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!


1 Answer 1


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

$$f({ad}_V)(W)$$ where $$f(x)=\frac{\exp(x)-1}{x}=\int_0^1\exp(xs)\,ds.$$

To have $t\neq1$, replace $V$ by $tV$ and $W$ by $tW$.

  • $\begingroup$ user8268, thanks for your answer! Could you please suggest a possible resource for this kind of computation? \\ In my case: $g$ is a space of sufficiently smooth vector field whose values and derivatives go to $0$ at infinity, considered with the Lie bracket of vector fields. Is $g$ nilpotent?\\ Important: could you explain more this part: $$f({ad}_V)(W)$$ where $$f(x)=\frac{\exp(x)-1}{x}=\int_0^1\exp(xs)\,ds.$$ Here you're basically substituting $x$ by an operator $ad_V$ on $g$, but then what's the meaning of your "division by $ad_V$ " when you substitute $ad_V$ for $x$ in denominator? $\endgroup$ May 5, 2015 at 23:31
  • $\begingroup$ Also, could you explain $f(ad_V)(W)$? Is it $\frac{exp[V,W]-W}{[V,W]}$ if I directly apply the definition? But then the term does not make sense, since we're diving by $[V,W]\in g$ and also we're subtracting $W\in g$ from an element $exp[V,W]\in G$. $\endgroup$ May 6, 2015 at 0:03
  • $\begingroup$ @Mathmath The Lie algebra of vector fields is not nilpotent; it doesn't mean that the formulas don't make sense, but one needs to check convergence. As for $f({ad_V})$: write $f$ as a Taylor series $f(x)=1+x/2+x^2/6+x^3/24\dots$, then substitute ${ad}_V$ for $x$. You'll get $f({ad_V})W=W+[V,W]/2+[V,[V,W]]/6+[V,[V,[V,W]]]/24+\dots$. Convergence might be an issue, so you may prefer the formula $\int_0^1 Ad_{\exp(sV)}(W)\,ds$ $\endgroup$
    – user8268
    May 6, 2015 at 10:11

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