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Let $R_n=\mathbb{C}[q_1, \ldots, q_n, p_1, \ldots p_n]$ be a polynomial algebra over complex numbers with even number of variables. Then $R$ admits Poisson bracket defined on linear functions as follows $$ \{p_i, q_j\} =\delta_{ij},\\ \{q_i, q_j\}=0,\\ \{p_i, p_j\}=0. $$ and extended to all polynomials by Leibniz's rule. Consider $R_n$ as (infinite dimensional) Lie algebra with such bracket.

I want to find non-trivial (and even interesting) examples of finite dimensional Lie subalgebras containing polynomials of degree at least 3.

Set $P_n=\mathbb{C}[q_, \ldots, q_n]$, if $v=\sum_{i=1}^n v_i \partial_i \in \operatorname{Der}(P_n)$ is a derivation with polynomial coefficients $v_i \in P_n$ then $\sigma(v)= \sum_{i=1}^n v_i p_i \in R_n$ and it is easy to check $$ \sigma([v_1, v_2])=\{\sigma(v_1), \sigma(v_2)\}. $$

So, a special case of the question is finding a finite dimensional Lie algebra of polynomial vector fields on an affine space, containing at least one vector field with at least one quadratic coefficient.

If $n=1$ then $L=<\partial, q \partial, q^2 \partial>$ seems to be a example of vector fields, giving the following 3-dimensional subalgebra $K=<p, qp, qp^2>$ of the Poisson algebra. Are there any other examples for $n=1$?

This example can be generalized to any $n$. The linear span of $\partial_i$, $q_i \partial_j$ and $q_i E$, where $E=\sum_i q_i \partial_i$, generate Lie algebra $sl(n+1, \mathbb{C})$. Moreover, this algebra is maximal: http://math.univ-lyon1.fr/~ovsienko/Publis/LecLMP.pdf

Hence, we can embed $gl(n, \mathbb{C})$ into $R_{n}$ by embedding it first in $sl(n+1, \mathbb{C})$.

Another way to embed $gl(n, \mathbb{C})$ is to embed it into $sp(2n, \mathbb{C})$ and notice that symplectic Lie algebra is realized inside $R_{2n}$ by quadratic polynomials. So, this not an example of what I'm looking for.

Alex
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  • Feels like cheating, but you can take any 1-dimensional subspace of $R$ and it will be a Lie subalgebra – user8268 Sep 10 '19 at 06:23
  • Ok, I want non-trivial examples. – Alex Sep 10 '19 at 07:01
  • You can generalize your last example to higher $n$'s and get $sl(n+1)$ (via infinitesimal projective transformations) – user8268 Sep 10 '19 at 16:52
  • Do you want to exclude things like $\langle p, qp, p^4 \rangle$? – Ricardo Buring Sep 15 '19 at 13:10
  • No, I don't. Looks like we can span an abelian Lie algebra by some power of $p$'s and then add $qp$, which acts as Euler derivative. – Alex Sep 15 '19 at 17:19
  • Also, when you fix the number of generators and bound their degrees, then this is a big system of polynomial equations for the coefficients of the generators and the structure constants of the subalgebra. You can produce these equations and find some solutions, e.g. with SageMath. – Ricardo Buring Sep 15 '19 at 17:37

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