More Facts about Finite Symmetry Groups

Transcription

1 More Facts about Fiite Symmetry Groups Last time we proved that a fiite symmetry group caot cotai a traslatio or a glide reflectio. We also discovered that i ay fiite symmetry group, either every isometry is direct or there is a equal umber of direct ad opposite isometries. Now we just eed a few more facts before we ca put the pieces together to obtai Leoardo s Theorem. Lemma. I a fiite symmetry group, every rotatio must have the same cetre. Proof. Suppose that our fiite symmetry group has a rotatio R 1 with cetre O 1 ad a rotatio R 2 with cetre O 2. Obviously, our goal is to prove that O 1 = O 2. Cosider the compositio R2 1 R1 1 R 2 R 1 it must be a direct isometry ad, due to the way that rotatios compose together, it must actually be a traslatio. 1 But we already kow that our group ca t cotai traslatios which are t the idetity obviously this meas that the compositio R2 1 R1 1 R 2 R 1 must i fact be the idetity. Therefore, we kow that R2 1 R1 1 R 2 R 1 (O 1 ) = O 1 from which it follows that R2 1 R1 1 R 2 (O 1 ) = O 1. If we apply R 2 to both sides, the R 2 ad the R2 1 o the left had side kock each other out ad the equatio simplifies to R1 1 R 2 (O 1 ) = R 2 (O 1 ). Now we apply R 1 to both sides so that the R 1 ad the R1 1 o the left had side kock each other out ad the equatio simplifies yet agai to R 1 R 2 (O 1 ) = R 2 (O 1 ). Aother way to say the same thig is that R 2 (O 1 ) is a fixed poit of R 1. However, sice R 1 is a rotatio, we kow that it has a uique fixed poit, which we have called O 1. So we ca deduce that R 2 (O 1 ) = O 1. Aother way to say the same thig is that O 1 is a fixed poit of R 2. However, sice R 2 is a rotatio, we kow that it has a uique fixed poit, which we have called O 2. So we ve deduced that O 1 = O 2. A simple cosequece of the previous result is the followig. Lemma. I a fiite symmetry group, every mirror of a reflectio passes through the same poit. Proof. If there is oly oe reflectio i the fiite symmetry group, the there is othig to prove. If there are at least two mirrors, the you ca compose the correspodig reflectios to obtai rotatios. By the previous lemma, we kow that all of these rotatios have the same cetre O. Now suppose that two mirrors meet at a poit P. The the compositio of the correspodig reflectios is a rotatio about P. But we ve already stated that this rotatio, sice it belogs to our fiite symmetry group, must have cetre O. Therefore, the two mirrors must have met at O ad it follows that every mirror passes through O. Oe fial result that we ll eed to use is the followig little lemma. We ll leave the proof of this as a exercise for the ethusiastic reader. Lemma. I a fiite symmetry group, the rotatios are of the form I, R, R 2, R 3,..., R 1 for some rotatio R. 1 Expressios like R2 1 R1 1 R 2 R 1 which take the form a b a 1 b 1 are kow as commutators ad are icredibly useful i group theory ad other areas of mathematics. 1

2 The Proof of Leoardo s Theorem Recall that Leoardo s theorem states that if a subset of the Euclidea plae has fiitely may symmetries, the its symmetry group must be the cyclic group C or the dihedral group D for some positive iteger. Most of the mathematical cotet i the proof of Leoardo s theorem is i the lemmas we proved above. All we eed to do ow is fit the pieces of the puzzle together as follows. Proof. Suppose that someoe gives you a fiite symmetry group. By the lemma proved earlier, we kow that there are o traslatios or are there glide reflectios. I other words, every elemet of the fiite symmetry group must be a rotatio or a reflectio. If there are oly rotatios, the the last lemma above guaratees that the fiite symmetry group must be C for some positive iteger. If there are oly reflectios as well as the idetity, the the fiite symmetry group must be D 1. This is because there must be a equal umber of direct ad opposite isometries. So we re left with the case that there are both rotatios ad reflectios, i which case there must be equal umbers of each. 2 The first lemma above states that all of the rotatios must share the same cetre O while the secod lemma above states that all of the mirrors of reflectios must pass through O. We kow that the rotatios are evely spaced by the previous lemma, so the rotatios must be by the agles 0 360,, ,..., ( 1), for some positive iteger. If the mirrors were ot evely spaced as well, the there must be two of them which create a agle strictly less tha 180. The compositio of the reflectios through these two mirrors will the yield a rotatio of strictly less tha 360 a cotradictio. Therefore, we ca coclude that the rotatios are equally spaced with cetre O ad that the mirrors are equally spaced ad pass through O. But this just meas that the fiite symmetry group is dihedral. Frieze Patters Let s tur our attetio ow to ifiite symmetry groups. Sice we ve already see that a fiite group of symmetries ca t cotai a traslatio, a easy way to create a ifiite symmetry group is to cosider a sigle traslatio T. A symmetry group which cotais T ecessarily cotais the ifiitely may elemets 3..., T 3, T 2, T 1, I, T, T 2, T 3,.... Keepig this i mid, we defie a frieze patter to be a subset of the Euclidea plae whose symmetry group cotais a horizotal traslatio T alog with ad o other traslatios...., T 3, T 2, T 1, I, T, T 2, T 3,..., 2 Here, we re cosiderig the idetity as a rotatio through a agle of zero. 3 Here, the otatio T stads for the compositio of copies of T, while T stads for the compositio of copies of T 1. 2

3 A frieze patter ca certaily have other symmetries which are t traslatios, but there are t too may possibilities for such other symmetries. Actually, it should t be too difficult to see that they ca oly come i four types. H = a reflectio i a horizotal mirror V = a reflectio i a vertical mirror R = a 180 rotatio G = a glide reflectio alog a horizotal axis So for each frieze patter, we ca give it a HVRG symbol, depedig o which of these four symmetries it possesses. For example, a frieze patter with symbol VRG would have symmetry group which cotais a reflectio i a vertical mirror, a 180 rotatio ad a glide reflectio alog a horizotal axis, but ot a reflectio i a horizotal mirror. If we decide to classify frieze patters i this way, the we obtai the followig result. Theorem. There are exactly seve types of frieze patter. Proof. There are at most sixtee possible HVRG symbols. oe, H, V, R, G, HV, HR, HG, VR, VG, RG, VRG, HRG, HVG, HVR, HVRG However, the followig four observatios allows us to exclude some of these examples. If you have H, the you have G. If you have V ad R, the you have G. If you have R ad G, the you have V. If you have G ad V, the you have R. You ca prove the first of these observatios, for example, by composig H ad T to obtai G. The other three facts ca be proved i a similar way. If you go ahead ad use these observatios to elimiate some of the possibilities, the you should fid that there are oly seve remaiig. oe, V, R, G, HG, VRG HVRG Of course, this oly shows that there are at most seve types of frieze. We still have to demostrate that each of these possibilities actually arises, which we accomplish i the followig diagrams. oe HG V VRG R G 3 HVRG

4 The mathematicia Joh Coway who ofte has his ow spi o certai mathematical theorems ad proofs has coied his ow set of ames for the types of frieze patters. HOP (oe) SIDLE (V) SPINNING HOP (R) STEP (G) JUMP (HG) SPINNING SIDLE (VRG) SPINNING JUMP (HVRG) The followig diagrams should hopefully explai Coway s rather strage omeclature for the frieze patters. 4

5 The Crystallographic Restrictio Cosider applyig two traslatios S ad T i differet directios to a poit P. By takig various combiatios of S ad T, we obtai a set of poits i the plae. Ay set of poits i the plae which you ca obtai i this way is called a lattice. You ca thik of a lattice as a equally spaced orchard of apple trees. Of course, the traslatios S ad T are symmetries of the lattice, as is ay combiatio ivolvig S ad T. Aother symmetry of the lattice is a rotatio by 180 about oe of the lattice poits. However, it could be possible that there are other rotatioal symmetries, although there is some restrictio as to what those rotatioal symmetries could be. I order to state the theorem more explicitly, let s defie the order of a rotatio R to be the smallest positive iteger such that R = I. Theorem (Crystallographic Restrictio). If the symmetry group of a lattice cotais a rotatio, the that rotatio must have order 2, 3, 4 or 6. Proof. First, we state a simple fact, which the ethusiastic reader is ecouraged to prove o their ow. Ay group which cotais a rotatio of order also cotais a rotatio by agle 360. Let s call a poit i the plae special if it s the cetre of a rotatioal symmetry of the lattice of order. Pick ay special poit O ad let P be the closest special poit to it. By the fact stated above, a rotatio by 360 about ay special poit must be a symmetry of the lattice. But, furthermore, ay such rotatio must be a symmetry of the set of special poits. So if we rotate O about P by 360, we obtai a poit Q which is special. For 7, this poit Q is closer to O tha P, which cotradicts our assumptio. So we ca deduce that there is o rotatioal symmetry of the lattice of order greater tha or equal to 7. Q Q P 360 O P O R If = 5, the we rotate P about Q by 360 to obtai aother special poit R. However, this poit R is ow closer to O tha P, which cotradicts our assumptio. So we ca deduce that there is o rotatioal symmetry of the lattice of order 5. Hece, we may coclude that ay rotatioal symmetry of the lattice must have order 2, 3, 4 or 6. This result derives its ame from the fact that it also holds for three-dimesioal lattices. I that case, the statemet is importat i crystallography the study of crystals ad guaratees that ay rotatioal symmetry of a crystal must have order 2, 3, 4 or 6. It would be ice if the theorem also held i greater tha three dimesios, but it just is t true. This is because rotatios i higher dimesios behave i a very differet way tha i two or three dimesios. 5

6 Wallpaper Patters If a subset of the Euclidea plae has a symmetry group whose traslatios form a lattice, the we call that subset a wallpaper patter. Simply put, they re patters that cover the whole plae ad are repetitive i two differet directios. Just as we did for friezes, we ca try to classify the types of wallpaper patters however, the game is much harder this time. It turs out that there are exactly sevetee differet types of wallpaper patters, although we wo t provide a proof here. We ll merely cotet ourselves with kowig how to idetify them. To each wallpaper patter, we associate a RMG symbol cosistig of the followig three umbers R, M ad G. R = the maximum order of a rotatioal symmetry M = the maximum umber of mirrors which pass through a poit G = the maximum umber of proper glide axes which pass through a poit By a proper glide axis, we mea the axis of a glide reflectio which is ot itself a mirror. The sevetee possible types of wallpaper patters are pictured below. For each wallpaper patter, make sure that you ca fid a cetre of a rotatioal symmetry of order R, a poit through which M mirrors pass, ad a poit through which G proper glide axes pass. Ufortuately, the RMG symbol does t quite distiguish all sevetee wallpaper patters there are two which are described by 332. So let s call them 332A ad 332B ad ote that they ca be distiguished usig the followig observatio. For the wallpaper patter 332A, it s possible to fid a cetre of rotatio which does t lie o a mirror. For the wallpaper patter 332B, every cetre of rotatio lies o a mirror

7 A 332B 7

8 Problems Problem. Prove that if a symmetry group of a frieze patter cotais a reflectio i a vertical mirror ad a rotatio by 180, the it must also cotai a glide reflectio. Proof. If we deote the reflectio i a vertical mirror by V ad the rotatio by 180 by R, the the symmetry group of the frieze patter must also cotai R V. This compositio is a opposite isometry ad it is easy to see that it is either reflectio i a horizotal mirror or a glide reflectio alog a horizotal axis. (Remember that the cetre of rotatio for R does ot have to lie o the mirror for V.) Sice the symmetry group of a frieze patter cotais a horizotal traslatio by defiitio, i either case, the symmetry group of the frieze patter must cotai a glide reflectio alog a horizotal axis. 8

9 So how do you kow that the compositio R V is a reflectio i a horizotal mirror or a glide reflectio alog a horizotal axis? Oe way to see this is to cosider what happes to the picture of a left footprit walkig right, above the cetre of rotatio of R. After applyig the reflectio V, this becomes a right footprit walkig left, above the cetre of rotatio of R. Ad the after applyig R, this becomes a right footprit walkig right below the cetre of rotatio of R. The oly way to start with a left footprit walkig right ad ed up wit ha right footprit walkig right is via a reflectio i a horizotal mirror or a glide reflectio alog a horizotal axis. Problem. For each wallpaper patter pictured above, if it has symbol RMG, fid o the diagram a poit which is the cetre of a rotatioal symmetry of order R, a poit where M mirrors meet, ad a poit where G proper glide axes meet. Proof. For example, the diagram below shows the wallpaper patter with RMG symbol equal to 423 with a poit which is the cetre of a rotatioal symmetry of order 4, a poit which 2 mirrors pass through, ad a poit which 3 proper glide axes pass through. You ca do somethig similar for the other wallpaper patters. 9

10 Abel Earlier we spoke about Galois, a mathematicia who proved that it was impossibile to write dow a formula to solve the quitic equatio, before dyig at the teder age of twety. I fact, the first perso to give a complete proof of this beat Galois by several years. He was a Norwegia mathematicia by the ame of Niels Herik Abel ad he was bor i 1802 before dyig i 1829, livig a whole six years loger tha Galois. Abel s proof of the impossibility of solvig the quitic equatio was ot as deep ad far-reachig as Galois proof, but was oetheless extremely ovel. Ufortuately, his work was extremely difficult to read, partly because he had to cut out all of the details to save moey o pritig. His success i mathematics gave him some fiaces to travel aroud Europe ad meet some of the more well-kow mathematicias. Ufortuately, he cotracted tuberculosis while i Paris ad became quite ill. Durig this time, a fried of his had foud a prestigious professorship for Abel i Berli ad wrote him a letter to tell him the good ews. Ufortuately, the letter arrived two days after Abel died. The early death of this extremely taleted mathematicia cut short a career of extraordiary brilliace ad promise. Abel had maaged to clear some of the prevailig obscurities of mathematics ad paved the way for several ew fields. His complete works were edited ad evetually published by the Norwegia govermet. The adjective abelia is derived from Abel s ame ad is so commoplace i mathematics that mathematicias do t eve bother to capitalise it ay more. Abel is still a relatively big deal i Norway there have bee stamps, bakotes ad cois bearig his portrait; there is a crater o the Moo amed after him; ad there is the prestigious Abel prize for mathematicias, preseted by the Kig of Norway ad worth about a millio dollars i prize moey. 10