Given in this exercise is the following set: $U = \{f(z) = z^TCz\ \vert\ C \in \textrm{Mat}_{2n}(\mathbb R),\ C^T=C\}$ is a Lie Algebra with $\left\{\cdot , \cdot \right\}$ where $$ \left\{f,g \right\} = \sum_{j=1}^n \frac{\partial f}{\partial q_j} \frac{\partial g}{\partial p_j} - \frac{\partial f}{\partial p_j} \frac{\partial g}{\partial q_j}. $$

First I computed $\left \{ f,g \right \}$ in this set as follows:

Using $ z=\left( \begin{array}{c} q \\ p \end{array} \right) \in \mathbb R^{2n}$ with $q,p\in \mathbb R^n$, I can write $f(z)= f(q,p) = \begin{pmatrix}q&p\end{pmatrix}\,C\,\begin{pmatrix}p\\q\end{pmatrix}$ and so the derivatives of $f$ are \begin{align*} \frac{\partial f}{\partial q_j} &= 2C_{j,k}q_k + 2 C_{j,k+n}p_k,\\ \frac{\partial f}{\partial p_j} &= 2C_{j+n,k}q_k + 2C_{j+n,k+n}p_k. \end{align*} (Einstein notation assumed!) Computing the bracket itself gives: \begin{align*} \left\{f,g \right\} =& 4\sum_{j=1}^n \bigg[\left(C_{j,k}q_k + C_{j,k+n}p_k\right) \left(D_{j+n,l}q_l + D_{j+n,l+n}p_l\right)\\ & - \left(C_{j+n,k}q_k + C_{j+n,k+n}p_k\right) \left(D-{j,l}q_l + D_{j,l+n}p_l\right)\bigg]. \end{align*}

I think I should be able to find a Matrix K such that, $z^T K z = \left\{f,g\right\}$ to show that the Poisson Bracket is in U... but I can't figure out how to that. Otherwise I could show that the Bracket fulfills the 3 properties of a Lie Algebra ($ \left\{f,f \right\} = 0, \left\{(a+b)f,g \right\} = \left\{af,g \right\} + \left\{bf,g \right\}$ and the Jacobi Identity. But managing those multiple sums is an incredible nuisance and I can't really manage to do that either. I hope someone can set me off in the right direction!



The symmetric matrix $K$ always exists since the Poisson bracket is again a homogeneous quadratic form.

  • $\begingroup$ Could you please elaborate on this? How is the Poisson bracket a homogeneous quadratic form? $\endgroup$ – Howdy Ho Jan 21 '13 at 17:37
  • $\begingroup$ It follows from the formula for the Poisson bracket. Since $f$ is homogeneous quadratic its partial is homogeneous linear. The same applies for $g$. Therefore the Poisson bracket is homogeneous quadratic. $\endgroup$ – PAD Jan 21 '13 at 17:41

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