Kostant-Kirillov form versus Fubini-Study form on Plucker embedding

Question

Let $G$ be a connected complex reductive group with a maximal compact subgroup $K$. Let $\lambda$ be a dominant weight in the interior of the positive Weyl chamber. Let $V_\lambda$ denote the irreducible representation with highest weight $\lambda$ and fix a highest weight vector $v_\lambda$. Then $gB \mapsto [g. v_\lambda]$ gives a $G$-equivariant embedding of the flag variety $G/B$ into the projective space $\mathbb{P}(V_\lambda)$.

Fix a $K$-invariant Hermitian product on $V_\lambda$. This defines a $K$-invariant Fubini-Study form on $\mathbb{P}(V_\lambda)$ and hence on the image of $G/B$ in there.

My question is whether this symplectic form on $G/B$ coincides (up to a scalar) with the Kostant-Kirillov form on the coadjoint orbit of $\lambda^*$ (dual weight)? (via the moment map $\mu: \mathbb{P}(V_\lambda) \to Lie(K)^*$) restricted to the image of $G/B$.)

Answer 1

The answer is yes, due to formal properties of moment maps (that depend on almost no details of your situation). Namely, suppose $K$ acts transitively on any symplectic manifold $(X,\omega_X)$, with equivariant moment map $\mu:X\to\mathfrak k^*$. Then $\mu(X)$ is a coadjoint orbit $\mathcal O$ and $\omega_X=\mu^*\omega_{\mathcal O}$.

Proof: Write $Z(x)$ for the infinitesimal action of $Z\in\mathfrak k$ on $x\in X$. The moment map definition $i_Z\omega_X=-d\langle\mu(\cdot),Z\rangle$ says $$ \omega_X(Z(x),\delta x)=-\langle D\mu(x)(\delta x),Z\rangle $$ for all tangent vectors $\delta x\in T_xX$. By transitivity, any such writes $Z'(x)$ for some $Z'\in\mathfrak k$, and equivariance then gives $D\mu(x)(Z'(x))=Z'(\mu(x))=\langle\mu(x),[\cdot,Z']\rangle$. So the above equality becomes $$ \begin{align} \omega_X(Z(x),Z'(x))&=\langle\mu(x),[Z',Z]\rangle\\ &=\omega_{\mathcal O}(Z(\mu(x)), Z'(\mu(x))), \end{align} $$ where the last equality is the definition of the Kirillov-Kostant-Souriau 2-form $\omega_{\mathcal O}$.