## Taylor’s series for Lie groups

Let $G_1$ and $G_2$ be two (matrix) Lie groups, with $L(G_1)$ and $L(G_2)$ their respective Lie algebras.

I am interested to know if there is a well developed theory to approximate a (sufficiently) smooth function $f:G_1 \rightarrow G_2$ using a "Taylor's series" expansion. That is, I'd like to know how I can compute the functions $a_i: L(G_1) \rightarrow L(G_2)$, $i = 1,2, \dots$ such that the following identity holds

$f(g \exp( \varepsilon \zeta)) = f(g) \exp( \varepsilon a_1(\zeta) + \frac{\varepsilon^2}{2!} a_2(\zeta) + \frac{\varepsilon^3}{3!} a_3(\zeta) + \dots)$

with $\varepsilon \in \mathbb{R}$ and $\zeta \in L(G_1)$.

Clearly, $a_1(\zeta) = f(g)^{-1} Df(g)\cdot g\zeta$...

Thanks.

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Is $f$ supposed to be a group homomorphism? – José Figueroa-O'Farrill Sep 15 2011 at 18:08
No, $f$ is a generic mapping. Also the dimensions of $G_1$ and $G_2$ are arbitrary. I am really looking for a general formula, if any exists, that agrees with Taylor's when $G_1 = (\mathbb{R}^n, +)$ and $G_2 = (\mathbb{R}^m, +)$. Does assuming $f$ a group homomorphism help? – Alessandro Saccon Sep 15 2011 at 19:00
The exponential map is a local diffeomorphism at the origin, so Taylor's theorem for multivariate functions applies. – Fernando Muro Sep 15 2011 at 19:29
If $f$ is not a homomorphism, then I don't see that $G_i$ being Lie groups is particular relevant. As Fernando Muro points out, this is just a (smooth) map between manifolds, so compose with local charts and it's just a smooth map from $\mathbb{R}^m$ to $\mathbb{R}^n$. – José Figueroa-O'Farrill Sep 15 2011 at 19:49
Sorry, I forgot to answer the question in your comment. If $f$ is a homomorphism then $f(g \mathrm{exp}(t\zeta)) = f(g) f(\mathrm{exp}(t\zeta) = f(g) \mathrm{exp}(t f_*(\zeta)$, where $f_*$ is the Lie map of $f$: the induced homomorphism of Lie algebras. – José Figueroa-O'Farrill Sep 15 2011 at 22:06