines a k-cycle in Sn,thepermutationthatactsinthefollowingwayontheset X =[1,n]. i i ... ik i (aone-step“cyclicshift”oflistentries) (29) σ maps 1 → 2 → → → 1 ! j j for all j not in the list i1,...,ik → { } A1-cycle(k) is just the identity map idX so we seldom indicate them explicitly, though it is permissible and sometimes quite useful to do so. The support of a k-cycle is the set of entries supp(σ)= i1,...,ik ,innoparticularorder.Thesupportofaone-cycle (k) is the one-point set {k . } { } The order of the entries in the symbol σ =(i1,...,ik)matters,butcyclenotationis ambiguous: k different symbols (i1,...,ik)=(i2,...,ik,i1)=(i3,...,ik,i1,i2)=...=(ik,i1,...,ik−1) obtained by “cyclic shifts” of the list entries in σ;alldescribethesameoperationin Sn.Thusak-cycle might best be descibed by a “cyclic list” of the sort shown below, rather than a linearly ordered list, but such diagrams are a bit cumbersome for the printed page. If we change the cyclic order of the indices we get a new operator. Thus (1, 2, 3) = (2, 3, 1) = (3, 1, 2) =(1, 3, 2) because (1, 2, 3) sends 1 2while(1, 3, 2) sends 1 3. ̸ → → Figure 4.1. Action of the k-cycle σ =(i1,,...,ik)onX = {1, 2,...n}.Pointsℓ not in the “support set” supp(σ)={i1,...,ik} remain fixed; those in the support set are shifted one step clockwise in this cyclically ordered list. This σ is a “1-shift.” (A 2-shift would move points 2 steps in the cyclic order, sending i1 → i3 to . etc. One (cumbersome) way to describe general elements σ Sn employs a data arrray to show where each k [1,n]endsup: ∈ ∈ 123... n σ = " j1 j2 j3 ... jn # More efficient notation is afforded by the fact that every permutation σ can be uniquely written as a product of cycles with disjoint supports, which means that the factors commute. 1.2. Exercise. If σ =(i1,...,ir), τ =(j1,..,js)actondisjoint sets of indices, show that these operators commute. This is no longer true if the sets of indices overlap. Check this by computing the effect of the following products στ(k)=σ(τ(k)) of permutations in S5. 1. (1, 2, 3)(2, 4); 69 2. (2, 4)(1, 2, 3). Is either product a cycle? Thus the order of factors in a product of cycles is irrelevant if the cycles are disjoint. The product of two cycles στ = σ τ is a composition of operators, so the action of στ = σ τ on an element k [1,n]isevaluatedbyfeeding◦ k into the product from the ◦ ∈ right as below. Taking σ =(1, 2), and τ =(1, 2, 3) in S5 we have (1,2,3) (1,2) στ : k (1, 2, 3) k (1, 2) ((1, 2, 3) k) = ((1, 2)(1, 2, 3) k) −→ · −→ · · · To determine the net effect we track what happens to each k: Action Net Effect (1,2,3) (1,2,3) 1 2 111 2 −→ 3 −→ 32→ 3 3 −→ 1 −→ 23→ 2 4 −→ 4 −→ 44→ 4 5 −→ 5 −→ 55→ 5 −→ −→ → Thus the product (1, 2)(1, 2, 3) is equal to (2, 3) = (1)(2, 3)(4)(5), when we include re- dundant 1-cycles. On the other hand (1, 2, 3)(1, 2) = (1, 3) which shows that cycles need not commute if their supports overlap. As another example we have (1, 2, 3, 4)2 =(1, 3)(2, 4) which shows that a power σk of a cycle need not be a cycle, although it is a product of disjoint cycles. We cite without proof the fundamental cycledecompositiontheorem. 1.3. Theorem (Cycle Decomposition of Permutations). Every σ Sn is a product of disjoint cycles. This decomposition is unique (up to order of the commuting∈ factors) if we include the 1-cycles needed to account for all indices k [1,n]. ∈ 1.4. Exercise. Write 123456 σ = " 246513# as a product of disjoint commuting cycles. Hint: Start by tracking 1 2 4 ... until a cycle is completed; then feed σ the first integer not included in→ the previous→ → cycle, etc. 1.5. Exercise. Evaluate the net action of the following products of cycles 1. (1, 2)(1, 3) in S3; 4. (1, 2, 3, 4, 5)(1, 2) in S5; 2 2. (1, 2)(1, 3) in S6 ; 5. (1, 2) in S5; 2 3. (1, 2)(1, 2, 3, 4, 5) in S5; 6. (1, 2, 3) in S5. Write each as a product of disjoint cycles. −1 1.6. Exercise. Determine the inverses σ of the following elements in S5 1. (1, 2); 3. Any 2-cycle (i ,i )withi = i ; 1 2 1 ̸ 2 2. (1, 2, 3); 4. Any k-cycle (i1,...,ik). 1.7. Exercise. Evaluate the following products in Sn as products of disjoint cycles 1. (1, 5)(1, 4)(1, 3)(1, 2); 70 2. (1, 2)(1, 3)(1, 4)(1, 5); 3. (1,k)(1, 2,...,k 1). − 1.8. Exercise. The order o(σ)ofapermutationσ is the smallest integer m 1such that σm = σ ... σ = e. ≥ · · 1. Prove that every k-cycle has order o(σ)=k. th r 2. Verify that the r power σ of a k-cycle σ =(i1,...,ik)isan“r-shift” that moves every entry clockwise r steps in the cyclically ordered list of Figure 4.1. 3. If σ is a 6-cycle its square σ2 = σ σ is a cyclic 2-shift of the entries (i ,...,i ). ◦ 1 6 What is the order of this element in Sn? Hint: By relabeling, it suffices to consider the standard 6-cycle (1, 2, 3, 4, 5, 6) in answer- ing (3.) The only element in Sn of order 1 is the identity e;two-cycleshaveorder2.Asnoted above, in (2.) the powers σr of a k-cycle need not be cycles (but sometimes they are). Parity of a Permutation. In a different direction we note that the 2-cycles (i, j) generate the entire group Sn in the sense that every σ Sn can be written as a product ∈ σ = τ1 ... τr of 2-cycles. However these factors are not necessarily disjoint and need not commute,· · and such decompositions are far from unique since we have, for example, e =(1, 2)2 =(1, 2)4 =(1, 3)2 etc.. Nevertheless an important aspect of such factorizations is unique, namely its parity sgn(σ)=( 1)r − where r =#(2-cyclesinthefactorizationσ = τ1,...,τr). That means the elements σ Sn fall into two disjoint classes: even permutations that can be written as a product of∈ an even number of 2-cycles, and odd permutations.Itisnotobviousthatall2-cycle decompositions of a given permutation have the same parity. We prove that next, and then show how to compute sgn(σ)effectively. We first observe that a decomposition into 2-cycles always exists. By Theorem 1.3 it suffices to show that any k-cycle can be so decomposed.
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