## Definition

The coadjoint orbit of a Lie group $G$ is an orbit of $G$ in the space $\mathfrak{g}^\ast$ dual to $\mathfrak{g}=\text{Lie}(G)$, where $G$ acts on $\mathfrak{g}^\ast$ by the coadjoint action.

The group $G$ acts on itself by inner automorphisms $A_g : G\to G : h\mapsto g h g^{-1}$. The neutral element $e\in G$ is a fixed point of the action, so upon differentiation there is a map $$\operatorname{Ad}_g := (A_g)_\ast (e):\mathfrak{g} \to \mathfrak{g}.$$ called the adjoint action of $G$ on $\mathfrak{g}$. The map $g\mapsto \operatorname{Ad}_g$ is the adjoint representation of $G$.

Remark. For matrix group $G$ the Lie algebra $\mathfrak{g}$ is a subspace of $\text{Mat}_n(\mathbb{R})$, and the adjoint action is simply the matrix conjugation $$\operatorname{Ad}_g X = g\cdot X\cdot g^{-1},\quad X\in\mathfrak{g},g\in G.$$

For any linear representation $(\pi,V)$ of a group $G$, there is a dual representation $(\pi^\ast,V^\ast)$ in the dual space $V^\ast$: $$\pi^\ast(g) := \pi(g^{-1})^\ast$$ where $(\cdot)^\ast$ here means the dual operator defined by the pairing $\langle\cdot,\cdot\rangle:V^\ast\times V\to \mathbb{R}$, $$\langle A^\ast \alpha,v\rangle :=\langle f,Av\rangle\quad \forall v\in V,\alpha\in V^\ast$$

The dual representation $(\pi^\ast,V^\ast)$ is also called the contragradient representation. For a representation of a Lie group $G$ in $\mathfrak{g}^\ast$ that is dual to the adjoint representation in $\mathfrak{g}$, this representation, denoted $\operatorname{Ad}^\ast$, is called coadjoint, i.e. $$\langle \operatorname{Ad}^\ast_g \xi, X\rangle = \langle \xi,\operatorname{Ad}_{g^{-1}} X\rangle.$$

All coadjoint orbits are symplectic manifolds, and each coadjoint orbit possesses a canonical $G$-invariant symplectic structure.

Definition(Kirillov-Kostant-Souriau symplectic structure) . Let $\lambda \in \mathfrak{g}^\ast$ and let $\mathcal{O}_\lambda$ be the coadjoint orbit through $\lambda$. The Kirillov-Kostant-Souriau canonical symplectic structure $\omega$ is defined pointwisely by $$\omega_\lambda(\operatorname{ad}^\ast_X \lambda,\operatorname{ad}^\ast_Y \lambda) = - \langle \lambda, [X,Y]\rangle$$ where $\operatorname{ad}^\ast$ is the derivative of the coadjoint action $\operatorname{Ad}^\ast$.

### The Kirillov-Kostant-Sourial form is nondegenerate and $G$-invariant.

Let $\lambda\in\mathfrak{g}^\ast$ and let $\mathcal{O}_\lambda$ be the coadjoint orbit. Let $\text{stab}(\lambda)$ be the Lie algebra of $\text{Stab}(\lambda)\subseteq G$, the stabilizer of $\lambda$ under the coadjoint action (i.e. those $g\in G$ with $\operatorname{Ad}^\ast_g \lambda = \lambda$). The group $G$ is then a fiber bundle $$p_\lambda:G\to \mathcal{O}_\lambda: g\mapsto \operatorname{Ad}^\ast_g \lambda$$ over the base $\mathcal{O}_\lambda \cong G/\text{Stab}(\lambda)$. Clearly the fiber over $\lambda$ is $\text{Stab}(\lambda)$. Then there is an exact sequence of vector spaces $$0\to \text{stab}(\lambda)\to \mathfrak{g}\xrightarrow{(p_\lambda)_\ast} T_\lambda(\mathcal{O}_\lambda) \to 0.$$ The tangent space $T_\lambda(\Omega)$ is identified with the quotient $\mathfrak{g}/\text{stab}(\lambda)$.

G-invariance: Over a point $\lambda\in\mathcal{O}_\lambda$, the form $\omega_\lambda$ is invariant under $\text{Stab}(\lambda)$, since $$\langle \lambda,[\operatorname{Ad}_h X,\operatorname{Ad}_h Y] \rangle = \langle \lambda,\operatorname{Ad}_h [X,Y]\rangle = \langle \operatorname{Ad}^\ast_h \lambda,[X,Y]\rangle = \langle \lambda,[X,Y]\rangle$$ for $h\in\text{Stab}(\lambda)$, and for general $g\in G$, $\omega$ is sent to $\omega_{\operatorname{Ad}^\ast_g \lambda}$. Hence $\omega$ is $G$-invariant.

Non-degeneracy: The kernel of $\omega_{\lambda}$ is $\text{stab}(\lambda)$. \begin{align*} \operatorname{Ker}(\omega_\lambda) &= \{ X\in \mathfrak{g} | \langle \lambda, \operatorname{ad}_X Y\rangle =0, \forall Y\in\mathfrak{g}\}\\ &=\{ X\in\mathfrak{g} | \langle \operatorname{ad}^\ast_X \lambda,Y\rangle = 0,\forall Y\in\mathfrak{g}\}\\ &=\{X\in\mathfrak{g} | \operatorname{ad}^\ast_X \lambda =0 \}\\ &= \text{stab}(\lambda) \end{align*} Now since $T_\lambda(\mathcal{O}_\lambda)\cong \mathfrak{g}/\text{stab}(F)$, the kernel of $\omega$ is trivial on $\mathcal{O}_\lambda$.

### The Kirillov-Kostant-Souriau form is closed

Recall that the Maurer-Cartan form $\theta$ is a $\mathfrak{g}$-valued 1-form, $\theta \in T^\ast G \otimes \mathfrak{g}$. Defined pointwisely by the maps $\theta_g : T_g G \to T_e G = \mathfrak{g}$ s.t. $$\theta_g(V) = (L_{g^{-1}})_\ast V$$ for $V\in T_g G$ where $L_g: G\to G$ is the left-translation by $g$. . The Maurer-Cartan equation $$d\theta + \frac{1}{2}[\theta,\theta] = 0$$ becomes, for $X,Y$ left-invariant vector fields on $G$, $$d\theta = - [\theta(X),\theta(Y)].$$ Now consider the bundle $p_\lambda: G\to \mathcal{O}_\lambda$, $\omega\in \Omega^2(\mathcal{O}_\lambda)$ can be pulled back to $\Omega^2(G)$.

Proposition. The 2-form $\Omega_\lambda = p^\ast_\lambda(\omega)$ on $G$ is the exterior derivative of the left-invariant real-valued 1-form $\Theta_\lambda$, given by $$\Theta_\lambda = -\langle \lambda,\theta\rangle.$$ where $\theta$ is the Maurer-Cartan form on $G$.
Proof:  Let $X,Y$ be left-invariant vector fields on $G$. For $$d\Theta_\lambda = X\Theta_\lambda (Y) - Y\Theta_\lambda (X) - \Theta_\lambda([X,Y])$$ since $\Theta_\lambda(Y)$ and $\Theta_\lambda(X)$ are constant functions of $G$, the 1st and 2nd term on the R.H.S. vanish. Thus $$d\Theta_\lambda(X,Y) = -\Theta_\lambda([X,Y]) = -\langle\lambda,[X,Y]_e\rangle = p^\ast_\lambda(\omega)(X,Y) = \Omega_\lambda(X,Y).$$ Here $[X,Y]_e$ is the value of left-invariant vector fields at the origin $e$.

Now $p_\lambda$ is a submersion, and thus $(p_\lambda)_\ast$ is injective, so $(p_\lambda)^\ast$ is injective. $p^\ast_\lambda d\omega = d p^\ast_\lambda(\omega) = d\Omega_\lambda = d^2 \theta_\lambda = 0$. So $\omega$ is closed.

### The Lie-Poisson structure and symplectic foliation

On any $n$-dimensional vector space $V$ there is a linear Poisson structure $\pi$ given by $$\pi = \frac{1}{2}\pi_{ij}\partial_{x_i}\wedge \partial_{x_j}, \pi_{ij} = \sum_k c^{k}_{ij} x_k$$ where $c^k_{ij} = -c^k_{ji}$ and $(x_i)$ are linear coordinates. Jacobi identity $[\pi,\pi]=0$ makes the components $c^k_{ij}$ to be structure constants of an $n$-dimensional Lie algebra. This Lie algebra can be realized as the dual space $V^\ast$ with the bracket $$[f,g] = -\{f,g\}\quad (f,g\in V^\ast).$$

Let a Lie algebra $\mathfrak{g}^\ast$ be finite dimensional. It is a vector space, and thus its dual $\mathfrak{g}^\ast$ has a natural Poisson structure given by $$\{\phi,\psi\}(\gamma) = -\gamma([d\phi,d\psi])$$ where $\gamma\in\mathfrak{g}^\ast$ and $\phi,\psi \in C^\infty(\mathfrak{g}^\ast)$. Here $d\phi,d\psi \in \mathfrak{g}$, since $d\phi: T_\gamma \mathfrak{g}^\ast = \mathfrak{g}^\ast \to \mathbb{R}$. This is the Lie-Poisson structure of $\mathfrak{g}^\ast$.

Let $X\in\mathfrak{g}$, then $\iota_X \in C^\infty(\mathfrak{g}^\ast)$ and is a linear function on $\mathfrak{g}^*$. For these linear functions, $$\{\iota_X,\iota_Y\}(\gamma) = -\gamma([d\iota_X,d\iota_Y]) = -\gamma(d\iota_{[X,Y]}),$$ i.e. the Lie-Poisson structure is the original Lie bracket.

Now every Poisson manifold is foliated by symplectic leaves. $\mathfrak{g}^\ast$ with the Lie-Poisson structure is a Poisson manifold.

Theorem. The symplectic leaves of the Poisson manifold $(\mathfrak{g}^\ast,\pi)$, where $\pi$ is the Lie-Poisson structure, are the coadjoint orbits.
Proof:  Let $S_\lambda$ be the leaf that contains the point $\lambda\in\mathfrak{g}^\ast$. The tangent space $T_\lambda S_\lambda$ is, by definition, spanned by vectors $v_i = \pi_{ij}\partial^j = c^k_{ij}X_k\partial^j$, while $v_i$ is at the same time the value of the fundamental vector field at $\lambda$ corresponding to the infinitesimal coadjoint action of $X_i \in\mathfrak{g}$. This means $\mathcal{O}_\lambda$ and the leaf $S_\lambda$ have the same tangent space on $\lambda$. This is true for every $\xi \in \mathfrak{g}^\ast$, so the orbit is open in the symplectic leaf $S_\lambda$. $S_\lambda$ must be the disjoint union of some coadjoint orbits that it contains, and thus should be closed since it is the complement of the union of the remaining orbits which are all open. $S_\lambda$ is connected by definition, and $\mathcal{O}_\lambda$ is a nonempty open and closed subset, meaning $S_\lambda = \mathcal{O}_\lambda$.

## Integrality

A coadjoint orbit $\mathcal{O}_\omega$ where $\omega$ is the canonical symplectic form is integral if $$\int_C \omega \in \mathbb{Z}$$ for every integral singular 2-cycle $C$ in $\mathcal{O}_\omega$. This actually means $[\omega] \in H^2(\mathcal{O}_\omega,\mathbb{Z})$. The integrality condition can be interpreted geometrically, or representation theoretically:

Proposition. Let $G$ be a simply connected Lie group, the following are equivalent
1. The coadjoint orbit $\mathcal{O}_\omega \subset \mathfrak{g}^\ast$ is integral.
2. There is a $G$-equivariant complex line bundle over $\mathcal{O}_\omega$ with a $G$-invariant Hermitian connection $\nabla$ s.t. $\operatorname{curv}\nabla = 2\pi i \omega$.
3. For any $\lambda \in\mathcal{O}_\omega$ there is a unitary 1-dimensional representation $\chi$ of the connected (0-component) Lie group $\text{Stab}_0(\lambda)$ s.t. $$$$\label{rep} \chi(\exp X) = e^{2\pi i \langle\lambda,X\rangle}$$$$ where $X \in \mathfrak{g}$.

Notice that the representation $\chi$ in $\ref{rep}$ is, rather, that of an abelian Lie group $A = \text{Stab}_0(\lambda)/[\text{Stab}_0(\lambda),\text{Stab}_0(\lambda)] \cong \mathbb{T}^k\times \mathbb{R}^l$ since $\text{stab}(\lambda) = \operatorname{Ker} \omega$. The representation corresponds to a unitary 1-dimensional representation of $A$ iff it takes integer values on the $k$-dimensional lattice $\Lambda = \exp^{-1}(e)$ determined by the compact torus.

## Universal property and central extensions

Any coadjoint orbit is a homogeneous symplectic manifold. The converse is almost true, in that up to some algebraic and topological corrections any homogeneous symplectic manifold is a coadjoint orbit.

Given a connected Lie group $G$, the collection of all Poisson $G$-manifolds form a category $\mathcal{P}(G)$ of which the morphisms are Poisson maps $\alpha$ that make the following diagram commute Here $\mu_\bullet$ are moment maps. Now clearly the Poisson $G$-manifold $(\mathfrak{g}^\ast,\pi)$ with $\pi$ the linear Lie-Poisson structure is a final object in the category $\mathcal{P}(G)$, the morphism being the moment map itself.

The image $\mu(M)$ of a moment map $\mu: M\to\mathfrak{g}^\ast$ is a homogeneous submanifold in $\mathfrak{g}^\ast$, and thus is a coadjoint orbit $\mathcal{O}_\omega \subset \mathfrak{g}^\ast$. The $G$-action on $M$ is transitive, so the rank of the Jacobi matrix for $\mu$ is $\text{dim}\ M$, and hence $\mu$ is locally a diffeomorphism. Globally $\mu$ is a covering of $\mathcal{O}_\omega$.

For a symplectic manifold $(M,\omega)$, the property that the transitive $G$-action is Poisson is more restrictive than it being symplectic, since this means the $G$-action is Hamiltonian. In other words, let $\pi$ be the Poisson structure associated to $\omega$, then the $G$-action being Poisson on $(M,\pi)$ means $G$ is Hamiltonian on $(M,\omega)$.

Upon modifications of $(M,\pi)$ (where $\pi$ comes from $\omega$) and $G$ to $M’$ and $G’$, the Poisson manifold $(M’,\pi)$ can be made to a Poisson $G’$-manifold.

1. To make the canonical vector field defined by the Lie derivative $L_X$, where $X\in\mathfrak{g}$, be written as $\{f_X,\cdot\}$ for a globally defined function $f_X$, go to the simply connected cover $\tilde{M}$ of $M$. Replace the group $G$ with its universal cover $\tilde{G}$.
2. To make the map $\mathfrak{g}\to C^{\infty} : X\mapsto f_X$ a Lie algebra homomorphism and hence a co-moment map (so that it defines a moment map), pass to a central extension $G’$ of $\tilde{G}$. This is because $f_{[X,Y]}$ and $\{f_X,f_Y\}$ might differ by a constant $\pi(X,Y)$.

Note that here the central extension $G’$ is given by the Lie algebra central extension of $\tilde{\mathfrak{g}}$. $\mathfrak{g}’ = \tilde{\mathfrak{g}}\oplus\mathbb{R}$ as a vector space, with the commutator given by $$[(X,a),(Y,b)] = ([X,Y],\pi(X,Y)).$$ The Poisson structure $\pi:C^\infty(\mathfrak{g}^\ast)\times C^\infty(\mathfrak{g}^\ast) \to \mathbb{R}$ should be seen as $\pi:\mathfrak{g}\times\mathfrak{g}\to\mathbb{R}$, and by Jacobi identity it is a Lie algebra 2-cocycle on $\mathfrak{g}$.

The infinitesimal Poisson action of $\mathfrak{g}’$ on $\tilde{M}$ is defined by $L_{(X,a)}:= L_X$ since the center acts trivially in the adjoint and coadjoint representations.

Theorem. Any symplectic action of a connected Lie group $G$ on a symplectic manifold $(M,\omega)$ can be modified to a Poisson action (on the associated Poisson manifold, or a Hamiltonian action) of a central extension $G'$ of the universal covering of $G$ on some universal covering $\tilde{M}$ of $M$, s.t.

## Coadjoint orbits of the symplectic group

Let $G=Sp(2n,\mathbb{R})$, the symplectic group. Write $J_n = \begin{pmatrix} 0_n & 1_n \\ 1_n & 0_n \end{pmatrix}$ of which $G$ is the isometry group. The Lie algebra $\mathfrak{g}$ consist of $J_n$-symmetric matrices $X$, i.e. $$X^\mathsf{t} J_n + J_n X =0.$$ This is equivalently $$(J_n X)^\mathsf{t} = J_n X.$$

The presence of a nondegenerate $\operatorname{Ad}$-invariant bilinear form $\langle A,B\rangle = \operatorname{tr}(AB)$ on $\mathfrak{g}$ ensures that the coadjoint representation is equivalent to the adjoint one. The coadjoint orbit is the $G$-conjugacy classes in $\mathfrak{g}$.

The coadjoint orbits are then classified by the action $$S\mapsto g^\mathsf{t} S g, g\in So(2n,\mathbb{R})$$ of $Sp(2n,\mathbb{R})$ on symmetric matrices, i.e. by the conjugacy class of symmetric matrices. This can be seen by looking at $-X^\mathsf{t} J_n$ and compute the conjugation $$-X^\mathsf{t} J_n \mapsto g^{-1} (-X^\mathsf{t} J_n) g$$ the L.H.S turns out to be $g^{-1}(-X^\mathsf{t}J_n)(g^\mathsf{t})^{-1}$. Taking inverse yields $$S = J_n X \mapsto g^\mathsf{t} S g.$$