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The Exponential Map & One-Parameter Subgroups

The Lie algebra is the group's tangent space at the identity; the exponential map is the bridge back, turning each algebra vector into a smooth one-parameter flow through the group. This guide builds that bridge concretely, computes it for matrix groups, and is honest about exactly where it does and does not behave.

From the algebra back to the group

In the previous guide we built the Lie algebra g of a Lie group G as its tangent space at the identity, T_e G, equipped with the bracket inherited from left-invariant vector fields. That was a one-way trip: differentiate the group near e and you fall into the algebra, a finite-dimensional vector space that is far easier to handle than the curved group. The natural question now is whether you can travel the other way — given a vector X in g, can you reconstruct a piece of the group it came from? The exponential map is exactly that return ticket.

Here is the construction. Each X in g is the value at e of a unique left-invariant vector field, call it X-tilde, defined everywhere on G. Like any vector field, X-tilde generates a flow; but because it is left-invariant, that flow is unusually rigid. Define gamma(t) to be the integral curve through the identity, so gamma(0) = e and gamma'(0) = X. Then set exp(X) = gamma(1). In words: travel along the flow of X-tilde starting at e, and after one unit of time you arrive at the group element we call the exponential of X.

One-parameter subgroups: the trajectories

The identity gamma(s + t) = gamma(s)·gamma(t) from the callout is not a side remark — it is the whole point. A smooth curve gamma: R -> G satisfying gamma(s + t) = gamma(s)·gamma(t) for all s, t (and automatically gamma(0) = e) is called a one-parameter subgroup of G. It is a group homomorphism from the additive line (R, +) into G: a perfectly straight, never-bending path threaded through the curved group, the closest thing G has to a geodesic line for free.

The link to the exponential is a clean dictionary. Every one-parameter subgroup is exactly t -> exp(tX) for one and only one X in g, namely its velocity at zero, X = gamma'(0). Feeding the scalar t into the algebra before exponentiating simply rescales which vector you flow along, and exp(tX) traces out the entire subgroup as t runs over R. So one-parameter subgroups and vectors of the Lie algebra are the same data in two costumes: a direction X in g, or the straight trajectory exp(tX) it launches in G.

A small but vital warning about that scaling rule. We have exp((s + t)X) = exp(sX)·exp(tX) for free, because both sides move along ONE trajectory. But exp(X)·exp(Y) is almost never exp(X + Y): that would require X and Y to point along the same trajectory, i.e. to commute. The genuine answer is the Baker-Campbell-Hausdorff formula, exp(X)·exp(Y) = exp(X + Y + (1/2)[X, Y] + ...), whose correction terms are built entirely from iterated brackets. The failure of exp to be a homomorphism in two variables is precisely the bracket talking — which is why the Lie algebra remembers so much of the group.

Matrix groups: exp is the matrix exponential

For a matrix Lie group — any of the classical groups like GL(n), O(n), SU(n), or Sp(n) — the abstract exponential becomes a hands-on computation you can actually do. The Lie algebra g sits inside the matrices, the flow equation is the linear ODE gamma'(t) = gamma(t)·X with gamma(0) = I, and its solution is the honest power series. In other words, for matrices the Lie-group exp coincides with the matrix exponential, the same series you already know from solving systems of linear differential equations.

exp(X)  =  I + X + X^2/2! + X^3/3! + ...          (converges for every matrix X)

gamma(t) = exp(tX)    solves    gamma'(t) = gamma(t) X,   gamma(0) = I

Example, the rotation group SO(2):

        [ 0  -theta ]                  [ cos theta   -sin theta ]
  X  =  [ theta   0 ]   =>   exp(X) =  [ sin theta    cos theta ]

  (a skew-symmetric X exponentiates to an orthogonal rotation)
For matrix groups exp is the convergent power series; a skew-symmetric algebra element flows to a rotation in SO(2).

That tiny SO(2) example carries the whole lesson in miniature. The algebra so(2) consists of 2x2 skew-symmetric matrices, a single real line spanned by the matrix that rotates by angle theta; its exponential is the rotation matrix itself. This is the general pattern for the classical groups: skew-symmetric matrices exponentiate to orthogonal ones, skew-Hermitian to unitary, trace-free to determinant-one. The algebraic constraint defining the algebra (skew, trace-zero, ...) is the infinitesimal shadow of the constraint defining the group (orthogonal, unimodular, ...), and exp converts one into the other.

How well-behaved is exp, really?

Near the identity, exp is as good as you could hope. Its differential at the origin, d(exp)_0: g -> T_e G, is the identity map, so by the inverse function theorem exp is a local diffeomorphism: there is a neighborhood of 0 in g mapped smoothly and invertibly onto a neighborhood of e in G. This is the engine behind normal coordinates on a group and the reason small group elements can always be written uniquely as exp of a small algebra vector — a fact guide 4 leans on heavily when it studies the adjoint representation.

Globally, though, exp can misbehave in two genuinely different ways, and honesty demands you keep them straight. First, exp need not be surjective: for the non-compact group SL(2, R), there are elements (with trace less than -2) that are simply not the exponential of anything in sl(2, R), so you cannot reach all of the group by flowing straight from e. Second, exp need not be injective: on the circle group SO(2) the curve exp(t·X) returns to the identity every full turn, so infinitely many algebra vectors share the same image. Local diffeomorphism does not promise either global property.

Computing exp, step by step

Abstractly exp is a flow; in practice, for a matrix group, you compute it by diagonalizing. The power series exp(X) = I + X + X^2/2! + ... respects any change of basis, so if you can put X into a form whose powers are transparent, the whole series collapses into closed form. Here is the standard recipe, walked through on a vector X you have already identified as living in the Lie algebra.

  1. Confirm X lies in the Lie algebra g: check the infinitesimal constraint (e.g. skew-symmetric for so(n), trace-zero for sl(n)). This guarantees exp(X) lands back in the group G, not merely in GL(n).
  2. Diagonalize (or Jordan-decompose) X = P D P^(-1). Then every power obeys X^k = P D^k P^(-1), so the powers no longer fight you.
  3. Exponentiate the diagonal: exp(X) = P · exp(D) · P^(-1), where exp(D) just exponentiates each eigenvalue on the diagonal — the infinite series becomes ordinary scalar exponentials.
  4. Read off the one-parameter subgroup: replace X by tX throughout to get exp(tX), the entire straight trajectory through e, and sanity-check that it satisfies the group constraint for every t.

One last caveat to keep you honest, especially when X has repeated or complex eigenvalues. If X is not diagonalizable you must use its Jordan form, where a nilpotent block contributes polynomial-in-t factors (this is exactly how a shear, not a rotation, shows up). And sign conventions differ across books: some authors define exp via gamma'(t) = X·gamma(t) (right-invariant) rather than gamma(t)·X, which can flip a sign in non-abelian computations. Pick the left-invariant convention we used here, state it, and you will stay consistent with the exponential map as guide 4 and 5 deploy it.