Timeline for Taylor's series for Lie groups
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 22, 2013 at 19:56 | answer | added | Peter Michor | timeline score: 4 | |
| Sep 16, 2011 at 9:02 | comment | added | Alessandro Saccon | (José Figueroa-O'Farrill) thanks for the answers. I like the idea that the group homomorphism can be thought as an "affine" transformation (i.e., only the term in $t$ is present in the expansion). | |
| Sep 16, 2011 at 9:00 | comment | added | Alessandro Saccon | (Fernando Muro) Thanks. I actually started from there. I defined $F(\zeta):=\log(f(g)^{−1}f(g\exp(\zeta))$ so that $F(t\zeta) = t a_1(\zeta)+t^2 a_2(\zeta)+...$ but, if I use coordinates and differentiate $F$, I have to differentiate $\log$ and $\exp$, obtaining derivatives that are not intrinsic. I′d like instead to obtain a "Leibniz s rule" to get simpler subexpressions. Actually,using covariant derivative of 2−point tensors (using the Cartan (0) connection), I get things like $a_2(\zeta) = f(g)^{-1} \mathbb{D}^2 f(g) \cdot (g\zeta, g\zeta)$... which is promising. Is this well known? | |
| Sep 16, 2011 at 7:06 | comment | added | Mark Grant | Is a formal group an example of what you want? en.wikipedia.org/wiki/Formal_group | |
| Sep 15, 2011 at 22:06 | comment | added | José Figueroa-O'Farrill | 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. | |
| Sep 15, 2011 at 19:49 | comment | added | José Figueroa-O'Farrill | 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 $\mathbb{R}^n$. | |
| Sep 15, 2011 at 19:29 | comment | added | Fernando Muro | The exponential map is a local diffeomorphism at the origin, so Taylor's theorem for multivariate functions applies. | |
| Sep 15, 2011 at 19:00 | comment | added | Alessandro Saccon | 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? | |
| Sep 15, 2011 at 18:08 | comment | added | José Figueroa-O'Farrill | Is $f$ supposed to be a group homomorphism? | |
| Sep 15, 2011 at 18:05 | history | asked | Alessandro Saccon | CC BY-SA 3.0 |