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I am studying the Drinfel'd-Sokolov hierarchies. And here is what confuses me. Let $\mathfrak{g}$ be a (simple) finite-dimensional Lie algebra, let $u(x), v(x) \in C^\infty(\mathbb{R},\mathfrak{g})$. We can define the exponential of $F\in\operatorname{End}(\mathfrak{g})$ as

$\exp(F) := \mathrm{id} + F + \frac{1}{2!}(F\circ F)+\cdots \in \operatorname{GL}(\mathfrak{g})$

Question: how to calculate $\frac{d}{dx} \exp(\operatorname{ad} u(x))(v(x))$?

I believe there is an explicit formula but after long time searching, I cannot find any. And I hope I can get some inspiration here.

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  • $\begingroup$ Yes, but in the old papers it appears like that $\endgroup$
    – Fred Wang
    Commented May 23, 2017 at 9:30
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    $\begingroup$ 4.5 years later, I realised my comment was wrong, and deleted it. $\endgroup$
    – LSpice
    Commented Oct 6, 2021 at 0:07

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Every Lie group is locally isomorphic to a matrix Lie group. So in calculations like these it is in practice sufficient to do them in matrix groups. Moreover, you defined the exponential using power series which kind of implies that you are actually taking the exponential to live not in $G$ but in $\mathrm{Aut}(\mathfrak{g})$.

The result is $$ e^{\mathrm{ad}_{u(x)}} ([u'(x), v(x)] + v'(x)). $$

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    $\begingroup$ I am wondering this formula requires some conditions. See my question in mathoverflow.net/questions/405552/… . $\endgroup$
    – Po C.
    Commented Oct 6, 2021 at 0:15
  • $\begingroup$ Prompted by @PoC.'s comment and question, I wonder if you're sure this is true? With $u(x) = \begin{pmatrix} 0 & x \\ & 1 \end{pmatrix}$ and $v(x) = \begin{pmatrix} 0 \\ & 1 \end{pmatrix}$, I find that $\exp(\operatorname{ad}(u(x))v(x) = \begin{pmatrix} 0 & (1 - e^{-1})x \\ & 0 \end{pmatrix}$ … $\endgroup$
    – LSpice
    Commented Oct 6, 2021 at 1:29
  • $\begingroup$ … but $\exp(\operatorname{ad}(u(x))(\operatorname{ad}(u'(x))v(x) + v'(x)) = \begin{pmatrix} 0 & e^{-1} \\ & 0 \end{pmatrix}$. Maybe I miscomputed. $\endgroup$
    – LSpice
    Commented Oct 6, 2021 at 1:30
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It looks as though this is a standard lemma: https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula.

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  • $\begingroup$ I am not sure if this can be utilized for a general Lie algebra rather than a matrix algebra....nevertheless, maybe i can just calculate like that. $\endgroup$
    – Fred Wang
    Commented May 23, 2017 at 9:27
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    $\begingroup$ @FredWang, every finite-dimensional Lie algebra embeds in a matrix algebra: en.wikipedia.org/wiki/Ado%27s_theorem . (I'm not sure if this is applicable, because I realise I don't know what you're looking for: do you really put a Lie-algebra structure on $C^\infty(\mathbb R, G)$ (and, if so, what is it?), or do you mean just germs of such functions (say, at $0$)?) $\endgroup$
    – LSpice
    Commented May 23, 2017 at 12:37
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From wikipedia we know that $$\frac{d}{dx} \exp(u(x)) = \exp(u(x)) \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)!} (\operatorname{ad} u(x))^k \frac{d u(x)}{dx}.$$

Here in my question we have a discussion on differentiating $\exp(\operatorname{ad} X)$. My calculation shows that we actually have similar formula: $$\frac{d}{dx} \exp(\operatorname{ad} u(x)) = \exp(\operatorname{ad} u(x)) \operatorname{ad} \left( \sum_{k=0}^\infty \frac{(-1)^k}{(k+1)!} (\operatorname{ad} u(x))^k \frac{d u(x)}{dx} \right).$$

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