This question is related to a previous [question](https://mathoverflow.net/questions/16207/why-can-the-dolbeault-operators-be-realised-as-lie-algebra-actions) of mine. More specifically, it results from my attempts to understand the simplest incarnation of a phenomenon mentioned therein.


Put a grading on the elements of the coordinate algebra of $SU(2)$ by setting deg$(\hat{a})$=deg$(\hat{c})=1$, and deg$(\hat{b})$=deg$(\hat{d})=-1$; where 
$$
\hat{a}\left(\array{a & b \\\ c & d} \right) = a
$$
and so on. We can identify the coordinate algebra of $\mathbb{CP}^1 = SU(2,\mathbb{C})/U(1)$, with the elements of degree $0$, $\Omega^{(0,1)}(\mathbb{CP}^1)$ with the elements of degree $2$. Does there exist an element $X$ of $\mathfrak {su}(2)$ (or its enveloping algebra) such that 
$$
X(f) = \overline{\partial}(f) \in \Omega^{(0,1)}(\mathbb{CP}^1), 
$$ 
for all coordinate functions $f$ on $\mathbb{CP}^1$, where $X$ is the canonical action of $\mathfrak {su}(2)$ on the coordinate algebra of $SU(2)$, that is, 
$$
X(f)(g) = \frac{d}{dt}\_{t=0}(f(\exp(-tX)g)).
$$
If so, can someone explain in a more general context why so? (Perhaps in the context of spaces of the form $G/T$, where $G$ is a compact Lie group, and $T$ is a maximal torus.)


EDIT: I've been doing some more reading, and it seems that the required element of $\mathfrak{su} (2)$ does exist. Moreover, there exists a formula for extending this to the general $\mathbb{CP}^n$ case, see page 32 of this [paper](http://arxiv.org/abs/0901.4735) by D'Andrea and Dabrowski. What I still don't see though is why.



Dolbeault Operators for $CP^1$  as $\mathfrak{su}(2)$ Actions.
Let $K$ be a compact connected Lie group, and $L$ a Levi subgroup of $K$ (the centralizer of an element of the Lie algebra). Then $X=K/L$ is a complex manifold (a coadjoint orbit). 
So one can ask about the representation theoretic interpretation of the Dolbeault operator
on $X$. 
We have a direct sum decomposition $Lie(K)\_{\Bbb C}=Lie(L)\_{\Bbb C}\oplus {\mathfrak u}\_+\oplus {\mathfrak u}\_-$, 
where ${\mathfrak u}\_\pm$ 
are the nilpotent subalgebras corresponding to $L$. 
The antiholomorphic tangent bundle $T^{0,1}$ is the bundle associated to the representation ${\mathfrak u}\_+$ of $L$ and the principal $L$-bundle $K\to X$. 


Now, by the Peter-Weyl theorem, the space of functions (of finite type) on $K$ is 
${\rm Fun}(K)=\oplus V\otimes V^\ast$, where $V$ runs over irreducible representations of $K$. 
Thus, ${\rm Fun}(X)=\oplus V^L\otimes V^\ast$, and 
$\Omega^{0,1}(X)=\oplus ({\mathfrak n}\_+^\ast\otimes V)^L\otimes V^\ast$.
The Dolbeault operator $\bar{\partial}: {\rm Fun(X)}\to \Omega^{0,1}(X)$ is thus an 
operator
$\oplus V^L\otimes V^\ast\to \oplus ({\mathfrak n}\_+^\*\otimes V)^L\otimes V^\ast$.
It commutes with the action of $K$, so comes from some operator 
$\overline{\partial}\_V: V^L\to ({\mathfrak n}\_+^\*\otimes V)^L$ for each $V$. 


I claim that this operator is just the dual to the action map ${\mathfrak n}\_+\otimes V\to V$ (restricted to $L$-invariants). 


Indeed, it is well known that the exterior differential of functions on $K$ is given by the formula $df=\sum L\_{a\_i}(f) \omega\_{a\_i^\ast}$, where $a\_i$ is a basis of $Lie(K)\_{\Bbb C}$, 
$L\_a$ is the left invariant vector field corresponding to $a$, and 
$\omega\_u$ is the left invariant differential 1-form corresponding to $u$. 
Choosing a basis corresponding to the decomposition $Lie(K)\_{\Bbb C}=Lie(L)\_{\Bbb C}\oplus {\mathfrak u}\_+\oplus {\mathfrak u}\_-$, and restricting to function on $X=K/L$, we see that the terms corresponding to $Lie(L)\_{\Bbb C}$ are zero, and the terms for ${\mathfrak n}\_+$ and ${\mathfrak n}\_-$ give $\bar\partial$ and $\partial$ respectively.


In your basic example $K=SU(2)$, $L=U(1)$, and the Dolbeault operator is defined by the action of the generator $E$ of $sl(2)$. 


