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In this set of notes, we describe the basic analytic structure theory of Lie groups, by relating them to the simpler concept of a Lie algebra. Roughly speaking, the Lie algebra encodes the “infinitesimal” structure of a Lie group, but is a simpler object, being a vector space rather than a nonlinear manifold. Nevertheless, thanks to the fundamental theorems of Lie, the Lie algebra can be used to reconstruct the Lie group (at a local level, at least), by means of the exponential map and the Baker-Campbell-Hausdorff formula. As such, the local theory of Lie groups is completely described (in principle, at least) by the theory of Lie algebras, which leads to a number of useful consequences, such as the following:

• (Local Lie implies Lie) A topological group ${G}$ is Lie (i.e. it is isomorphic to a Lie group) if and only if it is locally Lie (i.e. the group operations are smooth near the origin).
• (Uniqueness of Lie structure) A topological group has at most one smooth structure on it that makes it Lie.
• (Weak regularity implies strong regularity, I) Lie groups are automatically real analytic. (In fact one only needs a “local ${C^{1,1}}$” regularity on the group structure to obtain real analyticity.)
• (Weak regularity implies strong regularity, II) A continuous homomorphism from one Lie group to another is automatically smooth (and real analytic).

The connection between Lie groups and Lie algebras also highlights the role of one-parameter subgroups of a topological group, which will play a central role in the solution of Hilbert’s fifth problem.

We note that there is also a very important algebraic structure theory of Lie groups and Lie algebras, in which the Lie algebra is split into solvable and semisimple components, with the latter being decomposed further into simple components, which can then be completely classified using Dynkin diagrams. This classification is of fundamental importance in many areas of mathematics (e.g. representation theory, arithmetic geometry, and group theory), and many of the deeper facts about Lie groups and Lie algebras are proven via this classification (although in such cases it can be of interest to also find alternate proofs that avoid the classification). However, it turns out that we will not need this theory in this course, and so we will not discuss it further here (though it can of course be found in any graduate text on Lie groups and Lie algebras).

Over the past few months or so, I have been brushing up on my Lie group theory, as part of my project to fully understand the theory surrounding Hilbert’s fifth problem. Every so often, I encounter a basic fact in Lie theory which requires a slightly non-trivial “trick” to prove; I am recording two of them here, so that I can find these tricks again when I need to.

The first fact concerns the exponential map ${\exp: {\mathfrak g} \rightarrow G}$ from a Lie algebra ${{\mathfrak g}}$ of a Lie group ${G}$ to that group. (For this discussion we will only consider finite-dimensional Lie groups and Lie algebras over the reals ${{\bf R}}$.) A basic fact in the subject is that the exponential map is locally a homeomorphism: there is a neighbourhood of the origin in ${{\mathfrak g}}$ that is mapped homeomorphically by the exponential map to a neighbourhood of the identity in ${G}$. This local homeomorphism property is the foundation of an important dictionary between Lie groups and Lie algebras.

It is natural to ask whether the exponential map is globally a homeomorphism, and not just locally: in particular, whether the exponential map remains both injective and surjective. For instance, this is the case for connected, simply connected, nilpotent Lie groups (as can be seen from the Baker-Campbell-Hausdorff formula.)

The circle group ${S^1}$, which has ${{\bf R}}$ as its Lie algebra, already shows that global injectivity fails for any group that contains a circle subgroup, which is a huge class of examples (including, for instance, the positive dimensional compact Lie groups, or non-simply-connected Lie groups). Surjectivity also obviously fails for disconnected groups, since the Lie algebra is necessarily connected, and so the image under the exponential map must be connected also. However, even for connected Lie groups, surjectivity can fail. To see this, first observe that if the exponential map was surjective, then every group element ${g \in G}$ has a square root (i.e. an element ${h \in G}$ with ${h^2 = g}$), since ${\exp(x)}$ has ${\exp(x/2)}$ as a square root for any ${x \in {\mathfrak g}}$. However, there exist elements in connected Lie groups without square roots. A simple example is provided by the matrix

$\displaystyle g = \begin{pmatrix} -4 & 0 \\ 0 & -1/4 \end{pmatrix}$

in the connected Lie group ${SL_2({\bf R})}$. This matrix has eigenvalues ${-4}$, ${-1/4}$. Thus, if ${h \in SL_2({\bf R})}$ is a square root of ${g}$, we see (from the Jordan normal form) that it must have at least one eigenvalue in ${\{-2i,+2i\}}$, and at least one eigenvalue in ${\{-i/2,i/2\}}$. On the other hand, as ${h}$ has real coefficients, the complex eigenvalues must come in conjugate pairs ${\{ a+bi, a-bi\}}$. Since ${h}$ can only have at most ${2}$ eigenvalues, we obtain a contradiction.

However, there is an important case where surjectivity is recovered:

Proposition 1 If ${G}$ is a compact connected Lie group, then the exponential map is surjective.

Proof: The idea here is to relate the exponential map in Lie theory to the exponential map in Riemannian geometry. We first observe that every compact Lie group ${G}$ can be given the structure of a Riemannian manifold with a bi-invariant metric. This can be seen in one of two ways. Firstly, one can put an arbitrary positive definite inner product on ${{\mathfrak g}}$ and average it against the adjoint action of ${G}$ using Haar probability measure (which is available since ${G}$ is compact); this gives an ad-invariant positive-definite inner product on ${{\mathfrak g}}$ that one can then translate by either left or right translation to give a bi-invariant Riemannian structure on ${G}$. Alternatively, one can use the Peter-Weyl theorem to embed ${G}$ in a unitary group ${U(n)}$, at which point one can induce a bi-invariant metric on ${G}$ from the one on the space ${M_n({\bf C}) \equiv {\bf C}^{n^2}}$ of ${n \times n}$ complex matrices.

As ${G}$ is connected and compact and thus complete, we can apply the Hopf-Rinow theorem and conclude that any two points are connected by at least one geodesic, so that the Riemannian exponential map from ${{\mathfrak g}}$ to ${G}$ formed by following geodesics from the origin is surjective. But one can check that the Lie exponential map and Riemannian exponential map agree; for instance, this can be seen by noting that the group structure naturally defines a connection on the tangent bundle which is both torsion-free and preserves the bi-invariant metric, and must therefore agree with the Levi-Civita metric. (Alternatively, one can embed into a unitary group ${U(n)}$ and observe that ${G}$ is totally geodesic inside ${U(n)}$, because the geodesics in ${U(n)}$ can be described explicitly in terms of one-parameter subgroups.) The claim follows. $\Box$

Remark 1 While it is quite nice to see Riemannian geometry come in to prove this proposition, I am curious to know if there is any other proof of surjectivity for compact connected Lie groups that does not require explicit introduction of Riemannian geometry concepts.

The other basic fact I learned recently concerns the algebraic nature of Lie groups and Lie algebras. An important family of examples of Lie groups are the algebraic groups – algebraic varieties with a group law given by algebraic maps. Given that one can always automatically upgrade the smooth structure on a Lie group to analytic structure (by using the Baker-Campbell-Hausdorff formula), it is natural to ask whether one can upgrade the structure further to an algebraic structure. Unfortunately, this is not always the case. A prototypical example of this is given by the one-parameter subgroup

$\displaystyle G := \{ \begin{pmatrix} t & 0 \\ 0 & t^\alpha \end{pmatrix}: t \in {\bf R}^+ \} \ \ \ \ \ (1)$

of ${GL_2({\bf R})}$. This is a Lie group for any exponent ${\alpha \in {\bf R}}$, but if ${\alpha}$ is irrational, then the curve that ${G}$ traces out is not an algebraic subset of ${GL_2({\bf R})}$ (as one can see by playing around with Puiseux series).

This is not a true counterexample to the claim that every Lie group can be given the structure of an algebraic group, because one can give ${G}$ a different algebraic structure than one inherited from the ambient group ${GL_2({\bf R})}$. Indeed, ${G}$ is clearly isomorphic to the additive group ${{\bf R}}$, which is of course an algebraic group. However, a modification of the above construction works:

Proposition 2 There exists a Lie group ${G}$ that cannot be given the structure of an algebraic group.

Proof: We use an example from the text of Tauvel and Yu (that I found via this MathOverflow posting). We consider the subgroup

$\displaystyle G := \{ \begin{pmatrix} 1 & 0 & 0 \\ x & t & 0 \\ y & 0 & t^\alpha \end{pmatrix}: x, y \in {\bf R}; t \in {\bf R}^+ \}$

of ${GL_3({\bf R})}$, with ${\alpha}$ an irrational number. This is a three-dimensional (metabelian) Lie group, whose Lie algebra ${{\mathfrak g} \subset {\mathfrak gl}_3({\bf R})}$ is spanned by the elements

$\displaystyle X := \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \alpha \end{pmatrix}$

$\displaystyle Y := \begin{pmatrix} 0 & 0 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$

$\displaystyle Z := \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ -\alpha & 0 & 0 \end{pmatrix}$

with the Lie bracket given by

$\displaystyle [Y,X] = -Y; [Z,X] = -\alpha Z; [Y,Z] = 0.$

As such, we see that if we use the basis ${X, Y, Z}$ to identify ${{\mathfrak g}}$ to ${{\bf R}^3}$, then adjoint representation of ${G}$ is the identity map.

If ${G}$ is an algebraic group, it is easy to see that the adjoint representation ${\hbox{Ad}: G \rightarrow GL({\mathfrak g})}$ is also algebraic, and so ${\hbox{Ad}(G) = G}$ is algebraic in ${GL({\mathfrak g})}$. Specialising to our specific example, in which adjoint representation is the identity, we conclude that if ${G}$ has any algebraic structure, then it must also be an algebraic subgroup of ${GL_3({\bf R})}$; but ${G}$ projects to the group (1) which is not algebraic, a contradiction. $\Box$

A slight modification of the same argument also shows that not every Lie algebra is algebraic, in the sense that it is isomorphic to a Lie algebra of an algebraic group. (However, there are important classes of Lie algebras that are automatically algebraic, such as nilpotent or semisimple Lie algebras.)

Let ${G}$ be a Lie group with Lie algebra ${{\mathfrak g}}$. As is well known, the exponential map ${\exp: {\mathfrak g} \rightarrow G}$ is a local homeomorphism near the identity. As such, the group law on ${G}$ can be locally pulled back to an operation ${*: U \times U \rightarrow {\mathfrak g}}$ defined on a neighbourhood ${U}$ of the identity in ${G}$, defined as

$\displaystyle x * y := \log( \exp(x) \exp(y) )$

where ${\log}$ is the local inverse of the exponential map. One can view ${*}$ as the group law expressed in local exponential coordinates around the origin.

An asymptotic expansion for ${x*y}$ is provided by the Baker-Campbell-Hausdorff (BCH) formula

$\displaystyle x*y = x+y+ \frac{1}{2} [x,y] + \frac{1}{12}[x,[x,y]] - \frac{1}{12}[y,[x,y]] + \ldots$

for all sufficiently small ${x,y}$, where ${[,]: {\mathfrak g} \times {\mathfrak g} \rightarrow {\mathfrak g}}$ is the Lie bracket. More explicitly, one has the Baker-Campbell-Hausdorff-Dynkin formula

$\displaystyle x * y = x + \int_0^1 F( \hbox{Ad}_x \hbox{Ad}_{ty} ) y\ dt \ \ \ \ \ (1)$

for all sufficiently small ${x,y}$, where ${\hbox{Ad}_x = \exp( \hbox{ad}_x )}$, ${\hbox{ad}_x: {\bf R}^d \rightarrow {\bf R}^d}$ is the adjoint representation ${\hbox{ad}_x(y) := [x,y]}$, and ${F}$ is the function

$\displaystyle F( t ) := \frac{t \log t}{t-1}$

which is real analytic near ${t=1}$ and can thus be applied to linear operators sufficiently close to the identity. One corollary of this is that the multiplication operation ${*}$ is real analytic in local coordinates, and so every smooth Lie group is in fact a real analytic Lie group.

It turns out that one does not need the full force of the smoothness hypothesis to obtain these conclusions. It is, for instance, a classical result that ${C^2}$ regularity of the group operations is already enough to obtain the Baker-Campbell-Hausdorff formula. Actually, it turns out that we can weaken this a bit, and show that even ${C^{1,1}}$ regularity (i.e. that the group operations are continuously differentiable, and the derivatives are locally Lipschitz) is enough to make the classical derivation of the Baker-Campbell-Hausdorff formula work. More precisely, we have

Theorem 1 (${C^{1,1}}$ Baker-Campbell-Hausdorff formula) Let ${{\bf R}^d}$ be a finite-dimensional vector space, and suppose one has a continuous operation ${*: U \times U \rightarrow {\bf R}^d}$ defined on a neighbourhood ${U}$ around the origin, which obeys the following three axioms:

• (Approximate additivity) For ${x,y}$ sufficiently close to the origin, one has

$\displaystyle x*y = x+y+O(|x| |y|). \ \ \ \ \ (2)$

(In particular, ${0*x=x*0=x}$ for ${x}$ sufficiently close to the origin.)

• (Associativity) For ${x,y,z}$ sufficiently close to the origin, ${(x*y)*z = x*(y*z)}$.
• (Radial homogeneity) For ${x}$ sufficiently close to the origin, one has

$\displaystyle (sx) * (tx) = (s+t)x \ \ \ \ \ (3)$

for all ${s,t \in [-1,1]}$. (In particular, ${x * (-x) = (-x) * x = 0}$ for all ${x}$ sufficiently close to the origin.)

Then ${*}$ is real analytic (and in particular, smooth) near the origin. (In particular, ${*}$ gives a neighbourhood of the origin the structure of a local Lie group.)

Indeed, we will recover the Baker-Campbell-Hausdorff-Dynkin formula (after defining ${\hbox{Ad}_x}$ appropriately) in this setting; see below the fold.

The reason that we call this a ${C^{1,1}}$ Baker-Campbell-Hausdorff formula is that if the group operation ${*}$ has ${C^{1,1}}$ regularity, and has ${0}$ as an identity element, then Taylor expansion already gives (2), and in exponential coordinates (which, as it turns out, can be defined without much difficulty in the ${C^{1,1}}$ category) one automatically has (3).

We will record the proof of Theorem 1 below the fold; it largely follows the classical derivation of the BCH formula, but due to the low regularity one will rely on tools such as telescoping series and Riemann sums rather than on the fundamental theorem of calculus. As an application of this theorem, we can give an alternate derivation of one of the components of the solution to Hilbert’s fifth problem, namely the construction of a Lie group structure from a Gleason metric, which was covered in the previous post; we discuss this at the end of this article. With this approach, one can avoid any appeal to von Neumann’s theorem and Cartan’s theorem (discussed in this post), or the Kuranishi-Gleason extension theorem (discussed in this post).