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Home , Adjoint representation, Diffeomorphism, Linear group, Manifold, Submanifold, Topological group

Chapter 4

Manifolds, Lie Groups, and Lie Algebras; “Baby Case”

In this section we define precisely embedded submani- folds, Lie groups,andtheirLie algebras.

One of the reasons that Lie groups are nice is that they have a di↵erential structure,whichmeansthattheno- tion of makes sense at any of the .

Furthermore, the at the identity happens to have some , that of a .

Roughly, the tangent space at the identity provides a “lin- earization” of the , and it turns out that many properties of a Lie group are reflected in its Lie algebra.

139 140 CHAPTER 4. , LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” Fortunately, most of the Lie groups that we need to con- sider are subspaces of RN for some suciently large N.

In fact, they are all isomorphic to of GL(N,R) for some suitable N,evenSE(n), which is isomorphic to asubgroupofSL(n +1).

Such groups are called linear Lie groups (or matrix groups).

Since the groups under consideration are subspaces of RN , we do not need immediately the definition of an abstract .

We just have to define embedded (also called submanifolds) of RN (in the case of GL(n, R), N = n2). 141 In general, the dicult part in proving that a of GL(n, R)isaLiegroupistoprovethatitisamanifold.

Fortunately, there is simple a characterization of the lin- ear groups.

This characterization rests on two theorems. First, a Lie subgroup H of a Lie group G (where H is an embedded of G)isclosedinG.

Second, a theorem of Von Neumann and Cartan asserts that a closed subgroup of GL(n, R)isanembeddedsub- manifold, and thus, a Lie group.

Thus, a linear Lie group is a closed subgroup of GL(n, R). 142 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” Asmallannoyingtechnicalarisesinourapproach,the problem with discrete subgroups.

If A is a of RN ,recallthatA inherits a from RN called the ,anddefinedsuch that a subset V of A is open if V = A U \ for some open subset U of RN .

Apointa A is said to be isolated if there is some open 2 subset U of RN such that a = A U, { } \ in other words, if a is an open in A. { } 143 The group GL(n, R)ofrealinvertiblen n matrices 2 ⇥ can be viewed as a subset of Rn ,andassuch,itisa under the subspace topology (in fact, a 2 dense open subset of Rn ).

One can easily check that multiplication and the inverse operation are continuous, and in fact smooth (i.e., C1- continuously di↵erentiable).

This makes GL(n, R)atopological group.

Any subgroup G of GL(n, R)isalsoatopologicalspace under the subspace topology. 144 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” AsubgroupG is called a discrete subgroup if it has some .

This turns out to be equivalent to the fact that every point of G is isolated,andthus,G has the discrete topology (every subset of G is open).

Now, because GL(n, R)isatopologicalgroup,itcan be shown that every discrete subgroup of GL(n, R) is closed,andinfactcountable.

Thus, discrete subgroups of GL(n, R) are Lie groups!

But these are not very interesting Lie groups so we will consider only closed subgroups of GL(n, R)thatarenot discrete. 145

We wish to define embedded submanifolds in RN .

For the sake of brevity, we use the terminology manifold (but other authors would say embedded submanifold,or something like that).

The intuition behind the notion of a smooth manifold in RN is that a subspace M is a manifold of m if every point p M is contained in some open subset U of M (in the subspace2 topology) that can be parametrized by some ':⌦ U from some open subset ⌦ ! of the origin in Rm,andthat' has some nice properties that allow: (1) The definition of smooth functions on M and (2) The definition of the tangent space at p.

For this, ' has to be at least a , but more is needed: ' must be smooth, and the '0(0m) at the origin must be injective (letting 0m =(0,...,0)). m

| {z } Manifolds

N 146 CHAPTER 4.Submanifolds MANIFOLDS, LIE GROUPS, embedded AND LIE ALGEBRAS; in R “BABY CASE”

EN

p m U E

M t0

The function Figure 4.1: A manifold in RN : U Definition 4.1. Given any ! N,m,with is called a (local) parametrization of M at p. If 0m and (0m)=p, we say that is N m 1, an m-dimensional2 smooth manifold in centered at p. RN , for short a manifold,isanonemptysubsetM of ComputationalRN such Manifolds that and for Applications every point (CMA) p- 2011,M IMPA,there Rio de are Janeiro, two RJ, open Brazil 3 2 ⌦ Rm and U M,withp U,andasmooth ✓ ✓ 2 function ':⌦ RN such that ' is a homeomorphism ! between ⌦and U = '(⌦), and '0(t0)isinjective,where 1 t0 = ' (p).

The function ':⌦ U is called a (local) parametriza- ! tion of M at p.If0m ⌦and'(0m)=p,wesaythat ':⌦ U is centered at2 p. ! 147

Recall that M RN is a topological space under the subspace topology,✓ and U is some open subset of M in the subspace topology, which means that U = M W \ for some open subset W of RN .

Since ':⌦ U is a homeomorphism, it has an inverse 1 ! ' : U ⌦thatisalsoahomeomorphism,calleda (local) chart! .

Since ⌦ Rm,foreveryp M and every parametriza- ✓ 2 1 tion ':⌦ U of M at p,wehave' (p)=(z ,...,z ) ! 1 m for some zi R,andwecallz1,...,zm the local coordi- 2 1 nates of p (w.r.t. ' ). 148 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” We often refer to a manifold M without explicitly speci- fying its dimension (the m).

Intuitively, a chart provides a “flattened” local of a region on a manifold.

Remark: We could allow m =0indefinition4.1.Ifso, amanifoldofdimension0isjustasetofisolatedpoints, and thus it has the discrete topology.

In fact, it can be shown that a discrete subset of RN is countable. Such manifolds are not very exciting, but they do correspond to discrete subgroups. 149

Example 4.1. The S2 in R3 defined such that 2 3 2 2 2 S = (x, y, z) R x + y + z =1 2 | is a smooth 2-manifold, because it can be parametrized using the following two '1 and '2:

2u 2v u2 + v2 1 ' :(u, v) , , 1 7! u2 + v2 +1 u2 + v2 +1 u2 + v2 +1 ✓ ◆ and 2u 2v 1 u2 v2 ' :(u, v) , , . 2 7! u2 + v2 +1 u2 + v2 +1 u2 + v2 +1 ✓ ◆ 150 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE”

The map '1 corresponds to the inverse of the stereo- graphic from the north pole N =(0, 0, 1) onto the z =0,andthemap'2 corresponds to the in- verse of the from the south pole S =(0, 0, 1) onto the plane z =0,asillustratedin Figure 4.2.

The reader should check that the map '1 parametrizes 2 2 S N and that the map '2 parametrizes S S (and that{ } they are smooth, , etc.).{ }

Using '1,theopenlowerhemisphereisparametrizedby the open of O and radius 1 contained in the plane z =0.

1 The chart '1 assigns local coordinates to the points in the open lower hemisphere. 151 1

N

O '1(u, v)

(u, v) z =0 S '2(u, v)

Figure 4.2: Inverse stereographic projections

We urge our readers to define a manifold structure on a . This can be done using four charts.

Every open subset of RN is a manifold in a trivial way. Indeed, we can use the inclusion map as a parametriza- tion. 152 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE”

2 In particular, GL(n, R)isanopensubsetofRn ,since its is closed (the set of invertible matrices is the inverse of the function, which is continuous).

Thus, GL(n, R)isamanifold.WecanviewGL(n, C)as 2 asubsetofR(2n) using the defined as follows:

For every complex n n matrix A,constructthereal 2n 2n matrix such⇥ that every entry a + ib in A is replaced⇥ by the 2 2block ⇥ a b ba ✓ ◆ where a, b R. 2 It is immediately verified that this map is in fact a group . 153 Thus, we can view GL(n, C)asasubgroupofGL(2n, R), 2 and as a manifold in R(2n) .

A1-manifoldiscalleda(smooth) ,anda2-manifold is called a (smooth) (although some authors re- quire that they also be connected).

The following two lemmas provide the link with the defi- nition of an abstract manifold.

Lemma 4.1. Given an m-dimensional manifold M in RN , for every p M there are two open sets O, W N 2 ✓ R with 0N O and p M W , and a smooth di↵eomorphism2 ': O W2, such\ that ! '(0N )=p and m '(O (R 0N m )) = M W. \ ⇥{ } \ 154 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” There is an open subset ⌦of Rm such that

m O (R 0N m )=⌦ 0N m , \ ⇥{ } ⇥{ } and the map :⌦ RN given by !

(x)='(x, 0N m) is an and a homeomorphism onto U = W M; so is a parametrization of M at p. \

We can think of ' as a promoted version of which is actually a di↵eomorphism between open subsets of RN ; see Figure 4.3. 155

W

p U

M

φ

Ω

ON

O

Figure 4.3: An illustration of Lemma 4.1, where M is a surface embedded in R3,namely m =2andN =3. 156 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” The next lemma is easily shown from Lemma 4.1. It is a key technical result used to show that interesting properties of maps between manifolds do not depend on parametrizations.

Lemma 4.2. Given an m-dimensional manifold M in RN , for every p M and any two parametrizations 2 '1 :⌦1 U1 and '2 :⌦2 U2 of M at p, if U1 U2 = ! 1 1! 1 \ 6 , the map '2 '1 : '1 (U1 U2) '2 (U1 U2) is a; smooth di↵eomorphism. \ ! \

1 1 1 The maps '2 '1 : '1 (U1 U2) '2 (U1 U2)are called transition maps. \ ! \

Lemma 4.2 is illustrated in Figure 4.4. 157 1

⌦1

1 ' (U U ) 1 1 \ 2 '1 U1

1 ' ' U1 U2 2 1 \

'2 U2 1 ' (U U ) 2 1 \ 2 ⌦2

Figure 4.4: Parametrizations and transition functions

Using Definition 4.1, it may be quite hard to prove that aspaceisamanifold.Therefore,itishandytohave alternate characterizations such as those given in the next Proposition: 158 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE”

Proposition 4.3. A subset, M Rm+k, is an m- dimensional manifold i↵either ✓ (1) For every p M, there is some open subset, W 2 ✓ Rm+k, with p W and a (smooth) , k 2 1 f : W R , so that W M = f (0), or ! \ (2) For every p M, there is some open subset, W 2 ✓ Rm+k, with p W and a (smooth) map, k 2 f : W R , so that f 0(p) is surjective and ! 1 W M = f (0). \ See Figure 4.5. 159

W

p

f M

-1 f (0) = W ∩ M

0

Figure 4.5: An illustration of Proposition 4.3, where M is the torus, m =2,andk = 1. Note 1 that f (0) is the pink patch of the torus, i.e. the zero of the open W . 160 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” Observe that condition (2), although apparently weaker than condition (1), is in fact equivalent to it, but more convenient in practice.

This is because to say that f 0(p)issurjectivemeansthat the Jacobian matrix of f 0(p)hasrankk,whichmeans that some determinant is nonzero, and because the de- terminant function is continuous this must hold in some open subset W W containing p. 1 ✓

Consequently, the restriction, f1,off to W1 is indeed a submersion and 1 1 f (0) = W f (0) = W W M = W M. 1 1 \ 1 \ \ 1 \

The proof is based on two technical lemmas that are proved using the theorem. 161 Lemma 4.4. Let U Rm be an open subset of Rm ✓ and pick some a U.Iff : U Rn is a smooth immersion at a, i.e.,2 df is injective! (so, m n), a  then there is an , V Rn, with f(a) V , ✓ 2 an open subset, U 0 U, with a U 0 and f(U 0) V , ✓ n m 2 ✓ an open subset O R , and a di↵eomorphism, ✓ ✓ : V U 0 O, so that ! ⇥

✓(f(x1,...,xm)) = (x1,...,xm, 0,...,0), for all (x1,...,xm) U 0, as illustrated in the diagram below 2

f U 0 U / f(U 0) V ✓ ✓ ✓ in1 ( ✏ U 0 O ⇥ where in1(x1,...,xm)=(x1,...,xm, 0,...,0); see Fig- ure 4.6. 162 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE”

V

Θ f(a) f(U ‘)

U’ x O

f

a U‘ U

Figure 4.6: An illustration of Lemma 4.4, where m =2andn = 3. Note that U 0 is the base of the solid and ✓ is the di↵eomorphism between the solid cylinder and the solid gourd shaped V . The composition ✓ f injects U 0 into U 0 O. ⇥ 163 Lemma 4.5. Let W Rm be an open subset of Rm ✓ and pick some a W .Iff : W Rn is a smooth submersion at a,2 i.e., df is surjective! (so, m n), a then there is an open set, V W Rm, with a V , ✓ ✓ 2 and a di↵eomorphism, , with domain O Rm, so that (O)=V and ✓

f( (x1,...,xm)) = (x1,...,xn), for all (x1,...,xm) O, as illustrated in the diagram below 2

m m O R / V W R ✓⇡ ✓ ✓ ✏ f Rn, u where ⇡(x1,...,xm)=(x1,...,xn); see Figure 4.7. 164 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE”

W

V f O

a f(a ) Ψ

x = Ψ -1(a)

Figure 4.7: An illustration of Lemma 4.5, where m =3andn = 2. Note that is the di↵eomorphism between the 0 and the solid purple ball V .Thecompositionf projects O onto its equatorial pink disk. 165

Theorem 4.6. A nonempty subset, M RN , is an m-manifold (with 1 m N) i↵any of✓ the following conditions hold:   (1) For every p M, there are two open subsets ⌦ 2 ✓ Rm and U M, with p U, and a smooth func- ✓ 2 tion ':⌦ RN such that ' is a homeomorphism ! between ⌦ and U = '(⌦), and '0(0) is injective, where p = '(0). (2) For every p M, there are two open sets O, W N 2 ✓ R with 0N O and p M W , and a smooth 2 2 \ di↵eomorphism ': O W , such that '(0N )=p and ! m '(O (R 0N m )) = M W. \ ⇥{ } \ (3) For every p M, there is some open subset, W 2 ✓ RN , with p W and a smooth submersion N2 m 1 f : W R , so that W M = f (0). ! \ 166 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” (4) For every p M, there is some open subset, W N 2 ✓ R , and N m smooth functions, fi : W R, ! so that the linear forms df 1(p),...,dfN m(p) are linearly independent and 1 1 W M = f1 (0) fN m(0). \ \···\ See Figure 4.8.

p

M W

-1 f1 (0) p

-1 f2 (0) 0

Figure 4.8: An illustration of Condition (4) in Theorem 4.6, where N =3andm =1.The manifold M is the helix in R3.ThedarkgreenportionofM is magnified in to show 1 1 that it is the intersection of the pink surface, f1 (0), and the blue surface, f2 (0). 167 Condition (4) says that locally (that is, in a small open set of M containing p M), M is “cut out” by N m 2 smooth functions, fi : W R,inthesensethatthe portion of the manifold M! W is the intersection of \1 the N m , f (0), (the zero-level sets of i the fi)andthatthisintersectionis“clean”,whichmeans that the linear forms df 1(p),...,dfN m(p)arelinearly independent. 168 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” As an illustration of Theorem 4.6, the sphere n n+1 2 S = x R x 1=0 { 2 |k k2 } is an n-dimensional manifold in Rn+1.

n+1 2 Indeed, the map f : R R given by f(x)= x 2 1 is a submersion, since ! k k n+1

df (x)(y)=2 xkyk. Xk=1

n(n 1) The rotation group, SO(n), is an 2 -dimensional 2 manifold in Rn .

+ 2 Indeed, GL (n)isanopensubsetofRn (recall, GL+(n)= A GL(n) det(A) > 0 )andiff is defined by { 2 | } f(A)=A>A I, where A GL+(n), then f(A)issymmetric,so 2 n(n+1) f(A) S(n)=R 2 . 2 169 We showed earlier that

df (A)(H)=A>H + H>A. But then, df (A)issurjectiveforallA SO(n), because if S is any , we see that2 AS df (A) = S. 2 ✓ ◆ 1 As SO(n)=f (0), we conclude that SO(n)isindeed amanifold.

AsimilarargumentprovesthatO(n)isan n(n 1) 2 -dimensional manifold.

Using the map, f : GL(n) R,givenbyA det(A), we can prove that SL(n)isamanifoldofdimension! 7! n2 1. 1 Remark: We have df (A)(B)=det(A)tr(A B), for every A GL(n). 2 170 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” Aclassofmanifoldsgeneralizingthespheresandtheor- thogonal groups are the Stiefel manifolds.

For any n 1andanyk with 1 k n,letS(k, n)be the set of all orthonormal k-frames ;thatis,of k-tuples n of orthonormal vectors (u1,...,uk)withui R . 2 n 1 Obviously S(1,n)=S ,andS(n, n)=O(n).

Every orthonormal k-frame (u1,...,uk)canberepre- sented by an n k matrix Y over the canonical ⇥ of Rn,andsuchamatrixY satisfies the

Y >Y = I.

Thus, S(k, n)canbeviewedasasubspaceofMn,k.We claim that S(k, n)isamanifold.

Let W = A Mn,k det(A>A) > 0 ,anopensubset of M such{ that2 S(k,| n) W . } n,k ✓ 171 Generalizing the situation involving SO(n), define the function f : W S(k)by !

f(A)=A>A I. Basically the same computation as in the case of SO(n) yields

df (A)(H)=A>H + H>A.

The proof that df (A)issurjectiveforallA S(k, n)is 2 the same as before, because only the equation A>A = I is needed.

1 As S(k, n)=f (0), we conclude that S(k, n)isasmooth manifold of dimension

k(k +1) k(k 1) nk = k(n k)+ . 2 2 172 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” The third characterization of Theorem 4.6 suggests the following definition.

Definition 4.2. Let f : Rm+k Rk be a smooth func- ! tion. A point, p Rm+k,iscalledacritical point (of 2 k f) i↵ df p is not surjective and a point q R is called a critical value (of f) i↵ q = f(p), for some2 critical point, p Rm+k. 2 Apointp Rm+k is a regular point (of f) i↵ p is not 2 k critical, i.e., df p is surjective, and a point q R is a regular value (of f) i↵it is not a critical value.2

In particular, any q Rk f(Rm+k)isaregularvalue m+k 2 1 and q f(R )isaregularvaluei↵every p f (q) is a regular2 point (but, in contrast, q is a critical2 value i↵ 1 some p f (q)iscritical). 2 173 Part (3) of Theorem 4.6 implies the following useful propo- sition:

Proposition 4.7. Given any smooth function, f : Rm+k Rk, for every regular value, q f(Rm+k), ! 1 2 the preimage, Z = f (q), is a manifold of dimension m.

Definition 4.2 and Proposition 4.7 can be generalized to manifolds

Regular and critical values of smooth maps play an im- portant role in di↵erential topology.

Firstly, given a smooth map, f : Rm+k Rk,almost ! every point of Rk is a regular value of f.

To make this statement precise, one needs the notion of a set of zero. 174 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” Then, Sard’s theorem says that the set of critical values of a smooth map has measure zero.

Secondly, if we consider smooth functions, f : Rm+1 R, m+1 ! apointp R is critical i↵ df p =0. 2 Then, we can use second order to further clas- sify critical points. The Hessian matrix of f (at p)is the matrix of second-order partials @2f H (p)= (p) f @x @x ✓ i j ◆ and a critical point p is a nondegenerate critical point if Hf (p)isanonsingularmatrix. 175 The remarkable fact is that, at a nondegenerate critical point, p,thelocalbehavioroff is completely determined, in the sense that after a suitable change of coordinates (given by a smooth di↵eomorphism) f(x)=f(p) x2 x2 + x2 + + x2 1 ··· +1 ··· m+1 near p,where called the index of f at p is an integer which depends only on p (in fact, is the number of negative eigenvalues of Hf (p)).

This result is known as Morse lemma (after Marston Morse, 1892-1977).

Smooth functions whose critical points are all nondegen- erate are called Morse functions.

It turns out that every smooth function, f : Rm+1 R, gives rise to a large supply of Morse functions by adding! alinearfunctiontoit. 176 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE”

More precisely, the set of a Rm+1 for which the function 2 fa given by f (x)=f(x)+a x + + a x a 1 1 ··· m+1 m+1 is not a Morse function has measure zero.

Morse functions can be used to study topological proper- ties of manifolds.

In a sense to be made precise and under certain technical conditions, a Morse function can be used to reconstuct a manifold by attaching cells, up to equiv- alence.

However, these results are way beyond the scope of these notes. 177 Let us now review the definitions of a smooth curve in a manifold and the at a point of a curve.

Definition 4.3. Let M be an m-dimensional manifold in RN .Asmooth curve in M is any function : I ! M where I is an open in R and such that for every t I,lettingp = (t), there is some parametrization ':2⌦ U of M at p and some open interval (t ✏, t + ! 1 m ✏) I such that the curve ' :(t ✏, t + ✏) R is smooth.✓ The notion of a smooth curve is illustrated! in Figure 4.9.

Using Lemma 4.2, it is easily shown that Definition 4.3 does not depend on the choice of the parametrization ':⌦ U at p. ! Lemma 4.2 also implies that viewed as a curve : I RN is smooth. ! 178 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE”

γ

t

γ(t) = p

p U M

φ φ -1 φ-1(p) = 0

0 Ω

Figure 4.9: A smooth curve in a manifold M. 1

0(t)

p M

Figure 4.10: Tangent vector to a curve on a manifold. 179

Then the tangent vector to the curve : I RN at t, ! denoted by 0(t), is the value of the derivative of at t (a vector in RN )computedasusual: (t + h) (t) 0(t)=lim . h 0 h 7! Given any point p M,wewillshowthatthesetof tangent vectors to all2 smooth in M through p is a isomorphic to the vector space Rm.

Given a smooth curve : I M,foranyt I,letting p = (t), since M is a manifold,! there is a parametriza-2 tion ':⌦ U such that '(0m)=p U and some open interval J ! I with t J and such that2 the function ✓ 2 1 m ' : J R ! is a smooth curve, since is a smooth curve.

1 Letting ↵ = ' ,thederivative↵0(t)iswell-defined, and it is a vector in Rm. 180 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” But ' ↵: J M is also a smooth curve, which agrees with on J,andbythechainrule,!

0(t)='0(0m)(↵0(t)), since ↵(t)=0m (because '(0m)=p and (t)=p).

N Observe that 0(t)isavectorinR .

Now, for every vector v Rm,thecurve↵: J Rm defined such that 2 ! ↵(u)=(u t)v for all u J is clearly smooth, and ↵0(t)=v. 2 181 This shows that the set of tangent vectors at t to all m smooth curves (in R )passingthrough0m is the entire vector space Rm.

Since every smooth curve : I M agrees with a curve of the form ' ↵: J M for! some smooth curve ! ↵: J Rm (with J I)asexplainedabove,andsince ! ✓ it is assumed that '0(0m)isinjective,'0(0m)mapsthe vector space Rm injectively to the set of tangent vectors to at p,asclaimed.

All this is summarized in the following definition. 182 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” Definition 4.4. Let M be an m-dimensional manifold N in R .Foreverypointp M,thetangent space TpM 2 N at p is the set of all vectors in R of the form 0(0), where : I M is any smooth curve in M such that p = (0). ! m The set TpM is a vector space isomorphic to R .Every vector v T M is called a tangent vector to M at p. 2 p

We can now define Lie groups.

Definition 4.5. A Lie group is a nonempty subset G of RN (N 1) satisfying the following conditions: (a) G is a group. (b) G is a manifold in RN . (c) The group operation : G G G and the inverse 1 · ⇥ ! map : G G are smooth. !

Actually, we haven’t defined yet what a smooth map be- tween manifolds is (in clause (c)). 183 This notion is explained in Definition 4.8, but we feel that most readers will appreciate seeing the formal definition aLiegroup,asearlyaspossible.

It is immediately verified that GL(n, R)isaLiegroup. Since all the Lie groups that we are considering are sub- groups of GL(n, R), the following definition is in order.

Definition 4.6. A linear Lie group is a subgroup G of GL(n, R)(forsomen 1) which is a smooth manifold 2 in Rn .

Let Mn(R)denotethesetofallrealn n matrices (in- vertible or not). If we recall that the exponential⇥ map exp: A eA 7! is well defined on Mn(R), we have the following crucial theorem due to Von Neumann and Cartan: 184 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” Theorem 4.8. (Von Neumann and Cartan, 1927) A nondiscrete closed subgroup G of GL(n, R) is a linear Lie group. Furthermore, the set g defined such that tX g = X Mn(R) e G for all t R { 2 | 2 2 } is a nontrivial vector space equal to the tangent space TIG at the identity I, and g is closed under the Lie bracket [ , ] defined such that [A, B]=AB BA for all A, B Mn(R). 2

Theorem 4.8 applies even when G is a discrete subgroup, but in this case, g is trivial (i.e., g = 0 ). { }

For example, the set of nonzero reals R⇤ = R 0 = { } GL(1, R)isaLiegroupundermultiplication,andthe subgroup n H = 2 n Z { | 2 } is a discrete subgroup of R⇤.Thus,H is a Lie group.

On the other hand, the set Q⇤ = Q 0 of nonzero { } rational numbers is a multiplicative subgroup of R⇤,but it is not closed, since Q is dense in R. 185 If G is closed and not discrete, we must have n 1, and g has dimension n.

The first step to prove Theorem 4.8 is this:

Proposition 4.9. Given any closed subgroup G in GL(n, R), the set

g = X Mn(R) X = 0(0),: J G { 2 1 | ! is a C curve in Mn(R) such that (0) = I } satisfies the following properties:

(1) g is a vector subspace of Mn(R). tX (2) For every X Mn(R), we have X g i↵ e G 2 2 2 for all t R. 2 (3) For every X g and for every g G, we have 1 2 2 gXg g. 2 (4) g is closed under the Lie bracket.

The second step to prove Theorem 4.8 is this: 186 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” Proposition 4.10. Let G be a subgroup of GL(n, R), and assume that G is closed and not discrete. Then, dim(g) 1, and the exponential map is a di↵eomor- phism of a neighborhood of 0 in g onto a neighborhood of I in G. Furthermore, there is an open subset ⌦ ✓ Mn(R) with 0n,n ⌦, an open subset W GL(n, R) with I W , and2 a di↵eomorphism :⌦✓ W such that 2 ! (⌦ g)=W G. \ \

With the help of Theorem 4.8 it is now very easy to prove that SL(n), O(n), SO(n), SL(n, C), U(n), and SU(n) are Lie groups. It suces to show that these subgroups of GL(n, R)(GL(2n, R)inthecaseofSL(n, C), U(n), and SU(n)) are closed.

We can also prove that SE(n)isaLiegroupasfollows. 187 Recall that we can view every element of SE(n)asareal (n +1) (n +1)matrix ⇥ RU 01 ✓ ◆ where R SO(n)andU Rn. 2 2 In fact, such matrices belong to SL(n +1).

This embedding of SE(n)intoSL(n +1)isagroupho- momorphism, since the group operation on SE(n)corre- sponds to multiplication in SL(n +1):

RS RV + U RU SV = . 01 01 01 ✓ ◆ ✓ ◆✓ ◆ Note that the inverse is given by

R 1 R 1U R R U = > > . 01 01 ✓ ◆ ✓ ◆ 188 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” Also note that the embedding shows that as a manifold, SE(n)isdi↵eomorphictoSO(n) Rn (given a manifold ⇥ M1 of dimension m1 and a manifold M2 of dimension m ,theproductM M can be given the structure of 2 1 ⇥ 2 amanifoldofdimensionm1 + m2 in a natural way).

Thus, SE(n)isaLiegroupwithunderlyingmanifold SO(n) Rn,andinfact,asubgroupofSL(n +1). ⇥

Even though SE(n)isdi↵eomorphictoSO(n) Rn  ⇥ as a manifold, it is not isomorphic to SO(n) Rn as a group, because the group multiplication on SE⇥(n)is not the multiplication on SO(n) Rn.Instead,SE(n) ⇥ is a of SO(n)byRn.

Returning to Theorem 4.8, the vector space g is called the Lie algebra of the Lie group G.

Lie algebras are defined as follows. 189 Definition 4.7. A (real) Lie algebra is a real vector space together with a bilinear map [ , A]: called the Lie bracket on such that· · theA following⇥A! twoA identities hold for all a, b, cA : 2A [a, a]=0, and the so-called Jacobi identity

[a, [b, c]] + [c, [a, b]] + [b, [c, a]] = 0.

It is immediately verified that [b, a]= [a, b].

In view of Theorem 4.8, the vector space g = TIG associ- ated with a Lie group G is indeed a Lie algebra. Further- more, the exponential map exp: g G is well-defined. ! 190 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” In general, exp is neither injective nor surjective, as we observed earlier.

Theorem 4.8 also provides a kind of recipe for “comput- ing” the Lie algebra g = TIG of a Lie group G.

Indeed, g is the tangent space to G at I,andthuswecan use curves to compute tangent vectors.

Actually, for every X T G,themap 2 I : t etX X 7! is a smooth curve in G,anditiseasilyshownthat

X0 (0) = X.Thus,wecanusethesecurves.

As an illustration, we show that the Lie algebras of SL(n) and SO(n)arethematriceswithnulltraceandtheskew symmetric matrices. 191 Let t R(t)beasmoothcurveinSL(n)suchthat R(0) =7!I.Wehavedet(R(t)) = 1 for all t ( ✏,✏). 2 Using the , we can compute the derivative of the function

t det(R(t)) 7! at t =0,andweget

detI0 (R0(0)) = 0.

It is an easy exercise to prove that

detI0 (X)=tr(X), and thus tr(R0(0)) = 0, which says that the tangent vec- tor X = R0(0) has null . 192 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” Another proof consists in observing that X sl(n, R)i↵ 2 det(etX)=1 for all t R.Sincedet(etX)=etr(tX),fort =1,weget tr(X)=0,asclaimed.2

Clearly, sl(n, R)hasdimensionn2 1. Let t R(t)beasmoothcurveinSO(n)suchthat R(0) =7!I.SinceeachR(t)isorthogonal,wehave

R(t) R(t)> = I for all t ( ✏,✏). 2 193 Taking the derivative at t =0,weget

R0(0) R(0)> + R(0) R0(0)> =0, but since R(0) = I = R(0)>,weget

R0(0) + R0(0)> =0, which says that the tangent vector X = R0(0) is skew symmetric. 194 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” Since the diagonal elements of a skew symmetric matrix are null, the trace is automatically null, and the condition det(R)=1yieldsnothingnew.

This shows that o(n)=so(n). It is easily shown that so(n)hasdimensionn(n 1)/2. As a concrete example, the Lie algebra so(3) of SO(3) is the real vector space consisting of all 3 3realskew symmetric matrices. Every such matrix is of⇥ the form

0 dc d 0 b 0 cb0 1 @ A where b, c, d R. 2 The Lie bracket [A, B]inso(3) is also given by the usual , [A, B]=AB BA. 195 We can define an isomorphism of Lie algebras :(R3, ) so(3) by the formula ⇥ ! 0 dc (b, c, d)= d 0 b . 0 cb0 1 @ A It is indeed easy to verify that

(u v)=[ (u), (v)]. ⇥ It is also easily verified that for any two vectors 3 u =(b, c, d)andv =(b0,c0,d0)inR ,

(u)(v)=u v. ⇥ In robotics and in , (u)isoftendenoted by u . ⇥ 196 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” The exponential map exp: so(3) SO(3) is given by Rodrigues’s formula (see Lemma 1.12):!

sin ✓ (1 cos ✓) eA =cos✓I + A + B, 3 ✓ ✓2 or equivalently by

sin ✓ (1 cos ✓) eA = I + A + A2 3 ✓ ✓2 if ✓ =0,where 6 0 dc A = d 0 b , 0 cb0 1 @ A

2 2 2 2 2 03 ✓ = pb + c + d , B = A + ✓ I3,andwithe = I3. 197 Using the above methods, it is easy to verify that the Lie algebras gl(n, R), sl(n, R), o(n), and so(n), are respec- tively Mn(R), the set of matrices with null trace, and the set of skew symmetric matrices (in the last two cases).

Asimilarcomputationcanbedoneforgl(n, C), sl(n, C), u(n), and su(n), confirming the claims of Section 1.5.

It is easy to show that gl(n, C)hasdimension2n2, sl(n, C) has dimension 2(n2 1), u(n)hasdimensionn2,and su(n)hasdimensionn2 1. For example, the Lie algebra su(2) of SU(2) (or S3)is the real vector space consisting of all 2 2(complex)skew Hermitian matrices of null trace. ⇥ 198 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” As we showed, SE(n)isaLiegroup,anditsliealgebra se(n)describedinSection1.7iseasilydeterminedasthe subalgebra of sl(n +1)consistingofallmatricesofthe form

BU 00 ✓ ◆ where B so(n)andU Rn. 2 2 Thus, se(n)hasdimensionn(n +1)/2. The Lie bracket is given by

BU CV CV BU 00 00 00 00 ✓ ◆✓ ◆ ✓ ◆✓ ◆ BC CB BV CU = . 00 ✓ ◆ 199 We conclude by indicating the relationship between ho- momorphisms of Lie groups and of Lie algebras.

Definition 4.8. Let M1 (m1-dimensional) and M2 (m2- dimensional) be manifolds in RN .Afunction f : M M is smooth if for every p M there are 1 ! 2 2 1 parametrizations ':⌦1 U1 of M1 at p and :⌦2 U of M at f(p)suchthat! f(U ) U and ! 2 2 1 ✓ 2

1 m2 f ':⌦1 R ! is smooth; see Figure 4.11.

Using Lemma 4.2, it is easily shown that Definition 4.8 does not depend on the choice of the parametrizations ':⌦ U and :⌦ U . 1 ! 1 2 ! 2 200 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE”

M2

U2 f f(p) U1 M 1 p

ψ -1 ψ φ

Ω2

m2 Ω1 m 1

Figure 4.11: An illustration of a smooth map from the torus, M1,tothesolidellipsoidM2. The pink patch on M1 is mapped into interior pink ellipsoid of M2. 201 Asmoothmapf between manifolds is a smooth di↵eo- 1 if f is bijective and both f and f are smooth maps.

Definition 4.9. Let M1 (m1-dimensional) and M2 (m2- dimensional) be manifolds in RN .Foranysmoothfunc- tion f : M M and any p M ,thefunction 1 ! 2 2 1 fp0 : TpM1 Tf(p)M2,calledthetangent map of f at p, or derivative! of f at p, or di↵erential of f at p,is defined as follows: For every v T M and every smooth 2 p 1 curve : I M such that (0) = p and 0(0) = v, ! 1

f 0(v)=(f )0(0). p See Figure 4.12.

The map fp0 is also denoted by df p or Tpf. 202 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE”

M1

p

Tp M1 γ(0) = p f

M2 γ

f(p)

0 I

Figure 4.12: An illustration of the tangent map from TpM1 to Tf(p)M2. 203 Doing a few calculations involving the facts that

1 1 f =(f ') (' )and = ' (' ) and using Lemma 4.2, it is not hard to show that fp0(v) does not depend on the choice of the curve .Itiseasily shown that fp0 is a .

Finally, we define homomorphisms of Lie groups and Lie algebras and see how they are related.

Definition 4.10. Given two Lie groups G1 and ,a (or map) of Lie groups is a function f : G G that is a homomorphism of groups and a 1 ! 2 smooth map (between the manifolds G1 and G2). 204 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE”

Given two Lie algebras 1 and 2,ahomomorphism (or map) of Lie algebras Ais a functionA f : that is A1 !A2 alinearmapbetweenthevectorspaces 1 and 2 and that preserves Lie brackets, i.e., A A

f([A, B]) = [f(A),f(B)] for all A, B . 2A1

An isomorphism of Lie groups is a bijective function f 1 such that both f and f are maps of Lie groups, and an isomorphism of Lie algebras is a bijective function 1 f such that both f and f are maps of Lie algebras.

If f : G G is a homomorphism of Lie groups, then 1 ! 2 fI0 : g1 g2 is a homomorphism of Lie algebras, but in order to! prove this, we need the Ad, so we postpone the proof. 205 The notion of a one-parameter group plays a crucial role in Lie group .

Definition 4.11. Asmoothhomomorphism h:(R, +) G from the additive group R to a Lie group G is called! a one-parameter group in G.

All parameter groups of a linear Lie group can be deter- mined explicitly.

Proposition 4.11. Let G be any linear Lie group. 1. For every X g, the map h(t)=etX is a one- parameter group2 in G. 2. Every one-parameter group h: R G is of the tZ ! form h(t)=e , with Z = h0(0). In summary, for every Z g, there is a unique one- 2 parameter group h such that h0(0) = Z given by h(t)= eZt. 206 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” The exponential map is natural in the following sense:

Proposition 4.12. Given any two linear Lie groups G and H, for every Lie f : G H, the following diagram commutes: ! f / GO HO exp exp g / h df I

Alert readers must have noticed that we only defined the Lie algebra of a .

In the more general case, we can still define the Lie alge- bra g of a Lie group G as the tangent space TIG at the identity I.

The tangent space g = TIG is a vector space, but we need to define the Lie bracket. 207 This can be done in several ways. We explain briefly how this can be done in terms of so-called adjoint represen- tations.

This has the advantage of not requiring the definition of left- vector fields, but it is still a little bizarre!

Given a Lie group G,foreverya G we define left 2 translation as the map La : G G such that La(b)=ab for all b G,andright translation! as the map R : G 2 G such that R (b)=ba for all b G. a ! a 2

The maps La and Ra are di↵eomorphisms, and their derivatives play an important role. 208 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” The inner

Ada = R 1 La (= R 1La) a a of G also play an important role.

Note that 1 Ad (b)=R 1L (b)=aba . a a a

The derivative

(Ad )0 : T G T G a I I ! I of Ad : G G at I is an isomorphism of Lie algebras, a ! and since TIG = g,ifwedenote(Ada)I0 by Ada,weget amap

Ad : g g. a ! 209 The map a Ad is a map of Lie groups 7! a Ad: G GL(g), ! called the adjoint representation of G (where GL(g) denotes the Lie group of all bijective linear maps on g).

In the case of a linear group, we have

1 Ad(a)(X)=Ada(X)=aXa for all a G and all X g. 2 2 We are now almost ready to prove that if f : G G 1 ! 2 is a homomorphism of Lie groups, then fI0 : g1 g2 is a homomorphism of Lie algebras. !

What we need is to express the Lie bracket [A, B]interms of the derivative of an expression involving the adjoint representation Ad. 210 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” For any A, B g,wehave 2

tA tA Ad tA(B) 0(0) = (e Be )0(0) = AB BA =[A, B]. e

Proposition 4.13. If f : G1 G2 is a homomor- phism of linear Lie groups, then! the linear map df : g g satisfies the equation I 1 ! 2

df I(Ada(X)) = Adf(a)(df I(X)), for all a G and all X g1; that is, the following diagram commutes2 2

df I g1 / g2 Ad Ada f(a) ✏ ✏ g1 / g2 df I

Furthermore, df I is a homomorphism of Lie algebras. 211

If some additional assumptions are made about G1 and G2 (for example, connected, simply connected), it can be shown that f is pretty much determined by fI0.

The derivative

Ad0 : g gl(g) I ! of Ad: G GL(g)atI is map of Lie algebras, and if ! we denote AdI0 by ad, it is a map

ad: g gl(g), ! called the adjoint representation of g.

Recall that Theorem 4.8 immediately implies that the Lie algebra gl(g)ofGL(g)isthevectorspaceHom(g, g)of all linear maps on g. 212 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” If we apply Proposition 4.12 to Ad: G GL(g), we obtain the equation !

adA Ad A = e for all A g, e 2 which is a generalization of the identity of Proposition 3.1.

In the case of a linear group, we have

ad(A)(B)=[A, B] for all A, B g. 2 This can be shown using Propositions 3.1 and 4.12 213 One can also check that the Jacobi identity on g is equiv- alent to the fact that ad preserves Lie brackets, i.e., ad is amapofLiealgebras:

ad([A, B]) = [ad(A), ad(B)] for all A, B g 2

(where on the right, the Lie bracket is the commutator of linear maps on g).

Thus, we recover the Lie bracket from ad.

This is the key to the definition of the Lie bracket in the case of a general Lie group (not just a linear Lie group).

We define the Lie bracket on g as

[A, B]=ad(A)(B). 214 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” To be complete, we still have to define the exponential map exp: g G for a general Lie group. ! For this we need to introduce some left-invariant vector fields induced by the derivatives of the left translations and curves associated with such vector fields.

We conclude this section by computing explicitly the ad- joint representations ad of so(3) and Ad of SO(3).

Recall that for every X so(3), ad is a linear map 2 X ad : so(3) so(3). X ! Also, for every R SO(3), the map Ad : so(3) so(3) 2 R ! is an invertible linear map of so(3).

Now, as we saw earlier, so(3) is isomorphic to (R3, ), ⇥ where is the cross-product on R3, via the isomorphism ⇥ :(R3, ) so(3) given by the formula ⇥ ! 0 cb (a, b, c)= c 0 a . 0 ba0 1 @ A 215 In robotics and in computer vision, (u)isoftendenoted by u . ⇥ 3 The image of the canonical basis (e1,e2,e3)ofR is the following basis of so(3):

00 0 001 E1 = 00 1 ,E2 = 000 , 0 001 01 0 1001 @ @ A @ A0 10 E3 = 100 . 000011 Observe that @ AA

[E1,E2]=E3, [E2,E3]=E1, [E3,E1]=E2. 216 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE” Using the isomorphism ,weobtainanisomorphism between Hom(so(3), so(3)) and M3(R)=gl(3, R)given by

1 (f)= f , where (f)isexpressedinthebasis(e1,e2,e3).

By restricting to GL(so(3)), we obtain an isomorphism between GL(so(3)) and GL(3, R).

It turns out that if we use the basis (E1,E2,E3)inso(3), for every X so(3), the matrix representing ad 2 X 2 Hom(so(3), so(3)) is X itself, and for every R SO(3), 2 the matrix representing Ad GL(so(3)) is R itself. R 2 217 Proposition 4.14. For all X so(3) and all R SO(3), we have 2 2

(adX)=X, (AdR)=R, which means that ad is the inclusion map from so(3) to M3(R)=gl(3, R), and that Ad is the in- clusion map from SO(3) to GL(3, R). Equivalently, for all u R3, we have 2

adX( (u)) = (Xu), AdR( (u)) = (Ru).

These can also be written as

1 [X, u ]=(Xu) ,RuR =(Ru) . ⇥ ⇥ ⇥ ⇥ 218 CHAPTER 4. MANIFOLDS, LIE GROUPS, AND LIE ALGEBRAS; “BABY CASE”