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

Question

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.

Answer 1

Here is a partial answer.

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{equation} \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} \end{equation}

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 \begin{equation} \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. \end{equation}

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.