Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $SU(n+1)/U(n)$.

share|improve this question
1  
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. –  Allen Knutson Feb 27 '10 at 2:58
    
Yes of course, I've edited and it should be ok now. –  Jean Delinez Feb 27 '10 at 19:15
    
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. –  algori Feb 28 '10 at 3:06
    
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 $\partial f= \sum_{i=1}^n \frac{\partial f}{\partial z_i} dz_i$, –  Jean Delinez Mar 1 '10 at 21:06
    
$\overline{\partial}f = \sum_{i=1}^n \frac{\partial f}{\partial \overline{z}_i}d\overline{z}_i$, 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. –  Jean Delinez Mar 1 '10 at 21:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.