Clever way to show a property of Lie transformation

Question

Given the following Lie transformation:

$$ \exp(\lbrace H, \cdot \rbrace):=\sum_{n=0}^{\infty} \frac{(\lbrace H, \cdot \rbrace)^n}{n!} $$ and apply it to a Poisson Bracket $\lbrace g_1, g_2 \rbrace$. I would like to show that $$ \exp(\lbrace H, \lbrace g_1, g_2 \rbrace\rbrace)=\sum_{n=0}^{\infty} \frac{(\lbrace H, \lbrace g_1, g_2 \rbrace \rbrace)^n}{n!}=\dots=\lbrace \sum_{n=0}^{\infty} \frac{(\lbrace H, g_1\rbrace)^n}{n!}, \sum_{n=0}^{\infty} \frac{(\lbrace H, g_2\rbrace)^n}{n!}\rbrace=\lbrace\exp(\lbrace H, g_1\rbrace),\exp(\lbrace H, g_2\rbrace)\rbrace $$ I put some dots in the Passage I am not able to do...(I tried expressing the power using the binomial theorem)

I am sure this property holds, since I found this here (page 25): http://www.aps.anl.gov/Science/Publications/lsnotes/content/files/APS_1418211.pdf

but I have difficulties in proving this in an elegant way. Could someone give me a good reference?

This is the way I tried: $$\exp(\lbrace H, \lbrace g_1, g_2 \rbrace\rbrace)=\sum_{n=0}^{\infty} \frac{(\lbrace H, \lbrace g_1, g_2 \rbrace \rbrace)^n}{n!}= \sum_{n=0}^{\infty} \frac{( \{\{H,g_1\},g_2\}+\{g_1,\{H,g_2\}\} )^n}{n!}= \sum_{n=0}^{\infty} \frac{1}{n!} \sum_{k=0}^n {n \choose k}(\{\{H,g_1\},g_2\})^{n-k}(\{g_1,\{H,g_2\}\})^k = $$ $$\sum_{n=0}^{\infty} \sum_{k=0}^n \frac{1}{(n-k)!}(\{\{H,g_1\},g_2\})^{n-k}\frac{1}{k!}(\{g_1,\{H,g_2\}\})^k $$

Answer 1

Although OP strictly speaking writes something different in the question formulation (v4), it seems OP essentially wants to prove the following Lemma.

Lemma. Given a (possibly infinite-dimensional$^1$) Lie algebra $L$ over a field $\mathbb{F}\supseteq\mathbb{R}$ with Lie bracket $[\cdot,\cdot]:L\times L \to L$ and Lie algebra derivation$^2$ $D:L\to L$, which satisfies Leibniz rule $$ \forall x,y\in L:~~ D[x,y]~=~ [Dx,y]+[x,Dy].\tag{1}$$ Assume that the exponential map $$e^D:= {\bf 1}_L+ D + \frac{1}{2}D^2+ \ldots ~:~L\to L\tag{2}$$ is pointwise convergent. Then $$ \forall x,y\in L:~~ e^D[x,y]~=~[e^Dx,e^Dy].\tag{3}$$

Proof of eq. (3): The identity (3) follows by setting $t=1$ in the following identity:

$$ \forall t\in [0,1]\forall x,y\in L:~~ e^{tD}[x,y]~=~[e^{tD}x,e^{tD}y].\tag{4}$$

To prove eq. (4), first note that eq. (4) is trivially true for $t=0$. Next show that the lhs. and the rhs. of eq. (4) independently satisfy the same first-order ODE $$ \frac{d}{dt}(\ldots)~=~D(\ldots). \tag{5}$$ Hence they must be equal.

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$^1$ Note that the Poisson algebra $(C^{\infty}(M);\{\cdot,\cdot\}_{PB})$ of smooth functions over a finite-dimensional phase space $M$ is an infinite-dimensional Lie algebra.

$^2$ In OP's case the derivation $D=[z,\cdot]$ is an inner derivation.