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Jay Taylor
  • Member for 14 years
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Richardson Classes and the Bala Carter Theorem
OK, well thanks for all your help and very useful comments Jim. I'll take a look at the Bala-Carter papers to see if I can sort out my labelling issues. Thanks again.
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Richardson Classes and the Bala Carter Theorem
So let's just ignore the parabolic coming from the Richardson theory altogether. Can one determine in a nice way specifically the Levi subgroup attached to the class in the Bala-Carter theory? I had the impression that as the partition determines the Jordan normal form of the matrix that this would determine the Levi subgroup. I guess in my mind in the example in $D_4$ the two Jordan blocks of size 4 fit very nicely in some $A_3$ Levi, which will look like $GL_4\times GL_4$ in $SO_8$, (except one $GL_4$ is determined by the other). Using matrices in $SO_{2n}$ would probably make this concrete.
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Richardson Classes and the Bala Carter Theorem
Thanks for your answer Jim. Just a couple of questions. Sticking with the example of $D_4$ and the partition $(4,4)$. Do you really mean to say that the Bala-Carter theory gives a parabolic with Levi complement $A_3$ and the Richardson parabolic is of type $A_2$? What I got the impression that Carter was saying was that the parabolic for which the class is a Richardson class has Levi complement $A_3$. This is mentioned in his book (pg. 423 to be precise) in his description of the Springer correspondence.
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Richardson Classes and the Bala Carter Theorem
Yes, you're right. I have made it a bit clearer now for $D_4$, especially as this is the running example in my post. Yeah, I always get a bit haphazard with this. I get so used to writing "connected reductive" or "connected semisimple" that I forget when referring to a connected simple algebraic group that this is redundant. Thanks for the suggestion. In fact, no. I'm actually interested in these classes in the half spin group $HSpin_{2n}(\mathbb{K})$.
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Richardson Classes and the Bala Carter Theorem
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Richardson Classes and the Bala Carter Theorem
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