3
$\begingroup$

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!

$\endgroup$
3
$\begingroup$

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 $tW$.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Mathmath May 5 '15 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 $exp[V,W]\in G$. $\endgroup$ – Mathmath May 6 '15 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 '15 at 10:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.