# Weighted projective space with rational or real weights

The most common formulation of the weighted projective space is perhaps the global quotient $$(\mathbb{C}^{n+1} \setminus \{(0,\ldots,0\}) / \mathbb{C}^\ast$$ with the $\mathbb{C}^\ast$ group action given by $$\lambda \cdot (x_0,\ldots,x_n) = (\lambda^{w_0} x_0, \ldots, \lambda^{w_n} x_n),$$ using positive integer weights $(w_0, \ldots, w_n)$. Of course, it can also be, equivalently, defined as a toric variety using an appropriate lattice.

My question is: (1) is there any classification on weighted projective space with rational weights? (i.e. $w_0, \ldots, w_n$ are nonzero rational numbers). It seems to me that many of the results should remain valid. In particular, all the singularities should still be of the cyclic quotient singularity type.

(2) How about using real weights? (let $w_0, \ldots, w_n$ be real numbers) With a properly chosen branch of logarithm, powers like $\lambda^{w_0}$ are still meaningful. Is there any classification results on such weighted projective spaces?

-
You need branches already for rational weights, but to define the action you need the power functions defined on all of $C^∗$ at the same time, no? For rational exponents, you can lift to a real action on the whole of $C^*$ by using a (finite) covering of the group, in order to make sense (and then you have reduced everything to the case of integer exponents!) (there are choices involved in this lifting, though...); for real exponents, you could proceed similarly using an infinite covering. –  Mariano Suárez-Alvarez Sep 29 '13 at 18:49
Yes. It seems with weights having common denominator $m$, an $m$-fold covering of the group $\mathbb{C}^\ast$ will do the job. But I couldn't find any reference. Will this approach break down in the case of infinite covering? –  ssquidd Sep 29 '13 at 19:50
Probably the easiest definition is using toric geometry in terms of the moment polytope. Real weights will correspond to a non-rational simplex. But it depends on your purpose, because with non-rational weights the space will not be Hausdorff. –  Craig Dec 3 '13 at 5:10