# Explicit local normal form symplectomorphism at torus fixed point of a coadjoint orbit

Let $K$ be a compact, connected (probably also simple) Lie group and with a maximal torus $T$. Regular coadjoint orbits $\mathcal{O}_{\lambda} \cong K/T$, parameterized by a regular element $\lambda \in \mathfrak{t}^*$, have several properties:

• They are symplectic, equipped with KKS forms.
• They are Hamiltonian $T$-manifolds with respect to the coadjoint action of $T$, generated by the canonical projection map $\mathcal{O}_{\lambda} \to \mathfrak{t}^*$.
• The fixed points of the $T$-action are precisely the elements of $\mathcal{O}_{\lambda} \cap \mathfrak{t}^*$ (where $\mathfrak{t}^*$ is identified with the subspace $(\mathfrak{k}^*)^T$).
• Using the killing form (or whatever scalar multiple of it that you like), we can make an identification $T_{\lambda}\mathcal{O}_{\lambda} \cong \mathfrak{t}^{\perp}$. One has that $\mathfrak{t}^{\perp}$ is invariant under the adjoint action of $T$ and coincides under this identification with the isotropy action of $T$ at the fixed point $\lambda$.

The local normal form theorem for Hamiltonian torus actions tells us that there exists a symplectomorphism $\varphi$ from a neighbourhood of 0 in the symplectic vector space $T_{\lambda}\mathcal{O}_{\lambda}$ to a neighbourhood of $\lambda$ in $\mathcal{O}_{\lambda}$ such that:

1. $\varphi$ is $T$-equivariant.
2. $\varphi$ intertwines the moment maps, i.e. if $\mu = \phi + \lambda$ denotes the quadratic moment map for the isotropy action of $T$ on $T_{\lambda}\mathcal{O}_{\lambda}$ shifted by $\lambda$, and $\text{pr}\colon \mathcal{O}_{\lambda} \to \mathfrak{t}^*$, then $$\text{pr} \circ \varphi = \mu.$$

Since everything in this example is fairly explicit, it seems reasonable to ask for an explicit formula for the map $\varphi$.

What is a general formula for $\varphi$?

There is an obvious candidate'' that looks good at first: one can easily define the map $$\psi\colon \mathfrak{t}^{\perp} \to \mathcal{O}_{\lambda}, \, \psi(X) = Ad^*_{e^X}\lambda.$$ This map is $T$-equivariant, but it fails to be symplectic (which can be demonstrated easily for coadjoint orbits of $SU(2)$).

(As a side note, the killing form restricted to $\mathfrak{t}^{\perp}$ and then extended $\mathcal{O}_{\lambda}$ by $K$-invariance makes $\mathcal{O}_{\lambda}$ a reductive homogeneous space, and $\psi$ is in fact the exponential map for this metric.)

I believe that, essentially, the map $\psi$ fails to be symplectic because $e^Xe^Y \neq e^{X+Y}$ if $[X,Y]\neq 0$ (which, if you are not careful means that you can accidentally prove'' this map is symplectic and spend an afternoon being very confused), which suggests that maybe there is some sort of CBH-related correction to $\psi$ that makes it symplectic (I am really just taking a guess here).

Edit: It will be sufficient to find a $T$-equivariant map $\rho\colon \mathfrak{t}^{\perp} \to K$ such that $$\varphi(Y) := Ad_{\rho(Y)}\lambda$$ satisfies the condition $$d(\varphi)_Y(ad_X\lambda) = ad_{Ad_{\rho(Y)}X}\varphi(Y).$$

Edit 2: Unpacking the proof of local normal forms, one sees that there exists a map $\phi\colon U \to \mathfrak{t}^{\perp}$ defined on a neighbourhood $U$ of the origin in $\mathfrak{t}^{\perp}$ that fixes the origin and is a diffeomorphism onto it's image, such that the map

$$\varphi(Y) := Ad_{e^{\phi(Y)}}\lambda$$

is a symplectomorphism. $\phi$ is the time-1 flow of a moser vector field. It's not clear from this perspective whether an explicit formula for $\phi$ is a reasonable thing to hope for.

Edit 3: I found a partial answer that I've posted below, but I am interested if anyone can expand on it.

The map $\mathfrak{t}^{\perp} \to T_{\lambda}\mathcal{O}_{\lambda}$, $X \mapsto ad_X(\lambda)$, endows $\mathfrak{t}^{\perp}$ with a linear symplectic form $$\omega_0(X,Y) = \langle \lambda, [X,Y]\rangle.$$ Based on edit 2 above, we know there exists a $T$-equivariant map $\phi\colon \mathfrak{t}^{\perp} \to \mathfrak{t}^{\perp}$ (at least, defined and invertible near 0) such that $\phi(0) = 0$ and the composition $$\varphi(Y) := Ad_{e^{\phi(Y)}}\lambda$$ is a symplectomorphism.

Example ($K = SU(2)$) Here we can compute using cylindrical coordinates on the coadjoint orbit and polar coordinates on $\mathfrak{t}^{\perp}$ to show that $$\phi\left(\left(\begin{array}{cc} 0 & \rho e^{i\theta} \\ -\rho e^{-i\theta} & 0 \end{array}\right)\right) = \left(\begin{array}{cc} 0 & \arcsin(\rho) e^{i\theta} \\ -\arcsin(\rho) e^{-i\theta} & 0 \end{array}\right).$$

Let $\pi\colon K \to \mathcal{O}_{\lambda}$ be the map $k \mapsto Ad_k\lambda$. We know that

• $d(\pi)_e Y = ad_Y\lambda$
• $d(\pi)_k = d(Ad_k)_{\lambda} \circ d(\pi)_e \circ d(\mathcal{L}_{k^{-1}})_k$ (where $\mathcal{L}_k\colon K \to K$ is left multiplication).
• $d(\mathcal{L}_{k^{-1}})_k$ is simply the (left-invariant) Maurer-Cartan form $\theta$ evaluated at $k$
• Because it is a linear map, $d(Ad_k)_{\lambda} = Ad_k$. We also know that $Ad_kad_X \lambda = ad_{Ad_kX}Ad_k\lambda$.

Thus:
$$\begin{split} d(\varphi)_Y & = d(\pi)_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\\ & = d(Ad_{e^{\phi(Y)}})_{\lambda} \circ d(\pi)_e \circ d(\mathcal{L}_{e^{-\phi(Y)}})_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\\ & = Ad_{e^{\phi(Y)}}\left(d(\pi)_e \circ \theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\right)\\ & = Ad_{e^{\phi(Y)}}\left(ad_{\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y}\lambda\right)\\ & = ad_{Ad_{e^{\phi(Y)}}\left(\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y\right)}\left(Ad_{e^{\phi(Y)}}\lambda\right). \end{split}$$

The map $\varphi$ is a symplectomorphism if $\varphi^* \omega = \omega_0$. Explicitly, for $X,Z \in \mathfrak{t}^{\perp}$, $$(\varphi^*\omega)_Y(X,Z) = \omega_{\varphi(Y)}\left(d(\varphi)_Y X,d(\varphi)_Y Z\right)$$ which by the previous equation $$= \langle Ad_{e^{\phi(Y)}}\lambda, [Ad_{e^{\phi(Y)}}\left(\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y X\right), Ad_{e^{\phi(Y)}}\left(\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y Z\right)]\rangle$$ $$= \langle \lambda, [\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y X, \theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)}\circ d\phi_Y Z]\rangle.$$ This simplifies by substituting the formula for the derivative of the exponential map: $$\theta_{e^{\phi(Y)}} \circ d(\exp)_{\phi(Y)} = e^{-\phi(Y)}e^{\phi(Y)}\left(\frac{1-e^{-ad_{\phi(Y)}}}{ad_{\phi(Y)}}\right) = \frac{1-e^{-ad_{\phi(Y)}}}{ad_{\phi(Y)}}$$ to give $$(\varphi^*\omega_{\lambda})_Y(X,Z) = \bigg\langle \lambda, \frac{1-e^{-ad_{\phi(Y)}}}{ad_{\phi(Y)}}[ d\phi_Y X, d\phi_Y Z]\bigg\rangle.$$

Thus, our equation for $\phi$ is $$\bigg\langle \lambda, \frac{1-e^{-ad_{\phi(Y)}}}{ad_{\phi(Y)}}[ d\phi_Y X, d\phi_Y Z]\bigg\rangle = \langle \lambda, [X,Z]\rangle.$$

The series expansion of the LHS will have some terms that disappear because they pair with $\lambda$ to zero, but I don't have a general formula for the ones that remain.