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Right-trivialized differential of the Lie-group exponential

Statement

Assume ${A C}_{ω}$ . Let $G$ be a finite-dimensional real Lie group with Lie algebra $g$ . For $Z, V \in g$ , right translation identifies the differential of the exponential with

$d (R_{\exp_{G} (- Z)})_{\exp_{G} Z} (d (\exp_{G})_{Z} V) = \sum_{n = 0}^{\infty} \frac{{ad}_{Z}^{n} (V)}{(n + 1)!} .$

The operator series converges absolutely in any norm and is denoted

$\frac{e^{{ad}_{Z}} - I}{{ad}_{Z}} (V),$

with the displayed power series—not division by a possibly singular ${ad}_{Z}$ —as its definition.

More generally, for every finite-dimensional real vector space and every endomorphism $B$ , the linear-ODE exponential used here satisfies

$e^{t B} = \sum_{n = 0}^{\infty} \frac{t^{n} B^{n}}{n!},$

with absolute convergence uniformly on compact $t$ -intervals.

Facts & Assumptions

Given: ${A C}_{ω}$ , a finite-dimensional real Lie group $G$ , and $Z, V \in g$ .

[F1]

Lie-group exponentials are smooth, and $t \mapsto \exp_{G} (t W)$ is the integral curve of $W^{L}$ through the identity. The Lie-group exponential map is smooth with identity differential at zero. One-parameter subgroups are integral curves of left-invariant fields.

[F2]

Left and right translations and their differentials give the standard tangent trivializations. Left and right translations on a Lie group.

[F3]

Assuming countable choice, ${Ad}_{\exp_{G} (t Z)} = e^{t {ad}_{Z}}$ , where the right side is the linear-ODE exponential. The Axiom of Countable Choice ( ${A C}_{ω}$ ). Adjoint exponential identity.

[F4]

Linear matrix initial-value problems have unique solutions on compact intervals. Linear matrix ODEs have unique global solutions on a fixed interval.

[F5]

A finite basis gives bounded coordinates; the scalar exponential series converges everywhere and real power series differentiate termwise inside their radii. A chosen algebraic basis identifies a finite-dimensional normed space with a coordinate space. The exponential series converges absolutely for every real argument. Inside its radius a real power series may be differentiated term by term, and the differentiated series has the same radius.

[F6]

The vector-valued fundamental theorem of calculus integrates a continuous derivative componentwise. If $f : [a, b] \to R^{m}$ is differentiable with integrable $f^{'}$ then $\int_{a}^{b} f^{'} = f (b) - f (a)$ ; and a bounded derivative makes $f$ Lipschitz.

[F7]

Differentials of smooth maps obey the chain rule. The chain rule for differentials of smooth maps.

Proof

technique · direct

1.1F4F5

Put $B = {ad}_{Z}$ . In one basis, the operator series $S (t) = \sum_{n \geq 0} t^{n} B^{n} / n!$ converges absolutely and uniformly on compact $t$ -intervals, since its norm is bounded by the scalar exponential majorant $\sum_{n \geq 0} (∣ t ∣ ∥ B ∥)^{n} / n!$ . Coordinatewise termwise differentiation gives $S^{'} (t) = B S (t)$ and $S (0) = I$ . By [F4]--[F5], uniqueness therefore identifies $S (t)$ with the linear-ODE exponential $e^{t B}$ .

1.2F1F2F7algebra

Consider the smooth variation $F (s, t) = \exp_{G} (t (Z + s V))$ and write $γ (t) = F (0, t) = \exp_{G} (t Z)$ . Right-trivialize its variational field by $ξ (t) = d (R_{γ (t)^{- 1}})_{γ (t)} (\partial_{s} F (0, t))$ . By [F1], $\partial_{t} F (s, t) = d (L_{F (s, t)})_{e} (Z + s V)$ . Differentiate this identity in $s$ , commute the two coordinate partial derivatives, and differentiate the right-trivialization using multiplication and inversion. The two terms containing $Z$ cancel, leaving $ξ^{'} (t) = {Ad}_{γ (t)} V$ and $ξ (0) = 0$ .

2.1F3step 1.1

Consequently [F3] gives ${Ad}_{\exp_{G} (t Z)} = S (t)$ for every real $t$ . This is the exact point where ${A C}_{ω}$ is used.

3.1F5F6step 2.1step 1.2

By steps 2.1 and 1.2, $ξ^{'} (t) = S (t) V$ . The uniformly convergent series $A (t) = \sum_{n \geq 0} t^{n + 1} B^{n} (V) / (n + 1)!$ has $A (0) = 0$ and, coordinatewise by [F5], $A^{'} (t) = S (t) V$ . Applying [F6] to $ξ - A$ on $[0, 1]$ gives $ξ (1) = A (1) = \sum_{n \geq 0} B^{n} (V) / (n + 1)!$ .

4.1F2F7step 3.1

By the definition of $F$ and the chain rule, $\partial_{s} F (0, 1) = d (\exp_{G})_{Z} V$ ; by the definition of $ξ$ , its value at one is the right translation of that vector by $\exp_{G} (- Z)$ . Step 3.1 is therefore exactly the asserted formula.

5.1F1F2F3F4F5F6F7step 1.1step 2.1step 1.2step 3.1step 4.1∎

A Lie group is nonempty and boundaryless. In dimension zero both sides are the unique zero vector; in dimension one the bracket vanishes and the formula reduces to $d R_{\exp (- Z)} d \exp_{Z} (V) = V$ . Singular ${ad}_{Z}$ is allowed because the quotient notation means its entire power series. The variation uses the compact interval $[0, 1]$ including both endpoints. Countable choice is inherited through [F1] and [F3]; one basis and fixed vectors add no choice. No metric dependence or biconditional is asserted.

Right-trivialized differential of the Lie-group exponential

Statement

Facts & Assumptions

Proof

Depends on

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