groups & symmetries notes

8/10/2019 Groups & Symmetries Notes

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Groups and Symmetries January - April 2013

Lecturer: Dr. Benjamin Doyon

Contents

1 Lecture 1 2

2 Lecture 2 3

3 Lecture 3 4

4 Lecture 4 7

5 Lecture 5 8

6 Lecture 6 11

7 Lecture 7 13

8 Lecture 8 17

9 Lecture 9 18

10 Lecture 10 23

11 Lecture 11 24

12 Lecture 12 27

13 Lecture 13 29

14 Lecture 14 30

15 Lecture 15 30

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16 Lecture 16 32

17 Lecture 17 36

18 Lecture 18 40

19 Lecture 19 42

20 Lecture 20 43

21 To be adjusted 45

1 Lecture 1

Definition:A group is a set Gwith

Closure: a product G GG denoted g1g2; that is, g1g2G for all g1, g2G.

Associativity: g1(g2g3) = (g1g2)g3 for all g1, g2, g3G

Identity: there exits a eG such that ge= eg = g for all gG

Inverse: for all g there exists a g1 G such that gg1 =g1g= e.

Easy consequences (simple proofs omitted):

The identity is unique

The inverse is unique

(g1g2)1 =g12 g11 (g1)1 =g.

Definition:The order of a group Gis the number of elements ofG, denoted|G|.Definition: Ifg1g2 = g2g1 for all g1, g2G, then G is a commutative, or abelian, group.

Definition:A subset HG is a subgroup ofG if

h1h2H for all h1, h2H

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eH

h1 H for all hH.

Easy consequence (simple proof omitted): a subgroup is itself a group.

Cyclic groups

Notation: gn (nZ) is then-times product ofg for n >0, the n-time product ofg1 forn , is the set of all elements h ofGof the form h = gn for some n Z. This can be written as

< g >={gn :n Z, gn G}={gn :n Z}.

(this is a set: elements dont have multiplicities).

Note: < g > is in fact a subgroup ofG.

Definition: A group G is called cyclic if there exists a g G such that G =< g >. Such anelement is called a generating element for G.

2 Lecture 2

Easy observation: a cyclic group is abelian.

Let us consider for a while groups of finite order only, that is,|G|,and g |G| =e for all gG.

Proof. SinceG is cyclic, there is ag0 such thatG =< g0>. First,gn0 , n= 0, 1, 2, . . . , |G| 1 are

all distinct elements. Indeed, if this were not true, say they were the same for n1 and n2, then

we would have that gn2n10 =e. Denote q= n2 n1 there are only q , a contradiction; and ifg|G|0 = gn0 for somen= 1, . . . , |G| 1, then we would have the situation that we had above with n2 n1= q


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Theorem 2.2

Every subgroup of a cyclic group is cyclic.

Proof. LetHG =< a >={e,a,a2, . . . , a|G|1}be a subgroup. Let qbe the smallest positiveinteger such that aq H. Consider c = an H, and write an = akq+r for k integer andr = 0, . . . , q 1. Since H is a subgroup, then akq H, so that cakq H. Hence,ar H.This is a contradiction unless r = 0. Hence, c= akq = (aq)k. That is, we have H=< aq >, soHis cyclic.

Examples of cyclic groups:

the integers Z =1. {e2ik/n : k = 0, 1, . . . , n 1} under multiplications for n a positive integer, the group is

given bye2i/n. the integers modulo n, Zn ={0, 1, 2, . . . , n 1}under addition modulo n, Zn =1(note:

same notation as for Z, but here the multiplication law is different).

3 Lecture 3

Symbols and relations, and another example of a cyclic group.

Consider an alphabet of two letters: {e, a}. Let us consider all words that we can form fromthese: e, a,ea, aaa2, a3e, aea, etc. This set of words does not quit form a group, but it is aset with a multiplication law: the concatenationof words, e.g. a multiplied with ea gives aea.

The multiplication law is automatically associative (easy to check).

Let us now impose some relations. The trivial ones that we will always impose when looking at

symbols and relations are those having to do with the identity element: e2 =e and ea = ae = a.

That reduces the words we can form: ea = a, aea = a2, etc. It also guarantees that we now

have a set with an associative multiplication law and an identity element, the word e.

The words we have now are e, a, a2, a3, a4, etc. Let us impose one more relation: an = e for

some fixed positive integer n (e.g. forn = 4). This unique additional relation reduces further

the number of words we can make. Take n = 4 for instance. We now have e, a, a2, a3 and

nothing else (any other word reduces to one of these four). This now forms a group: we can

check that every element has an inverse a1 = a3, (a2)1 = a2, and (a3)1 = a. It is a cylic

group, generated by a, so it is the group < a >.

Maps

Consider two sets X, Y and a map f : X Y. We write y = f(x) for the value in Y that ismapped from xX.

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Definition:y is the image ofxunder f.

Definition:f(X) ={yY :x|y = f(x)} Y is the image ofXunderf.

Definition:The map fis onto (or surjective) if every y in Y is the image of at least one x in

X, i.e. iff(X) =Y (denoted f :X Y).

Definition: The map f is one-to-one (or injective) if for all y f(X) there exists a uniquexXsuch thaty = f(x). That is, if the following proposition holds:f(x1) =f(x2)x1= x2.

Definition:A map fthat is one-to-one and onto is called bijective

Observation: iff is bijective, then there is a unique correspondance betweenXand Y, and the

inverse map f1 : Y X exists and is itself bijective (easy proof omitted). The inverse mapf1 is defined by f1(y) =x iff(x) =y.

Given two maps f, g from X to X, we can form a third by compsition: h = f g given byh(x) = f(g(x)). Consider the set M ap(X, X) all such maps. Then: 1) if f and g are such

maps, then f g also is; 2) given f ,g ,h M ap(X, X), we have that (f g) h = f (g h);3) M ap(X, X) contains the identity id(x) = x for all x X. Hence we only need one moreproperty, the existence of inverse, to have a group.

Theorem 3.1

The set of all bijective maps of a finite set X forms a group under compositions, called the

permutation group,Perm(X).

Proof. Just need to check inverse. Since we have bijectivity, then f1 exists and is bijective

itself by the comments above. Hence f1 is an element of the group. It has the property that

f1

f=f f1

= id because f1

(f(x)) is the element x

such that f(x

) =f(x), hence suchthat x =x by bijectivity off, and because f(f1(x)) is f(x) where x is such that f(x) =x.

IfX is a finite set, label by integers 1, 2, . . . , n. An element Perm(X) is a mapping kik fork= 1, 2, . . . , n, where all integers ik are distinct (so that the map is bijective).

Definition:The permutation group ofnobjects is called the symmetric group Sn.

We can denote elements ofSn by

1 2 ni1 i2 in

. The product is easy to work out in this

notation.

S3 has 6 elements: e the identity, a the shift to the right by one, b the inversion

1 2 3

3 2 1

,

and then the shift to the right by 2 (or equiv. to the left by 1) a2, the inversion plus shift to the

right by 1 ab and by 2 a2b. It is easy to see

a3 =e, b2 =e, ab= ba2, a2b= ba.

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Hence, we have S3 ={e,a,a2,b,ab,a2b} subject to these relations. So we can say thatS3 isgenerated bya and b, and we writeS3 =< a, b >(with in mind that a and b satisfy the relations

above).

Note: S3 is nonabelian.

Isomorphisms

Definition:A map f :G1 G2 is an isomorphism if it is bijective and if it satisfies f(gh) =f(g)f(h) for all g , hG1 (in other words: fpreserves the multiplication rule).

Definition:Two groupsG1andG2 are isomorphic if there exists an isomorphism f :G1G2.

Theorem 3.2

The isomorphic relation is an equivalence relation.

Proof. We need to check that the relation is reflective, symmetric and transitive. Reflexive: G1

is isomorphic toG1 because the identity map id is an isomorphism. Symmetric: iff :G1

G2

is an isomorphism, than so is f1 : G2 G1. Indeed f1 is a bijection. Further, let g =f(h), g = f(h) G2 with h, h G1 (h and h always exist because f is a bijection). Thenf1(gg) =f1(f(h)f(h)) =f1(f(hh)) where in the last step we have used that f preserves

the multiplication law. Then this equals hh = f1(g)f1(g) by definition of f1, so f1

preserves the multiplication law. Finally, transitive: letf andfbe isomorphism. Certainlyffis bijective. Alsof f(gg) =f(f(gg )) =f(f(g)f(g)) =f(f(g))f(f(g)) =f f(g) f f(g)sof fpreserves the multiplication law.

We have:

f(e1) =e2: f(g) =f(ge1) =f(g)f(e1) hence f(e1) =f(g)1f(g) =e2.

f(g1) =f(g)1: e2= f(gg1) =f(g)f(g1) hence f(g1) =f(g)1.

Theorem 3.3

Two cyclic groups of the same order are isomorphic

Proof. (proof done in class).

Example

Let G1 ={1, 1} under multiplication of integers, and G2 = Perm({1, 2}) = 1 21 2 , 1 22 1 .Consider the function f(1) =

1 2

1 2

and f(1) =

1 2

2 1

. The map is clearly bijec-

tive. Also,f((1) (1)) = f(1) =

1 2

1 2

=

1 2

2 1

1 2

2 1

= f(1)f(1). Also,

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f(1) =

1 2

1 2

hence f maps identity to identity. This is sufficient to verify that f is an

isomorphism.

Example

Let G1 = Zn and G2 ={e2ik/n

: k = 0, 1, . . . , n 1}. Consider the map f : G1 G2 givenbyf(k) =e2ik/n, k= 0, 1, 2, . . . , n 1. It is well-defined: since we restrict to k between 0 andn 1, they are all different integers when taken mod n. It is surjective: just by definition, weget all e2ik/n : k = 0, 1, . . . , n 1. It is injective: iff(k) = f(k) then e2pii(kk)/n = 1 hencekk = 0 modnso thatk = k modn. Further, we havef(k+k mod n) =e2i(k+k mod n)/n =e2i(k+k

)/n =e2ik/ne2ik/n =f(k)f(k).

Cosets and Lagranges theorem

Let Hbe a subgroup ofG, define equivalence relation a

b iffab1

H. It is an equivalence

relation because:

reflexive: aa

symmetric: abba

transitive: ab, bc ac.

(simple proofs omitted).

The equivalence class ofa is denoted [a] :=

{b

G : a

b

}.

Definition:The set of such classes is the set of right cosets ofGwith respect to H.

We have [a] =H a. Clearly, by definition, ifab then [a] = [b].

4 Lecture 4

Theorem 4.1

Two right cosets ofG with respect to Hare either disjoint or identical.

Proof. Leta, bG. If [a] and [b] have no element in common, then they are disjoint. Ifc[a]and c[b], then then ac and bc, hence ab hence [a] = [b].

Note: the latter is true for equivalent classes in general, not just cosets of groups.

Theorem 4.2

All cosets ofGw.r.t. Hhave the same number of elements.

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Proof. Consider the map Ma : Ha H defined by b ba1. It is a bijection. Indeed, ifMa(b) =h then b = ha is unique and exists.

Definition:the number of cosets ofG wrt H is called the index ofH inG, which we will denote

i(H, G).

Theorem 4.3

(Lagrange Theorem) Let Hbe a subgroup ofG. The order of H divides the order ofG. More

precisely,|G|=|H| i(H, G).

Proof. The right cosets ofG w.r.t. Hdivide the group intoi(H, G) disjoint sets with all exactly

|H|elements.

Definition:A proper subgroup of a group G is a subgroup Hthat is different from the trivial

group{e}and from G itself.

Definition:The order of an elementa in a group G is the smallest positive integer k such that

a

k

=e.

Note: ifH is a subgroup ofGand|H|=|G|, then H=G.

Corollary (i): If|G| is prime, then the group G has no proper subgroup.

Proof. IfH is a proper subgroup ofG, then|H| divides|G| and|H| is a number not equal to 1or|G|. Contradiction.

Corollary (ii): Let aG and let k be the order ofa. Then k divides|G|.

Proof. Let us look at the cyclic subgroup < a > ofG generated by a. This has order k. Hence

k divides|G|.Corollary (iii): If|G| is prime then G is a cyclic group.

Proof. Given any a G, consider the subgroup < a >. Since|G| is prime, it has no propersubgroup. Since the order of< a > is greater than 1, < a > must be G itself.

Hence: Any group of prime order is unique up to isomorphisms.

5 Lecture 5

Example

Examples of cosets. Take the group G = S3 ={e,a,a2,b,ab,a2b} with the relations writtenpreviously, a3 =e, b2 = e, ab=ba2, a2b = ba. Take the subgroup H={e,a,a2}. Let uscalculate all the right cosets associated to this subgroup. We have

He= H, Ha={a, a2, a3 =e}= H, Ha2 ={a2, a3 =e, a4 =a}= H

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and

Hb={b,ab,a2b}, Hab={ab,a2b, a3b= b}= Hb, Ha2b={a2b, a3b= b, a4b= a2b}= H b.

Hence we find that any two cosets Hg1 and Hg2 for g1, g2 G are either disjoint or identical.We find that there are exactly 2 different cosets occurring, which are the sets HandH b. This is

in agreement with Lagranges theorem, because|G|= 6,|H|= 3 so that the number of differentcosets should be i(H, G) =|G|/|H|= 2, which it is.

Example

Consider S3 ={e,a,a2,b,ab,a2b} with the relations as before. Consider the subgroup H ={e, b} =< b >. We haveHa ={a, a2b}, Ha2 ={a2, ab}. We have 3 cosets, each 2 elements,total 6 elements.

Groups of low order

|G|= 1: G={e}.

|G| = 2: G ={e, a} with a= e. It must be that a2 = a or a2 = e. In the former case,a2a1 =a so that e=a, a contradiction. Hence only the latter case is possible: a2 =e.

The group is called Z2. Here, we could simply have used the results above: 2 is prime,

henceG must be cyclic, G =< a > with a2 =e.

|G| = 3: since 3 is prime, G must be cyclic, so we can always write G ={e,a,a2} witha3 =e.

|G|= 4: G ={e,a,b,c} (all distinct). < a > is a subgroup, so its order divide 4. Hencethere are 2 or 4 elements in< a >. Hencea2 =e or a4 =e. In the latter case, < a >=G,

soG is cyclic. Let us assume thatG is not cyclic, in order to see what other group we can

have. Soa2 =e. We can then do the same for b and c, so we have a2 =b2 =c2 =e. Then

considerab, and check if associativity holds.

1. ab= a: (a2)b= b anda(ab) =a2 =e contradiction.

2. ab= b: similarly, a = e contradition.

3. ab= e: a2b= a

b = a contradiction.

Hence if group exists, must have ab = c. Similarly, must have ba = c, bc = cb = a and

ca= ac = b. To show existence of the group, must check associativity in all possible triple

products abc, a2b, etc. (left as exercise).

The above arguments show:

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Theorem 5.1

Every group of order 4 is either cyclic, or has the rules (this is a Cayley table)

e a b c

e e a b c

a a e c b

b b c e ac c b a e

Definition:The group with the rules above is denoted V4, and called Klein four-group (Vier-

ergruppe).

Notes:

Both the cyclic group and the group V4are abelian. Hence there are no non-abelian groupsof order less then or equal to 4.

The group V4 is the smallest non-cylic group.

V4 is such that all elements different form ehave order 2.

V4 has 5 subgroups: the trivial one and V4 itself, as well as the 3 proper subgroups< a >,< b > and < c >.

V4 can be seen as a subgroup ofS4:

e, 1 2 3 4

2 1 4 3 ,

1 2 3 4

3 4 1 2 ,

1 2 3 4

4 3 2 1 .

V4 can also be described using symbols and relation. It is generated by the symbolsa, b,with the relations a2 =b2 =e and ab = ba. We have V4=< a,b >={e,a,b,ab}.

Example

Consider the 2 2 matrices

e=

1 0

0 1

, a=

1 0

0 1

, b=

1 0

0 1

, c=

1 0

0 1

and the product rules on these matrices given by the usual matrix multiplication. These formthe group V4.

Example

Take V4 ={e,a,b,ab} with the relations shown above. A (cyclic) subgroup is of courseH ={e, a} =< a >. One coset isH = Ha, the other is Hb ={b,ab}. They have no element in

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r and t are positive integers. The smallest values for t and r are: t= all prime factors ofp not

included in q, and r= all prime factors ofqnot included in p. Ifp and q are relatively prime,

then we find t= p and r=q. Hence, n= pq. That is, the subgroup ofZpZq hasorder pq=|ZpZq|. Hence we have found Zp Zq =: it is cyclic, hence isomorphicto Zpq. For the opposite proposition, we show that if p and q are not relatively prime, then

Zp

Zq is not isomorphic to Zpq. In the argument above, in place of (a, b) we take an arbitrary

element of the formc = (av, bw) forv, w0, and look for some n >0 such that cn =e. Clearly,(av)p =e and (aw)q =e. Hence the argument above witha replaced byav andb replaced by bw

shows that we can taket < p and r < q, so thatn < pq. This means that any element ofZpZqhas order less than pq. But in Zpq there is at least one element that has order pq. Hence, we do

not have an isomorphism.

Matrices, symmetry transformations and dihedral groups

Let us consider the natural transformations of the Euclidean plane: translations, rotations and

reflections. Consider the subset of the plane formed by a circle centred at the origin and of

radius 1:{(x, y) :x2 + y2 = 1}. (1)

What are its symmetries? The only transformations that preserve this circle are the rotations

w.r.t. the origin, and the reflections w.r.t. any axis passing by the origin. We can describe these

by matrices, acting on the coordinates

x

y

simply by matrix multiplication. Rotations are

A =

cos sin

sin cos

.

For the reflections, one is that through the x axis (i.e. inversing the y coordinate):

B=

1 0

0 1

.

Reflections through any other axis can be obtained from these two transformations: if we want

reflection through the axis that is at angle , we just need to construct

B :=ABA

(i.e. first rotate the angle- axis to the x axis by a rotation A, then do a reflection, then

rotate back by an angle ). Hence the set of all symmetry transformations is{

A :

[0, 2)}{ABA :[0, )}(note that we only need to consider axes of reflection of angles between 0

and , because a rotation of an axis gives back the same axis). In general, a multiple rotation

lead to a single rotation with the sum of the angles:

AA =A+

(check by matrix multiplication).

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In order to check mathematically that some matrix multiplication gives rise to a symmetry of

the subset (1), we can proceed as follows: we start with the expression for transformed subset,

and make a change of variable. Notice thatA,B and ABA all are orthogonal matrices, i.e.

matrices M satisfyingMTM=M MT =I. Let us then consider such an orthogonal matrix for

the transformation:

M

xy

R2 :x2 + y2 = 1

=

M

xy

R2 :

xy

T xy

= 1

=

x

y

R2 :

MT

x

y

TMT

x

y

= 1

=

x

y

R2 :

x

y

TM MT

x

y

= 1

=

x

y

R2 : x

y

T

x

y

= 1

=

x

y

R2 : (x)2 + (y)2 = 1

Hence we get back to the set (1). We will discuss later on what group this form. But for now,

let us consider simpler groups formed by subsets of the transformations discussed here. Let us

consider, instead of the circle, less symmetric objects: polygons.

Definition: Let n 2 be an integer. The set of rotations and reflections that preserve theregular polygon Pn formed by successively joining the points

cos

2kn

sin

2k

n

, k= 0, 1, 2 . . . , n 1

is called the dihedral group Dn. This is the symmetry group of this polygon.

Note that the case n= 2 does not quite form a polygon - it is rather just a segment, and it is

not actually closed. But it does fit well in the considerations below, so it is appropriate to add

it here.

7 Lecture 7

n= 2. From geometric considerations, the symmetries of the segment are the rotations byangle 0 (the identity) and , as well as the reflections w.r.t. axes at angles 0 and /2, i.e.

thex and y axes. These are the matrices, in the notation A (rotation by an angle) and

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B (reflection w.r.t. the axis at angle from the x axis, with B0= B) introduced above:

1=

1 0

0 1

, B=

1 0

0 1

, B/2=

1 0

0 1

, A =

1 0

0 1

To show that, e.g., A is a symmetry, we proceed as follows: the segment is described by

{(x, y)

R2 :y = 0,

1< x


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Conjugations and normal subgroups

Definition:Given a group G, we say that a is conjugate to b if there exists a gG such thata= gbg1.

Theorem 7.1

The conjugacy relation is an equivalence relation.

Proof. (i) a is conj. to itself: a= eae1. (ii) Ifa is conj. tob, then a= gbg1 b = g1ag =g1a(g1)1 henceb is conj. to a. (iii) Ifa is conj. to b and b is conj. to c, then a = gbg1 and

b= g c(g)1 (for some g, g G), hence a= ggc(g)1g1 =gg c(gg)1.

Hence, the group G is divided into disjoint conjugacy classes, [a] ={gag1 : g G}. Theseclasses cover the whole group,aG[a] =G, hence form a partition ofG. Further remarks:

[e] ={e}. Hence no other class is a subgroup (i.e. e[a] for any a=e).

All elements of a conjugacy class have the same order. Because ( gag1)n =gag1 gag1

gag1

(n factors) = gang1 (using g1g = e). Hence if an = e (i.e. a has order n), then

(gag1)n =geg1 =e. Hence alsogag1 has order n.

IfG is abelian, then [a] ={a}for all aG.

Definition:A subgroup His called normal or invariant ifgHg1 H for all gG. (note: asusual,gHg1 ={ghg1 :hH}). That is,His normal if for everyhHand everygG, wehave g hg1 H.

Notes:

A simple consequence of normality is that gHg1 =H. Indeed, we have HgH g1: forevery hH, there is a h =g1hgH (by normality) such that ghg1 =h.

Let H be a normal subgroup. Ifh H then [h] H. That is,H is composed of entireconjugacy classes. In fact: a subgroup is normal if and only if the subgroup is the union

of conjugacy classes.

{e}is a normal subgroup.

Every subgroup of an abelian group is normal.

Definition:A group is simple if it has no proper normal subgroup. A group is semi-simple if

it has no proper abelian normal subgroup.

Definition: The center Z(G) of a group G is the set of all elements which commute with all

elements ofG:

Z(G) ={aG : ag = gagG}

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Theorem 7.2

The center Z(G) of a group is a normal subgroup.

Proof. Subgroup: closure: let a, b Z(G) and g G then abg = agb = gab hence ab Z(G).identity: e Z(G). inverse: leta Z(G) and g G then ag1 = g1a hence ga1 = a1ghencea1 Z(G). Normal: letaZ(G) and gG then gag1 =gg1a= aZ(G).

Note: IfG is simple, then Z(G) ={e} or Z(G) =G.

Quotients

LetG be a group and Ha subgroup ofG.

Definition:A left-coset ofG with respect to His a class [a]L ={bG : a1bH}for aG.That is, [a]L = aH. (Compare with right-cosets same principle: a

1bHis an equivalencerelation between a and b).

Hence we have two types of cosets (right and left), with two types of equivalence relations. We

will denote the equivalence relations respectively byR andL, and the classes by [a]R=H aand [a]L= aH.

Definition: The quotient G/H={[a]L : aG} is the set of all left-cosets; that is, the set ofequivalence classes underL. The quotient H\G={[a]R] :aG}is the set of all right-cosets;that is, the set of equivalence classes underR.

We only discuss G/H, but a similar discussion hold forH\G.

Given two subsets A and B ofG, we define the multiplication law AB ={ab: aA, bB}.

Theorem 7.3IfHis normal, then the quotient G/H, with the multiplication law on subsets, is a group.

Proof. We need to check 4 things.

Closure: We have [g1]L[g2]L =g1Hg2H= g1g2g12 Hg2H =g1g2HHwhere we used H =g12 Hg2 (because H is normal). Since e H (because H is a subgroup), we have thatHH H. Clearly HH H by closure. Hence HH = H. Hence we find [g1]L[g2]L =g1g2Hso that

[g1]L[g2]L = [g1g2]L.

Associativity: This follows immediately from associativity ofG and the relation found inthe previous point.

Identity: Similarly it follows that [e]L is an identity.

Inverse: Similarly it follows that [g1]L is the inverse of [g]L.

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We callG/Hthe (left-)quotient group ofGwith respect to H.

Homomorphisms

Definition: Let G1 and G2 be groups. A map : G1 G2 is a homomorphism if(gg ) =(g)(g) for all g , g G1.

Note:

(e1) =e2 (g1) =(g)1

Example

HomomorphismS2S3:

1 21 2

1 2 3

1 2 3 , 1 2

2 1

1 2 3

2 1 3

Note: this is not an isomorphism though S2 and S3 are not isomorphic (in particular, they

dont even have the same number of elements).

A homomorphism which is bijective is an isomorphism.

8 Lecture 8

Definition:Letbe a homomorphism ofG1 ontoG2. Then the kernel of1 is

ker1 ={gG1: (g) =e2}

Note: e1ker1.

Theorem 8.1

The kernel is a normal subgroup.

Proof. Let : G G and H = ker G. We have: 1) if h1, h2 H then (h1h2) =(h1)(h2) = ee = e hence h1h2 H; 2) (e) = e hence eH; 3) ifhH then (h1) =(h)1 = e hence h1 H. Hence H is a subgroup. Also: if h G and g G then(ghg1) =(g)(h)(g1) =(gg1) =(e) =e hence ghg1 H. Hence His normal.

Theorem 8.2

The image of a homomorphism : G1G2 is a subgroup ofG2.

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Proof. Let g2, g2 Im G2. Closure: g2 = (g1), g2 = (g1) with g1, g1 G1. So g2g2 =

(g1)(g1) =(g1g

1)Im. Identity: we know that(e1) = e2, hence e2 Im. Inverse: we

know that (g1)1 =(g11 ). We have g2= (g1) sog

12 =(g1)

1 =(g11 )Im.

Note: Imin general is notnormal.

The homomorphism theorem

Theorem 8.3

(Homomorphism theorem) Let : GG be a homomorphism. Then, G/ker=im.

Proof. Suppose without loss of generality that is onto G, i.e. im= G. Let H= ker (this

is a normal subgroup of G) and G = G/H. Let us first find a homomorphism : G G.Define (gH) = (g). This is well defined because if g1H = g2H (recall, g1H and g2H are

either equal or disjoint) then g1L g2 hence g1 =g2h for some hHhence (g1) =(g2h) =(g2)(h) = (g2) because H is the kernel of . Also, is a homomorphism: (gHg

H) =

(gg H) =(gg ) =(g)(g) = (gH)(gH).

to be continued...

9 Lecture 9

...continuation

Second, we show that is bijective. It is clearly surjective (onto G), because is onto G:

the set of all left-cosets is the set of all gH for all g G, and ({gH : g G}) = (G) = G.

Hence we only need to show injectivity. Suppose (gH) = (g

H). Then(g) =(g

). Hence(g)1(g) =e hence (g)1(g) =e hence (g1g) =e hence g1g H hence g =g h forsome hH, hence gLg , so that gH=g H.

Example

Take S3 ={e,a,a2,b,ab,a2b} (with the usual relations). Choose H ={e,a,a2}. Check thatit is a normal subgroup. Clearly it is a subgroup, as it is H =< a >. We need to check that

gH g1 H; it is sufficient to check for g = b,ab,a2b (i.e. g H). We have bab1 = bab =a2bb = a2 H, and ab a (ab)1 = abab1a1 = ababa2 = aa2bba2 = a3a2 = a2 H anda2b a (a2b)1 = a2bab1(a2)1 = a2baba= a2a2bba= a4a = a5 = a2 H. Using these results,the rest follows: ba2b1 = (bab1)2 = a4 = a H, etc. Hence H is normal. Interestingly,we also obtain from these calculations the conjugacy classes of a and a2 (because the other

things to calculate are trivial since H is abelian: aaa1 =a, etc.): we have [a]C ={a, a2} and[a2]C={a, a2}. Along with [e]C={e}, we see indeed thatH= [e]C [a]Cso it contains wholeconjugacy classes.

We have two left-cosets: H= [e]L andbH= [b]L={b,ab,a2b}. Hence,S3/Hhas two elements,

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[e]L and [b]L. Explicitly multiplying these subsets we find:

[e]L[e]L= [e]L, [e]L[b]L= [b]L, [b]L[b]L= [e]L

Indeed, this forms a group, and is in agreement with the relation found (using b2 = e). We

have [e]L = [a]L = [a2]L and [b]L = [ab]L = [a

2b]L. The multiplication law is also in agreement

with other choices of representatives, because e.g. [a]L[ab]L= [a2

b]L and [ab]L[ab]L = [abab]L =[a3b2]L = [e]L, etc. In the end, we find that the multiplication law is that ofZ2, i.e.

S3/H= Z2

Example

Consider the above example again. We have the following homomorphism:: S3={e,a,a2,b,ab,a2b} Z2 =

{e, b

}given by (an) = e and (anb) = b. This is a homomorphism: (anam) = e =

(an)(am), (anbamb) = (anmb2) = (anm) = e = b2 = (anb)(amb), (anbam) =

(anmb) = b = be = (anb)(am) and likewise (anamb) = b = (an)(amb). We see the

following: given the group H, we have found a homomorphism : S3S3/H(onto) such thatker= H. (And put differently, we also see that, as in the previous example, S3/ker= im.)

Example

Let R be the nonzero reals. This is a group under multiplication of real numbers (e = 1,

x1 = 1/x). Let Z2 be the group{1, 1} under multiplication. Let R+ be the group of positivereal numbers (it is a normal subgroup ofR, but this doesnt matter). Define : R

R

+

(onto) by (x) =|x|. This is a homomorphism: (xx) =|xx|=|x| |x|= (x)(x). Its kernelis ker ={x R :|x| = 1} ={1, 1} = Z2. Hence Z2 is a normal subgroup ofR. Further,let us calculate R/Z2. This is the set{xZ2 : x R} ={{x, x} : x R} ={{x, x} :x R+}. That is, it is the set of pairs of number and its negative, and each pair can becompletely characterised by a positive real number. These pairs form a group (the quotient

group):{x, x} {x, x}={xx, xx}.

We note that there is an isomorphism between R/Z2 and R+. Indeed, define : R/Z2 R+

as the bijective map ({x, x}) =|x| (for x R). It is clearly onto, and it is injectivebecause given a value of

|x|> 0, there is a unique pair

{x,

x}

. Also, it is a homomorphism:

({x, x} {x, x}) =({xx, xx}) =|xx|=|x| |x|= ({x, x} {x, x}). Hence, we havefound that R/Z2= R+, that is,

R/ker=im

(where im= (G) is the image of).

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Example

not done in class

Consider the group

K2=

1 0

: C, C

L2=

1 0

1

: C

We can check that K2 is a group, and that L2 is a normal subgroup ofK2.

K2 is a group: 1) closure: 1 0

1 0

=

()1 0

1 +

2) associativity: immediate from matrix multiplication;

3) identity: choose = 1 and = 0;4) inverse:

1 0

1=

0

1

(just check from the multiplication law above).

L2 is a subgroup: just choose = 1, this is a subset that is preserve under multiplication (check

multiplication law above), that contains the identity (= 0) and that contain inverse of every

element (check form of inverse above).

L2 is a normal subgroup: the argument is simple: under the multiplication rule, elements of

the diagonal get multiplied directly. Hence, with element g =

1 0

K2 and h =

1 0

1

L2, we have that ghg1 is a matrix with on the diagonal 11 = 1 and

1 1 = 1, hence matrix of the form

1 0 1

L2.

So, we can form the quotient group K2/L2: this is the group of left-cosets, under element-wise

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multiplication of left-cosets. The set of left cosets is

{gL2: gK2} =

1 0

1 0

1

: C

:, C; = 0

=

1 0

+

: C

:, C; = 0

=

1 0

: C

:, C; = 0

=

1 0

C

:, C; = 0

(2)

where in the third step, we changed variable to = + (and then renamed to ) which

preserves C, i.e. {+ : C} = C, because = 0. That is, a left-coset is a subset

1 0

C .

The multiplication law is 1 0

C

1 0

C

=

()1 0

C1 + C

=

()1 0

C

hence clearly the identity in the quotient group is 1 0

C 1

=L2

There exists a bijective map : K2/L2 C ={ C := 0}, given by

1 0

C

=

This is bijective. Indeed, it is surjective: given C, there is the element

1 0

C

that maps to it; and it is injective: if both

11 0

C 1

and

12 0

C 2

map to , then

1= 2= , hence

1

1 0C 1

=

1

2 0C 2

.

The mapis also a homomorphism, hence it is an isomorphism. Indeed, using the multiplication

law of the quotient group above, we see that

(

1 0

C

1 0

C

) =(

()1 0

C

) == (

1 0

C

)(

1 0

C

).

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Example

LetaG. Define a: GG by a(g) =aga1. Then a is an automorphism:

Homomorphism: a(g1g2) =ag1g2a1 =ag1a1ag2a1 =a(g1)a(g2).

Onto: given gG, there exists g G such that a(g) =g): indeed, take g =a1ga.

kera={e}: indeed ifa(g) =e then aga1 =e then g= a1ea= e.

10 Lecture 10

Example

Consider G = Z2 Z2, with Z2 ={e, a}, a2 = e. Define : G G by ((g1, g2)) = (g2, g1).This is an nontrivial (i.e. different from identity map) automorphism. Indeed: 1) it is bi-

jective, 2) it is nontrivial: ((e, a)) = (a, e)= (e, a), 3) it is a homomorphism: (g)(g) =((g1, g2))((g

1, g

2)) = (g2, g1)(g

2, g

1) = (g2g

2, g1g

1) =((g1g

1, g2g

2)) =(gg

). But there is no

g Z2 Z2 such that = g. Indeed, we would have g((e, a)) = (g1, g2)(e, a)(g11 , g12 ) =(e, g2ag

12 )= (a, e) for any g2.

Definition:An inner automorphism is an automorphism such that = a for some aG.If is not inner, it is called outer. The set of inner automorphisms of a group G is denoted

Inn(G).

Definition:the set of all automorphisms of a group G is denoted Aut(H).

Note: IfG is abelian, then every inner automorphism is the identity map.

Theorem 10.1

The set of all autmomorphisms Aut(G) is a group under composition. The subset Inn(G) is a

normal subgroup.

Proof. First, we know that composing bijective maps we get a bijective map, that composition

is associative, and that there is an identity and an inverse which are also bijective. Hence

we need to check 3 things: composition is homomorphism, identity is homomorphism, and

inverse is homomorphism. Let 1, 2 Aut(G). Closure: (1 2)(gg) = 1(2(gg )) =1(2(g)2(g)) =1(2(g))1(2(g)) = (1 2)(g)(1 2)(g) so indeed the composed mapis a homomorphism (hence automorphism because bijective). Also, the identity is obviously a

homomorphism. Finally, let Aut(G). If (g1) = g1, then g1 is the unique one such thatthis is true, and the inverse map is defined by 1(g1) = g1. Let also (g2) = g

2. Then,

1(g1g2) =

1((g1)(g2)) = 1((g1g2)) = g1g2 =

1(g1)1(g2) so that indeed

1 is

homomorphism (hence, again, automorphism).

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Second, the subset of inner automorphisms is a subgroup. Indeed, a b = ab becausea(b(g)) =abgb

1a1 = (ab)g(ab)1. Also, the identity automorphism is e and the inverse of

a is a1 .

Third, the subset of inner automorphisms is normal. Let be any automorphism. Then a 1 =(a), so it is indeed an inner automorphism. This is shown as follows: a 1(g) =(a(1(g))) =(a1(g)a1) =(a)g(a1) =(a)g(a)1.

Note on notation: be careful of the meaning of where we put the1 in the exponent: (g1) =(g)1 on the r.h.s. we take the inverse of the element(g). But in a1(g) =

1a (g), on the

r.h.s. we take the inverse 1 of the map , and then apply it to g .

Note the important formulae:

a 1 =(a), a b= ab,

11 Lecture 11

Theorem 11.1

G/Z(G)=Inn(G).

Proof. We only have to realise that the map : GAut(G) given by (g) =g is a homomor-phism. Indeed, the calculation above have showed that (g1g2) =g1g2 =g1g2 =(g1)(g2).Hence, we can use the homomorphism theorem, G/ker=im. Clearly, by definition, im =Inn(G). Let us calculate the kernel. We look for allg G such that (g) = id. That is, allgsuch that g(h) = h

h

G. That is,ghg1 = h

h

G so that gh = hg

h

G. Hence,

gZ(G), so that indeed ker= Z(G).

Basics of matrices

We denote byMN(C) andMN(R) the set of allN Nmatrices with elements in Cand Rresp.

We may:

add matrices C=A + B

multiply matrices by scalars C=A,

C or

R,

multiply matrices C=AB

The identity matrix is I with elements Ijk = jk = 1(j = k), 0(j= k). Note: 1)IA = AI; 2)matrix multiplication is associative.

Given any matrix A, we may

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Take its complex conjugate A: (A)jk =Ajk Take its transpose AT : (AT)jk =Akj Take its adjoint A = AT =AT.

Definition:A matrix A is

self-adjoint ifA =A

symmetric ifAT =A

unitary ifA =A1

diagonal ifAjk = 0 for j=k

Definition:The trace of a matrix is Tr(A) = Nj=1 Ajj .

Note: Tr(A1A2 Ak) = Tr(AkA1 Ak1), Tr(I) =N(simple proofs omitted).

Definition:A matrixA is invertible if there exists a matrix A1 such thatAA1 =A1A= I.

Definition: The determinant of a matrix is det(A) =N

j1=1 NjN=1 j1jNA1j1 ANjN

where123N= 1 andj1jNis completely anti-symmetric: it changes its sign if two indices are

interchanged. Example: 12= 1, 21=1, 11= 22= 0.

Properties:

det(AB) = det(A) det(B) (hence, in particular, ifA1

exists, then det(A1

) = 1/ det(A))

det(A)= 0 if and only ifA is invertible (recall Kramers rule for evaluating the inverse ofa matrix)

det(I) = 1

det(A) = det(A)

det(AT) = det(A)

det(A) =Ndet(A)

ifAis diagonal, then det(A) = Nj=1 Ajj .Note: det(SAS1) = det(A), and also Tr(SAS1) = Tr(A), by the properties above.

Note: the determinant can also be written as det(A) =

SNpar()

Nk=1 Ak,(k). Here, par()

is the parity of a permutation element. The parity of a permutation is defined as the number

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Theorem 11.3

SL(N, C) is a normal subgroup ofGL(N, C).

Proof. By definition, we have SL(N, C) = ker det, where by det we mean the map det :

GL(N, C)C. Hence, SL(N, C) it is a normal subgroup ofGL(N, C) (so in particularit is a group.)

Similarly for S L(N, R).

Theorem 11.4

GL(N, C)/SL(N, C)= C. AlsoGL(N, R)/SL(N, R)= R.Proof.Again, consider the homomorphism det : GL(N, C) C, whose kernel is SL(N, C),and which is onto C. By the homomorphism theorem, the theorem immediately follow.

Idem for the real case.

3. Unitary group

U(N) ={AMN(C) :A =A1}

Note: the conditon A = A1 automatically implies that A1 exists, because of courseA exists for any matrix A; hence it implies that det(A)= 0. The condition can also bewritten AA= AA =I.

Group: 1) closure: ifA1, A2U(N) then (A1A2) =A2A1=A12 A11 = (A1A2)1 henceA1A2 U(N). 2) associativity: from matrix multiplication. 3) identity: I = I = I1hence I U(N). 4) inverse: ifA = A1 then using (A) = A we find (A1) = A =(A1)1, hence that A1 U(N).

12 Lecture 12

Note: U(1) ={z C : zz = 1}. Writingz = ei we see that the condition zz implies R; we may restrict to [0, 2). In terms of, the group is addition modulo 2.Theorem 12.1

The map det :U(N)U(1) is onto and is a group homomorphism.Proof. Onto: for z C with|z| = 1, we can construct A diagonal with A11 = z andAjj = 1 for j >1. Homomorphism: because of property of det as before.

4. Special unitary group

SU(N) ={AU(N) : det(A) = 1}Clearly SU(N) is a subgroup of U(N). It is SU(N) = kerdet, hence it is a normal

subgroup. Since det : U(N)U(1) maps onto U(1), we find that

U(N)/SU(N)=U(1)

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5. Orthogonal group

O(N) ={AMN(R) :AT =A1}This is the group of orthogonal matrices. Note: O(N) U(N). If A O(N) thendet(A) =1. Hence, here det : O(N) Z2 is onto.

6. Special orthogonal group

SO(N) ={AO(N) : det(A) = 1}

As before, SO(N) = ker det for det :O(N) Z2. Hence as we have again

O(N)/SO(N)= Z2

Centers of general linear and special linear groups

Theorem 12.2

Z(GL(N, C))= C. AlsoZ(GL(N, R))= R.

Proof. Suppose AB=BA for all BGL(N,C). That is,k

AjkBkl =k

BjkAkl.

Since this holds for all B , choose B diagonal with diagonal entries all different from each other.

Then we have

AjlBll =BjjAjl(Bll Bjj )Ajl = 0henceAjl = 0 for j=l. Hence, we find that Amust be diagonal. Further, consider

0 1 0

0

1 0 0 00 0 1 0...

. . .

0 0 0 1

Check that det(B) = 1, so B GL(N,C). The equation with j = 1 and l = 2 then gives usA11B12 =B12A22 hence A11 =A22. Similarly, Ajj =A11 for all j. So A= I for C. Thegroup of such diagonal matrices is obviously isomorphic to C. Idem for the real case.

Theorem 12.3

Z(SL(N, C))= ZN. AlsoZ(SL(N,R))= Z2 ifN is even, and Z(SL(N, R))={1} ifN is odd.Proof. The proof for GL(N, C) above goes through to show that the center must be matrices

proportional to the identity I. Hence we look for all a C such that det(aI) = 1. Butdet(aI) = aN hence a must be a Nth root of unity. This group of these roots of unity under

multiplication is isomorphic to the group ZN. For the case a R: ifN is even, then a=1, ifN is odd, then a = 1.

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13 Lecture 13

Structures of some groups

Theorem 13.1

IfN is odd, O(N)= Z2 SO(N).

Proof. Let us construct an isomorphism that does the job. We define it as

: O(N) Z2 SO(N)A(A) =

det(A),

A

det(A)

All we have to show is that this maps into the right space as specified (because this is not

immediately obvious), and then that it is indeed an isomorphism.

First: that it maps into Z2SO(N) is shown as follows. 1) Clearly det(A) Z2by the discussionabove. 2) Also (A/ det(A))T = AT/ det(A) = A1/ det(A) and (A/ det(A))1 = det(A)A1 =

det(A)2A1/ det(A) = (1)2A1/ det(A) =A1/ det(A). Hence indeed A/ det(A)O(N). 3)Further, det(A/ det(A)) = det(A)/ det(A)N = det(A)1N = (1)1N so ifN is odd, thenN 1is even, so (1)1N = 1. Hence indeed det(A/ det(A)) = 1 so A/ det(A)S O(N).

Second: that it is a homomorphism:

(AB) =

det(AB),

AB

det(AB)

=

det(A)det(B),

A

det(A)

B

det(B)

=

det(A), A

det(A)

det(B), B

det(B)

= (A)(B)

Third: that it is bijective. Injectivity: if(A1) =(A2) then det(A1) = det(A2) and A1/ det(A1) =

A2/ det(A2) hence A1= A2 so indeed it is injective. Surjectivity: take a Z2 andBS O(N).We can always find a matrix A O(N) such that (A) = (a, B). Indeed, just takeA = aB.This has determinant det(A) = det(aB) = aNdet(B) = a det(B) (since N is odd) = a (since

BSO(N)). Hence, (A) = (a, A/a) = (a, B) as it should.

We see that the inverse map is 1(a, B) =aB: simply multiply theS O(N) matrix by the sign

a. But of course this only works for N odd. For N even, there is the concept of semi-direct

product that would work...

Semi-direct products

The semi-direct product is a generalisation of the direct product. Take two groups G and

H, and consider the Cartesian product of these sets, GH ={g, h) : g G, h H}.

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This new set can be given the structure of a groups simply by taking the multiplication law

(g, h)(g, h) = (gg , hh). But there is another way of defining the multiplication law.

Definition:Given a homomorphism : GAut(H), where we denote(g) =:g, gG, wedefine the semi-direct product G Hby the group with elements all those of the set G H,and with multiplication law

(g, h)(g

, h

) = (gg

, hg(h

)).

14 Lecture 14

To check that this defines a group, we must check associativity,

(g, h)((g, h)(g, h)) = (g, h)(gg, hg(h)) = (ggg, hg(h

g(h)))

where the second member in the last term can be writtenhg(h)g(g(h

)) =hg(h)gg(h

).

On the other hand,

((g, h)(g, h))(g, h) = (gg , hg(h))(g, h) = (gg g, hg(h

)gg(h)

which is in agreement with the previous result. We must also check the presence of an iden-

tity, id = (id, id) (obvious using id = id and g(id) = id, recall the general properties of

homomorphisms). Finally, we must check that an inverse exists. It is given by

(g, h)1 = (g1, g1(h1))

because we have

(g, h)1(g, h) = (id, g1(h1)g1(h)) = (id, id)

and

(g, h)(g, h)1 = (id, hg(g1(h1))) = (id, hid(h

1)) = (id, id).

15 Lecture 15

Theorem 15.1

Consider Z2=

{+1,

1

}and the isomorphism : Z2

{I, R

},R = diag(

1, 1, 1, . . . , 1), given

by(1) =I and (1) =R. IfN is even, then

O(N)= Z2 S O(N),

where(s) =s is given by s(g) =(s)g(s) fors Z2 and gSO(N).

Proof. 1)(done in Lecture 14) We first check that: Z2Aut(SO(N)) is a homomorphism.

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a) We check that s is an automorphism ofSO(N) for any s. This is clear, using(s)2 = I:

we have s(gg) = (s)gg (s) = (s)g(s)(s)g(s) = s(g)s(g

). It is also onto be-

cause s((s)g(s)) = g for any g SO(N), and if det(g) = 1 then det((s)g(s)) =det((s))2 det(g) = det(g) = 1. Further, its kernel is the identity: ifs(g) = id then g = id.

Hence it is an isomorphism from S O(N) onto S O(N).

b) Then, we check that is a homomorphism. Indeed, (s)(s) = ss which acts as(s s)(g) =(s)(s)g(s)(s) =(ss)g(ss) =(ss)g(ss) =ss(g).

2) Second, we construct an isomorphism that mapsO(N) onto Z2SO(N). We define it as

: O(N) Z2 SO(N)A(A) = (det(A), A(det(A)))

Again all we have to show is that this maps into the right space as specified (because this is not

immediately obvious), and then that it is indeed an isomorphism.

First: that it maps into Z2SO(N) is shown as follows. a) Clearly det(A) Z2. b) Also((det(A))A)T = AT(det(A))T = A1(det(A)) and ((det(A))A)1 = A1(det(A))1 =

A1(det(A)1) =A1(det(A)). Hence indeed A(det(A))O(N). c) Further, det(A(det(A))) =det(A)det((det(A))) = det(A)2 = 1. Hence indeedA(det(A))SO(N).

Second: that it is a homomorphism:

(AB) = (det(AB),AB(det(AB)))

= (det(A) det(B),AB(det(A)det(B)))

= (det(A) det(B),AB(det(A))(det(B)))

= (det(A) det(B), A(det(A))(det(A))B(det(A))(det(B)))

= (det(A) det(B), A(det(A))(det(A))B(det(B))(det(A)))

=

det(A)det(B), A(det(A))det(A)(B(det(B)))

= (det(A), A(det(A))) (det(B), B(det(B)))

= (A)(B)

Third: that it is bijective. Injectivity: if(A1) =(A2) then det(A1) = det(A2) and A1(det(A1)) =

A2(det(A2)) henceA1= A2so indeed it is injective. Surjectivity: takes Z2and BS O(N).We can always find a matrix A

O(N) such that (A) = (s, B). Indeed, just take A = B(s).

This has determinant det(A) = s det(B) =s (since B S O(N)). Further, A(s) =B (s)2 =B. Hence, (A) = (det(A), A(s)) = (s, B) as it should.

The semi-direct product decomposition makes very clear the structures involved in the quotient,

e.g. O(N)/SO(N)= Z2; see below.

Theorem 15.2

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The subset{(e, h) :hH} G His a subgroup ofG Hthat is isomorphic toHand thatis normal. The subset{(g, e) :gG} is a subgroup ofG Hthat is isomorphic to G.

Proof. For the first statement: it is a subgroup because it contains the identity, it is closed

(e, h)(e, h) = (e,he(h)) = (e,hh), and it contains the inverse, (e, h)1 = (e, h1) by the

multiplication rule just established. It is also clearly isomorphic to H, with (e, h)h, thanksagain to the multiplication rule. Further, it is normal:

(g, h)1(id, h)(g, h) = (g1, g1(h1h))(g, h) = (id, g1(h

1hh)).

For the second statement, the subset contains the identity, is closed (g, e)(g, e) = (gg , g(e)) =

(gg , e), and by this multiplication law, it contains the inverse. Clearly again, it is isomorphic

to G.

A special case of the semi-direct product is the direct product, whereg = id for all gG (thatis, : GAut(H) is trivial, (g) = id). In this case, both G and H are normal subgroups.

Note: we usually denote simply by G and Hthe subgroups{(g, e) :gG}and{(e, h) :hH}ofG H(as we did for the direct product).

Theorem 15.3

The left cosets ofGHwith respect to the normal subgroupHare the subsets{(g, h) :hH}for allgG. Also, (G H)/H=G.

Proof. For the first statement: the left cosets are (g, h)(e, H) = (g,hg(H)) = (g,hH) = (g, H)

sinceg is an automorphism. For the second statement: the isomorphism is (g, H)g. This isclearly bijective, and it is a homomorphism, because (g, H)(g, H) = (gg, Hg(H)) = (gg

, H).

16 Lecture 16

In general, ifHis a normal subgroup of a groupJ, it is not necessarily true that J is isomorphic

to a semi-direct product G H with G= J/H. But this is true in the case where 1) His the kernel of some homomorphism (actually this is always true, but the inverse of the

homomorphism theorem, Theorem 10.3), and 2) there is a subgroup G in J on which is an

isomorphism.

Coming back to our example: SO(N) is indeed a normal subgroup ofO(N)= Z2 SO(N),but the Z2 of this decomposition, although it is a subgroup, is not normal. The Z2 of this

decomposition can be obtained as an explicit subgroup of O(N) by the inverse map 1 of

Theorem 15.1: 1((s, I)) = (s) for s =1. Hence the subgroup is{I, diag(1, 1, . . . , 1)}.Here, we indeed have that SO(N) is the kernel of det, and that{I, R} is a subgroup on whichdet is an isomorphism.

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Note: Clearly, there are many Z2 subgroups, for instance{I, I}; this one is normal. But itdoes not take part into any decomposition ofO(N) into Z2 andSO(N).

Extra material

Continuous groups

SN

: finite number of elements. Z: infinite number of elements, but countable. But other groups have

infinitely many elements forming a continuum. Ex: C = C{0}(group under multiplication of complexnumbers); or all the classical matrix groups.

More precisely: a group, as a set, may be a manifold; i.e. it may be locally diffeomorphic to open subsets

ofRN, and have a continuous, differentiable structure on it.

Manifold:

A topological space(M, J)(M: the space,J: the open sets) that is Hausdorff (every distinct pointspossess distinct neighbourhoods);

an atlas=

{(U, U) : U

I

}, I

J, where(U, U) is a chart;

UIU=M; U :U Rn homeomorphism; ifU V= forU, V I, thenV 1U , which mapsU(U V) Rn into Rn, is smooth.

The coordinates onUare the components of the map U, that is, in a neighbourhoodU, the coordinates

of a pointpM arex1 =1U(p), x2 =2U(p), etc.

Then above is the dimension of the manifold.

IfG is a manifold, then clearly also G

G, the cartesian product, is a manifold, with the cartesian product

topology, etc. The dimension is2n ifn is the dimension ofG.

Definition: A Lie group G is a manifoldG with a group structure, such that the group operations are

smooth functions: multiplicationG GG, and inverseGG.

Essentially, for the classical matrix groups above, the number of dimensions is the number of free, con-

tinuous parameters.

Example

The (real) dimensions for some of these Lie groups are

C

: 2 dimensions.

GL(N, R): N2 dimensions. SL(N, R): N2 1 dimensions due to the one conditiondet(A) = 1. SO(N): N(N1)/2dimensions. Indeed: there areNNreal matrices so there areN2 parameters.

There is the condition ATA = I. This is a condition on the matrix ATA, which contains N2

elements. But this matrix is symmetric no matter whatA is, because(ATA)T =ATA. Hence, the

constraint ATA = I in fact has1 + 2 +... + N constraints only (looking at the top row withN

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elements, then the second row withN1 elements, etc.). That is,N(N+ 1)/2constraints. Theseare independent constraints. Hence, the dimension isN2 N(N+ 1)/2 = N(N 1)/2.

Euclidean group

We now describe the Euclidean group EN. We start with a formal definition, then a geometric

interpretation.

Formal definition

Consider the groups O(N) and RN. The former is the orthogonal group ofN byN matrices.

The latter is the (abelian) group of real vectors in RN under addition. The Euclidean group is

EN=O(N) RN

where: O(N)Aut(RN), (A) =A is a homomorphism, with A defined byA(b) =Ab.

To make sure this is a good definition: we must show that is a homomorphism and that that

is an automorphism. First the latter. A is clearly bijective because the matrixA is invertible

(work out injectivity and surjectivity from this). Further, it is a homomorphism because A(x +

y) =A(x + y) =Ax + Ay= A(x) + A(y). Second, is a homomorphism because AA(b) =

AAb= A(Ab) =A(A(b)) = (A A)(b) so that(AA) =(A)(A) (the multiplicationin automorphisms is the composition of maps). Hence, this is a good definition of semidirect

product.

Explicitly, the multiplication law is

(A, b)(A, b) = (AA, b + Ab).

Geometric meaning

Consider the space of all translations in RN. This is the space of all mapsxx + bfor bRN:maps that take each point x to x + bin RN. Denote a translation by Tb, so thatTb(x) =x + b.

The composition law is obtained from

(Tb Tb)(x) =Tb(Tb(x)) =Tb(x + b) =x + b + b =Tb+b(x).

Hence,Tb Tb =Tb+b , so that compositions of translations are translations. Further, there isa translation that does nothing (choosingb = 0), one can always undo a translation, Tb Tb=T0 = id. Hence, the set of all translations, with multiplication law the composition, is a group

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TN. Note also that Tb=Tb ifb=b and for any bRN there is a Tb. Clearly, then, this is agroup that is isomorphic to RN, by the map

Tbb

Consider now another type of operations on RN: the orthogonal transformations (those that

preserve the length of vectors in RN), forming the group O(N) as we have seen. For a matrix

AO(N), we will denote the action on a point x in RN byA(x) :=Ax.

Consider the direct product of the sets of orthogonal transformations and translations, with

elements (A, Tb) for A O(N) and Tb TN. Suppose we define the action of elements ofthis form on the space RN by first an orthogonal transformation, then a translation. That is,

(A, Tb) :=Tb A, i.e.(A, Tb)(x) =Tb(A(x)) =Ax + b.

Then, let us see what happen when we compose such transformations. We have

((A, Tb) (A, Tb))(x) =A(Ax + b) + b= AAx + Ab + b= (AA, TAb+b)(x).

That is, we obtain a transformation that can be described by first an orthogonal transformation

AA, then a translation by the vector Ab +b. Combined with the definition of the Euclidean

group above and the fact that Tbb is an isomorphism, what we have just shown is that theset of all transformations orthogonal transfo followed by translation is the same set as the

setEN, and has the same composition law hence the first is a group, and the two groups are

isomorphic. That is, the Euclidean group can be seen as the group of such transformations.

Note how the semi-direct multiplication law occurs essentially because rotations and translations

dont commute:

A(Tb(x)) =A(x + b) =Ax + Ab, Tb(A(x)) =Ax + b

so thatTbATbA =TbTbAA. We rather haveTbATbA =TbATbA1AA,and we find the conjugation law

A Tb A1(x) =A(A1x + b) =x + Ab =TAb(x)

That is: the conjugation of a translation by a rotation is again a translation, but by the rotated

vector, and this is what gives rise to the semi-direct product law. This is true generally: if two

type of transformations dont commute, but the conjugation of one by another is again of thefirst type, then we have a semi-direct product. Recall also examples ofSO(2) and Z2 in their

geometric interpretation as rotations and reflections.

There is more. We could decide to try to do translations and rotations in any order that is,

we can look at all transformations ofRN that are obtained by doing rotations and translations

in any order and of any kind. A general transformation will look like A1 A2 Tb1 Tb2

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A1 A2 etc. But since orthogonal transformations and translations independently formgroups, we can multiply successive orthogonal transformations to get a single one, and like wise

for translations, so we get something of the formA Tb A . . .etc. Further taking into accountthat we can always put the identity orthogonal transformation at the beginning, and the identity

translation at the end, if need be, we always recover something of the form ( A, Tb)(A, Tb) .

Hence, we recover a Euclidean transformation. Hence, the Euclidean group is the one generated

by translations and orthogonal transformations. We have proved:

Theorem 16.1

The Euclidean groupEN is the group generated by translations and orthogonal transformations

ofRN.

Euclidean symmetries of subsets

Consider a subset of R2. In order to find the Euclidean transformations of R2 that preserve

it, follow these steps. 1) Try to see if translations are symmetries. 2) Then, try to see if there

is one or many points about which rotations are symmetries, whether by all possible angles or

by some discrete set of angles. 3)Then, try to see if there are one or many axes about which

reflection is a symmetry.

Translations form the group R or R2 (depending in how many directions you can translate -

here, for simplicity, were in 2-d only). Rotations about one given fix point (any point) and by

all possible angles form the group S O(2); if its not all possible angles that we have, but just n

of them (including the 0 angle), then this is the group Zn. Reflections about any given fix axis

form the group Z2 (including the identity transformation).

When we put things together: if the group actions commute, then its a direct product. If

not, its a semi-direct product. In semi-direct products, we always have the order reflections

to the left of rotations to the left of translations. Remember that Z2 SO(2)= O(2), thatO(n) Rn=E(n), that Z2 Zn=Dn (in the latter one the case n= 2 is Z2 Z2= Z2 Z2:the semi-direct product becomes a direct product). If there is translation, then there is a factor

Rn. If there is rotation as well, then pick one center, this corresponds to the factorSO(n) in

the semi-direct product. If there is reflection as well, then pick one axis, this corresponds to the

factor Z2 in the semi-direct product.

(Examples...)

17 Lecture 17

The group SO(2)

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Let us explicitly construct the group SO(2).

A=

a b

c d

, ATA= I , det(A) = 1.

We have 4 conditions:

ATA= Ia2 + c2 = 1, b2 + d2 = 1, ac + bd= 0.

The first two conditions imply thata = cos ,c = sin (here we put a minus sign for convenience)

andb = cos ,d = sin for some, in [0, 2) (or, R mod 2). The second condition thenis cos( ) = 0. Hence, = /2 mod 2. This givesb = sin , d = cos or b = sin ,d= cos . The last condition

det(A) = 1ad bc= 1

then implies that the first choice must hold:

A= A() :=

cos sin sin cos

, [0, 2)

This clearly is of dimension 1 (there is one real parameter remaining) as it should.

Explicit multiplications of matrices give

A()A() =A(+ ), A()1 =A()

In particular, S O(2) is abelian.

Here, the group manifold is the circleS1

. Indeed, we have one parameter, and the matrix elements varycontinuously for[0, 2) and also going back to 0 instead of2. Every point onS1 corresponds to aunique group element A(), and all group elements are covered in this way. Note: geometrically,the manifold is indeedS1 rather than the interval [0, 2) (which strictly speaking wouldnt be a manifold

anyway) because of periodicity, i.e. continuity from the endpoint2 back to the starting point 0.

The interpretation of SO(2) is that of rotations about the origin: we act with SO(2) on the

plane R2 by

v =Av, ASO(2), v=

x

y

, (x, y) R2.

This givesx =x cos y sin , y =x sin + y cos

All this makes it clear that SO(2)= U(1). Indeed, just put the points x, y into the complexplane via x + iy, and consider action ofU(1) as ei. Hence the isomorphism is

SO(2)U(1) :A()ei

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The group O(2)

Two disconnected sets: O(2) =SO(2) CwhereC= orthogonal matrices with determinant -1.Note: C is not a group. The condition of det = -1 implies from the previous analysis that

C=

B() :=

cos sin

sin cos

, [0, 2)

Note that, with R =

1 0

0 1

, we have

A()R= B()

for all . Note also that RO(2), in fact RC. The group elementR is the reflection w.r.t.the y axis, and we have

O(2) ={group generated by rotations about origin and reflections about y axis}

O(2) is not abelian, in particular A()R

=RA() in general.

Clearly, then,O(2) cannot be isomorphic to Z2SO(2) because both Z2andSO(2) are abelian.The semi-direct product Z2 SO(2), however, is not abelian. We see that the subgroupZ2 in

Z2 SO(2) corresponds to the group composed of the identity and the reflection, {I, R}. Wesee also that det is the isomorphism of this subgroup onto Z2.

The geometric interpretation of the semi-direct product structure is as follows: we can always

represent an element ofO(2) AB whereA is a rotation and B is the reflection w.r.t. y axis (for

instance). Indeed, ifC O(2) then C = C(det C)(det C) and we can take A = C(det C)and B = (det C). Then let us perform in succession C2 then C1 on the vector v. The

result is C1C2v. The transformationC3 = C1C2 can also be written as C3 = A3B3. We haveC1C2= A1B1A2B2= A1(B1A2B

11 )B1B2and we can identifyA3= A1B1A2B

1

1and B3= B1B2.

So if we use the notation in paris C= (B, A), we obtain (B1, A1)(B2, A2) = (B1B2, A1B1A2B11),

which is exactly the semi-direct product structure with B(A) =BAB1, as we had done (here

we dont use explicitly the , as we use directly the matrices B , which are I orR; we could set

Bi= (gi) and use the pairs C= (g, A) instead).

Note: we have the following relation:

RA()R= A()

But then,A()RA() =A()RA()RR= A()2R= A(2)R= B(2)

The quantityA()RA() has a nice geometric meaning: reflection w.r.t. axis rotated by angle fromy axis. In order to cover all B() for[0, 2), we only needA()RA() for[0, ).This indeed gives all possible axes passing by the origin. Hence:

O(2)={group of all rotations about origin and all reflections about all different axes passing by the origi

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The group SU(2)

Let us look at SU(2). We will use the Pauli matrices: along with the identity I, the Pauli

matrices form a basis (over C) for the linear space M2(C).

1= x:=

0 1

1 0

, 2= y :=

0 ii 0

, 3= z :=

1 0

0

1

.

We will denote a general complex 2 by 2 matrix by

A= aI+ b where we understand as a vector of Pauli matrices, so that b = bxx+ byy+ bzz.

Some properties:

2i =I , ij =ji(i=j), xy =iz, yz =ix, zx= iy, i =i, det(A) =a2||b||2.From these, we find:

x

y

= (x

y)I+ i(x

y)

where x y is the vector product. We also find(aI+ b )(aI b ) =a2I (b b)I i(b b) = (a2 ||b||2)I= det(A)I.

Hence,

(aI+ b )1 = aI b det(A)

All these properties point to a nice analogy with complex numbers. Define, instead of one, three imaginary

numbers, =ix, = iy, k =iz. They satisfy 2 = 2 =k2 =1.

Definition: The division algebraH of quaternions is the non-commutative field of all real linear combi-

nationsz = a +bx +by +bzk (a, bx, by, bz R), with the relations 2 = 2 =k2 =1 and = =kand cyclic permutations. It is associative.

We define the quaternion conjugate byz = abx by bzk(from the point of view of 2 by 2 matrices, thisisz= z), and we havez z = zz = a2 + ||b||2 0, with equality iffz = 0. Hence we can define|z|= zz.We also havez1 = z/|z|2, as for complex numbers. An important identity is|z1z2| =|z1| |z2| for anyz1, z2 H, which follows from|z1z2|2 = z1z2z2z1 = z2z2z1z1 where we used the fact thatz2z2 R Hhence commutes with everything. Any quaternionz has a unique inverse, except for 0. This is what makes

the quaternions a division algebra: we have the addition and the multiplication, with distributivity and

with two particular numbers, 0 and 1, having the usual properties, and we have that only 0 doesnt have

a multiplicative inverse. Note that there are no other division algebras that satisfy associativity, besides

the real numbers, the complex numbers and the quaternions. There is one more division algebra, which

is not associative in addition to not being commutative: the octonions, with 7 imaginary numbers.

Group S U(2): with A= aI+ x , we require det(A) = 1 hence a2 ||x||2 = 1, and A =A1hence aI+ x = aI x , so that a R and x =x. Writingx = ib we have b R3.That is,

SU(2) ={A= aI+ ib : a R, b R3, a2 + ||b||2 = 1}

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Note that, in terms of quaternions, this is:

SU(2) ={z H :|z|= 1}.

Note the similarity with

U(1) ={z C :|z|= 1}Note that in both cases we have a group because in both cases

|z1z2

|=

|z1

| |z2

|, so the condition

|z

|= 1is

preserved under multiplication.

Geometrically, the condition|z| = 1 for quaternions is the condition for a 3-sphere in R4. This is themanifold ofS U(2).

18 Lecture 18

Invariance of physical laws: scalars and vectors

Invariance of mathematical objects under transformations is the main concept that leads to thestudy of groups. A proper definition of an invariance requires us to say three things: 1) what

is the family of mathematical objects that we consider, 2) what is the family of transformations

that we admit, and 3) what particular object do we claim to be invariant. We have seen the

examples of subsets in the plane, and briefly of vector fields.

We now consider other examples. First, the equations of motion of physics. Suppose the

3-dimensional vectorsx(t) andy(t), as functions of time t, satisfy the following equations (New-

tons equations):

md2x(t)

dt2 =GmM x(t)

y(t)

||x(t) y(t)||3 , Md2y(t)

dt2 =GmM y(t)

x(t)

||x(t) y(t)||3 . (4)

Then clearly the new vectors x(t) = Ax(t) + b and y(t) = Ay(t) + b, where A O(3) is anorthogonal matrix and b is a constant vector, also satisfy the same equations, because

d2

dt2(Ax(t) + b) =A

d2

dt2x(t)

and

GmM Ax(t) Ay(t)||Ax(t) Ay(t)||2 =GmM A x(t) y(t)||x(t) y(t)||3

so that for instance

md2x(t)

dt2 + GmM

x(t) y(t)||x(t) y(t)||3 =A

d2

dt2x(t) + GmM

x(t) y(t)||x(t) y(t)||3

= 0

The transformations (x, y)(Ax + b, Ay + b) satisfy the rules of the Euclidean group: hence,Newtonss equations are invariant under the Euclidean group.

Let us concentrate on the O(3) transformations for a while, forgetting about the translations.

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Definition:An object V that transforms like V =AV under the transformation AO(3) issaid to be a vector under O(3). If it transforms like that only underSO(3) then it is a vector

under SO(3). An objectSthat transforms like S = Sunder the same transformation is said

to be a scalar.

The basic vectors in our example arex(t) andy(t): 3-components functions oft. They transform

as vectors by the very definition of how we want to act on space. But then, the objectx(t)y(t)is a new object, form out of the previous ones. It is also a vector. Likewise, d2x(t)/dt2 is a

vector. Further, the object||x(t) y(t)|| is also a new object, and it is a scalar. All in all, theleft- and right-hand sides of (4) are all vectors. This fact, that they are all vectors, is why the

equation is invariant under O(3).

Other examples: x(t) y(t) = is a scalar because the inner product is invariance under O(3),x(t + t0) is a vector, etc.

Another example that can be worked out: ifx and y are vectors underS O(3), then also is x y(the vector product).

A more precise definition of vectors/scalars would require the theory of representation. For our

purposes, we may do as follows. Letf(x1, x2, . . .) be a function of many variables x1, x2, . . ..

Consider a transformation of variable, with x1 = Ax1, x2 = Ax2, . . ., where A O(3) (resp.

SO(3)). Then, we say that f(x1, x2, . . .) is a vector under O(3) (resp. SO(3)) iff(x1, x

2, . . .) =

Af(x1, x2, . . .) for all AO(3) (resp. SO(3)). If there is just one variablex1, then we can seethis variable as a position variable in R3; then a vector according to this definition is just a O(3)

(or SO(3)) invariant vector field.

More complicated example (which doesnt quite fall into the more precise definition above, but

which follows the general principle): The 3-component differential operator

/x1

/x2

/x3

is also a vector. This is because by the chain rule,

xi=j

xjxi

xj=j,k

Ajkxkxi

xj=j

Aji

xj

so that (using the vector notation)

x=AT

x

x =A

x

where we used AT =A1.

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19 Lecture 19

Inner product

LetVbe a finite-dimensional vector space over C. A map V V C, x, y(x, y) is an innerproduct if it has the properties:

(x, y) = (y, x), (x, ay+ bz) =a(x, y) + b(x, z), (x, x)0, (x, x) = 0x = 0

(z = z is the complex conjugate). This makes V into a Hilbert space. Note that the first and

second property imply

(ay+ bz, x) =a(y, x) + b(z, x).

This along with the second property is called sesquilinearity. The restriction over the real-vector

space (real restriction: CN becomes RN; same basis, but only consider real coefficients) then

gives

(x, y) = (y, x), (ay+ bz, x) =a(y, x) + b(z, x).

This restriction is bilinear and symmetric. The only example we will use is:

(x, y) =i

xi yi.

In particular, using matrix and column vector notation, this is

(x, y) =xy.

This implies that ifA is a linear operator, then

(x, Ay) = (Ax, y).

The norm of a vector is defined by

||x||=

(x, x)

(positive square-root).

Structure ofO(N)

Theorem 19.1

A real-linear transformation A ofRN is such that||Ax||=||x|| for all x RN iffAO(N).

Proof. IfA

O(N), then AT =A1 so that

||Ax||2 = (Ax, Ax)= (Ax)TAx

= xTATAx

= xTx

= ||x||2 (5)

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where in the 2nd step we use reality, so that =T. Also: if||Ax|| =||x||, then replace x byx + y and use bilinearity and symmetry:

(A(x+y), A(x+y)) = (x+y, x+y)(Ax, Ax)+(Ay, Ay)+2(Ax, Ay) = (x, x)+(y, y)+2(x, y).

Using||Ax||=||x|| and||Ay||=||y||, the first and last terms cancel out, so

(Ax, Ay) = (x, y).

Hence (ATAx, y) = (x, y) so that (ATAx x, y) = 0. This holds for all y, hence it must bethat ATAx x = 0 (obtained by choosing y = ATAx x, so that ATAx = x. This holds forall x, hence ATA= I. Hence AO(N).

Note: the same hold if we say A RN and ask for||Ax|| =||x|| for all x CN. Indeed,this includes x RN, so in one direction this is obvious; in the other direction we use again(Ax, Ax) = (ATAx, x) which holds for x CN as well.

Theorem 19.2IfxCN is an eigenvector ofAO(N) with eigenvalue , then||= 1.

Proof. Ax= x with x= 0, hence (Ax, Ax) = (x, x) hence (x, x) =||2(x, x) hence||2 = 1since x= 0.

Theorem 19.3

IfASO(3) then there exists a vector nR3 such that An= n (i.e. with eigenvalue 1).

20 Lecture 20

Proof. Consider P() = det(AI). We know that P(0) = 1 because A SO(3). Also,P() =3 +. . .+ 1 because the only order-3 term is from theI part. Hence, P() =( 1)( 2)( 3) with 123= 1.

There are 2 possibilities: First, at least one of these, say 1, is complex. It must be a pure

phase: 1=ei for (0, 2) {}. Ifx is the associated eigenvector, then Ax= eix hence

Ax =eix so that x is a new eigenvector with a different eigenvalue, say 2 =ei. But

since 123 = 1, it must be that 3= 1.

Second, all i are real, so are1. Since 123= 1, not the three are1, hence at least one is1.

Finally, ifAn= n, and ifn is not real, then n is another eigenvector with eigenvalue 1. Hence

the thirds eigenvalue also must be 1: all eigenvalues are 1. Since the space of all eigenvectors

span C3, thenAx= x for anyx C3 so thatA = I, so we may choosen real. Hence, ifAn= n,then n is real or may be chosen so.

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We may normalise to||n||= 1. Suppose n= (unit vector in x direction). Then A = implies

A=

1 0 0

Further,A

T

A= I implies

A=

1 0 0

0 0

The 2 by 2 matrix in the bottom-right, which we denote a, has the property

aTa= I , det(a) = 1

hence it is an S O(2) matrix. Hence,

A=

1 0 0

0 cos sin 0 sin cos

. (6)That is, A is a rotation by around the x axis. In general, A will be a rotation about the

axis spanned by n, i.e. the axis{cn: c R}, ifAn= n. Hence, an SO(3) matrix is always arotation w.r.t. a certain axis. Clearly, is A is a rotation, then (Ax, Ax) = (x, x) for any x R3,so A O(3). Also, An = n for any vector n along the rotation axis, and in a perpendicular,right-handed basis that includes n,Ahas the form (6), hence ASO(3). That is:

SO(3) ={all rotations about all possible axes in R3}.

Geometrically: we may parametrise the space of axes times direction of rotation by the space

of unit vectors in R3: the rotation will be about the axis{cn: c R} and the rotation in theright-handed direction w.r.t. n. This space is the unit sphere in R3. We may then characterise

the space of all rotations by angles in [0, ], in any direction, by giving a non-unit length to

the vectors, equal to the angle of rotation. Hence, this space is the closed ball of radius in

R3. But we are over-counting: we need to take away exactly half of the sphere of radius (so

take away the upper closed hemisphere, and the closed half of the equator, and one of the two

points joined by the equator). This is the space of all S O(3). To make it a manifold: we simply

have to connect points on the sphere of radius that are diametrically opposed. These pointsare indeed very similar rotations: a rotation in one direction by , or in the other direction by

, is the same rotation. This is the manifold ofSO(3), and it is such that multiplications (i.e.

composition) of rotations by small angles correspond to a small motion on the manifold.

For O(3), we simply need to add the reflections, as before. The setO(3) is all rotations, and

all reflections about all possible planes . A reflection about the three axes is described by the

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matrixI, and the group structure isO(3)= Z2SO(3) where the Z2 is the subgroup{I, I}.But there is another way of representing O(3): we could also consider reflection about the y zplane, for instance. This is the matrix R = diag(1, 1, 1), and the subgroup{I, R} also hasthe structure ofZ2. Using this Z2, we then can write O(3)= Z2 SO(3): the same semidirectproduct that we used in the case ofO(n) with n even.

21 To be adjusted

We can now construct particular elements of the Euclidean group that are of geometric interest,

in 3 dimensions (and also in 2) for simplicity: rotations or reflections about an axis passing by

a translated point b. These are described by, for AO(3) (or AO(2)),RA,b(x) :=Tb A Tb(x) =Ax Ab + b.

Indeed, they preserve the point b: with x= b we obtain back b. Moreover, they preserve the

length of vectors starting at band ending at x:||RA,b(x) RA,b(b)||=||Ax Ab + b b||=||A(x b)|| =||x b||. (More generally, of course, this is an orthogonal transformation withrespect to the point b.) TakeAS O(3): this is a rotation, and there is a vector n such thatAn = n (this is the vector in the axis of rotation). We could modify b by a vector in the

direction of n without modifying RA,b. That is RA,b+cn(x) = AxA(b+ cn) +b + cn =Ax Ab + b= RA,b(x).

Structure ofU(N)

Theorem 21.1

A complex linear transformation A on CN preserves the norm,||

Ax||

=||

x||

for all xC

N, iff

AU(N).

Proof. 1) If A U(N) then||Ax||2 = (Ax, Ax) = xAAx = xx = (x, x) =||x||2 hence||Ax||=||x||.

2) Consider x = y+ az. We have||Ax||2 = (Ay+ aAz, Ay+ aAz) =||Ay||2 +|a|2||Az||2 +a(Ay, Az) +a(Az, Ay). Likewise,||x||2 =||y||2 +|a|2||z||2 +a(y, z) +a(z, y). Then, using||Ay||2 =||y||2 and||Az||2 =||z||2, we have

||Ax||= xa(Ay, Az) + a(Az, Ay) =a(y, z) + a(z, y).

Choosinga = 1 this is (Ay, Az)+(Az, Ay) = (y, z)+(z, y) and choosinga = i this is (Ay, Az)(Az, Ay) = (y, z) (z, y). Combining these two equations, we find (Ay, Az) = (y, z), for ally, z CN. Then we can use similar techniques as before: using (Ay, Az) = (AAy, z) we get((AA I)y, z) = 0 for all y, z CN, which implies AA= I.

A 3-sphere in R4 can be imagined by enumerating its slices, starting from a pole (a point on the

3-sphere. The slices are 2-spheres, much like the slices of a 2-sphere are 1-sphere (i.e. circles).

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Starting from a point on the 3-sphere, we enumerate growing 2-spheres until we reach the single

one with the maximal radius (this is the radius of the 3-sphere, i.e. 1 in our case), then we

enumerate shortening 2-spheres back to a point. Putting all these objects together, we see that

the set is a double-cover of an open unit ball in R3 plus a single unit 2-sphere. This looks very

similar to the manifold ofSO(3), but there we had exactly half of that (an open ball plus half

of a 2-sphere). There is a group structure behind this:

A nice theorem

Theorem 21.2

SU(2)/Z2=SO(3)

Proof. This proof is not complete, but the main ingredients are there. The idea of the proof is

to use the homomorphism theorem, with a homomorphism : SU(2) SO(3) that is onto,such that ker = Z2.

Proposition: There exists a linear bijective map : R3 between the real-linear space ofself-adjoint traceless 2 by 2 matrices, and the real-linear space R3.

Proof. Note that the Pauli matrices are self-adjoint traceless 2 by 2 matrices. Take a general 2

by 2 matrix A = aI+ b for a, bx, by, bz C. The condition of tracelessness imposesa = 0.The condition of self-adjointness imposes b =b hence bR3. Hence, the real-linear space of self-adjoint traceless 2 by 2 matrices is the space of real linear combinations A = b . GivenA such a matrix, we have a unique (A) = b R3, hence we have injectivity. On the otherhand, given any b R3, we can form A= b , hence we have surjectivity. Finally, it is clearthat given A, A and cR, we have (cA) = c(A) and (A+A) = (b +b ) =((b + b)

) =b + b = (A) + (A), so is linear.

Given anyUS U(2), we can form a linear bijective map U : as follows:

U(A) =U AU.

This maps into because ifA, then 1) Tr(U(A)) = Tr(U AU) = Tr(UUA) = Tr(A) = 0,and 2) (U AU) = U AU = U AU. Hence, U(A) . Moreover, it is bijective because 1)injectivity: ifU AU =U AU thenUU AUU=UU AUUhenceA = A, and 2) surjectivity:

for any B , we have that UBU (by the same arguments as above) and we haveU(U

BU) =U UBUU =B so we have found a A= UBU that maps to B. Finally, it

is linear, quite obviously.

The map U induces a map on R3 via the map . We define RU : R

3 R3 by

RU= U 1

for any U S U(2). By the properties of Uand of derived above we have that RU is linearand bijective.

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We now want to show that RUS O(3).

1) from the properties of Pauli matrices, we know that det(b ) =||b||2. Hence, we have foranyA that det(A) =||(A)||2, or in other words det(1(b)) =||b||2. Hence,

||RU(b)||2 =||(U(1(b)))||2 =det(U(1(b))) = det(U1(b)U) = det(1(b)) =||b||2.

That is, RU is a real-linear map on R3

that preserves lengths of vectors. By the previoustheorems, it must be that RUO(3).

2) Further, the map g : SU(2) Rgiven by g(U) = det(RU) (where we see the linear map RUas a 3 by 3 real orthogonal matrix). This is continuous as a function of the matrix elements

ofU. Indeed, we can calculate any matrix element ofRUby choosing two basis vectors x and

y in R3 and by computing xRU(y). This is x(U1(y)U). The operation U Uand the operations of matrix multiplications are continuous in the matrix elements, hence the

map UU1(y)U is, for any matrix element of the resulting 2 by 2 matrix, continuous inthe matrix elements ofU. Since is linear, it is also continuous, and finally the dot-product

operation is continuous. Hence, all matrix elements of RU are continuous functions of thematrix elements ofU, so that det(RU) is also a continuous function of the matrix elements of

U. Moreover, we know that with U = I, we find RU = I (the former: identity 2 by 2 matrix,

the latter: identity 3 by 3 matrix). Hence, g(I) = 1. But sinceg(U) {1, 1} (because thedeterminant of an O(3) matrix is1), it must be that g (U) = 1 for all U SU(2) that can bereached by a continuous path from I (indeed, if : [0, 1] SU(2) is such a continuous path,(0) =Iand(1) =Uand(t) a continuous function oft, theng((t)) is a continuous function

of t with g((t)) {1, 1} and g((0)) = 1; the only possibility is g((1)) = 1 by continuity).SinceS U(2) is connected, then all USU(2) can be reached by continuous path from I, henceg(U) = 1 for all U

SU(2), hence det(R

U) = 1 for all U

SU(2) hence R

USO(3).

We have shown that RU SO(3). Hence, we have a map : SU(2)SO(3) given by

(U) =RU.

We now want to show that is a homomorphism. We have

1(RU1U2(b)) = U1U2(1(b)) =U1U2

1(b)U2U1 = U1(U2(

1(b)))

hence

(U1U2) =RU1U2 = 1RU1U2 = U1U21 = U11U21 =RU1RU2which is the homomorphism property.

Then, we would have to prove that is onto this requires more precise calculation of what

is as function of the matrix elements ofU. We will omit this step.

Finally, we can use the homomorphism theorem. We must calculate ker . The identity inO(3)

is the identity matrix. We have(U) =I O(3) iff (U(1(b))) =b for all bR3, which

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Proof. First in the opposite direction: ifQ is a Euclidean transformation, Q = (A, Tb), then

||Q(x) Q(y)||=||Ax + b Ay bb||=||A(x y)||=||x y||

so indeed it preserves the distance function.

Second the opposite direction: assumeQ preserves the distance function. Let b = Q(0). Define

Q =Tb Q. Hence, we have Q(0) =0: this preserves the origin. Hence, Q preserves lengthsof vectors:||Q(x)|| =||Q(x) Q(0)|| =||x 0|| =||x||. More than that: Q also preservesthe inner product:

(Q(x), Q(y)) = 1

2(Q(x), Q(x)) +

1

2(Q(y), Q(y)) 1

2(Q(x) Q(y), Q(x) Q(y))

= 1

2||Q(x)||2 +1

2||Q(y)||2 1

2||Q(x) Q(y)||2

= 1

2||x||2 +1

2||y||2 1

2||x y||2

= (x, y)

Now we show that Q is linear. Consider ei, i = 1, 2, . . . , N the unit vectors in orthogonal

directions (standard basis for RN as a vector space). Letei:= Q(ei). Then, (e

i, e

j) = (ei, ej) =

i,j . Hence ei, i= 1, 2, . . . , N also form an orthonormal basis. Now letx=

i xiei for xi R,

and writeQ(x) =

i xie

i (this can be done because e

i form a basis). We can find x

i by taking

inner products with e i: xi= (Q

(x), ei). But thenxi= (Q

(x), ei) = (Q(x), Q(ei)) = (x, ei) =

xi. Hence,Q(x) =

i xie

i, which means that Q

is a linear transformation. Hence, we have

found that Q is a linear transformation which preserves length of vectors in RN, so it must be

in O(N). Hence,Q has the form Tb A for AO(N), so that QEN.

The poincare group

groups & symmetries notes

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