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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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  • $\begingroup$ Is $f$ supposed to be a group homomorphism? $\endgroup$
    – José Figueroa-O'Farrill
    Commented Sep 15, 2011 at 18:08
  • $\begingroup$ 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? $\endgroup$
    – Alessandro Saccon
    Commented Sep 15, 2011 at 19:00
  • $\begingroup$ The exponential map is a local diffeomorphism at the origin, so Taylor's theorem for multivariate functions applies. $\endgroup$
    – Fernando Muro
    Commented Sep 15, 2011 at 19:29
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    $\begingroup$ 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 $\mathbb{R}^n$. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Sep 15, 2011 at 19:49
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    $\begingroup$ 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. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Sep 15, 2011 at 22:06

1 Answer 1

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Let me sketch a solution as a three step process:

For smooth function $f:M\to G_2$ consider its left logarithmic differential $\delta^l f\in \Omega^1(M,\mathfrak g_2)$ which satisfies the right Maurer-Cartan equation $d(\delta^l f) +\frac12 [\delta^l f,\delta^lf]=0$. It can be reconstructed on simply connected domains in $M$ uniquely up to constant right translation from $\delta^l f$. This called the Cartan development. See 4.2 of here, e.g., for a detailed proof.

Thus for $f:G_1\to G_2$ we have $\delta^l f\in \Omega^1(G_1,\mathfrak g_2)$. By left trivializing $TG_1$ we can view $\delta^l f$ as an element of $C^\infty (G_1, L(\mathfrak g_1,\mathfrak g_2))$.

Thm 2.6 in the following paper is the Taylor theorem with remainder term for functions on a Lie group $G_1$ (with values in a vector space, here $L(\mathfrak g_1,\mathfrak g_2))$. The infinite Taylor series factors to a linear functional on the universal enveloping algebra of the Lie algebra $\mathfrak g_1$.

  • Peter W. Michor: The cohomology of the diffeomorphism group is a Gelfand-Fuks cohomology. Rendiconti del Circolo Matematico di Palermo, Serie II, Suppl. 14 (1987), 235-- 246 (pdf)

Putting all together again, we get a Taylor series with remainder term for a smooth mapping $G_1\to G_2$.

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