most recent 30 from http://mathoverflow.net 2013-05-24T20:13:34Z http://mathoverflow.net/feeds/user/31094 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/120550/are-two-nice-transformation-groupoids-with-the-same-coarse-moduli-and-isomorphi Ana 2013-02-01T21:37:52Z 2013-02-01T21:37:52Z

<p>Hi!</p> <p>I am stuck with the following question: suppose we have a semisimple connected algebraic group acting on a quasi-affine variety X by closed orbits, and suppose that inertia is flat. Suppose we have another quasi-affine variety Y which embeds inside X, and assume the normalizer in G of Y, say N, is a reductive algebraic group such that Y has closed orbits under its action. Assume that </p> <p>$Y//N\cong X//G$</p> <p>and that the restriction of the inertia stack from $[X/G]$ to $[Y//N]$ gives the inertia of the latter stack. Is it true that </p> <p>$[Y/N]\cong [X/G]$</p> <p>via the obvious morphism induced by the embedding $Y\hookrightarrow X$?</p> <p>To fix ideas, one could consider </p> <p>$X=\mathfrak{g}_{reg,ss}$</p> <p>the regular semisimple points of the Lie algebra of $G$, the action being the adjoint, $Y=\mathfrak{t}_{reg}$ the regular points of a Cartan subalgebra of $\mathfrak{g}$ and $N=N(T)$ the normalizer of the corresponding torus. This example may be somehow misleading, as in this case inertia is abelian, so it descends uniquely to the coarse moduli giving a gerbe structure, but this may be false in the more general situation.</p> <p>Thanks!</p>