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Given a Lie group $G$, it acts smoothly on the dual $\mathfrak g^*$ of its Lie algebra $\mathfrak g$ by the coadjoint action. The orbits of that action are called coadjoint orbits.

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Can all hermitian symmetric spaces be realised as coadjoint orbits?

This is true. One can use a few facts from Helgason's Differential Geometry, Lie Groups, and Symmetric Spaces to show that, indeed, $K = \mathrm{Stab}_G(Z)$. Since $M=G/K$ is an irreducible Hermitian …
Robert Bryant's user avatar