# Why can the Dolbeault Operators be Realised as Lie Algebra Actions

I've been looking at an example in the non-commutative geometry literature and I'm having trouble figuring out what the classical motivation is. I'll just describe the classical case here: Recall that $\mathbb{CP}^2 = SU(\mathbb{C},3)/U(\mathbb{C},2)$. Recall also that since $T_\mathbb{C}^*(\mathbb{CP}^2)$ can be viewed as a vector bundle associated to the $SU(2)$-bundle $SU(\mathbb{C},3) \to \mathbb{CP}^2$, we can view $\Omega_\mathbb{C}^1(\mathbb{CP}^2)$ as a subset of $\mathcal{O}(\mathbb{CP}^2) \oplus \mathcal{O}(\mathbb{CP}^2)$. Now in the example, the action of each of the Dolbeault operators $\partial,\overline{\partial}$ is given in terms of two mappings, $\partial_1,\partial_2$ and $\overline{\partial_1},\overline{\partial_2}$, such that $$\partial(f) = (\partial_1(f),\partial_2(f)) \in \mathcal{O}(\mathbb{CP}^2) \oplus \mathcal{O}(\mathbb{CP}^2),$$ and similarly for $\overline{\partial}$. The mappings ${\partial}_i, \overline{\partial}_i$ are constructed using an action of the Lie algebra $\mathfrak{su}(3)$ on $\mathcal{O}(SU(3))$ constructed using the canonical pairing between Lie algebras and coordinate algebras.

I have a feeling this is a complicated incarnation of a simple classical object. Does any of this ring any bells with anyone?

Edit: This question has been superseded by this question and I am voting to close. I would ask others to do likewise.

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This sounds pretty cool, but I'm tired and your LaTeX didn't compile. Could you please put dollar-signs around the equations? – userN Feb 24 '10 at 0:17
There are dollar signs around the equations and no problems on my side. Are there problems with all the equations for you? – Jean Delinez Feb 24 '10 at 0:22
Yep. I'll check my browser. Thanks! – userN Feb 24 '10 at 1:34
Are the Dolbeault operators you are referring to the usual ones defined in terms of the de Rham differential and the complex structure? – William Slofstra Mar 1 '10 at 19:00
Yes, exactly. – Jean Delinez Mar 1 '10 at 20:30

I might be misunderstanding the question, but 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 $C^{\infty}(G) \otimes V)^{\mathfrak{p}}$).

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Thanks a lot for your answer. It may contain what I'm looking for, but I'm not familiar enough with the material to see if this is the case. I've asked a new, and much more direct, question entitled "Dolbeault Operators for $CP^1$ and the Lie Algebra of $SU(2)$", perhaps you'd like to take a look at this and see how your answer would look in that context? Thanks again. – Jean Delinez Mar 2 '10 at 17:38
No problem... I like Pavel's answer to the new question. It gives all the details of how to do the translation I mentioned. – William Slofstra Mar 4 '10 at 8:24

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.)

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For more on the Hodge-Lefschetz operators and the $\mathfrak{sl}_2$ action they give, there's Tim Perutz's great answer to my (not-as-great) question: mathoverflow.net/questions/14667/… – Charles Siegel Feb 24 '10 at 3:02