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Let $G$ be a Lie Group and $H$ a closed subgroup such that $G/H$ (the set of right cosets) is a complex manifold manifold. Now $\Omega^1(G/H)$, the space of complex one forms, is a $H$-equivariant bundle (with respect to left multiplication of $H$). What I want to know is: Are the Dolbeault maps $\partial, \overline{\partial}$, $H$-equivariant, that is, is it true that $$ \partial(f)(h.\overline{g}) = h.(\partial(f)(\overline{g})), $$ and $$ \overline{\partial}(f)(h.\overline{g}) = h.(\overline{\partial}(f)(\overline{g})), $$ for all $f \in \mathbb{C}[G/H]$, $h \in H$, $g \in G$.

The example I have in mind $SU(n+1)/U(n)$.

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    $\begingroup$ You've already lost me at "complex (compact) Lie group", and given SU(n+1) as an example. That is compact and not complex, e.g. for n=1 it is odd-real-dimensional. I'm pretty sure what you want, in order to have Dolbeault maps, is complex but not compact. Take G = SL(n+1), H = 1+n block upper triangular matrices (more often denoted P for parabolic). Also, Omega^1 is G-equivariant, not just H-equivariant. $\endgroup$
    – Allen Knutson
    Commented Feb 27, 2010 at 2:58
  • $\begingroup$ Yes of course, I've edited and it should be ok now. $\endgroup$
    – Jean Delinez
    Commented Feb 27, 2010 at 19:15
  • $\begingroup$ Jean -- your "closed subspace" $H$ is presumably a closed subgroup. Your phrase that $\Omega^1(G/H)$ is an $H$-equivariant bundle is also mysterious: $H$ does not act on $G/H$ (but $G$ does, with conjugates of $H$ as stabilizers). And it would be nice if you explained the notation in your formulae. $\endgroup$
    – algori
    Commented Feb 28, 2010 at 3:06
  • $\begingroup$ Yes I mean $H$-equivariant, as now explained in the question. Yes, I am assuming that $H$ is a subgroup. Finally, the "bar" in $\overline{g}$ indicates the coset containing $g$, and $\partial f= \sum_{i=1}^n \frac{\partial f}{\partial z_i} dz_i$, $\endgroup$
    – Jean Delinez
    Commented Mar 1, 2010 at 21:06
  • $\begingroup$ $\overline{\partial}f = \sum_{i=1}^n \frac{\partial f}{\partial \overline{z}_i}d\overline{z}_i$, where $\frac{\partial}{\partial z_i} = \frac{\partial}{\partial {x}_i} + i\frac{\partial}{\partial y_i}$, and $\frac{\partial }{\partial \overline{z}_i} = \frac{\partial }{\partial {x}_i} - i\frac{\partial}{\partial {y}_i}$, and $dz_i$ and $d\overline{z_i}$ are their respective duals. $\endgroup$
    – Jean Delinez
    Commented Mar 1, 2010 at 21:06

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