Let $G$ be a (not necessarily compact, probably even infinite dimensional) Lie group, and $g$ be its Lie algebra. Let $V,W\in g$. Consider $J(t):=(Dexp)_{tV}(tW)$ be the result of differential of the Lie group exponential map $exp:g\to G$ at $tV$ acting on $tW$.
Can we express $J(t)$ in terms of the adjoint operator $ad$ on $g$, defined by $ad_X(Y)=[X,Y]$.
I'd really appreciate a detailed answer if possible. In my case $G$ is a certain subgroup of diffeomorphism group of a manifold, and $g$ is a subspace of vector fields on that manifold. But I don't think this piece of information will be necessary.
Also, how can we relate the derivative of the Lie group exponential map $exp:g\to G$ in terms of the above adjoint operator?
Thank you!
Best Answer
Let us set $t=1$. There is a nice formal calculation of $J(1)$ which makes perfect sense if $\mathfrak g$ is nilpotent and thus $G$ sits in the enveloping algebra $U\mathfrak g$ (otherwise take it as a formal calculation that has to be made rigorous): $$J(1)=\frac{d}{d\epsilon}\exp(V+\epsilon W)|_{\epsilon=0}=\lim_{n\to\infty}\frac{d}{d\epsilon}(1+\frac{V+\epsilon W}{n})^n|_{\epsilon=0}$$ $$=\lim_{n\to\infty}\sum_{k=0}^n(1+\frac{V}{n})^k\, W\, (1+\frac{V}{n})^{n-k-1} =\int_0^1\exp(sV)W\exp((1-s)V)\,ds$$ $$=\Bigl(\int_0^1\exp(sV)W\exp(-sV)\,ds\Bigr)\exp(V).$$
This gives us that $J(1)\exp(-V)$, i.e. $J(1)$ translated by the right translation to the origin, is $$\int_0^1\exp(sV)W\exp(-sV)\,ds=(\int_0^1 Ad_{\exp(sV)}\,ds)(W)$$ which can be rewritten in the nilpotent/formal case as
To have $t\neq1$, replace $V$ by $tV$ and $W$ by $tW$.