Be $T$ a complex torus (which is not necessarily am abelian variety) of complex dimension $n \geq 2$. On $T$ we have an involution corresponding to the $(-1)$ application (i.e. passing to the inverse for the group law on $T$). Quotient $T$ by the action of such involution: we have a complex analytic space $\tilde{T}$ which is singular exactly at those points where the action of the involution is not free, i.e. $2-$torsion points. If we blow-up $\tilde{T}$ at these points we obtain again a complex manifolds and the exceptional divisors dominating the $2-$torsion points are all $\mathbf{P}^{n-1}$. Now is such a desingularization of the quotient of $T$ again Kaehler? It is clear to me that on the non-singular points of $\tilde{T}$ there are Kaehler forms coming from $T$, since invariant closed $2-$forms on $T$ are fixed for the $(-1)$ involution.

More generally, given a Kaehler manifold $X$, given a group $G$ acting on $X$ (not necessarily in a free, proper discountinous way), under what restrictions on $G$ and its action we will have that the desingularization of $X/G$ is still Kaehler?


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In the case of the torus the desingularisation will be Khaler (for any dimension). In order to prove this it is sufficinet to show that if you consider an ivolution on C^n, then on the blow up there is a Khaler metric that coincides with the quotient one at infinity. This will give you a local model and then you can scale it down to glue on the quotient torus.

The a similar thing should happen if the fixed points of your action are isolated. The point is that in this case locally in a neighborhood of a fixed point the action is algebraic (can be modelled on the action on C^n) and you can always find a (locally) algebraic resolution that automatically admits a Kahler form (an I guess this form can be chosen to coincide with the quoteint form outside of a little resolved neighborhood). In the case when the singularities are not isolated you can not apply this reasoning but I would still guess that a Khaler resolution should exist (but I never saw this statement written down)


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