Are the canonical actions on Schubert Cells Linearizable? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T19:27:58Z http://mathoverflow.net/feeds/question/111126 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/111126/are-the-canonical-actions-on-schubert-cells-linearizable Are the canonical actions on Schubert Cells Linearizable? P Vanchinathan 2012-11-01T10:01:43Z 2012-11-29T23:03:17Z <p>G. Schwarz constructed a (counter)example for an action of a simple algebraic group on an affine space that is not linearizable (i.e., it is not a representations). </p> <p>Natural examples of affine spaces that are not readily vector spaces are Schubert cells. So it was tempting to look for reductive group actions on them and see if they can lead to more counter-examples.</p> <p>For a parabolic subgroup \$P\$ of a linear algebraic group \$G\$, (say \$G\$ semi-simple) we can take a Schubert cell \$C\subset G/P\$. By definition \$C\$ is the orbit for a (maximal) solvable subgroup<br> which is far from a reductive group. However one can look at the largest parabolic subgroup of \$G\$ acting on \$C\$, and restrict the action to a Levi part \$L\$ and ask if the action of \$L\$ on \$C\$ is linearizable. (Easy to see examples where \$L\$ is more than a maximal torus of \$G\$).</p> <p>In a 5-minute meeting with M. Brion I asked this question and he said `Yes, it would follow from the slice theorem'. Can any one elaborate on his brief answer?</p>