User ana - MathOverflow 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 Are two "nice" transformation groupoids with the same coarse moduli and isomorphic inertia isomorphic? 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>