Why can the Dolbeault Operators be Realised as Lie Algebra Actions

These are the bells this rings, at least for the case of a Kähler manifold. It's perhaps not what you are thinking of, but it seems to provide an answer to the question in the title.

Among the many equivalent definitions of Kähler manifolds is one coming from physics, which says that

supersymmetry of a non-linear sigma model in four dimensions requires a Kähler target space.

Perhaps one should elaborate: a four-dimensional sigma model is described by an action functional for maps $$X : \mathbb{R}^{3,1} \to M,$$ where $\mathbb{R}^{3,1}$ is Minkowski space-time and $(M,g)$ is a riemannian manifold. The supersymmetric extension consists of adding fermionic fields $\psi$, which are sections of $S \otimes X^*TM$, where $S$ is a spinor bundle on $\mathbb{R}^{3,1}$, in such a way that the resulting action is invariant under the Poincaré superalgebra.

This is a Lie superalgebra $\mathfrak{p} = \mathfrak{p}_0 \oplus \mathfrak{p}_1$, where $\mathfrak{p}_0$ is the Poincaré algebra and $\mathfrak{p}_1$ transforms as a spinor representation of the Lorentz subalgebra (with translations acting trivially). In four-dimensions, the Lorentz subalgebra is isomorphic to $\mathfrak{so}(3,1)$.

There is a well-defined procedure of dimensional reduction by which you can take a four-dimensional action and construct a one-dimensional action by simply declaring that the fields do not depend on three of the Minkowski coordinates. If one does this to the above four-dimensional supersymmetric sigma model, one obtains a much studied supersymmetric quantum mechanical system. The canonical quantisation of this systems results in a Hilbert space which is isomorphic to the $L^2$ complex differential forms on $M$ and a Hamiltonian which is the Hodge laplacian. Under this identification, the Dolbeault operators correspond to the "supercharges" in the Poincaré superalgebra; i.e., the action of $\mathfrak{p}_1$.

Further the dimensional reduction (which has to choose a direction) breaks the Lorentz symmetry to an $\mathfrak{so}(2,1)$ subalgebra whose action on the Hilbert space corresponds, under the above identification, to the action of the Hodge-Lefschetz operators $L,\Lambda,H$. The Hodge identities are what is left of the Poincaré supersymmetry after dimensional reduction. The action of $\mathfrak{sl}(2)$ on the cohomology of a Kähler manifold is nothing else but the action of the residual Lorentz symmetry on the ground states of the quantum-mechanical system.

If I may be forgiven for pointing to my own work, this is explained in some detail in Supersymmetry and the cohomology of (hyper)Kähler manifolds written with Bill Spence and Chris Köhl back in 1997. (There we also treat the case of hyperkähler manifolds, which are the targets of six-dimensional supersymmetric sigma-models. Under dimensional reduction to one dimension, there is a similar story with a residual Lorentz symmetry which acts on the cohomology of a hyperkähler manifold, "explaining" an earlier result of Misha Verbitsky's.)

The Dolbeault operators are usually defined in terms of the de Rham operator and the complex structure (see e.g. Wells' book or Griffith and Harris). The example you outline generalizes to the situation $G_{\mathbb{C}} / P = G / G_0$, where $G$ is compact, $G_{\mathbb{C}}$ is the complexification, $P$ is a parabolic subgroup, and $G_0 = G \cap P$. The holomorphic tangent bundle is the homogeneous vector bundle $G_{\mathbb{C}} \times_P \mathfrak{g}\_{\mathbb{C}} / \mathfrak{p}$, and the cotangent bundle is a homogeneous vector bundle in similar fashion.

In this case, the Dolbeault complex with coefficients in a homogeneous vector bundle $G_{\mathbb{C}} \times_P V$ translates to the Koszul complex for the relative Lie algebra cohomology $H^*(\mathfrak{g},\mathfrak{g}\_0,V \otimes C^{\infty}(G))$. The Dolbeault operator $\overline{\partial}$ translates to the boundary operator for Lie algebra cohomology, which of course involves the action of $\mathfrak{g}$. I haven't worked out what happens to $\partial$ in this situation, but most likely a similar expression in terms of the Lie algebra can be derived. The translation works by thinking about the smooth sections as elements of $C^{\infty}(G) \otimes V)^{\mathfrak{g}\_0}$ (for holomorphic sections use $C^{\infty}(G) \otimes V)^{\mathfrak{p}}$).