Let $\mathfrak{g}$ be the Lie algebra of a Lie group $G$, and $exp:\mathfrak{g}\to G$ be its exponential map. The group $G$ could be finite or infinite dimensional. Let $G$ have the property that

$\bullet$ For each smooth curve $X\in C^{\infty}(\mathbb R,\mathfrak g)$ there exists a curve $g\in C^{\infty}(\mathbb R,G)$ whose right logarithmic derivative is $X$, i.e., $$ g(0) = e, \qquad \partial_t g(t) = T_e(\mu^{g(t)})X(t) = X(t).g(t),\quad\text{where } \mu(a,b)=\mu_a(b)=\mu^b(a) = a.b $$ The curve $g$ is uniquely determined by its initial value $g(0)$, if it exists.

$\bullet$ Put $\operatorname{evol}^r_G(X)=g(1)$ where $g$ is the unique solution required above. Then $\operatorname{evol}^r_G: C^{\infty}(\mathbb R,\mathfrak g)\to G$ is required to be $C^{\infty}$ also.

Note that for $X$ constant in time, $\operatorname{evol}^r_G(X)=\exp(X)$. So each regular Lie group admits an exponential mapping.

My questions are:

1) Is there any expression for the derivative of $\operatorname{evol}^r_G$. It's motivated by the fact that that there's an expression of $(Dexp)_{\mathfrak g}$ using the adjoint action on $\mathfrak g$ and derivative of the left translation in $G$.

2) Can we construct a new Lie group $H\subset = C^{\infty}(\mathbb R,G)$ with Lie algebra $C^{\infty}(\mathbb R,\mathfrak g)$ so that $\operatorname{evol}^r$ becomes the exponential map $C^{\infty}(\mathbb R,\mathfrak g)\to H$?

I must admit that I got the idea of the question from Peter Michor's answer to this question:

Exponential map


1 Answer 1


Answer to 1:

$TG$ is again a Lie group, semidirect product go $G$ over the normal $\mathfrak g$. See 6.7 of here. $TC^\infty(\mathbb R, \mathfrak g) = C^\infty(\mathbb R, \mathfrak g\times \mathfrak g)$. Then $$ T(evol_G^r) = evol^r_{TG}. $$ Be careful with the identifications. This is treated in detail in 38.10 of this book.

Answer to 2:

See 38.11 and 42.21 of this same second book.


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