MATH 47A FALL 2008 INTRO TO MATH RESEARCH 1 1. Permutations or: group theory in 15 minutes For those of you who already took a course in group theory, you probably learned about “abstract groups” which are sets with binary operations satisfying a list of conditions. I want to talk about permu- tations. Those who already know group theory can think about the question: Why is every group isomorphic to a permutation group? Definition 1.1. A permutation on a set X is a bijection f : X X. Recall that a bijection is a mapping which is → (1) 1-1 (injective) and (2) onto (surjective). If X = 1, 2, , n , a permutation of X is called a permutation on n letters. {The set· · · of p}ermutations of X will be denoted by P erm(X). I discussed three notations for permutations: 1 2 3 4 (1) σ = means σ(1) = 3, σ(2) = 4, σ(3) = 2, σ(4) = 3 4 2 1 1 or σ!(x) = y where" x is given in the first row and y is given in the second row. (2) (cycle form) σ = (1324). This means σ sends 1 to 3, 3 to 2, 2 to 4 and 4 to 1: 1 3 2 4 1. (3) (graphic notation).→Here→ you→ put→ the numbers x = 1, , n vertically on the right, put y = 1, , n vertically on the· · ·left and connect each x to each y = f· ·(x· ). For example, if f = (1423), you connect 1 on the right to f(1) = 4 on the left with a straight line, etc. In cycle notation, f = (1423). y x f x 1 3 M 1 f MM z MMM zz MM zz MMM zz MMM zz 2 4 MM zz 2 fMM MMz hhhh MM zzMMhhhh MMM zhzhhhMMM MM hhhzh MM hhhMMh zz MM hhhh zMzMM MM 3 2 thh zz MM 3 zz MMM zz MM zz MMM zz MMM zz MM 4 1 } 4 2 NOTES Definition 1.2. If f, g P erm(X), fg = f g is the composition of f and g. This is the perm∈ utation defined by ◦ fg(x) = f(g(x)) It means you do g first and then f. Question: If g = (34) what is gf? Write the answer in all three nota- tions and demonstrate the composition in the notation. (1) Stack up f, g, putting the first operation f on top and the second g underneath. Then cross out the second line: 1 2 3 4 1 2 3 4 4 3 1 2 = 3 4 1 2 3 4 1 2 ! " (2) Write the cycles next to each other and compute the image of each x by applying the cycles one at a time going from right to left: gf = (34)(1423) = (13)(24) for example: gf(1) =3 (34)4(1423)1 = 3 (3) Draw the diagrams next to each other, putting the first permu- tation f on the right: y g f x o 1 3 3 fMM 1 MM zz MMM zz MMM zz MM zz MMM zz 2 4 o 4 fM MM z hh 2 MM MzMz hhhh MMM zzhhMMhMh MM hzhzhh MM MhMhhhz MM hhhh MMzz MM thhhh zzMMM M 3 1 gN 2 zz MM 3 NNN ppp z MM NN pp zz MMM pNpNNp zz MM pp NN zz MMM ppp NNN zz MM 4 2 wp 1 } 4 Note that g = (34) switches the “letters” in locations 3 and 4. We discussed the fact that two of the crossings cancel when we redraw the picture: MATH 47A FALL 2008 INTRO TO MATH RESEARCH 3 y gf x 1 3 iSSS kk 1 SSS kkk SSS kkk SSS kkk SSS kkk SSS kkk SSS kkk 2 4 iSS kkkSSS k 2 SSS kkk SSS kkk SSS kkk SSS kkk SSSkkk SSkSkk kkk SSS kkk SSS kkk SSS kkk SSS kkk SSS kkk SSS 3 1 uk kSkSSk 3 kkk SSS kkk SSS kkk SSS kkk SSS kkk SSS kkk SSS 4 2 ukk S 4 Then we discussed the number of crossings. 4 NOTES 2. Transpositions On Day 2, I tried to formalize things we talked about at the end of Day 1, namely, in what way do the crossings in the diagram give transpositions and: What is the “longest word” ? Definition 2.1. A transposition is a 2-cycle (ab). For example 1 2 3 4 (14) = 4 2 3 1 ! " is a transposition. A simple transposition (also called a simple reflec- tion) is a transposition of two consecutive letters: si := (i, i + 1) In the group of permutations of n + 1 letters there are n simple reflec- tions: s = (12), s = (23), , s = (n, n + 1). 1 2 · · · n We talked about the following vaguely phrased theorem. Theorem 2.2. A permutation f can be written as a product of simple reflections, one for every crossing in the diagram for f. I drew a bunch of diagrams to illustrate this. The first was: y g x 1 3 fM hh 1 MM hhhh MM hhhh MMM hhhh MhMhhh hhhh MM thhhh MMM 2 1 MMM hhh 2 (12) MM hhhh hhhMhM hhhh MM hhhh MMM hhhh MM 3 2 thhh (23) 3 The crossing on the right is s2 = (23). The one on the left if s1 = (12). Reading these from right to left we get: g = (123) = (12)(23) = s1s2. When two crossing lie above each other (so that you don’t know which one is on the left and which is on the right) you can write them in MATH 47A FALL 2008 INTRO TO MATH RESEARCH 5 either order. For example: y (14) x 1 4 1 cFF x FF xx FF xx FF xx FF(12) (12) xx FF xx 2 2 o FF xx 2 FF xx FF xx (23) xxFF xx FF xx FF 3 3 o xx FF 3 xx FF xx(34) (34) FF xx FF xx FF xx FF 4 1 {x 4 This gives: (14) = (12)(34)(23)(12)(34) = s1s3s2s1s3 But the crossings labeled s1 = (12) and s3 = (34) lie above each other. So we can write them in either order: (14) = s3s1s2s3s1 In general, the rule is s s = s s if j i 2 i j j i | − | ≥ Definition 2.3. Suppose that f is a permutation of the letters 1, 2, , n+ 1. Then the length "(f) of f is defined to be the number of pairs· · ·of integers i, j so that (1) 1 i < j n + 1 (2) f(≤i) > f(j≤). In other words, it is the number of pairs of numbers whose order is switched by f. Corollary 2.4. Any permutation f can be written as a product of "(f) simple reflections. Proof. Draw the diagram of the permutation. Move the lines slightly up or down so that there are no triple crossings of lines. Lines i and j will cross if and only if the are in one order on the right (i < j) and in the other order on the left (f(i) > f(j)). Therefore, the number of crossings is equal to "(f) as defined above. The theorem says that each crossing gives one simple reflection and that f is a product of those simple reflections. Therefore, f is a product of "(f) reflections. ! 6 NOTES How large can "(f) be? The maximum possible number occurs when every single pair of in- tegers i < j gets reversed. This is the permutation called w0: 1 2 3 n + 1 w = 0 n + 1 n n 1 · · · 1 ! − · · · " The length of w is the number of all pairs 1 i < j n + 1 which is 0 ≤ ≤ n + 1 n(n + 1) "(w ) = = = 1 + 2 + + n 0 2 2 · · · ! " This is a triangle number. As Tri pointed out in class, w0 can be written as a product of n s ’s, n 1 s ’s, etc. For example, for n = 4, 1 − 2 s4 s3 s3 w0 = s2 s2 s2 s1 s1 s1 s1 = s1s2s1s3s2s4s1s3s2s1 If we want to draw the diagram with 1 at the top and n + 1 = 5 at the bottom, we could also write: s1 s2 s2 w0 = s3 s3 s3 s4 s4 s4 s4 In this method the simple reflection si occurs i times. And this is the method I used on Day 0. There are many different ways to write w0 as n+1 a product of 2 simple reflections. We will study this in detail later. ' ( MATH 47A FALL 2008 INTRO TO MATH RESEARCH 7 3. more group theory Today we talked about the main properties of the elements of a group (although I didn’t define a group yet). I just said that these are all permutations. I changed the format of the class to problem solving, similar to problem sessions, instead of lecturing. 3.1. inverse. Definition 3.1. The inverse of a permutation σ is given by 1 σ− (x) = y where σ(y) = x 1 a) Show that this defines a permutation σ− . b) Find a formula for the inverse of τ = (a1, a2, , ak). 1 1 1 · · · c) Show that (στ)− = τ − σ− . Students found Question (a) confusing so we first did (b) and (c). 3.1.1. inverse of a k-cycle. The inverse of a cycle is given by writing the cycle backwards: 1 τ − = (ak, ak 1, , a2, a1) − · · · This is supposed to obvious, but a proof would go like this: 1 Proof. Let x = ai. Then τ(ai 1) = ai = x.