Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Given a group $(G,*)$ there is no candidate for what can be understood as a derivative of a function $$f:G\rightarrow\mathbb{R}.$$ However, for the special case of Carnot groups there is the so-called Pansu derivative and one can get surprisingly far with this notion (of course for any Lie group one can speak about the pushforward but that is not what I am interested in). As a recall, Carnot groups are particular examples of Subriemannian manifolds: by definition a Carnot group $G$ of step $n$ is a simply connected Lie group with a stratified Lie algebra $\mathfrak{g}$, i.e. $$\mathfrak{g}=V_1\oplus\cdots\oplus V_n$$ with $[V_i,V_1]=V_{i+1}$. Due to this stratification there exists a natural dilation $$\delta_{\epsilon}:G\rightarrow G$$ One says that $f$ is Pansu differentiable at $g\in G$ if the maps $$\frac{f(g*\delta_{\epsilon}(.))-f(g)}{\epsilon}$$ converge (locally uniformly) as $\epsilon\rightarrow 0$ to a homomorphism $$Df(g):(G,*)\rightarrow (\mathbb{R},+)$$ (in fact, one take any other Carnot group instead of $\mathbb{R}$ but let's keep things simple). My question is if whether this can be extended to higher derivatives. More precisely in linear spaces $U,V$ the $k$th derivative of $F:U\rightarrow V$ is a map $$D^k F:U\rightarrow L(U^{\otimes k},V)$$ but in the setting above I already do not know how to replace $U^{\otimes k}$ (e.g. what are $k$-linear maps on the direct products of $G$). If there is such a notion, can one also make sense of Taylor approximations to $f$?

share|improve this question
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.