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In quantum case, we have commutators. In classical case, we have Poisson bracket. Let $T$ be a Poisson group, $a, b \in \mathbb{C}_q[T].$ It seems that we have $$ [a, b]=(q-1)\{a,b\}+o((q-1)^2). $$ Is this true? When $q\to 1$, we have $[a,b]/(q-1)\to \{a,b\}$.

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    $\begingroup$ What is $\mathbb{C}_q[T]$? $\endgroup$ – Qiaochu Yuan Jan 26 '15 at 5:56
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Your notation entails undefined terms and peculiar notation. I will, instead, use the conventional ones.

Using the invertible Wigner map from linear operators Φ in Hilbert space to functions in phase space f(x,p), one has $$ f(x,p)= 2 \int_{-\infty}^\infty \text{d}y~e^{-2ipy/\hbar}~ \langle x+y| \Phi [f] |x-y \rangle. $$ In phase space, one may now compare apples with apples, and indeed, the commutator of two operator maps to the bilinear of the corresponding two functions f and g, $$ f \star g - g \star f = f \, \exp{\left( \frac{i \hbar}{2} \left(\overleftarrow {\partial }_x \overrightarrow{\partial }_p-\overleftarrow{\partial }_p \overrightarrow {\partial }_x \right) \right)} \, g - g \, \exp{\left( \frac{i \hbar}{2} \left(\overleftarrow {\partial }_x \overrightarrow{\partial }_p-\overleftarrow{\partial }_p \overrightarrow {\partial }_x \right) \right)} \, f \\ = 2i~~ f \sin \left ( {\frac{\hbar }{2} (\overset{\leftarrow}{\partial_x} \overset{\rightarrow}{\partial_p}-\overset{\leftarrow}{\partial_p}\overset{\rightarrow}{\partial_x})} \right ) \ g =i\hbar \{f,g\} + O(\hbar^3), $$ with the classical $\hbar\to 0$ limit you are evoking.

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