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The components of the classical angular momentum $L_i$, satisfy the Poisson bracket relation $$\{L_i,L_j\}=\epsilon_{ijk}L_k,\tag{1}$$ and forms a Lie algebra (i.e, anti-symmetric, obeys the Jacobi identity etc).

In the theory of Lie groups, the commutator $$[\hat{L}_i,\hat{L}_j]=i\epsilon_{ijk}\hat{L}_k\tag{2}$$ is definitely called a $\mathfrak{so}(3)$ Lie algebra, because the $\hat{L}_i$'s generate ${\rm SO}(3)$ group elements $\left(\exp\left(-i\theta_i\hat{L}_i\right)\right)$.

I wonder whether the PB Lie algebra $(1)$ is also called a $\mathfrak{so}(3)$ Lie algebra. And if so, why? What is the relation of $(1)$ to the ${\rm SO}(3)$ group, if any?

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    $\begingroup$ Yes, it is the $\mathfrak{so}(3)$ Lie algebra. physics.stackexchange.com/q/532270 and links therein might help $\endgroup$ – Nihar Karve Dec 29 '20 at 12:59
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    $\begingroup$ In fact, the linked question (with its most popular answer) appears as a near duplicate. $\endgroup$ – Cosmas Zachos Dec 29 '20 at 16:31
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    $\begingroup$ Note the linear operators $\{L_i, \bullet \}\equiv \partial L_i /\partial q^j \partial \bullet /\partial p^j - \partial L_i /\partial p^j \partial \bullet /\partial q^j $ furnish a faithful realization of the elements $\hat L_i$ of (2). Verify this by acting on an arbitrary phase-space function f(x,p). $\endgroup$ – Cosmas Zachos Dec 29 '20 at 16:46
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    $\begingroup$ Near duplicate. $\endgroup$ – Cosmas Zachos Dec 29 '20 at 18:59
  • $\begingroup$ I have voted to reopen because the duplicate does not address the issue of the finite transformation $\in SO(3)$, which seems core to this question. $\endgroup$ – ZeroTheHero Dec 31 '20 at 15:44