I have the following system of ODEs:
\begin{align*}
\dot{x}_1 &= x_2x_3A_{32} + x_5 x_6 A_{65}\\
\dot{x}_2 &= x_1 x_3 A_{13} + x_4 x_6 A_{46}\\
\dot{x}_3 &= x_1 x_2 A_{21} + x_4 x_5 A_{54}\\
\dot{x}_4 &= x_3x_5 A_{35} + x_2 x_6 A_{62}\\
\dot{x}_5 &= x_3 x_4 A_{43} + x_1 x_6 A_{16}\\
\dot{x}_6 &= x_2 x_4 A_{24} + x_1 x_5 A_{51}
\end{align*}
where $A_{ij} = \frac{1}{I_i} – \frac{1}{I_j}$.
Questions: What is the Poisson bracket associated with the above system? If it is as I describe, then why is $\{H,H\} \neq 0$? Am I doing something wrong?
My Work: For a smaller system:
$$
\begin{cases}
\dot{x}_{1} &= \alpha_{1} x_2 x_3,\\
\dot{x}_{2} &= \alpha_{2} x_3 x_1,\\
\dot{x}_{3} &= \alpha_{3} x_1 x_2
\end{cases}
$$
where $\alpha_i = \frac{1}{I_k} – \frac{1}{I_j}$s are some physical constants.
I had been using the following definition
$$
\{\mathbf{x},\mathbf{y\}} = (\nabla f(\mathbf{x}))^{T} \omega(\mathbf{x}) \nabla \mathbf{y}
$$
where $\omega$ is a $n \times n$ matrix called the Poisson matrix. For this small system $$\omega(x) = \begin{pmatrix}0 & -x_{3} & x_{2} \\ x_{3} & 0 & -x_{1} \\ -x_{2} & x_{1} & 0 \end{pmatrix} \in so(3)$$ Using this definition we can confirm that the function $H = \frac{1}{2}\frac{x_1^2}{I_1} + \frac{1}{2}\frac{x_2^2}{I_2} + \frac{1}{2}\frac{x_3^2}{I_3}$ is the Hamiltonian of this system. This is done by recovering the system from the Poisson bracket, e.g, $$\dot{x}_1 = \{x_1, H\}$$.
There are two ways I done this computation. The first way keeping everything as matrices and taking the trace at the end: $(-,-) = tr(x^{T}\omega y)$ BUT I DONT KNOW WHY THIS WORKS. Why the trace?
The basis for $so(3)$ are the three matrices
$e_1 = \begin{pmatrix}
0 & 0 & 0\\
0 & 0 & -1\\
0 & 1 & 0
\end{pmatrix}$
$e_{2} = \begin{pmatrix}
0 & 0 & 1 \\ 0 & 0 & 0 \\ -1 & 0 & 0
\end{pmatrix}$
$e_{3} = \begin{pmatrix}
0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0
\end{pmatrix}$
Then by definition $\nabla x_1 = \sum_{i=1}^{3} \frac{\partial x_1}{x_i} \cdot e_i = e_1 = \begin{pmatrix}
0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0
\end{pmatrix}$. Similarly, $\nabla H = \begin{pmatrix}
0 & -\frac{x_3}{I_3} & \frac{x_2}{I_2}\\
\frac{x_3}{I_3} & 0 & -\frac{x_1}{I_1}\\
-\frac{x_2}{I_2} & \frac{x_1}{I_1} & 0
\end{pmatrix}$
And so I then do the computation \begin{align*}
\{x_1, H\} &= \underbrace{\begin{pmatrix}
0 & 0 & 0\\
0 & 0 & 1\\
0 & -1 & 0
\end{pmatrix}}_{(\Delta_x f)^T} \underbrace{\begin{pmatrix}0 & -x_{3} & x_{2} \\ x_{3} & 0 & -x_{1} \\ -x_{2} & x_{1} & 0 \end{pmatrix}}_{\omega(x)} \underbrace{\begin{pmatrix}
0 & -\frac{x_3}{I_3} & \frac{x_2}{I_2}\\
\frac{x_3}{I_3} & 0 & -\frac{x_1}{I_1}\\
-\frac{x_2}{I_2} & \frac{x_1}{I_1} & 0
\end{pmatrix}}_{\Delta_x H}\\
&= \begin{pmatrix}
0 & 0 & 0 \\
-x_2 & x_2 & 0\\
-x_3 & 0 & x_1
\end{pmatrix}\begin{pmatrix}
0 & -\frac{x_3}{I_3} & \frac{x_2}{I_2} \\
\frac{x_3}{I_3} & 0 & -\frac{x_1}{I_1}\\
-\frac{x_2}{I_2} & \frac{x_1}{I_1} & 0
\end{pmatrix}\\
&= \underbrace{\begin{pmatrix}
0 & 0 & 0 \\
\frac{x_1 x_3}{I_3} & \frac{x_2 x_3}{I_3} & -\frac{x_2^2}{I_2} – \frac{x_1^2}{I_1} \\
-\frac{x_1x_2}{I_2} & \frac{x_3^2}{I_3} + \frac{x_1^2}{I_1} & -\frac{x_2 x_3}{I_2}
\end{pmatrix}}_{A}
\end{align*}
Taking the trace of the matrix above yields:
\begin{align*}
tr(A) &= \frac{x_2x_3}{I_3} – \frac{x_2x_3}{I_2}\\
&= (\frac{1}{I_3} – \frac{1}{I_2})x_2 x_3
\end{align*} which is exactly the first equation of our little system! We can also recover $\dot{x}_2$ and $\dot{x}_3$. But, again, I don't really know why we have to take the trace other than it works.
The function $f = x_1^2 + x_2^2 + x_3^2$ is also a conserved quantity and is verified by showing $\{f,H\} = 0$. So $f$ and $H$ are in involution.
The second way to do the computation is a little different. I compute with: $(-,-) = (\nabla x)^T B (\nabla y)$ where $(x), (y)$ are written in terms of the basis and $B$ is made up of $b_{ij} = \{e_i, e_j\}$. Turns out that $B = \omega$ here. Then $(\nabla x) = \begin{pmatrix}1 \\ 0 \\ 0\end{pmatrix}$ and $(\nabla y) = \begin{pmatrix}\frac{x_1}{I_1} \\ \frac{x_2}{I_2} \\ \frac{x_3}{I_3} \end{pmatrix}$. So then doing the computation $$\begin{pmatrix}1; 0; 0\end{pmatrix} B \begin{pmatrix}\frac{x_1}{I_1} \\ \frac{x_2}{I_2} \\ \frac{x_3}{I_3} \end{pmatrix}$$
yields what we desire: $x_2x_3\alpha_1$.
So we see these two computations are equivalent. But why are they equivalent? The trace seems kinda arbitrary to me.
Now, for the big system I have found three suspected conserved quantities. One of them being the Hamiltonian $H = \frac{1}{2}\frac{x_1^2}{I_1} + \frac{1}{2}\frac{x_2^2}{I_2} + \frac{1}{2}\frac{x_3^2}{I_3} + \frac{1}{2}\frac{x_4^2}{I_4} + \frac{1}{2}\frac{x_5^2}{I_5} + \frac{1}{2}\frac{x_6^2}{I_6}
$. I won't write the computations here, but I confirmed that we recover the big system. That is, I get $$\dot{x}_1 = \{x_1, H\}$$ where $\omega \in so(4)$. HOWEVER, I have seen that
$$\{H,H\} \neq 0$$
This is bad since it should be zero. I do not understand why it is not zero since I did the calculation showing we recover the system so that means the system is Hamiltonian doesn't it?
Questions: What is the Poisson bracket associated with the big system? If it is as I describe, then why is $\{H,H\} \neq 0$? Am I doing something wrong?
Best Answer
Let me state from the beginning that I'm only answering this question because I found it interesting, which is why I read up slightly on the subject (thanks to @Ricardo Buring for pointing me in the right direction).
So, the first thing is that I'm not exactly sure why taking the trace gives the right answer. I suspect it has something to do with the theory of Lie algebras and how one chooses the basis of $\mathfrak{so}(3)$ to construct an isomorphism $\mathfrak{so}(3) \cong \Bbb{R}^3$... so at this point it just seems like a mystery to me which just "happens to work". Hopefully someone else can shed some light on the matter.
Here's what I do understand. You finding $\{H,H\} \neq 0$ means you've definitely done something wrong, because the bracket must be skew-symmetric, hence it should be $0$. Some possible places you could have gone wrong is that the bracket you defined is not skew-symmetric? Or perhaps you just made an arithmetic error somewhere along the calculation of $\{H,H\}$ which is why you're getting a non-zero result.
The second thing I observed while studying this is that your question
is not a well posed question, in the sense that there are several answers. If you find one pair of bracket and hamiltonian, then you can find infinitely many simply by an overall rescaling. But more than that, I found two different ways of describing the bracket and Hamiltonian respectively; i.e I found two different Poisson brackets and Hamiltonian pairs $(\{\cdot, \cdot\}_1, H_1)$ and $(\{\cdot, \cdot\}_2, H_2)$, which are not simply a constant multiples of one another, but such that they reproduce the same system of ODEs. (this seems like a pretty cool thing though).
Here's how I approached the problem after acquainting myself with some of the material.
With this in mind, what I did is write down the system of equations you provided and I explicitly plugged in $A_{ij} = \frac{1}{I_i} - \frac{1}{I_j}$. Then, note that if we want the equations to be described by a Hamiltonian and a Poisson bracket, we must have \begin{align} \dot{x}^a = \{x^a,H\} = \pi^{ab} \dfrac{\partial H}{\partial x^b} \end{align}
So, I first considered the Hamiltonian \begin{align} H_1 = \sum_{a=1}^6 \dfrac{(x^a)^2}{2I_a} \end{align} Then, by simply computing the partial derivatives, and plugging them into the system of equations, and pattern matching, I was able to obtain the coefficients $(\pi_1)^{ab}$, which I organize into a $6 \times 6$ matrix as follows: \begin{align} [(\pi_1)^{ab}] &= \begin{bmatrix} \begin{pmatrix} 0 & -x^3 & x^2\\ x^3 & 0 & -x^1 \\ -x^2 & x^1 & 0 \end{pmatrix} & \begin{pmatrix} 0 & -x^6 & x^5\\ x^6 & 0 & -x^4 \\ -x^5 & x^4 & 0 \end{pmatrix} \\\\ \begin{pmatrix} 0 & -x^6 & x^5\\ x^6 & 0 & -x^4 \\ -x^5 & x^4 & 0 \end{pmatrix} & \begin{pmatrix} 0 & -x^3 & x^2\\ x^3 & 0 & -x^1 \\ -x^2 & x^1 & 0 \end{pmatrix} \end{bmatrix} \end{align}
This $6 \times 6$ matrix has a lot of structure. First of all, the skew symmetry is clear: $(\pi_1)^{ab} = -(\pi_1)^{ba}$. Next, notice how it is a block matrix of the type $\begin{pmatrix} P & Q \\ Q & P\end{pmatrix}$, where both $P,Q \in \mathfrak{so}(3)$, but $P,Q$ are "independent" in the sense that $P$ only depends on the first three coordinates $x^1, x^2, x^3$, while $Q$ only depends on $x^4, x^5, x^6$. Because of this, it is much easier to verify that the Jacobi identity holds. Thus, $\pi_1$ does indeed define a Poisson bracket, via \begin{align} \{f,g\}_1 := (\pi_1)^{ab} \dfrac{\partial f}{\partial x^a} \dfrac{\partial g}{\partial x^b} \end{align} ($a,b \in \{1, \dots, 6\}$). So, by plugging in everything, you'll get an explicit formula in terms of the various $x$'s.
However, I got the above Poisson bracket only because I started out with the assumption that the Hamiltonian is given by $H_1 = \sum_{a=1}^6 \dfrac{(x^a)^2}{2I_a}$. If however, I consider a different Hamiltonian function, say \begin{align} H_2 &= \dfrac{1}{2}\sum_{a=1}^6 (x^a)^2, \end{align} then by again writing out the equations explicitly and pattern matching properly, we get a new collection of functions $(\pi_2)^{ab}$, given as \begin{align} [(\pi_2)^{ab}]&= \begin{bmatrix} \begin{pmatrix} 0 & \frac{x^3}{I_3} & -\frac{x^2}{I_2}\\ -\frac{x^3}{I_3} & 0 & \frac{x^1}{I_1}\\ \frac{x^2}{I_2} & -\frac{x^1}{I_1} & 0 \end{pmatrix} & \begin{pmatrix} 0 & \frac{x^6}{I_6} & -\frac{x^5}{I_5}\\ -\frac{x^6}{I_6} & 0 & \frac{x^4}{I_4}\\ \frac{x^5}{I_5} & -\frac{x^4}{I_4} & 0 \end{pmatrix} \\\\ \begin{pmatrix} 0 & \frac{x^6}{I_6} & -\frac{x^5}{I_5}\\ -\frac{x^6}{I_6} & 0 & \frac{x^4}{I_4}\\ \frac{x^5}{I_5} & -\frac{x^4}{I_4} & 0 \end{pmatrix} & \begin{pmatrix} 0 & \frac{x^3}{I_3} & -\frac{x^2}{I_2}\\ -\frac{x^3}{I_3} & 0 & \frac{x^1}{I_1}\\ \frac{x^2}{I_2} & -\frac{x^1}{I_1} & 0 \end{pmatrix} \end{bmatrix} \end{align}
All I did was "transfer" the $I_a$'s from the Hamiltonian to the Poisson matrix (along with a few minus signs), so this will indeed yield the same system of differential equations. However, this gives us a completely different Poisson bracket: \begin{align} \{f,g\}_2 &= (\pi_2)^{ab}\dfrac{\partial f}{\partial x^a} \dfrac{\partial g}{\partial x^b} \end{align}
So, hopefully, I've convinced you that there is no "the" Poisson bracket (i.e just specifying the ODEs doesn't uniquely determine the Poisson bracket... you also have to specify a Hamiltonian).
Anyway, regardless of which function you choose as your Hamiltonian, and which Poisson bracket you choose, you should always get that $\{H,H\} = 0$, simply because the bracket is skew-symmetric, and indeed the brackets I constructed are valid Poisson brackets, so we have \begin{align} \{H_1, H_1\}_1 = \{H_2, H_2\}_2 = 0. \end{align}