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Do there exist results towards answering the following question?

Consider the Poisson algebra of regular functions $A=\mathbb R[V]$ on the symplectic vector space $V:=T^* \mathbb R^n$. Using canonical coordinates we can write $A=\mathbb R[q^1,\dots,q^n,p_1,\dots,p_n]$ with Poisson bracket being $\{q^i,p_j\}=\delta_j^i$ (all other brackets between coordinates being zero).

Question: What are the finite dimensional Lie subalgebras of the Poisson algebra $(A,\cdot,\{\:,\:\})$?

Partial answer: 1. There are subalgebras in degree 2 coming from moment maps of cotangent lifted representations. 2. There are a couple of Heisenberg subalgebras. 3. There are plenty of abelian subalgebras.

I wonder if one can give a complete classification using these building blocks.

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  • $\begingroup$ The Poisson bracket is supposed to be a Lie bracket (valued in the vector space), but the formula you gives yields a scalar-valued bracket. What is the Lie bracket? $\endgroup$
    – YCor
    Commented Jul 25, 2015 at 8:49
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    $\begingroup$ This problem is at least as hard as classifying the finite dimensional Lie subalgebras of the vector fields in $\mathbb{R}^n$. Even when one identifies two such algebras when they differ by a diffeomorphism, this latter problem is quite hard for $n=2$, and a complete list for $n=3$ (with various 'regularizing' assumptions) is still not done. As far as I know, no one has seriously attempted $n=4$. $\endgroup$
    – Robert Bryant
    Commented Jul 25, 2015 at 10:20
  • $\begingroup$ Hello YCor, a scalar is a constant regular function, i.e. $\in A$. One can use the relation $\{q^i,p_j\}=\delta_j^i$ together with the Leibniz rule $\{a,bc\}=b\{a,c\}+\{a,b\}c$ to define $\{\:,\:\}:A\times A\to A$ since $A$ is generated as an algebra by the coordinate functions. By definition, the bracket of a Lie subalgebra arrises as the restriction of the Lie bracket of the ambient space, in this case $\{\:,\:\}$. $\endgroup$
    – HCH
    Commented Jul 25, 2015 at 18:53

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Using the Leibniz rule and the induction on $deg(f) + deg (g)$, we can see that the braket on $A$ is exactely the Poisson bracket defined by him in 1809 :) That is: $$ \{f, \, g\} = \sum_{i=1}^r \, \Bigl({\frac {\partial f} {\partial p_i}} {\frac {\partial g} {\partial q_i}} - {\frac {\partial f} {\partial q_i}} {\frac {\partial g} {\partial p_i}} \Bigl) $$ for all $f$, $g \in A$.

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