Nilpotent Lie Groups

Chapter 1

Nilpotent Lie groups

In this chapter, we deﬁne the basic notions from Lie theory and explain the relations between Lie groups and Lie algebras.

1.1 Basics of Lie groups and Lie algebras

Deﬁnition 1.1.1. A Lie group G is a group that is at the same time a C∞ diﬀerentiable manifold, such that the group operations are C∞ maps1.

The identity element of G is denoted by e. Write Lg and Rg respectively for the left and right translations by g ∈ G, i.e. Lgh = gh and Rgh = hg for all g ∈ G. For example, for a ﬁnite dimensional vector space V ,

GL(V ) = {linear automorphism A of V with det A 6= 0}

d is a Lie group. GL(V ) is written as GL(d, R) if V = R . The Lie algebra g of G is the tangent space TeG of G at e. For example, the Lie algbera of GL(V ) is the space of endomorphisms gl(V ) = End(V ) = {linear isomorphisms of V }. The Lie algebra is a vector space. However, to be a Lie algebra, the vector spaces needs to be endowed with an additional structure called the Lie bracket, which we will deﬁne now. −1 Given an element g ∈ G, the map Ψg : h → ghg is a continuous automorphism of G. In particular, it sends e to e and is diﬀerentiable.

1In fact, G being a topological manifold with continuous group operations would guar- antee the existence of such a desired C∞ structure, or even real analytic structure. This was Hilbert’s Fifth Problem and was known by the works of Gleason, Montgomery and Zippin in the 1952.

1 CHAPTER 1. NILPOTENT LIE GROUPS 2

Moreover, g → Ψg is a group morphism from G to Aut(G). Let Adg = DeΨg be the derivative at e, then it is a linear automorphism of TeG = g. In ∼ particular, it can be regarded as an element of GL(g) = GL(dim g, R). If G is a linear group, i.e. a subgroup of GL(d, R), it is not hard to see −1 that Adg X = gXg for g ∈ G and X ∈ g ⊂ gl(d, R) = Matd×d(R). Ad : g → Adg is called the Adjoint action, as it is a group action of G on g by automorphisms. Φ : g → Adg is a group morphism from G to GL(g). Let us diﬀerentiate one more time at e: DeΦ is now a map from g to ∼ gl(g) = Matdim G×dim G(R). For X ∈ g, we denote its image DeΦ(X) by adX , called the adjoint map. For X,Y ∈ g, adX Y is also denoted by [X,Y ], called the Lie bracket.

Deﬁnition 1.1.2. The Lie algebra of G is the tangent space g = TeG, equipped with the Lie bracket [·, ·]: g × g → g. Some times g is also denoted as Lie(G).

Example 1.1.3. If G ⊂ GL(d, R) is a linear group, then [X,Y ] = XY −YX.

Proof. To calculate the image adX ∈ gl(g), we diﬀerentiate the smooth curve Adg(t) ∈ GL(g) at t = 0 for a smooth curve g : R → G with g(0) = e and g0(0) = X. Indeed:

0 −1 0 adX (Y ) =(Adg(t) Y ) |t=0 = (g(t)Y g(t) ) |t=0 0 −1 0 −1 =(g (t)Y g(t) + g(t)Y (g(t) g (t)g(t) ))|t=0 =g0(t)Y − Y g0(t) = XY − YX.

In this example, it is clear that when G is a linear Lie group, [X,Y ] = adX Y satisﬁes: • (biliearity) [X, aY + bZ] = a[X,Y ] + b[X,Z] and [aX + bZ, Y ] = a[X,Y ] + b[Z,Y ];

• (skew-symmetricity) [X,Y ] = −[Y,X];

• (Jacobi identity) [X, [Y,Z]] + [Y, [Z,X]] + [Z, [X,Y ]] = 0.

In fact, these also hold for general Lie groups.

Proposition 1.1.4. If G is a Lie group, then the Lie bracket on its Lie algebra g is a bilinear form and satisﬁes skew-symmetricity and the Jacobi idenity. CHAPTER 1. NILPOTENT LIE GROUPS 3

Proof of skew-symmetricity. Bilinearity of [·, ·] is clear from the deﬁnition. We give the proof of skew-symmetricity, while that of the Jacobi identity is left as Exercise 1.1.1. For X,Y ∈ G, let g(t), h(t) be smooth curves such that g(0) = h(0) = e, g0(0) = X, h0(0) = Y . The ﬁrst claim is that

0 (g(t)h(t)) |t=0 = X + Y. (1.1)

To see this, it suﬃces to note that 0 ∂ ∂ (g(t)h(t)) |t=0 = (g(r)h(s)) + (g(r)h(s)) ∂r ∂s r=s=0 ∂ ∂ −1 = (g(r)h(s)) + (g(r)h(s)g(r) g(r)) ∂r ∂s r=s=0 0 0 = Rh(s)g (r)|r=0 s=0 + (Rg(r))∗ Adg(r) h (s)|s=0 r=0 =X + Y.

An immediate consequence is the special case h(t) = g(t)−1, which yields that g0(t) + (g(t)−1)0 = 0 for all smooth curve g(t) with g(0) = 0. ∂ ∂ −1 −1 Now we may consider (∗) = ∂t ∂s g(t)h(s)g(t) h(s) . On the one hand, by earlier claims, ∂ (∗) = (Ad Y − Y ) = ad Y = [X,Y ]. ∂t g(t) X On the other hand, by the same argument, ∂ ∂ ∂ ∂ (∗) = g(t)h(s)g(t)−1h(s)−1 = − h(s)g(t)h(s)−1g(t)−1 ∂s ∂t ∂s ∂t = − [Y,X].

The skew-symmetricity is hence established.

Lie algebras can be abstractly deﬁned using these properties.

Deﬁnition 1.1.5. A Lie algebra is a vector space equipped with a Lie bracket [·, ·], which is a bilinear form that satisfy both skew-symmetricity and the Jacobi identity.

Deﬁnition 1.1.6. A Lie algebra g is abelian if its Lie bracket is trivial, i.e. [X,Y ] = 0 for all X, Y . CHAPTER 1. NILPOTENT LIE GROUPS 4

Observe that the skew-symmetricity guarantees that [X,X] = 0.

Deﬁnition 1.1.7. A Lie group morphism is a smooth map between two Lie groups that is at the same time a group morphism. A Lie algebra morphism is a linear transform ψ between two algebras such that [ψX, ψY ] = ψ[X,Y ] for all X, Y .

Note that ad : X → adX is a map from a Lie algebra to gl(g), which is a Lie algebra itself.

Lemma 1.1.8. ad is a Lie algebra morphism, i.e. [adX , adY ] = ad[X,Y ] for all X,Y ∈ g.

Proof. It suﬃces to show that

[adX , adY ]Z = adX adY Z − ad Y adX Z = [X, [Y,Z]] − [Y, [X,Z]] =[X, [Y,Z]] + [Y, [Z,X]] and ad[X,Y ] Z = [[X,Y ],Z] = −[Z, [X,Y ]] are equal, which follows from the Jacobi identity.

The left translation Lg by an element g ∈ G sends h to hg. It induces a pushforward (Lg)∗ from ThG to Thg(G). In particular, for all X ∈ g = TeG, (Lg)∗X ∈ TgG defines a smooth vector field on G as g varies in G. Lemma 1.1.9. The ordinary differential equation

0 g (t) = (Lg(t))∗X (1.2) with initial data g(0) = e has a unique solution g(t), which is itself a group morphism from R to G.

Proof. Since the right hand side (Lg(t))∗X of the ODE depends Lipschitz continuously (which is because the group operations are C∞), by the basic theory of differential equations, the equation, conditional to the initial con- dition g(0) = e, has a unique solution g(t) at least locally near t = 0, i.e. in t ∈ (−τ, τ) for some τ. We now show that if t, s, t + s ∈ (−τ, τ), then g(t)g(s) = g(t + s). In fact, ∂ (g(t)g(s)) = (L ) g0(s) = (L ) (L ) X = (L ) X, (1.3) ∂s g(t) ∗ g(t) ∗ g(s) ∗ g(t)g(s) ∗ CHAPTER 1. NILPOTENT LIE GROUPS 5 and ∂ g(t + s) = g0(t + s) = (L ) X. ∂s g(t+s) ∗ 0 So g(t)g(s) and g(t+s) solve the same ODE h (s) = (Lh(s))∗X in s, with the same initial data g(t)g(s)|s=0 = g(t + s)|s=0 = g(t), and therefore coincide τ τ by uniqueness. In particular, g(t) commutes with g(s) for all t, s ∈ (− 2 , 2 ). Given this property, one can easily extend the map g :(−τ, τ) → G to n τ τ (−∞, ∞) by g(nt) = g(t) for all n ∈ N and s ∈ (− 2 , 2 ). One can check that the definition is self consistent, as if nt = ms for n, m ∈ N, then they both t s n m n n m m equal mnr where r = m = n . Then g(t) = (g(r) ) = (g(r) ) = g(s) . Moreover, for all t, s ∈ R, there exists n such that t = nt1 and s = ns1 τ τ n n n for some t1, s1 ∈ (− 2 , 2 ), and hence g(t)g(s) = g(t1) g(s1) = g(t1 + s1) = g(n(t1 + s1)) = g(t + s). So t → g(t) is a group morphism from R. At the same time, observe g is differentiable at every t. This is because for a fixed t n large value of n, g(t1) is differentiable at t1 = n , and g(t) = g(t1) depends differentiably on g(t1). It remains to prove that g(t) is a global solution to (1.2) . Notice that as g(s) is a solution to the equation (1.2) near s = 0, (1.3) still applies. 0 ∂ ∂ Thus g (t) = ∂s g(t + s) s=0 = ∂s (g(t)g(s)) s=0 = (Lg(t)g(0))∗X = (Lg(t))∗X. Hence g(t) is a solution for all t ∈ R. In particular, it is unique by the fundamental theorem of ordinary differential equations.

Deﬁnition 1.1.10. Given X ∈ g, the element g(1) is denoted by exp X, where g(t) is the solution in Lemma 1.1.9. The map exp : g → G is called the exponential map. Obviously, exp 0 = e, since for X = e, the equation (1.2) has constant solutions. Another easy corollary to Lemma 1.1.9 is that {exp(tX)}t∈R is a one-parameter subgroup, i.e. a continuous group morphism from R.

Lemma 1.1.11. For X ∈ g, {exp(tX)}t∈R is the unique one-parameter subgroup g(t) satisifying g0(0) = X. Proof. If g(t) is such a group, then g(0) = e and ∂ g0(t) = (g(t)g(s))| = (L ) g0(s)| = (L ) X. ∂s s=0 g(t) ∗ s=0 g(t) ∗ That is, g(t) satisﬁes (1.2). Hence by Lemma 1.1.9, exp(tX) is the unique solution.

In particular, exp(X)−1 = exp(−X). CHAPTER 1. NILPOTENT LIE GROUPS 6

We remark that exp is locally a diﬀeomorphism near 0 ∈ g. This follows from the following

Lemma 1.1.12. The map exp is diﬀerentiable at 0, and D0 exp = Id.

0 Proof. (D0 exp)X = (exp(tX)) |t=0 = (Lexp(tX))∗X|t=0 = (Le)∗X = X. In general, exp does not have to be either injective or surjective, even if G is connected. Though we will see later in this chapter that, for the main object of these notes, namely connected nilpotent Lie groups, exp is a diﬀeomorphism between g and G. We now state two consequences of Lemma 1.1.12.

Corollary 1.1.13. If Ψ: G → H is a Lie group morphism between two Lie groups G and H. Then DeΨ is a Lie algebra morphism and the following diagram commutes: g −−−−→DeΨ h   exp exp y y G −−−−→Ψ H where g = Lie(G), h = Lie(H).

Proof. To show DeΨ is a Lie algebra morphism, observe that for X,Y ∈ g, ∂ ∂ D Ψ([X,Y ]) =D Ψ( Ad Y ) = D Ψ(Ad Y ) e e ∂t exp tX ∂t e exp tX ∂ ∂ = Ψ(exp tX exp sY exp(−tX)) ∂t ∂s ∂ ∂ = Ψ(exp tX)Ψ(exp sY )Ψ(exp(−tX)) (1.4) ∂t ∂s ∂ = AdΨ(exp tX) Y = ad ∂ Ψ(exp tX) DeΨ(Y ) ∂t ∂t

= adDeΨ(X) DeΨ(Y ) = [DeΨ(X),DeΨ(Y )].

Furthermore, exp(tX) is a one-parameter subgroup in G, and thus so is 0 Ψ◦exp(tX) in H. Thanks to Lemma 1.1.12, the derivative (Ψ◦exp(tX)) |t=0 is given by DeΨ ◦ D0 exp(X) = DeΨ(X). By Lemma 1.1.11, exp(DeΨ(X)) is the only one-parameter subgroup with this property. So Ψ ◦ exp(tX) = exp(DeΨX), which is the statement of the corollary.

Corollary 1.1.14. If two connected Lie subgroups H1, H2 of G have the same Lie algebra h, then H1 = H2. CHAPTER 1. NILPOTENT LIE GROUPS 7

Proof. Let U be a small neighborhood of 0 ∈ h, then by Lemma 1.1.12, exp is a diﬀeomorphism between U and exp U, which is a neighborhood of identity in both H1 and H1. Let H be the group generated by exp(U). 0 Then H ⊆ H1 ∩ H2. Moreover, H1 is both open and closed in H1, as both 0 H1 and its complement in H1 are invariant under translations by elements from exp U and are thus open. Since H1 is connected, H = H1. Similarly H = H2. On the other hand, one may wonder whether there is a Lie subgroup H corresponding to each Lie subalgebra h. It turns out this is the case.

Proposition 1.1.15. Suppose G is a Lie group and h is a Lie subalgebra of g = Lie(G). Then there is a unique connected subgroup H (not necessarily closed) such that Lie(H) = h.

While we will omit the proof of Proposition 1.1.15, let us remark that the idea of it. Every vector X ∈ h can be pushed forward by left translations to get a left invariant vector field FX . All such vector fields span a distribution F of subspaces in the vector space bundle TG. The key observation is the Lie bracket between X, Y coincides with the notion of Lie brackets in differential geometry between the two vector fields FX , FY . And that h being a Lie algebra makes the distribution F integrable; that is, there is a submanifold Mg through any given point g such that the tangent bundle TMg coincides with F everywhere on Mg. It can then be proved that Me is the desired subgroup H. For the detailed proof, see e.g. [Hel01, Theorem II.2.1].

Corollary 1.1.16. Suppose G, H are Lie groups and g, h are their respec- tive Lie algebras. Suppose G is simply connected, then for any Lie algebra morphism ψ : g → h, there is a unique Lie group morphism Ψ: G → H such that ψ = DeΨ. Proof. Since ψ interwines the Lie brackets on g and h, the linear graph g4 = (Id × ψ)g4 = {(X, ψX) is a Lie subalgebra of g × h. By Proposition 1.1.15, there is a Lie subgroup G4 in G×H which has g4 as the Lie algebra. 4 Note that the derivative Deπ1 of the projection π : G → G is the projection from g4 to g, which is an Lie algebra isomorphism. Hence π is locally a diﬀeomorphism. By Lemma 1.1.17 below, we know that π is a covering map. Because G is simply connected and G4 is connected, π must be bijective. So the composition π0 ◦ π−1 is a group morphism from G to H, where π0 stands fro the projection from G4 to H. CHAPTER 1. NILPOTENT LIE GROUPS 8

Observe that by Corollary 1.1.13, Ψ is completely determined on the image exp g of exp. Since Ψ is a group morphism and exp g generates G,Ψ is unique.

Lemma 1.1.17. Suppose a group morphism π : G → H between two topological groups is also a local homeomorphism near e ∈ G; i.e. there exists an open neighborhood U of e such that π|U is a homeomorphism between U and its image π(U). Then π is a covering map. Proof. For every g ∈ G, gU is an open neighborhood of g, and for a ∈ −1 gU, π(a) = π(g)π(g a). So π|gU is a homeomorphism between gU and π(g) · π(U). Hence π is a covering map.

We ﬁnish this section by stating the following well-known theorem of Cartan. Theorem 1.1.18. (Cartan) All closed subgroups of a Lie group are Lie subgroups, i.e. submanifolds that are closed under group operations. For the proof, follow Exercise 1.1.4. Notation 1.1.19. In the rest of this book, we shall denote the Lie algebras of Lie groups G, H, K, N, etc. by the corresponding calligraphic lower case letters g, h, k, n, etc. without further explanation.

Exercises

Exercise 1.1.1. Prove the Jacobi identity for the Lie algebra g of a Lie group G. (Hint: use Corollary 1.1.13 to show that for all X,Y ∈ g, [adX , adY ]gl(g) = ad[X,Y ], then apply Example 1.1.3.) Exercise 1.1.2. Prove that: the solution g(t) in Lemma 1.1.9 coincides with exp(tX).

Exercise 1.1.3. Prove that: if G ⊆ GL(d, R) is a linear Lie group, then P∞ (tX)n exp(tX) = n=0 n! for all t ∈ R and X ∈ g ⊆ gl(d, R). Exercise 1.1.4. Assume that H is a connected closed subgroup of a Lie group G. Let h = {X ∈ g : exp(tX) ∈ H, ∀t ∈ R}. (1) Show that h is a vector subspace of g. (2) Show that the projective space Ph is closed in Pg. (3) Show that exp is surjective from a small neighborhood of 0 ∈ h to a suﬃciently small neighborhood of e ∈ H. (5) Show that H is a submanifold and hence a Lie group. CHAPTER 1. NILPOTENT LIE GROUPS 9

1.2 Lie’s Theorems and Ado’s Theorem

In this part, we present three theorems due to Sophus Lie, which claim the correspondence between Lie groups and their Lie algebras are, in the weaker sense of local structures near identity, bijective.

Deﬁnition 1.2.1. Two Lie groups G, G0 are locally isomorphic if there are open neighborhoods of identity U, U 0 respectively in G, G0 and a smooth diﬀeomorphism Ψ: U → U 0, such that for all g, h ∈ U such that gh ∈ U, Ψ(gh) = Ψ(g)Ψ(h).

Theorem 1.2.2. (Lie) If Lie groups G, G0 are locally isomorphic, then g and g0 are isomorphic.

Theorem 1.2.3. (Lie) For Lie groups G, G0, if their Lie algebras g, g0 are isomorphic, then G and G0 are locally isomorphic.

Theorem 1.2.4. (Lie) For every ﬁnite dimensional Lie algebra g, there exists, up to Lie group isomorphisms, a unique simply connected Lie group G such that Lie(G) =∼ g.

Another important theorem is Ado’s theorem:

Theorem 1.2.5. (Ado) Every ﬁnite dimensional Lie algebra g admits a faithful linear representation, i.e. an injective Lie algebra morphism ψ : g → gl(d, R) for some d. We ﬁrst provide the proof of Theorem 1.2.2. The other theorems, though valid for general Lie groups, are easier to prove in the nilpotent case, which is our main focus. These proofs in the nilpotent case will appear in later sections.

Proof of Theorem 1.2.2. Suppose U, V , Ψ are as in Deﬁnition 1.2.1. Then DeΨ is a Lie algebra morphism. The proof of this fact is the same as in (1.4), while one should note in addition that for all X,Y ∈ g, there exists such that for t, s ∈ (−, ), all the group multiplications involved in (1.4) take place in either U or U 0, depending on the context, in which case Ψ behaves like a group morphism. Since Ψ is a diﬀeomorphism, DeΨ is bijective and hence a Lie algebra isomorphism.

Lemma 1.2.6. If G is a connected topological group, then the universal cover Ge is also a topological group and the projection π : Ge → G is a group morphism. CHAPTER 1. NILPOTENT LIE GROUPS 10

Proof. Consider the multiplication map τ :(a, b) → ab from G×G → G. Be- cause Ge×Ge is the universal cover of G, τ 0 lifts to a smooth map τ : Ge×Ge → Ge such that τ(e, e) = e. The two elements τ(a, τ(b, c)) and τ(τ(a, b), c) of Ge both project to a0b0c0 ∈ G, where a0, b0, c0 respectively denote the projections of a, b, c to G. Hence τ(a, τ(b, c)) is the translate of τ(τ(a, b), c) by some γ(a, b, c) ∈ Γ, where Γ = π1(G) is the deck tranformation group associated to the projection Ge → G. Since Γ is discrete and γ is continuous in a, b, c, we know that γ must be a constant, and in fact be the identity element by checking (a, b, c) = (e, e, e). So τ(a, τ(b, c)) = τ(τ(a, b), c). Similarly, one can ﬁnd a smooth map ι : Ge → Ge that lifts the inverse map ι0 : a → a−1 on G. such that ι(e) = e. Using similar arguments one can show τ(a, ι(a)) = e. Thus Ge is a simply connected group that covers G.

Proof of Theorem 1.2.3. Let Ge, Ge0 respectively be the universal covers of G and G0, then by Lemma 1.2.6 they are both simply connected Lie groups whose Lie algebras are both isomorphic to g. Let ψ : Lie(Ge) → Lie(Ge0) be a Lie algebra isomorphism. Then by Corollary 1.1.16, there exist Lie 0 0 group morphisms Ψ : Ge → Ge and Φ : Ge → Ge such that DeΨ = ψ and −1 DeΦ = ψ . Then De(Ψ ◦ Φ) = Id. By uniqueness in 1.1.16,Ψ ◦ Φ: Ge → Ge must be the identity map on G. Similarly, so Ψ ◦ Φ is the identity map on Ge0. So Ge =∼ Ge0 as Lie groups. Therefore, G and G0 are both covered by the same Lie group Ge and thus locally isomorphic. Proof of Theorem 1.2.4 assuming Theorem 1.2.5. By Theorem 1.2.5, there 0 ∼ 0 is a Lie algebra g ∈ gl(d, R) for some d such that g = g . By Proposition 0 1.1.15, one can ﬁnd a Lie group G ⊂ GL(d, R) such that the Lie algebra of G is g0. (It should be remarked here that the topology of G0 is not inherited from GL(d, R). Instead, the topology is the one determined by path connectivity if G is not closed.) One then take the universal cover G of G0, which is by Lemma 1.2.6 has a group structure that lifts that of G0. In particular, since G0 is a Lie group in the current setting, G is a simply connected Lie group with the same Lie algebra g.

The proof of Ado’s theorem (Theorem 1.2.5) is not a simple one for general Lie algebras, however considerably simpler for nilpotent Lie algebras. The proof in this special case will be presented in the appendix. While the general proof is omitted, the special case is going to be suﬃcient for our purposes.

Exercises CHAPTER 1. NILPOTENT LIE GROUPS 11

Exercise 1.2.1. Prove that all: connected abelian Lie groups are isomorphic k l to R × T for some k, l ≥ 0.

1.3 Baker-Campbell-Hausdorﬀ formula

For real numbers x, y, the equality exp x exp y = exp(x+y) always holds. Nevertheless, this property fails for the exponential map on Lie algebras, as a consequence of the non-commutativity represented by [·, ·]. The Baker- Campbell-Hausdorﬀ formula characterizes how the Lie brackets determines the multiplicative operations of the Lie group G.

Theorem 1.3.1 (Baker-Campbell-Hausdorﬀ formula). Suppose X, Y are elements of the Lie algebra g of a Lie group G, then

exp X exp Y = exp Z (1.5) formally holds where Z = Z(X,Y ) is an inﬁnite series whose all terms are of the form b[X1, [X2, [··· , [Xm−1,Xm]] ··· ], where b ∈ Q and Xi ∈ {X,Y }. When X, Y are in a suﬃciently neighborhood of 0, the equality actually holds.

The ﬁrst a few terms of Z(X,Y ) reads: 1 1 1 Z = X + Y + [X,Y ] + [X, [X,Y ]] − [Y, [X,Y ]] + ··· . (1.6) 2 12 12 The omitted terms all have at least three layers of Lie brackets. Here, the meaning of “formally holds” is that exp X exp Y and exp Z are both real analytic functions of (X,Y ) ∈ g2 and their power series agree. In general, the power series does not always converge. Before going further, we ﬁrst make the observation that if [X,Y ] = 0, then Z = X + Y . In this case, exp X exp Y = exp Y exp X = exp(X + Y ). In other words, the Lie bracket is responsible for the non-commutativity in G. We will discuss the proof of Theorem 1.3.1 only in the case of linear Lie groups. Indeed, this special case implies the theorem for general Lie groups thanks to Ado’s theorem (Exercise 1.3.3). In the proof, we will omit certain details (Exercise 1.3.1). CHAPTER 1. NILPOTENT LIE GROUPS 12

Lemma 1.3.2. For any Lie algebra g and X,Y ∈ g,

∞ X adn Y Ad Y = exp(ad )Y = X (1.7) exp X X n! n=0 Proof. Applying Corollary 1.1.13 to Ψ = Ad : G → GL(g), we see that ∼ Adexp X = exp(adX ), where adX is a matrix in gl(g) = Matdim G×dim G R. Thus by Exercise 1.1.3,

∞ ∞ X adn X adn Y Ad Y = exp(ad )Y = X Y = X exp X X n! n! n=0 n=0 for all X,Y ∈ g.

When g = gl(d, R) = Matd×d(R), both sides in (1.7) can be explicitly written as inﬁnite series, using Exercise 1.1.3 and the relations

−1 AdX Y = XYX , adX Y = XY − YX. (1.8)

Since X, Y are arbitrarily chosen matrices in arbitrary dimensions, it is not hard to believe that (1.7) holds formally for abstract symbols X, Y . In particular, we may express this relation in the setting where X = X(t) d is a smooth curve in gl(d, R), and Y represents the diﬀerential operator dt . While this discussion is not a rigorous proof, we obtain the statement below:

Lemma 1.3.3. For a smooth curve X(t) ∈ gl(d, R), adopting the conven- tions (1.8), we have the equality

∞ n d d d X adX(t)( dt ) Ad = exp ad = (1.9) exp X(t) dt X(t) dt n! n=0

d between linear diﬀerential operators acting on smooth curves γ : R → R . The rigorous proof of Lemma 1.3.3 is left as Exercise 1.3.1. One can derive from this lemma the diﬀerentiation formula of exp at any base point, which generalizes Lemma 1.1.12.

Corollary 1.3.4. For X,Y ∈ gl(d, R),

∞ X (−1)n−1 adn−1 (D exp)Y = (exp X) X Y. X n! n=1 CHAPTER 1. NILPOTENT LIE GROUPS 13

d 0 Proof. First of all, observe the fact that adX(t) dt = −X (t) as an operator. d To see this, observe that for all smooth curves ν(t) in R , d d d ad (ν(t)) =(X(t) ◦ − ◦ X(t))ν(t) = X(t)ν0(t) − (X · ν)0(t) X(t) dt dt dt = − X0 · ν(t).

Suppose γ(t) is a constant function in t. Applying both sides of (1.9) to −X(t) and γ = γ(t), we respectively get: d d Ad (γ) = exp(−X(t)) (exp X(t)γ) exp(−X(t)) dt dt = exp(−X(t))(exp X(t))0γ, and ∞ n d ∞ adn−1 (X0(t)) X ad−X(t)( dt ) X −X(t) (γ) = · γ n! n! n=0 n=1 ∞ n−1 n−1 0 X (−1) adX(t)(X (t)) = · γ. n! n=1 Here we used the fact above. Comparing both equalities, we can conclude the proof.

Proof of Theorem 1.3.1 for linear Lie groups. Assume that G ⊆ GL(d, R) and hence g ⊆ gl(d, R). We will prove the Baker-Campbell-Hausdorff formula holds as an actual equality for matrices X, Y from a sufficiently small neighborhood of 0 in g. Since exp tX exp tY is real analytic in t, this would prove the formula as a formal equality. When X, Y have small norms, exp X exp tY is also close to 0. For t ∈ (−2, 2), there is a unique Z(t) chosen from a small neighborhood of 0 such that exp Z(t) = exp X exp tY . This is because exp is a local diffeomorphism at 0. Moreover, Z depends differentiably on t. Applying Corollary 1.3.4 to exp X exp tY = exp Z(t) yields

∞ n−1 n−1 X (−1) adZ(t) (exp X exp tY )0 = exp X exp tY Z0(t). n! n=1 As the derivative on the left hand side equals (exp X exp tY )Y . So

∞ n−1 n−1 X (−1) adZ(t) Y = Z0(t). n! n=1 CHAPTER 1. NILPOTENT LIE GROUPS 14

(−1)n−1 adn−1 P∞ Z(t) Note that the ﬁrst term in the power series n=1 n! is the identity transform Id ∈ GL(g). Therefore, for small values of Z(t), which is being (−1)n−1 adn−1 P∞ Z(t) assumed, the linear transform n=1 n! has an inverse transform φ(adZ(t)) ∈ GL(g), where φ is the MacLaurin series determined by

∞ !−1 X (−1)n−1zn−1 z z z2 φ(z) = = = 1 + + + ··· . n! 1 − e−z 2 12 n=1 Then, 0 Z (t) = φ(adZ(t))Y. Furthermore, remember that for t ∈ (−2, 2), X,Y,Z(t) are in a small neighborhood U of 0 in g, and thus so are adX , ad Y, adZ(t) in Endg. When U is small, exp |U is a diﬀeomorphism, and by Lemma 1.3.2, −1 −1 −1 adZ(t) = exp (Adexp Z(t)) = exp (Adexp X exp tY ) = exp (Adx AdtY ) −1 = exp (exp(adX ) exp(t adY )). Since exp is, in this context, from End(g) to GL(g), it is given by the A2 −1 power series exp A = Id + A + 2! + ··· . The inverse exp takes place in a neighborhood of Id ∈ GL(g), and is given by the series exp−1(Id + A) = A2 A3 A − 2 + 3 − · · · . It follows that adZ(t) can be written as an inﬁnite power series ψ(adX , t adY ) in adX and t adY , whose terms have the form i1 j1 ik i1 j1 ik jk c adX ◦(t adY ) ◦ · · · ◦ adX or c adX ◦(t adY ) ◦ · · · ◦ adX ◦(t adY ) , where c is always rational. Hence φ ◦ ψ is a power series of similar form, and 0 Z (t) = φ ◦ ψ(adX , adY )Y . Integrating on t ∈ [0, 1], we know that Z 1 Z 1 Z(1) = Z(0) + φ ◦ ψ(adX , adY )Y dt = X + φ ◦ ψ(adX , adY )Y dt. 0 0 Remark that this is a series of the form in the statement of Theorem 1.3.1. Because exp Z(1) = exp X exp Y , the theorem is proved.

The constant and ﬁrst order terms in the series of exp(adX ) exp(t adY ) are Id+adX +t adY , thus ψ(adX , adY ) = adX +t adY +(higher order terms). 1 And φ ◦ ψ(adX , adY ) = Id + 2 (adX +t adY ) + (higher order terms) as well. So Z 1 1 1 Z(1) = X + (Y + ([X,Y ] + t[Y,Y ] ··· )dt = X + Y + [X,Y ] + ··· , 0 2 2 where “··· ” represents higher order terms involving two or more Lie brackets. CHAPTER 1. NILPOTENT LIE GROUPS 15

Exercises

Exercise 1.3.1. Prove Lemma 1.3.3. (Hint: Compute all the partial deriva- −sX(t) ∂ sX(t) tives in s of the real analytic expression e ∂t (e γ(t)) at s = 0.) Exercise 1.3.2. Verify that the third order terms at the beginning of the Baker-Campbell-Hausdorﬀ formula are those in (1.6). Exercise 1.3.3. Assuming Ado’s theorem, prove that: if the Baker-Campbell- Hausdorﬀ formula holds for all linear Lie groups, then it holds for all Lie groups. (Hint: use Ado’s theorem and Proposition 1.1.15 to prove that given a Lie group G, there is a linear Lie group G0 locally isomorphic to G.)

1.4 Nilpotent Lie algebras

The Baker-Campbell-Hausdorff formula tells us that how the local commutator relations determine the structure of a Lie groups. However, in practice, it is difficult to do so, because of the following reasons. First, the coefficients in the formula, though can be made explicit with some efforts, are highly complicated. Second, there can be infinitely many terms, because the repeated Lie brackets can be infinitely long. The group relations are analytic, however not polynomial in general. Nilpotent Lie groups are Lie groups whose Baker-Campbell-Hausdorff formula has only finitely many terms, or equivalently in whose Lie algebra repeated brackets of sufficiently high orders all vanish. In this case, the group structure is characterized by polynomial equations. We will give more precise definitions of such Lie algebras in this section. 0 0 If h, h ⊂ g is a Lie subalgebra. Denote [h, h ] = R-span{[X,Y ]: X ∈ h,Y ∈ h0}. It can be easily verified to be a vector subspace of g. The subalgebra h is called an ideal of g if [g, h] ⊆ h. Lemma 1.4.1. If h, h0 are ideals of g, then [h, h0] is an ideal of each of g, h and h0. Proof. Let X, Y , Z be respectively vectors from g, h and h0, then

[X, [Y,Z]] = − [Y, [Z,X]] − [Z, [X,Y ]] ∈ [h, [h0, g]] + [h0, [g, h]] =[h, h0] + [h0, h] = [h, h0].

Hence [g, [h, h0]] ⊆ [h, h0], or in other words [h, h0] is an ideal in g. Furthermore, because h and h0 are ideals, [h, h0] ⊆ h ∩ h0. It then follows that [h, h0] is also an ideal in both h and h0. CHAPTER 1. NILPOTENT LIE GROUPS 16

[h, [g, h]] ⊆ [g, [g, h]] ⊆ [g, h]. In particular, [[g, h], [g, h]] ⊆ [g, h], so h is a subalgebra, and furthermore and ideal.

If h is an ideal of g, then the quotient vector space g/h can be equipped with a Lie bracket descending from that of g. Namely, if X1,Y1 ∈ h are respectively represented by X,Y ∈ g, then [X1,Y1] ∈ g/h is the equivalence 0 0 class represented by [X,Y ]. If X ,Y also represent X1,Y1 respectively, then X0 = X + A, Y 0 = Y + B where A, B ∈ h. So [X0,Y 0] = [X,Y ] + [X,B] + [A, Y 0] ∈ [X,Y ] + h. Thus the Lie bracket on g/h is independent of the choice of representatives X, Y , and hence well-defined. Definition 1.4.2. Given a Lie algebra g, the lower central series (or derived series) is the sequence of Lie subalgebras g = g(1) Bg(2) Bg(3) B··· inductively defined by g(i+1) = [g, g(i)].

Lemma 1.4.1 guarantees that the g(i)’s are ideals of G and g(i) C g(j) if i ≥ j. The word “central” in the name comes from the fact that g(i)/g(i+1) is in the center of g/g(i+1), i.e. [X,Y ] = 0 for all X ∈ g(i)/g(i+1) and Y ∈ g/g(i+1).

Lemma 1.4.3. [g(i), g(j)] ⊆ g(i+j). Proof. We prove the statement inductively as i increases. For i = 1 and all j, this is the definition of lower central series. Suppose i ≥ 2 and [g(i−1), g(j)] ⊆ g(i+j−1), then it suffices to show [[X,Y ],Z] ∈ gi+j−1 for all X ∈ g, Y ∈ g(i−1) and Z ∈ g(j) because [g(i), g(j)] is linearly spanned by vectors of this form. By Jacobi identity, it suffices to show [X, [Y,Z]] and [Y, [X,Z]] are both in g(i+j). By inductive assumption, [X, [Y,Z]] ∈ [g, g(i+j−1)] ⊆ g(i+j), and [Y, [X,Z]] ∈ [g(i−1), g(j+1)] ⊆ g(i+j). The proof is completed.

Deﬁnition 1.4.4. A Lie algebra g is nilpotent if g(i+1) = {0} for some i ∈ N. The smallest such i is called the step of nilpotency, and n is called s-step nilpotent.

If the step of nilpotency is s, then g(n) = 0 for all n ≥ s + 1. Lemma 1.4.5. g is nilpotent with step of nilpotency at most s, if and only if [X1, [X2, [··· , [Xs,Xs+1] ··· ]] = 0 for all X1, ··· ,Xs+1 ∈ g.

Proof. All such iterated Lie brackets belong to g(s+1), and conversely g(s+1) is generated by these special vectors.

If the step of nilpotency is s, then g(n) = 0 for all n ≥ s + 1. CHAPTER 1. NILPOTENT LIE GROUPS 17

Lemma 1.4.6. If g is nilpotent, with step s, then all of its Lie subalgebras and quotient Lie algebras are also nilpotent and their step of nilpotencies are at most s.

Proof. Let h ⊆ g be a subalgebra and f ⊆ g be an ideal, then one can easily prove by induction that h(i) ⊆ g(i) and (g/f)(i) = g(i)/(f ∩ g(i)). Both of these become trivial when i = s + 1.

Deﬁnition 1.4.7. A matrix A ∈ Matd×d(R) is nilpotent (resp. unipotent) if all of its eigenvalues are equal to 0 (resp. 1).

Lemma 1.4.8. If g is nilpotent, then for every X ∈ g, adX ∈ End(g) is a s nilpotent matrix and adX = 0. Proof. Using Jordan canonical form, one can see that a matrix A is nilpotent n if and only if A = 0 for suﬃciently large n. This is true for adX with n = s.

Example 1.4.9. By deﬁnition, a Lie algebras is 1-step nilpotent if and only if it is abelian.

Example 1.4.10. Let n ⊆ gl(d, R) denote the vector subspace of consisting of strictly upper triangular matrices   0 a12 a13 ··· a1,d−1 a1,d  0 a23 ··· a2,d−1 a2,d     .. . .   . . .     .. . .   . . .     0 ad−1,d 0

Then n is closed under [·, ·]; that is, for all X,Y ∈ n,[X,Y ] = XY − YX is still in n. This is because n is closed under matrix multiplication and both XY and YX are also in n. d Furthermore, n can be written as {(aij)i,j=1 : aij = 0 unless j ≥ i + 1}. For X = (xij),Y = (yij) ∈ n, we now calculate the entries of [X,Y ]. Pd The entry of [X,Y ] with index ij is given by k=1(xikykj − yikxkj). In order for any term of the form xikykj or yikxkj to be non-zero, one needs CHAPTER 1. NILPOTENT LIE GROUPS 18

k ≥ i+1 and j ≥ k +1, so j ≥ i+2. Thus n(2) = [n, n] is inside the subspace d {(aij)i,j=1 : aij = 0 unless j ≥ i + 2}. Indeed, n(2) coincides with this space (Exercise 1.4.3). Inductively, we can show n(k) ⊆ [n, n] is contained in (in fact, equal to) d the vector subspace {(aij)i,j=1 : aij = 0 unless j ≥ i+k}. Starting at k = d, this space is trivial. Hence n is (d − 1)-step nilpotent.

Example 1.4.11. The (2n+1)-dimensional Heisenberg Lie algebra h2n+1 is the vectors space spanned by 2n+1 linearly independent vectors X1, ··· ,Xn, Y1, ··· ,Yn, Z. All the Lie brackets among these base vectors are trivial, except: [Xi,Yi] = −[Yi,Xi] = Z. Then all three-fold iterated Lie brackets of the form [∗, [∗, ∗]] vanish, hence (h2n+1)(3) = {0}. But (h2n+1)(2) is non-trivial because the Lie bracket is non-trivial. Hence g2n+1 is two-step nilpotent.

We ﬁnish this chapter by a useful fact. To state it, ﬁrst let g(2) = [g, g] be the commutator subalgebra of a nilpotent Lie algebra g. Then g(2) is an ideal. Suppose in addition that h is a Lie subalgebra of g.

Lemma 1.4.12. For a Lie subalgebra h of g, if h + g(2) = g then h = g.

Proof. We shall inductively prove that h+g(i) = g for every g(i) in the lower central series with i ≥ 2. As the lower central series eventually becomes trivial, this shows h = g. Suppose the inductive step is valid for index i. Then every X ∈ g can be written as Y +Z where Y ∈ h and Z ∈ g(i). The claim for i+1 would follow if we could further show that for every Z ∈ g(i), Z ∈ h + g(i+1). Without loss of generality we may assume Z = [U, V ] where U ∈ g and V ∈ g(i−1) as g(i) is spanned by such vectors. By inductive hypothesis, U = YU + ZU and V = YV + ZV , where YU ,YV ∈ h and ZU ,ZV ∈ g(i). Then

[U, V ] =[YU ,YV ] + ([YU ,ZV ] + [ZU ,V ])

∈h + ([g, g(i)] + [g(i), g]) ⊆ h + g(i+1).

The proof is complete.

Exercises

Exercise 1.4.1. Show that in the lower central series, [g(i), g(j)] C g(i+j). CHAPTER 1. NILPOTENT LIE GROUPS 19

Exercise 1.4.2. The upper central series {0} = g(0) ⊆ g(1) ⊆ g(2) ⊆ · · · of a Lie algebra g is the sequnce of subalgebras inductively deﬁned by

g(i+1) = {X ∈ g :[X,Y ] ∈ g(i), ∀Y ∈ g}.

Show that: (n+1) (1) g is nilpotent if and only if g = g for some n ∈ N; (2) The smallest such n equals the step of nilpotency s of g. (s−n+1) (3) g(n) ⊆ g for all n ≤ s. Exercise 1.4.3. In Example 1.4.10, show that

d g(k) = {(aij)i,j=1 : aij = 0 unless j ≥ i + k}.

Exercise 1.4.4. Show that the Lie algebra n of strictly upper triangular 3 × 3 matrices is isomorphic to the 3-dimensional Heisenberg Lie algebra.

Exercise 1.4.5. If A ∈ Matd×d(R) is a nilpotent matrix, then

d−1 X (−1)n−1 A = (exp A − Id)n. n n=1

1.5 Nilpotent Lie groups

Nilpotent Lie groups are deﬁned in a similar way as in the Lie algebra category. For subgroups H, H0 of a G, let the commutator group [H,H0] be the subgroup generated by {aba−1b−1 : a ∈ H, b ∈ H0}. Note that H is a normal subgroup of G (denoted as H C G) if and only if [G, H] ⊆ H. It is easy to check that for H ⊆ G,[G, H] is a normal subgroup of G. Similary to Lemma 1.4.1, if in addition H C G is a normal subgroup, then [G, H] ⊆ H, and hence is also normal in H. If we start with H = G, this deﬁnes a sequence of decreasing normal subgroups:

Deﬁnition 1.5.1. Given a group G, its lower central series (or derived series) is the sequence of normal subgroups G = G(1) B G(2) B G(3) B ··· inductively deﬁned by G(i+1) = [G, G(i)].

Deﬁnition 1.5.2. A Lie group G is nilpotent if G(i+1) = {0} for some i ∈ N. The smallest such i is called the step of nilpotency, and G is called s-step nilpotent. CHAPTER 1. NILPOTENT LIE GROUPS 20

A group is abelian if and only if it is 1-step nilpotent. Similar to Lemma 1.4.5, we have Lemma 1.5.3. If G is nilpotent, with step s, then all of its subgroups and quotient groups are also nilpotent and their step of nilpotencies are at most s. Proof. Let H ⊆ G be a subgroup and F ⊆ G be a normal subgroup, then induction shows that H(i) ⊆ G(i) and (G/F )(i) = G(i)/(F ∩ G(i)). Both series become trivial when i = s + 1. Lemma 1.5.4. If G is a connected Lie group, H is a connected Lie subgroup and g, h are the corresponding Lie algebras, then H C G if and only if h C g. Proof. Suppose H CG, then for all differentiable curves g(t) ∈ G and h(s) ∈ H with g(0) = h(0) = e, g(t)h(s)g(t)−1 ⊆ H. Differentiating in s, we see 0 0 that Adg(t) h (0) ∈ h. Differentiating again in t yields adg0(0) h (0) ∈ h. As g0(0) ∈ g and h0(0) ∈ h can be arbitrarily chosen, it follows that [g, h] ⊆ h, or equivalently h C g. −1 Conversely, if h C g, we wish to show that ghg ∈ H for all g ∈ G, h ∈ h. Let U, V be neighborhoods of 0 respectively in g and h such that exp |U : U → exp U and exp |V : V → exp V are diffeomorphisms. We first show that Adg h ⊆ h. First, suppose that g = exp X for some X ∈ U. Because adX preserves the Lie subalgebra h, so does Adg = exp(adX ). The claim follows for general g, because Ad : g → Adg is a group morphism and G, as a connected topological group, is generated by the neighborhood exp U of identity. We then prove ghg−1 ∈ H for all h ∈ H. If h = exp Y for some Y ∈ V , then h = h(1) where h = h(1) where h(t) = exp tY is the one parameter subgroup with initial velocity h0(0) = Y . Hence gh(t)g−1 is the one-parameter subgroup with initial veclocity Adg Y ∈ h. As such a subgroup exists in H and is unique in G, we obtain that gh(t)g−1 ∈ H. So ghg−1 ∈ H. To generalize this to all h ∈ H, it suffices to note that H is connected and thus generated by exp V . This completes the proof.

Deﬁnition 1.5.5. We deﬁne G(i)-admissible curves γ : R → G(i) recur- sively as follows: For i = 1, a G(1) admissible curve is a one-parameter subgroup {exp tY } in G. For i ≥ 1, a G(i)-admissible curve is a curve of the form −1 −1 exp(tY1)γ1(t) exp(−tY1)γ1(t) exp(tY2)γ2(t) exp(−tY2)γ2(t) −1 (1.10) ··· exp(tYn)γn(t) exp(−tYn)γn(t) , CHAPTER 1. NILPOTENT LIE GROUPS 21

where 1 ≤ n ≤ dim g(i), Yj ∈ g, and the γj’s are G(1) admissible curves. Lemma 1.5.6. Suppose G is a connected Lie groups, then for all i ≥ 1 1 and Z ∈ g(i), there exists a C diﬀerentiable curve γ : R → G(i) such that 0 g(0) = e, g (0) = Z. Moreover, γ is G(i)-admissible.

Proof. We prove by induction. For i = 1, G(i) = G, it suffices to take g(t) = exp tZ. Supppse i ≥ 2 and the statement holds for level i − 1. Suppose X ∈ g, Y0 ∈ g(i). By inductive hypothesis, there is a G(i−1)- 0 admissible curve h : R → G(i−1) such that h(0) = e, h (0) = Y0. Because h is differentiable at 0, one can modify its values on (−∞, 0) by setting h(−t) = h(t)−1 for all t > 0. Then the resulting new curve is still continuous at 0. Moreover, by (1.1), d d d d h(t)| = − h(−t)| = − h(t)−1| = h(t)| = Y . dt t=0− dt t=0+ dt t=0+ dt t=0+ 0 1 0 So the new curve is also C and h (0) = Y0. Thus we may assume h(−t) = −1 h(t) for all t ∈ R. Define a G(i)-admissible curve g(t) = (exp X)h(t) exp(−X)h(−t), which lies in [G, G(i−1)] = G(i). By applying (1.1) and Lemma 1.3.2, 0 d d g (0) = (exp Xh(t) exp(−X)) + h(−t) dt dt t=0 = Adexp X Y0 − Y0 = (exp(adX ) − Id)Y0 ∞ n ∞ n X ad Y0 X ad = X = X, Id + X Y n! (n + 1)! 0 n=1 n=1

n P∞ adX By Lemma 1.4.1, (Id + n=1 (n+1)! )Y0 ∈ g(i−1) + [g, g(i−1)] ⊆ g(i−1). Fur- thermore, when X is in a small neighborhood of 0, adX ∈ gl(g) has small n P∞ adX matrix norm, and hence so does the linear transform n=1 (n+1)! . In con- n P∞ adX sequence, Id + n=1 (n+1)! is an invertible linear transform of g(i−1). So we can conclude that: for vectors X ∈ g that are suﬃciently close to 0, and all vectors Y ∈ g(i−1), there exists Y0 ∈ g(i−1) such that the curve g(t) inside 0 the subgroup G(i) has value g(0) = e and derivative g (0) = [X,Y ] at 0. By replacing the pair (X,Y ) by (RX,R−1Y ) where R > 1, one can drop the assumption that X is close to 0 and assert that for all X ∈ g, Y ∈ g(i−1), 0 there exists a G(i)-admissible curve g in G(i) with g(0) = e, g (0) = [X,Y ]. CHAPTER 1. NILPOTENT LIE GROUPS 22

By applying (1.1) again, we can then, by taking an n-fold product where 0 n ≤ dim g(i), construct an admissible curve g such that g(0) = e and g (0) is any linear combination Z of vectors of the form [X,Y ], X ∈ g, Y ∈ g(i−1). This is what we need, since g(i) is by deﬁnition the collection of all such Z’s.

Theorem 1.5.7. For a connected Lie group G, G is nilpotent if and only if g = Lie(G) is a nilpotent Lie algebra. Moreover, their steps of nilpotency are equal, and exp : g → G is a covering map.

Proof of the “only if” part. Suppose G is s-step nilpotent, then G(s+1) is trivial. Therefore, according to Lemma 1.5.6, g(s+1) cannot contain any non-zero vectors. So g is nilpotent, and its step of nilpotency is at most s.

We start the proof of the “if” part by declaring the following fact, which does not necessarily hold for general Lie groups. Proposition 1.5.8. If the Lie algebra g of a simply connected Lie group G is nilpotent, then: (1) exp : g → G is a diffeomorphism between g and G. (2) The Baker-Campbell-Hausdorff formula (1.5) holds for all X,Y ∈ g. Proof. Suppose g is s-step nilpotent. Then for sufficiently short vectors X,Y ∈ g, exp X exp Y = exp Z where Z = Z(X,Y ) is, by the Baker- Campbell-Hausdorff formula, a sum 1 Z(X,Y ) = X + Y + [X,Y ] + ··· 2 in which each term is a repeated Lie bracket between X and Y of at most s layers. There are only finitely many such combinations of Lie brackets, and (X,Y ) → [X,Y ] is a bilinear map. Therefore, Z(X,Y ) is a finite sum, more precisely a polynomial map from g×g to g whose linear part is X +Y . In particular, Z(X,Y ) is defined for all X,Y ∈ g, not only those from a neighborhood of 0. We claim that g has a group structure, where the identity map is 0, the multiplication is given by

X Y = Z(X,Y ) (1.11) and inversion is given by X → −X. For this, one needs to verify

Z(X,Z(Y,W )) = Z(Z(X,Y ),W ) (1.12) CHAPTER 1. NILPOTENT LIE GROUPS 23 and Z(0,X) = Z(X, 0) = X,Z(X, −X) = 0 (1.13) Since (1.13) follows immediately from the fact that [X, −X] = [X, 0] = [0,X] = 0, we focus on (1.12). Notice that for X, Y , W of small norms, all intermediate expressions in (1.12) takes place on a small neighborhood U of 0 ∈ g on which exp is a diffeomorphism. In this case Z(X,Z(Y,W )) = exp−1(exp X · exp Z(Y,W )) = exp−1(exp X · exp Y exp W ) = exp−1(exp Z(X,Y ) · exp W ) = Z(Z(X,Y ),W ). So Z(X,Z(Y,W ))−Z(Z(X,Y ),W ) vanishes when (X,Y,W ) lies in a certain neighborhood of the origin in g3. On the other hand, Z(X,Z(Y,W )) − Z(Z(X,Y ),W ) is a polynomial map as Z is. This forces the polynomial to vanish on the entire g3. So (1.12) holds for all X, Y , Z and we obtain the desired group structure on g. Moreover, the Lie bracket g determined by the group multiplication coincides with the original bracket on g, because exp : g → G behaves like a group morphism on U and De exp = Id. Notice that in the Lie group (g, ·), {tX} is a one-parameter subgroup 0 and (tX) |t=0 = X. By the uniqueness in Lemma 1.1.11, the exponential map exp from g to the Lie group (g, ) sends tX to tX for all t ∈ R and X ∈ g. In other words, exp = Id. By Corollary 1.1.16 there is a unique Lie group morphism Ψ from g to G, such that DeΨ = Id. By Corollary 1.1.13,Ψ ◦ exp = exp ◦Id, or equivalently Ψ = exp. Thus we can regard exp as a Lie group morphism from (g, ) and (G, ·), with De exp = Id. So, exp is a covering map by Lemma 1.1.17, and then must be bijective because g and G are both simply connected. In conclusion, exp : g → G is a diffeomorphism. This shows part (1). To prove part (2), notice that by the definition of (1.11)and the fact that exp is a group morphism, exp X · exp Y = exp(X Y ) = exp Z(X,Y ) for all X, Y , which is the Baker-Campbell-Hausdorff formula.

In fact, for the Baker-Campbell-Hausdorﬀ formula to hold for all Lie algebra elements, we do not need the simply connectedness assumption on G. (Excercise 1.5.1) Corollary 1.5.9. If G is a simply connected Lie group whose Lie algebra g is nilpotent, then for all ideals h ⊆ g, H = exp h is a simply connected Lie subgroup of G and exp is a diﬀeomorphism between h and H. CHAPTER 1. NILPOTENT LIE GROUPS 24

In particular, all connected closed subgroups of G are simply connected.

Proof. Note that h is invariant under the group multiplication by Baker- Campbell-Hausdorﬀ formula and the inversion X → −X. Hence h is a subgroup in (g, ). The corollary is a direct sequence from the fact that exp : (g, ) → (G, ·) is a Lie group isomorphism.

In order to show the “if” direction in Theorem 1.5.7, we further need the following lemma:

Lemma 1.5.10. Assume that G is a simply connected Lie group whose Lie algebra g is nilpotent. Assume h C g is an ideal, then [G, exp h] ⊆ exp[g, h]. Proof. We first show that [G, exp h] ⊆ exp[g, h]. For this, it suffices to prove that exp X exp Y exp(−X) exp(−Y ) ∈ exp[g, h] for all X ∈ g, Y ∈ h. By the Baker-Campbell-Hausdorff formula, this expression is given by exp(Z(Z(X,Y ),Z(−X, −Y )). However,

Z(Z(Z(X,Y ),Z(−X, −Y )) 1 1 =Z(X + Y + [X,Y ] + ··· , −X − Y + [X,Y ] + ··· ) 2 2 1 1 =[X,Y ] + [X + Y + [X,Y ] + ··· , −X − Y + [X,Y ] + ··· ] 2 2 =[X,Y ] + [X + Y, −X − Y ] + ··· = [X,Y ] + ··· ,

where all the omitted terms are repeated Lie brackets that have at least three layers, and has at least one component that is equal to Y . Each of these terms belong to [g, h] as well as [X,Y ]. Therefore, we obtain that exp X exp Y exp(−X) exp(−Y ) ∈ exp[g, h]

Proof of the “if” part of Theorem 1.5.7. Suppose G is a connected Lie group whose Lie algebra g is s-step nilpotent. Then, since its universal cover Ge is also a connected Lie group endowed with the same Lie algebra, by Proposition 1.5.8, exp : g → Ge is a diﬀeomorphism and indeed, by the proof of Proposition 1.5.8, an isomorphism between the Lie group (g, ) and Ge. By Lemma 1.5.10,[G,e exp g(i)] ⊆ exp g(i+1) for all i. Because exp g(1) = exp g = Ge, it follows by induction that Ge(i) ⊆ exp g(i). As g(s+1) = {0}, G(s+1) is a trivial group. So Ge is nilpotent and its step of nilpotency is s. Lemma 1.5.3 implies that G is nilpotent with step bounded by s. This completes the proof of Theorem 1.5.7. CHAPTER 1. NILPOTENT LIE GROUPS 25

The two lower central series, in Lie algebra and in Lie group, correspond to each other for simply connected nilpotent Lie groups.

Corollary 1.5.11. For a simply connected nilpotent Lie group G, for all indices i, G(i) = exp g(i).

Proof. By the proof of the “if” part above, G(i) ⊆ exp g(i). It suffices to show that exp g(i) ⊆ G(i). Because exp g(i) is a connected Lie group and is generated by any open neighborhood of its identity, it suffices to show that for a sufficiently small neighborhood U ⊆ G(i) of e, U ⊆ G(i). Note that in the recurvsive Definition 1.5.5 we can make n = dim g(i) in (1.10) by adding trivial components when n < dim g(i). So Lemma 1.5.6 tells us that for every 1 ≤ i ≤ s, there exist constants N = N(i), M = M(i) and an expression γ(t, Y1, ··· ,YN ) ∈ G(i) representing a G(i)- admissive curve that can be decomposed into a product exp ··· exp tYh tYh1 M of exp(tY1), ··· , exp(tYN ), where h1, ··· , hM ∈ {1, ··· ,N} such that the ∂ N map L(Y1, ··· ,YN ) = ∂t γ(t, Y1, ··· ,YN ) t=0 from g to g(i) is surjective. The product P (X1, ··· ,XN ) = exp ··· exp Xh defines a differen- Xh1 M N tiable map from g to G(i). Observe that L = (∇P )(Y1, ··· ,YN ), and thus N ∇P is a surjective linear map from g to g(i). By implicit function theorem, the image of P contains a small neighborhood U of the identity in G(i). This completes the proof.

Example 1.5.12. The family of upper triangular unipotent d × d matrices, i.e. upper triangular matrices whose diagonal entries are all 1, form a subgroup N of SL(d, R). This group is d − 1 step nilpotent as its Lie algebra is the nilpotent Lie algebra n in Example 1.4.10.

Example 1.5.13. The (2n+1) dimensional Heisenberg Lie group is H2n+1 = 2n+1 R equipped with the group multiplication

(x, y, z)(x0, y0, z0) = (x + x0, y + y0, z + z0 + x · y0)

−1 n and group inversion (x, y, z) = (−x, −y, −z + x · y), where x, y ∈ R and z ∈ R. Its Lie algebra is the Lie algebra h2n+1 from Example 1.4.11. There- fore H2n+1 is 2-step nilpotent. The center of H2n+1 is the one-parameter subgroup {(0, 0, z)}.

We proved in Corollary 1.5.9 that every connected closed subgroup H of a simply connected nilpotent Lie group G is simply connected. In fact, so is the quotient G/H if H is a normal subgroup. CHAPTER 1. NILPOTENT LIE GROUPS 26

Corollary 1.5.14. If H is a connected closed normal subgroup of a simply connected nilpotent Lie group G, then G/H is a simply connected nilpotent group which is diﬀeomorphic to its Lie algebra g/h through the exponential map. Proof. Because H is closed, G/H has a manifold structure and hence is a Lie group. Its Lie algebra can be naturally identiﬁed with g/h. By Corollary 1.1.13 we have the commutative diagram

g −−−−→π g/h   exp exp (1.14) y y G −−−−→π G/H We know that exp : g/h → G/H is a local diffeomorphism at the identity. By this diagram, it is also surjective because exp : g → G is a diffeomorphism. It suffices to show its injectivity. Assume for the sake of contradiction that Z,X ∈ g satisfies exp Z ∈ exp X · H. We want to show that Z − X ∈ h. Because H = exp h, exp Z = exp X exp Y for some Y ∈ h. Hence Z = Z(X,Y ) is given by the Baker- Campbell-Hausdorff formula. More precisely, Z = X + Y + ··· , where all the omitted terms are repeated Lie bracktet involving Y at least once. Since Y ∈ h and h is an ideal (due to the normality of H), these omitted terms all belong to h. Thus Z − X ∈ h.

From Lemma 1.4.8 and Theorem 1.5.7 we deduce: Corollary 1.5.15. If G is a connected Lie group of step s, then for every s g ∈ G, Adg ∈ GL(g) is a unipotent matrix and (Adg −Id) = 0. Proof. By Theorem 1.5.7, g = exp X for some X. By Corollary 1.1.13, P∞ 1 n Adg = exp(adX ) = n=0 n! adX . Because adX is a nilpotent matrix by Lemma 1.4.8, Adg is unipotent as a matrix on g. In fact, Adg −Id = P∞ 1 n adX ·B where B = n=0 (n+1)! adX is a matrix commuting with adX . So s s s s (Adg −Id) = adX B = 0 as adX = 0. Thus Adg −Id is nilpotent and Adg is unipotent.

Finally, we state another fact about connected closed subgroups in simply connected nilpotent Lie groups. Lemma 1.5.16. Let H is a closed subgroup of a simply connected nilpotent Lie group G. If H has ﬁnitely many connected components then it is connected and simply connected. CHAPTER 1. NILPOTENT LIE GROUPS 27

Proof. Simply connectedness follows from connectedness so it suffices to show that there is only one connected component. Let H0 be the connected component of H containing identity, then it is a connected closed subgroup and hence a Lie subgroup. Assume there is an element h ∈ H that is not in 0 n 0 H , we claim that h ∈/ H for all n ∈ Z\{0}. In fact, h = exp X for some X ∈ g as exp : g → G is a diffeomorphism, and hn = exp(nX). If hn ∈ H0, then nX belongs to the Lie algebra h of H0 and thus so does X. This would contradict the fact that h∈ / H0. It follows that hnH0 are different connected components of H for all 0 n ∈ Z and H has infinite index in H. This contradicts the assumption. The proof is completed.

Exercises

Exercise 1.5.1. Show that for all connected nilpotent Lie group G, the Baker-Campbell-Hausdorﬀ formula is true for all pairs of elements from g.

Exercise 1.5.2. Show that h2n+1 from Example 1.4.11 is isomorphic to the Lie algebra of H2n+1 from Example 1.5.13. Moreover, the identity map

n n X X xiXi + yiYi + zZ → (x1, ··· , xn, y1, ··· , yn, z) i=1 i=1 from h2n+1 to H2n+1 is the exponential map. 1 t Exercise 1.5.3. Let u = . t 0 1 (1) Prove that t → ut is a group morphism. In particular, u :(t, v) → 2 2 utv, where t ∈ R, v ∈ R , is a left R-action on R . 2 (2) Prove that the semi-product group R nu R is a nilpotent Lie group. 2 (3) Prove that R nu R is isomorphic to the 3-dimensional Heisenberg group.

Exercise 1.5.4. Show that for all a ∈ SL(d, R), the group

{a ∈ SL(d, ) : lim gnag−n = Id} R n→∞ is a nilpotent Lie group and describe its Lie algebra. CHAPTER 1. NILPOTENT LIE GROUPS 28

1.6 Mal’cev basis

From previous discussions, we know that simply connected Lie groups are diﬀeomorphic to their Lie algebras, and that in this case the Lie group structure and the Lie algebra structure determine each other. Which data do we need in order to completely describe these structures?

Deﬁnition 1.6.1. A ﬁltration of a nilpotent Lie algebra g is a sequence of Lie subalgebras g = g1 B g2 B ··· B gr = {0} such that [g, gi] ⊆ gi+1.

In particular, each gi is an ideal of g. Moreover as in Lemma 1.4.3, [gi, gj] ⊆ gi+j. By definition, the lower central series g(i) of a nilpotent Lie algebra g is a filtration. For a filtration, gi/gi+1 is an abelian Lie algebra; in other words, di Pr it is isomorphic to R for some di. Denote mi = dim gi, then mi = j=i dj and is decreasing in i. Write m = m1 = dim g.

Definition 1.6.2. A Mal’cev basis X adapted to the filtration {gi} is a basis X1, ··· ,Xm of g such that Xm−mi+1, ··· ,Xm spans gi for each i. Notice that given the filtration, a Mal’cev basis always exists.

Example 1.6.3. The basis {X1, ··· ,Xn,Y1, ··· ,Yn,Z} in Example 1.4.11, in that order, form a Mal’cev basis adapted to the lower central series ﬁl- tration of the Heisenberg Lie algebra.

Suppose Xj, Xk are from the Mal’cev basis X and j ≤ k, then there exists 1 ≤ i ≤ r such that m − mi + 1 ≤ k ≤ m − mi+1. Hence Xk ∈ gi and l [Xj,Xk] ∈ gi+1 because {gi} is a ﬁltration. So there are constants cjk for all m − mi+1 + 1 ≤ l ≤ m, such that

m X l [Xj,Xk] = cjkXl. (1.15) l=m−mi+1+1

l The constants cjk, called structural constants, completely determines the Lie bracket on g, and hence, via Baker-Campbell-Hausdorﬀ formula, the multiplication on the simply connected Lie group G associated to g as well. d Once the Mal’cev basis X is chosen, the map ψ : R → G given by

ψ(u1, ··· , um) = exp(u1X1 + ··· + umXm) (1.16) is a diﬀeomorphism by Corollary 1.5.9. The following fact is evident: CHAPTER 1. NILPOTENT LIE GROUPS 29

Lemma 1.6.4. In the coordinate system (1.16), the multiplication on G is given by the polynomial map in (1.11). And a group morphism Ψ: G → G of the Lie group G is simply given by the linear transform DeΨ thanks to the diagram (1.14).

In this coordinate system we will write u = (u1, ··· , um) and, for j =

1, ··· , s, uj = (um−mj +1, ··· , um−mj+1 ). Then u = (u1, ··· , ur).

m−m 2 d Lemma 1.6.5. There exist polynomial functions θi :(R i ) → R i for all 1 ≤ i ≤ r, such that ψ(u)ψ(v) = ψ(w) if and only if

wi = ui + vi + θi(u1, ··· , ui−1, v1, ··· , vi−1). (1.17)

Moreover, θ1 = 0. Proof. Write U = Pm−mi+1 u X and similarly deﬁne V , W . By con- i j=m−mi+1 j j i i struction of ψ, m r r X X X ( Uj) ( Vi) = Wi. (1.18) i=r i=1 i=1 By Baker-Campbell-Hausdorﬀ formula, (1.18) is equal to

r X X (Ui + Vi) + (bracket terms), i=1 where each bracket term has the form b[Y1, [Y2, ··· [Yt,Yt+1] ··· ]] with b ∈ Q and Y1, ··· ,Yt+1 ∈ {U1, ··· ,Ur,V1, ··· ,Vr}. If at least one of these is from {Ui, ··· ,Ur,Vi, ··· ,Vr} ⊂ gi, then the bracket term will be in gi+1 as [g, gi] ⊆ gi+1. In this case, that bracket term belongs to the span of Xm−mi+1 , ··· ,Xm, and therefore does not contribute to the component Wi in (1.18). Therefore, Wi − Ui − Vi is a ﬁntie sum consisting of the Lm−mi+1 X -components of the bracket terms who only in- j=m−mi+1 R j volve U1, ··· ,Ur−1,V1, ··· ,Vr−1. Because this set of vectors are linearly parametrized by u1, ··· , ui−1, v1, ··· , vi−1 and the Lie bracket is bilinear, Wi −Ui −Vi is a polynomial in u1, ··· , ui−1, v1, ··· , vi−1. This is equivalent to (1.17). When i = 1, there are no such bracket terms as i − 1 = 0, and thus θ1 = 0. Instead of the coordinate change ψ, we can also deﬁne another map d φ : R → G by

φ(v1, ··· , vm) = exp(v1X1) ··· exp(vmXm). (1.19) CHAPTER 1. NILPOTENT LIE GROUPS 30

The map is differentiable and a local diffeomorphism near 0, because it Pm is easy to check that D0φ(u1, ··· , um) = j=1 umXm. Indeed, we will see that it is a global diffeomorphism. This follows from the following

m m Proposition 1.6.6. There is a bijective polynomial map f : R → R such m−mi 2 di that φ = ψ ◦ f. Moreover, there are polynomials ζi, ζbi :(R ) → R , such that if f(u) = v, then

vi = ui + ζi(u1, ··· , ui−1), (1.20)

ui = vi + ζbi(v1, ··· , vi−1). (1.21) ˆ Moreover, ζ1 = ζ1 = 0. Proof. The proof is by induction in r, the step of the filtration with respect to which X is defined. When r = 1, g is abelian, and therefore φ = ψ by Baker-Campbell- Hausdorff formula. So it suffices to take f = id and ζ1 = ζb1 = 0. 0 Assume the r − 1 case is known. Note that X = {Xd1+1, ··· ,Xm} is 0 0 a Mal’cev basis of g2. Let ψ , φ be the corresponding maps defined for 0 m m this basis. Then there is a bijective polynomial f : R 2 → R 2 satisfying φ0 = ψ0 ◦ f 0. Suppose φ(u) = ψ(v). Then

0 0 ψ(v) = φ(u1, 0)φ (u ), (1.22)

0 where u stands for (u2, ··· , ur). Then φ0(u0) = ψ0(v0) = ψ(0, v0) where v0 = f 0(u0). More precisely, according to the inductive hypothesis, for all 2 ≤ i ≤ r,

0 0 vi = ui + ζi(u2, ··· , ui−1), ∀2 ≤ i ≤ r (1.23)

0 0 0 0 ui = vi + ζbi(v2, ··· , vi−1), ∀2 ≤ i ≤ r, (1.24) 0 0 where ζi and ζbi are polynomial maps. Furthermore, by Baker-Campbell-Hausdroﬀ formula,

d1 X φ(u1, 0) = exp(u1X1) ··· exp(ud1 Xd1 ) = exp ujXj + η(u1) . j=1

Here η is a ﬁnite sum of iterated Lie brackets among u1X1, ··· , ud1 Xd1 , and therefore belongs to [g, g] = g2 and is polynomial in u1. In other words,

φ(u1, 0) = ψ(u1, η(u1)). (1.25) CHAPTER 1. NILPOTENT LIE GROUPS 31

Combining (1.22)-(1.25) with Lemma 1.6.5, we obtain that

v1 = u1 + 0 = u1 (1.26) and for 2 ≤ i ≤ r,

0 vi =(η(u1))i + vi 0 0 + θi u1, (η(u1))2, ··· , (η(u1))i−1, 0, v2, ··· , vi−1 h 0 (1.27) =ui + (η(u1))i + ζi(u2, ··· , ui−1) 0 0 i + θi u1, (η(u1))2, ··· , (η(u1))i−1, 0, v2, ··· , vi−1

Note that the sum in square brackets is a polynomial of u1, ··· , ui−1 0 0 0 because η, ζi and v2, ··· vi−1 are all polynomial. This proves (1.20) for index i. The array (u1, ··· , ui−1) is a polynomial function of (v1, ··· , vi−1), because (1.21) is assumed for all indices up to i − 1. Thus the function ζi(u1, ··· , ui−1) is a polynomial, which we denote by −ζbi, in (v1, ··· , vi−1). The equality (1.21) for i then follows from (1.20).

Exercises

Exercise 1.6.1. Using the Mal’cev basis {X1, ··· ,Xm,Y1, ··· ,Ym,Z} of the (2n+1)-dimensional Heisenberg Lie algebra h2n+1, express the multiplication rule on the corresponding Lie group H2n+1 in the coordinate system (1.19).

Exercise 1.6.2. Show that all simply connected 3-dimensional nilpotent 3 Lie groups are isomorphic to either H3 or R .