Welcome to MathOverflow

Question

2

1

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.

flag	asked Sep 15 2011 at 18:05 Alessandro Saccon 21●2

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

1

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

1

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

show 3 more comments

Taylor’s series for Lie groups

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.