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Nov 2, 2014 at 14:24 comment added Daniel Litt Fair enough; the root stacks need not be DM in char p, though, as I'm sure you know.
Nov 2, 2014 at 9:05 comment added Niels About your first comment: it seems better to me to give the simplest possible example than a class including pointless complications such as non-generic stabilizers. About your second comment : I was careful about what you mention. Your quotient stack is interesting but is definitely not a stack of roots.
Nov 1, 2014 at 21:06 comment added Daniel Litt Oh, and one must be careful in characteristic $p$; for example, $\mathbb{P}^1/(\mathbb{Z}/p\mathbb{Z})$ where $\mathbb{Z}/p\mathbb{Z}$ acts by translations is manifestly not simply connected, even though it has a single orbifold point at $\infty$.
Oct 31, 2014 at 15:41 comment added Daniel Litt Isn't this the special case of my answer, $\mathcal{P}(1, n)$ (at least in characteristic zero)? (Indeed, it must be, since Behrend-Noohi show that the only smooth proper DM curves are $\mathcal{P}(m, n)$ in characteristic zero.)
Oct 31, 2014 at 13:56 history answered Niels CC BY-SA 3.0