Lie Groups

A Lie Group, M, is a smooth manifold equipped with a smooth multiplication μ = (: (M: x*y ←y :M) ←x :M) under which it forms a group, whose inversion mapping is also smooth.

That μ is smooth says that, for each x in M, (M: x*y ←y :M) is a smooth mapping (M: :M) and that, when we consider the induced smooth manifold of such smooth mappings, μ is a smooth mapping from M to this induced manifold. In practice this just means that, for any x in M, both (M: x*y ←y :M) and (M: y*x ←y :M) are smooth.

That M forms a group under μ says that * is both cancellable and complete; that is

or equally (and more orthodoxly) as: there is some e in M for which e*x = x = x*e for every x in M; and there is a mapping (M: inv |M) for which inv(x)*x = e = x*inv(x) for every x in M. The specification of a smooth manifold requires that inv be a smooth mapping (M: |M).

That M is a smooth manifold implies that we have mutually dual linear spaces of tangents and gradients at every point of M; we get mappings (:T:M) and (:G:M) for which, for each x in M, T(x) is a linear space of tangents and G(x) is its dual, a linear space of gradients. For any smooth scalar field ({scalars}:h:M) whose (:h|) subsumes a neighbourhood of some x in M, we get a gradient dh(x) in G(x); for any smooth trajectory (M:f:{scalars}) whose (:f|) subsumes some neighbourhood of some scalar t, we get a tangent f'(t) in T(f(t)); and, when (:h|) subsumes some neighbourhood of f(t), dh(f(t))·f'(t) = (h&on;f)'(t), the (scalar) derivative at t of the smooth composite ({scalars}: h&on;f :{scalars}).

Whenever we have smooth manifolds M, N and a smooth mapping (N:S:M) from some neighbourhood, in M, of some member p of M, we obtain natural mappings from M's trajectories through p to N's trajectories through S(p) and from N's scalar fields defined in neighbourhoods of S(p) to M's scalar fields defined in neighbourhoods of p. These in turn induce natural linear maps from M's tangents at each p to N's tangents at the corresponding S(p) and from N's gradients at S(p) to M's gradients at p:

(The latter's name is here taken from discussion of more general alternating forms on smooth manifolds.) When S is locally iso (that is, each p in (:S|) has an open neighbourhood U subsumed by (:S|) for which (:S:U) is monic) these derived linear maps are also monic, so their reverses are mappings.

Left Transport

For any x in M we have μ(x) = (: x*y ←y :) as a smooth monic mapping (M||M) inducing

whose restriction, for each y in M, to T(y) is a linear map (T(x*y): Dμ(x) |T(y)). I'll describe this as a left transport of T, since it works by left-multiplying. Note that we can form an analogous linear map (T(y*x):|T(y)) by applying D to transpose(μ) = (: (: y*x ←y :) ←x :); this could sensibly be construed as a right transport. By construction, we may observe that Dμ(x)&on;Dμ(y) = Dμ(x*y); while Dμ(e) is necessarily the identity T(e).

Now, for any X in T(e), (: Dμ(x, X) ←x |M) is a vector field on M; its value at each x in M is a member of T(x). It is described as left invariant because it is a fixed point of Dμ(y), for each y in M. As a vector field, it defines a flow which we can integrate up to find curves always tangent to the vector field.

Now, for any trajectory (M:f:{scalars}) we can ask whether its tangent varies along its length the same way that Dμ&on;f would make it vary; that is, given scalars s and t of which (:f|) subsumes some connected neighbourhood,

or, to put it another way, whether Dμ(inv(f(t)), f'(t)) in T(e) is independent of t along the curve. If it is, we have some X in T(e) for which f'(t) = Dμ(f(t), X) along the curve; this is a linear first-order partial differential equation specifying the trajectory.

The exponential function

Because the solution of linear first order differential equations with well-specified boundary conditions is unique, any f and g with equal constant Dμ(inv(f(t)), f'(t)) = X = Dμ(inv(g(t)), g(t)) and passing through a common point, f(a) = g(b), must necessarily be related by f(t+a) = g(t+b), i.e. f = (: g(s) ←s−a+b :). For any u in M, consider the trajectory g = μ(u)&on;f; this yields

making g also a solution of the same differential equation. Now, if we pick u = f(a)*inv(f(b)) for some a, b in (:f|), we'll obtain g(b) = f(a), yielding g(s) = f(a−b+s) = f(a)*inv(f(b))*f(s). If we now consider such a trajectory passing through e at parameter 0, i.e. f(0) = e, f'(t) = Dμ(f(t), X) for some X in T(e), we obtain b = 0 and f(s+t) = f(s)*f(t) for all s, t in (:f|), making f a group homomorphism from the additive structure on {scalars} to the *-structure on M.

I can now define a relation ({trajectories}: p :T(e)) by: p relates f to x precisely if (:f|) is a connected open neighbourhood of 0, f(0) = e and, for each scalar t in (:f|), f'(t) = Dμ(f(t),x). For given x, this specification of f is a well-defined first-order partial differential equation with unambiguous initial condition, so various theorems of differential analysis imply uniqueness of solution; thus, if p relates f and g to the same x in T(e) then f and g agree in the intersection of (:f|) and (:g|). We may thus interpret p(x) as the union of all left values which p relates to x; this will, indeed, be one of the left values p does relate to x. Thus p is effectively a mapping ({trajectories}: |T(e)). In particular, for each x in T(e), we obtain p(x)'(0) = x and p(x,s+t) = p(x,s)*p(x,t) for all s, t in (:p|).

Now, for any trajectory (M:f:{scalars}) and any scalar k, we can construct g = (: f(k.t) ←t :) and infer g'(t) = k.f'(k.t) from the original construction of T and G (because the same holds true for the composite mappings from {scalars} to {scalars} that results from composing the a scalar field after the trajectories). If f is a solution to our differential equation, for some given X in T(e), this gives us

by linearity of the outputs of Dμ. We can thus infer that p(x,s) = p(s.x,1) so that it makes sense to define

characterized by:

Note that the last doesn't equate exp(x+y) to exp(x)*exp(y) except when x and y are parallel. Note, also, that (|exp:) need not be all of M – indeed, since (: exp(t.x) ←t; 0≤t≤1 :{scalars}) is a trajectory from e to exp(x), for each x in T(e), each output of exp is in the same path-connected component of M as e. In the Lie group of geometry-preserving linear maps in even just two real dimensions, the reflections are a separate connected component from the rotations, so no reflection lies in (|exp:). Indeed, in the multiplicative group of non-zero real scalars, only positive values are outputs of the standard exp function; the negatives are a separate connected component.

All the same, exp is manifestly smooth and monic on at least some neighbourhood of zero in T(e), with the result that (|exp:) contains some neighbourhood of e. (I'm sure this implies that it contains the whole connected component containing e, but I can't for the moment see how to prove it.)

The other thing to note about exp is its natural relationship with Lie group homomorphisms. Suppose (N:S|M) is a smooth homomorphism of Lie groups; let U be T(e) in M and V be the corresponding space of tangents at N's identity; then we have (V:DS|U) and both (N:exp|V) and (M:exp|U) yielding S&on;(M:exp|U) and (N:exp|V)&on;DS as mappings (N:|U). Since U's exp(0) is the identity in M and S is a group homomorphism, S(exp(0)) is the identity in N; while DS is linear, so maps 0 to 0 so that exp&on;DS also maps 0 in U to the identity in N. … Now, consider x in U; it's the tangent at 0 of (: exp(s.x) ←s :) so DS maps it to DS(x) = (: S(exp(s.x)) ←s :)'(0) in V which exp maps to a member of N, which we'll want to compare to S(exp(x)). So consider (: exp(t.DS(x)) ←t :)'(r) = Dν(exp(r.DS(x)), DS(x)) …

naturality will let us exploit the adjoint representation, in which exp looks like the familiar power series !

The adjoint representation

Any u in M yields γ(u) = (M| u*x*inv(u) ←x |M), which preserves the group structure since u*x*inv(u)*u*y*inv(u) = u*x*y*inv(u); in particular, e is a fixed point. Consequently this induces a linear map Adjoint(u) = (T(e)| (: u*f(t)*inv(u) ←t :)'(s) ←f'(s) |T(e)) which is necessarily iso. We thus obtain ({linear iso (T(e)||T(e))}: Adjoint |M), which is known as the adjoint representation of M. (A representation of a Lie group is a group monomorphism from it into the Lie group of auto-isomorphisms of some linear space; this effectively represents the Lie group as a family of matrices.) By looking at this embedding's derivative at e we get a linear map DAdjoint(e) embedding T(e) in tangents to {linear iso (T(e)||T(e))} at its identity, which is just (the identity on) T(e).

Now, {linear (T(e):|T(e))} is a linear space and its iso (T(e)||T(e)) members form an open sub-set of it, whose tangent space at any point – including its identity element – is simply the linear space, of which (|DAdjoint(e):T(e)) will necessarily be a linear sub-space. Thus DAdjoint(e) is linear and each of its outputs is a linear map (T(e):|T(e)), which makes DAdjoint(e) a bilinear multiplication on T(e), combining two members of T(e) to produce a third. Hereafter I'll denote it as DAdjoint(e) = (: (T(e): [x|y] ←y |T(e)) ←x |T(e)). (Orthodoxy writes [x,y] but I'm following some schools of computer programming and using this to denote the list x←0, y←1, so it would be confusing to also have it denote something else.)

Now, let us examine DAdjoint(e) and discover its properties. First, we need to characterize Adjoint(u), at least for u near e. We start with γ(u), which is a group homomorphism (M||M) and has at least u as a fixed point; its fixed points commute with u.

The requirement u near e can be expressed as u = exp(x) for some x in a (suitably small) neighbourhood of zero in T(e). This yields

Furthermore, we can replace f, s with f'(s) in T(e) using f'(s) = y, s = 0, f = (: exp(t.y) ←t :) to obtain:

The anti-symmetric product

On the way to defining exp I established that a general trajectory, f, with constant Dμ(inv(f(t)), f'(t)) in T(e) must satisfy

for all scalar r, s and t; and that, for any u in M, μ(u)&on;f is another trajectory with the same constant result when its tangent is transformed back to the identity using Dμ. Applying the latter with inv(u) a point on the trajectory gives us a trajectory through e; shifting the trajectory's input parameter by a constant then makes it pass through e at parameter 0, putting it in the standard form exp expresses above. The parameter shift can be un-done to put the trajectory through e in the form (: exp(k.x)*exp(t.x) ←t :) and applying μ(u) to this yields the same as applying μ(u*exp(k.x)) to the standard trajectory. Consequently the general trajectory, f, with constant Dμ(inv(f(t)), f'(t)) in T(e) is given by f(t) = u*exp(t.x) for some u in M and x in T(e). If f falls within the same path-connected component of M as e, then so does u and we may suppose u = exp(y) for some y; to discover now exp(x+y) and exp(x)*exp(y) are related, in general, we must now examine the case where y is not parallel to x.

For given x, y in T(e), consider Q = (: (: exp(s.x)*exp(t.y) ←t :) ←s :) – for each scalar s, Q(s) is a trajectory of the above type; and, for each scalar t, (: Q(s,t) ←s :) is also a trajectory, though not necessarily of the above type.

old mumblings

As a topological manifold, it's described by a collection of mutually isomorphic open sets which collectively generate (by taking finite intersections and arbitrary unions) an open topology. As a smooth manifold, it's described by an atlas consisting of a family of mutually smooth charts: a chart being an isomorphism between topologically trivial opens sets, one in the manifold and another in a linear space. The manifold is covered by (i.e. is (subsumed by) the union of) the open sets on its ends of the charts in its atlas. If we compose one chart with the reverse of another, we get an isomorphism between open sets – one way round these are open in the smooth manifold; the other way round, they are open in the linear space. [Note that composing a&on;b implicitly restricts a's right values to (|b:) and b's left values to (:a|), so the composite is as small as the intersection of (:a|) with (|b:); when this is empty, so is the composite; but, indeed, the empty mapping (construed as a mapping from the empty open subset of any linear space to that of any (optionally other) linear space) is smooth, albeit fatuously so.] For the latter case, linear analysis gives us a notion of differentiation, hence of smoothness: the atlas is smooth precisely if all composites it generates between neighbourhoods in the linear space are smooth. A mapping between neighbourhoods of the smooth manifold is then described as smooth precisely if all its appropriate composites between one chart and the reverse of another are smooth.

As a group, it supports a binary operator which we can construe as a (very well behaved, algebraically) mapping from the group to the collection of mappings from the group to itself. To be a Lie group, this mapping must be a smooth embedding of the manifold in the collection of smooth mappings from the manifold to itself.

I need to transcribe from lecture notes the explanations, but: there arises a natural mapping from the tangent bundle (canonically at the identity) to the group, called exp, for which, for any tangent T, (: exp(t.T) ←t :{scalars}) respects the group structures (so exp((t+s).T) = exp(t.T)*exp(s.T), with * as the binary operator in our group) and is a smooth trajectory, passing through the identity at t = 0 with tangent T. If the Lie group is compact, expect each such trajectory to be cyclic – i.e. there's some non-zero t for which exp(t.T) is the identity. We can then look at ({identity}: exp |), the collection of tangents, T, for which exp(T) is the identity, which should say interesting things about the Lie group; the zero tangent is in it and its other members describe the periodicities of exp. If we take exp's restriction, to a bubble about the tangent zero that's small enough to not get near the rest of ({identity}: exp |), it's a chart identifying this bubble with a neighbourhood of the identity in the Lie group.

Now exp maps tangents to members of the Lie group, among which we have our binary operator * and an inversion for which inv(a)*a and a*inv(a) are the identity, for any a in the group; this enables us to construct a*inv(b)*inv(a)*b = a*inv(a*b)*b from any a, b in the group; I think it's known as their commutator. We can apply this with a and b as exp(T) and exp(S), for some tangents T and S. We can expect inv(exp(T)) to be exp(−T) and likewise for S, so we'll be looking at exp(T)*exp(−S)*exp(−T)*exp(S). At least for T and S within some neighbourhood of the zero tangent, we can then ask for the tangent which exp maps to this product. Expect thereby to obtain an antisymmetric multiplication ^ on tangents for which exp(2.T^S) = exp(T)*exp(−S)*exp(−T)*exp(S); and expect that this multiplication is a quadratic form on tangents (i.e. T^S ← S is linear for any given T as is T^S←T for any given S; though anti-symmetry, T^S = −S^T, will induce linearity of either of these from that of the other).

Now T^S will be zero precisely if exp(T) and exp(S) commute with one another (which arises iff either commutes with the inverse of the other); in particular, this happens when T and S are parallel (because (: exp(t.T) ←t :) embeds the additive group structure of scalars faithfully in our group). Expect exp(T+S) to deviate from exp(T)*exp(S) in manners depending on T^S.

Expect ({identity}: exp |) to include a surface, among tangents at the identity, bounding a neighbourhood of the zero tangent whose intersection with ({identity}: exp |) is simply {zero} itself. IWBNI this surface were an ellipsoid: we could then use it to infer a natural symmetric positive definite quadratic form on tangents at the identity – i.e. a natural metric.

See also: someone else's page.

See also

Unitary transformations and rotations.

Lie Algebra

The space of tangents at the identity of a Lie Group supports an anti-symmetric bi-linear product, yielding its own members as values; the algebraic structure thus formed is described as the Lie Algebra of the given Lie Group. Distinct Lie Groups may have coincident Lie Algebras – notably, SU(2) and SO(3), or so I'm told.

For given tangents S, T at the identity and scalars s, t we can form the product exp(s.S)*exp(t.T)*exp(−s.S)*exp(−t.T) in the Lie Group and show that it differs from the identity by an amount proportional to s.t, essentially because, if we expand exp as a power series and ignore terms with more than two factors of s or t:

= (1 +s.S +s.s.S*S/2)*(1 +t.T +t.t.T*T/2)*(1 −s.S +s.s.S*S/2)*(1 −t.T +t.t.T*T/2)
= 1 +s.S +t.T −s.S −t.T +s.s.S*S +t.t.T*T +s.t.S*T −s.s.S*S −s.t.S*T −t.s.T*S −t.t.T*T +s.t.S*T
= 1 + s.t.(S*T −T*S)

Re-writing s.S as U and t.T as W, this is approximately exp(U)*exp(W)*exp(−U)*exp(−W) = exp(U*W −W*U). Thus U^W, alluded to above, is approximated by (U*W −W*U)/2; hence we see why it is anti-symmetric and bilinear.

Next look at how exp(s.S+t.T) relates to exp(s.S)*exp(t.T):

= (1 +s.S +s.s.S*S/2)*(1 +t.T +t.t.T*T/2)
= 1 +s.S +t.T +(s.s.S*S +2*s.t.S*T +t.t.T*T)/2
= 1 +s.S +t.T +s.t.(S*T −T*S)/2 +(s.S +t.T)*(s.S +t.T)/2

which is approximately exp(s.S +s.T +s.t.S^T), suggesting that we could alternatively have defined ^ by exp(U)*exp(V) = exp(U +V +U^V).

Natural Constancy

A smooth doesn't generally have a well-defined meaning for the notion constant for any other rank of its tensor bundle than the scalar fields. However, a Lie group naturally supports two families of automorphisms that map the tensors, of any given rank, at one point to those at another point. These provides meaningful notions of constancy and, hence, natural differential operators.

For any two members, h and t, of the Lie Group, L, (using x/y\z to denote the composite of x after y-inverse after z) the mappings

are permutations and map h←t. Composing either after a trajectory through t yields a trajectory at h, so each induces a natural isomorphism (T(h)||T(t)); and composing either before a scalar field at h yields a scalar field at t, so each induces a natural isomorphism (G(t)||G(h)). From these and their inverses we may induce natural isomorphisms, for each tensor rank, between the rank's spaces at t and h. Any value for a tensor at the identity (or, indeed, any other point of our Lie Group) can then be turned into a constant tensor field on the whole Lie Group by applying the relevant isomorphism for one or other of the above two permutations.

In principle this will give us two differential operators, one from each of the two permutations. They will necessarily differ by, in each rank at each point of the Lie Group, a linear map: we can either pick one of them, or take their average to obtain a standard differential operator. The first thing I must then do is discover the linear map that describes the difference. I have my suspicions that this will be interesting in its own right…

It should suffice to examine the question in a neighbourhood of the origin, since the structure of the question implies that we can just permute this to any other member of the Lie Group. So let us consider the case where t is the identity and h is of form exp(a.H) for some member, H, of the tangent bundle at the origin and some small real a. For any tangent K at the identity, we have a matching trajectory (: exp(b.K) ←b :) through the identity (at b = 0) and our permutation maps this trajectory to (: exp(a.H)·exp(b.K) ←b :) or (: exp(b.K)·exp(a.H) ←b :). Following out-bound along one trajectory and back along the other gets us the trajectory (: exp(a.H)·exp(b.K)·exp(−a.H) ←b :) at the origin.

Use exp as chart of a neighbourhood of the identity. We have tangents T(p) and gradients G(p) at each p in our neighbourhood (|exp:U) for some open U within T(1) with (:exp|U) as our chart and zero in U. For any w in G(1) we have W = (: w(x) ← exp(x) :) as a scalar field on (|exp:U) and can look at dW, its gradient, which is independent of our choice of differential operator. If we examine any trajectory C = (: exp(t.x) ←t :) through the identity, 1 = C(0), with tangent C'(0) = x, we find W&on;C = (: t.w(x) ←t :) so dW(C(0))·C'(0) = (W&on;C)'(0) = w(x), so dW(1)·x = w·x; as this is true for any tangent x we infer dW(1) = w. Thus a basis of G(1) yields a family of scalar fields whose gradients are that basis at 1 and, by continuity, must provide a basis of the gradient bundle in some neighbourhood of 1. The derivatives of such a basis fully describe any differential operator; so examining the derivatives the gradients of scalar fields (: w(x) ← exp(x) :) with w in G(0) suffices to fully characterize our differential operators.

First, then, let us look at the gradient of W = (: w(x) ← exp(x) :) away from the identity; we know dW(1) = w in G(1). Given H in T(1) and a small scalar h, consider dW(exp(h.H)) in G(exp(h.H)). A typical tangent at exp(h.H) is C = (: exp(h.H +t.x) ←t :) for some x in T(1), with C(0) = exp(h.H) and W&on;C = (: h.w·H +t.w·x ←t :); as before, dW(exp(h.H))·C'(0) = (W&on;C)'(0) = w·x, but now C'(0) is different. There will be a linear map (T(exp(h.H))| q |T(1)) which maps x to C'(0) with transpose (G(1)| |G(exp(h.H))) which maps dW(exp(h.H)) to w. We then need to characterize these two linear maps suitably to enable us to compute derivatives.


The useful course from my lecture notes was Adams, the Exceptional Lie {Groups,Algebras}.

Apparently there are four infinite (:|{naturals}) families of Lie groups – A, B, C and D – and five sporadic Lie groups, G2, F4, E6, E7 and E8. I'm guessing the last five are the exceptional ones; apparently they're all sub-groups of E8.

Any compact connected abelian Lie group is a torus, T(n) = Rene({T(1)}:|n) with T(1) = {phases} or {rotations in 2 real dimensions}. Any compact Lie group has maximal tori as sub-groups. The interesting thing about a Lie group is its quotient by a maximal torus, since this encodes is non-commutative aspects.

Classification of Simple Lie Groups.

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