Poisson bracket and co-adjoint orbits for $sl(2)$

adjoint-operatorsdifferential-geometrylie-algebraspoisson-geometrysymplectic-geometry

So I am trying to do this problem from Peter Olver's book Application of Lie groups to differential equations and I am wondering if somebody could check my work because I am not really sure about it and I am trying to really understand this material by testing myself.

enter image description here

So to start off for the Lie-Poisson bracket I am pretty convinced this is not correct because it just doesn't look right. So for $sl(2)$ I used the basis matrices

$$ A_1 = \begin{bmatrix} 0 \;\; 1
\\ 0 \;\; 0
\end{bmatrix}
A_2 = \begin{bmatrix} 1 \;\;\;\;\;\; 0
\\ 0 \;\; -1
\end{bmatrix}
A_3 = \begin{bmatrix} 0 \;\; 0
\\ 1 \;\; 0
\end{bmatrix}$$

Then I used this definition for the Poisson bracket
enter image description here

So I calculated all the structure constants with respect to my basis by using the Lie-brack of the matrices and I got

$$\{F,H\} = (2x^3,x^2,2x^1) \cdot \nabla F \times \nabla H$$

where $\nabla F = (\frac{dF}{dx^1}, \frac{dF}{dx^2}, \frac{dF}{dx^3})$ now this just feels very wrong… perhaps what is meant by the question is the poisson bracket on $sl(2)^*$? which might be what I calculated… I am not sure.

Now for the co-adjoint orbits, I first computed the adjoint map and then looked at the dual of that map to find the co-adjoint map and orbits. So what I did was took $X = \begin{bmatrix} a \;\; b
\\ c \;\; d
\end{bmatrix} \in SL(2) \implies ad-bc = 1$
then

$$Ad_X(A_1) = XA_1X^{-1} = a^2 \cdot A_1 -ac \cdot A_2 -c^2\cdot A_3$$
$$ Ad_X(A_2) = XA_2X^{-1} = -2ab \cdot A_1 + (ad+bc) \cdot A_2 + 2dc\cdot A_3 $$
$$Ad_X(A_3) = XA_3X^{-1} = -b^2 \cdot A_1 +bd \cdot A_2 +d^2\cdot A_3$$

So the matrix representation of $Ad_X$ is
$$ R= \begin{bmatrix}a^2 \;\;\;\; -2ab \;\;\;\; -b^2 \\
-ac \;\;\;\; ab+bc \;\;\;\; bd \\
-c^2 \;\;\;\; 2dc \;\;\;\; d^2
\end{bmatrix}$$

And so then the matrix representation $Ad_X^*$ is $(R^{-1})^T$ which I got to be
$$ (R^{-1})^T= \begin{bmatrix}d^2 \;\;\;\; cd \;\;\;\; -c^2 \\
2bd \;\;\;\; ab+bc \;\;\;\; -2ac \\
-b^2 \;\;\;\; -ab \;\;\;\; a^2
\end{bmatrix}$$

but then I am kind of confused as to how to even think about the orbits here. If I take $A \in sl(2) \implies A = xA_1 + y A_2 + zA_3$ so then
$$Ad_X^*(A) = \begin{bmatrix}d^2 \;\;\;\; cd \;\;\;\; -c^2 \\
2bd \;\;\;\; ab+bc \;\;\;\; -2ac \\
-b^2 \;\;\;\; -ab \;\;\;\; a^2
\end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z\\
\end{bmatrix} = \begin{bmatrix} xd^2 +ycd -zc^2 \\ x2bd + yad+ybc -2zac \\ -xb^2 -yab + za^2
\end{bmatrix}$$

this is the basis representation with respect to $A_1,A_2,A_3$ but I am not really sure what this gives me… it doesn't look obvious to me, and I cannot seem to find any solution on the web so I am wondering if I made a mistake or if this is just the final answer and I cannot see what it represents.

Best Answer

You have alreadey determined the coadjoint action $\mathrm{Ad}_X^*$ and the coadjoint orbits. Just replace $ab+bc$ by $ad+bc$ in the matrix.

Yes, the orbit of $(x,y,z)$ is the set of points of $\mathrm{sl}(2)^*\cong\mathbb{R}^3$

$$O((x,y,z))=\{Ad_X^*\left(\begin{smallmatrix}x\\y\\z\end{smallmatrix}\right),\ X\in\mathrm{SL}(2)\}.$$

These orbits are symplectic leaves of the Lie-Poisson structure. The Lie bracket in your basis : $$[A_1,A_2]=-2A_1,\ [A_1,A_3]=A_2, [A_2,A_3]=-2A_3.$$ The Poisson bracket is given by (see [linear Poisson][1]): $$\{F,H\}(A):=<A,[dF,dH]>$$ where $A=xA_1^*+yA_2^*+zA_3^*$, $dF=\frac{\partial F}{\partial x}A_1+\frac{\partial F}{\partial y}A_2+\frac{\partial F}{\partial x}A_3$... $$[dF,dH]=-2\left(\frac{\partial F}{\partial x}\frac{\partial H}{\partial y}-\frac{\partial F}{\partial y}\frac{\partial H}{\partial x}\right)A_1+\left(\frac{\partial F}{\partial x}\frac{\partial H}{\partial z}-\frac{\partial F}{\partial z}\frac{\partial H}{\partial x}\right)A_2-2\left(\frac{\partial F}{\partial y}\frac{\partial H}{\partial z}-\frac{\partial F}{\partial z}\frac{\partial H}{\partial y}\right)A_3$$ So the linear Poisson bracket is given by: $$\pi_\ell=-2x\,\partial_x\wedge\partial_y+y\,\partial_x\wedge\partial_z-2z\,\partial_y\wedge\partial_z.$$

You can check that $F=4xz+y^2$ is a Casimir function, so the orbits (symplectic leaves) are the connected components of the fibers of $F$. [1]: https://en.wikipedia.org/wiki/Poisson_manifold#Linear_Poisson_structures