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This question is related to a previous question 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 by D'Andrea and Dabrowski. What I still don't see though is why.

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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 $sl(2)$.

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