Matrix representation of adjoint & co-adjoint orbit of $so(3)$

Question

So I am trying to find the co-adjoint orbits of the lie algebra $so(3)^*$ from this example but I am stuck with a very trivial linear algebra property

now I found the adjoint orbits and I know the matrix representation of the adjoint action/map is $R$ as said above in the first highlighted passage. However, the co-adjoint action I thought was the dual map of the adjoint action so I figured its matrix representation would be $R^T$ as it normally is for dual maps however the author here says it is $(R^{-1})^T$ and so I am kind of confused why the inverse is in there when this is the matrix representation of the dual map.

EDIT: Here is the definition of $Ad^*R$

Answer 1

I would need to see the definition the author uses to be sure, but probably the inverse is included in the definition of $\text{Ad}^*R$ in order to make it a left action (satisfying $(\text{Ad}^*R_1)(\text{Ad}^*R_2) = \text{Ad}^*R_1R_2$). In other words, $\text{Ad}^*R:=(\text{Ad}\,R^{-1})^*$.

Answer 2

The Lie group $SO(3)$ has a Lie algebra $so(3)$ which can be defined as an algebra generated by $(X,Y,Z)$ with the brackets $$[Y, Z] = X, \hspace 1em [Z, X] = Y, \hspace 1em [X, Y] = Z.$$ The dual algebra $so(3)^*$ can be thought of as the algebra, itself, except that now $(X,Y,Z)$ play the role of coordinates, instead of vector basis elements. On it, (more precisely, over the function space $C^∞\left(so(3)^*\right)$) the Lie bracket becomes a Poisson bracket, whose fundamental brackets are just those corresponding to the Lie brackets, suitably rewritten: $$\{Y, Z\} = X, \hspace 1em \{Z, X\} = Y, \hspace 1em \{X, Y\} = Z.$$ The abuse of notation is, of course, legit since the $C^∞$ function space over $so(3)^*$ includes the linear functions; i.e. the space $so(3)^{**}$ which is, itself, equivalent to $so(3)$, as a vector space. So, this is the non-linear extension of the $so(3)$ Lie bracket.

The co-adjoint orbits are the symplectic leaves of the Poisson manifold given by this Poisson bracket. Each symplectic leaf is identified by the set of all functions invariant on it, and the values of those invariants. Therefore, they are determined by solving the equations $\{⋯, F\} = 0$, as "⋯" ranges over all the coordinates - here: $(X, Y, Z)$.

I'm going to take this one step further. Consider, instead, the one parameter family of Lie algebras, given by the brackets $$[Y, Z] = X, \hspace 1em [Z, X] = Y, \hspace 1em [X, Y] = α Z,$$ with the parameter $α ∈ ℝ$. All of the Poisson manifolds for these Lie algebras can be combined into a single Poisson manifold, with coordinates $(X, Y, Z, α)$ and Poisson bracket: $$\{F,G\} = X \frac{∂(F,G)}{∂(Y,Z)} + Y \frac{∂(F,G)}{∂(Z,X)} + α Z \frac{∂(F,G)}{∂(X,Y)},$$ where I'm using the abbreviation $$\frac{∂(w,x)}{∂(y,z)} = \frac{∂w}{∂y} \frac{∂x}{∂z} - \frac{∂w}{∂z} \frac{∂x}{∂y}.$$

In this Poisson manifold, the coordinate $α$ is a linear invariant. So, the symplectic leaves of the combined manifold together comprise the symplectic leaves of the Poisson manifolds of associated with the Lie algebras in the family, for each $α ∈ ℝ$.

So, we're finding the co-adjoint orbits for them all, all at once.

Let the total differential of $F$ be given by $dF = x dX + y dY + z dZ + A dα$. Then upon substitution of each coordinate into the equation $\{⋯, F\} = 0$, we find the following linear system: $$0 = \{X, F\} = α y Z - Y z, \hspace 1em 0 = \{Y, F\} = z X - α x Z, \hspace 1em 0 = \{Z, F\} = x Y - y X.$$

A constraint relation, denoted $U ≡ V$, will be defined as a relation that holds on a symplectic leaf. A condition for this to be true is that it be closed under application of the Poisson bracket, e.g. $\{X, U\} ≡ \{X, V\}$. New constraints that arise that way will be called secondary constraints. Thus, for instance, if we impose the constraint $X ≡ 0$, then upon application of the Poisson bracket with $Y$ and $Z$, we get $α Z ≡ 0$ and $Y ≡ 0$, respectively, as secondary constraints. Except for the symplectic leaves where $α ≡ 0$, this yields $(X, Y, Z, α) ≡ (0, 0, 0, α)$. For $α ≡ 0$, we get only that $(X, Y, Z, α) ≡ (0, 0, Z, 0)$ as $Z$ continues to be free to vary and - indeed - turns out to be an invariant on these symplectic leaves.

So, already, those are two sets of solutions:

• $(X, Y, Z, α) = (0, 0, 0, α)$, for $α ∈ ℝ$,
• $(X, Y, Z, α) = (0, 0, Z, 0)$, for $Z ∈ ℝ$.

Away from these symplectic leaves, we can solve $x Y - y X = 0$ under the assumption that $(X, Y) ≢ (0, 0)$, yielding the general solution: $(x, y) = λ (X, Y)$. Upon substitution into the remaining equations, we get: $$0 = α y Z - Y z = Y (α λ Z - z), \hspace 1em 0 = z X - α x Z = X (z - α λ Z).$$ Thus $z = α λ Z$. Upon substitution into the differential equation for $F$, we have $$dF = x dX + y dY + z dZ + A dα = λ (X dX + Y dY + α Z dZ) + A dα,$$ and, with integration by parts, we can write: $$α Z dZ = ½ α d\left(Z^2\right) = d\left(½ α Z^2\right) - ½ Z^2 dα,$$ thus allowing us to write: $$dF = λ d\left(½\left(X^2 + Y^2 + α Z^2\right)\right) + \left(A - ½ λ Z^2\right) dα,$$ which shows that $F = F\left(α, Φ_2\right)$ reduces to a function of the linear invariant $α$ and a quadratic invariant: $$Φ_2 = X^2 + Y^2 + α Z^2.$$

Thus, the remaining symplectic leaves are surfaces of constant $Φ₂$.

• If $α > 0$ and $Φ_2 > 0$, then they are ellipsoids.
• If $α = 0$ and $Φ_2 > 0$, then they are cylinders.
• If $α < 0$ and $Φ_2 > 0$, then they are one-sheeted hyperboloids.
• If $α < 0$ and $Φ_2 = 0$, then they are double cones.
• If $α < 0$ and $Φ_2 < 0$, then they are two-sheeted hyperboloids.

The cases for $Φ_2 = 0$, when $α ≥ 0$ have already been accounted for by the cases previously noted above, while $Φ_2 < 0$ is not possible with $α ≥ 0$. Since there are 4 coordinates and two invariants, then these symplectic leaves have 1+1 freely-running coordinates, while the other two families of symplectic leaves are each single points, with 0+0 freely-running coordinates.

For your case, $so(3)$, with $α = 1$, that includes the spheres of positive radius and the "0-radius" sphere: a single point.