Skip to main content
MathOverflow

Return to Question

edited tags
Link
YCor
YCor
  • 63.9k
  • 5
  • 187
  • 286
formatting
Source Link
YCor
YCor
  • 63.9k
  • 5
  • 187
  • 286

maximal Maximal dimension of abelian subalgebra of exceptional simple Lie algebra in positive characteristic

For complex semisimple Lie algebras, the maximal dimension of an abelian subalgebra was determined by Mal'cev in 1945. For E7$E_7$, for example, it is 27$27$, and is the radical of the E6$E_6$ parabolic.

What about in characteristic p$p$ for p>0$p>0$? I suspect the answer is nearly, but not quite, the same. Maybe you have that it is bounded by 29$29$ or something. Is there any literature on this? All of the papers and references I have found so far are in characteristic 0 $0$.

I don't need the exact bound, just something close will do. (I have a Lie algebra L$L$ with an abelian subalgebra of dimension, say, 43$43$, and I want to know that L$L$ cannot be embedded in E7$E_7$, and in particular is not E7$E_7$. But I'm in characteristic 7$7$, for example, or worse, 2$2$ or 3$3$.)

maximal dimension of abelian subalgebra of exceptional simple Lie algebra in positive characteristic

For complex semisimple Lie algebras, the maximal dimension of an abelian subalgebra was determined by Mal'cev in 1945. For E7, for example, it is 27, and is the radical of the E6 parabolic.

What about in characteristic p for p>0? I suspect the answer is nearly, but not quite, the same. Maybe you have that it is bounded by 29 or something. Is there any literature on this? All of the papers and references I have found so far are in characteristic 0.

I don't need the exact bound, just something close will do. (I have a Lie algebra L with an abelian subalgebra of dimension, say, 43, and I want to know that L cannot be embedded in E7, and in particular is not E7. But I'm in characteristic 7, for example, or worse, 2 or 3.)

Maximal dimension of abelian subalgebra of exceptional simple Lie algebra in positive characteristic

For complex semisimple Lie algebras, the maximal dimension of an abelian subalgebra was determined by Mal'cev in 1945. For $E_7$, for example, it is $27$, and is the radical of the $E_6$ parabolic.

What about in characteristic $p$ for $p>0$? I suspect the answer is nearly, but not quite, the same. Maybe you have that it is bounded by $29$ or something. Is there any literature on this? All of the papers and references I have found so far are in characteristic $0$.

I don't need the exact bound, just something close will do. (I have a Lie algebra $L$ with an abelian subalgebra of dimension, say, $43$, and I want to know that $L$ cannot be embedded in $E_7$, and in particular is not $E_7$. But I'm in characteristic $7$, for example, or worse, $2$ or $3$.)

Source Link
David A. Craven
David A. Craven
  • 1.1k
  • 7
  • 13

maximal dimension of abelian subalgebra of exceptional simple Lie algebra in positive characteristic

For complex semisimple Lie algebras, the maximal dimension of an abelian subalgebra was determined by Mal'cev in 1945. For E7, for example, it is 27, and is the radical of the E6 parabolic.

What about in characteristic p for p>0? I suspect the answer is nearly, but not quite, the same. Maybe you have that it is bounded by 29 or something. Is there any literature on this? All of the papers and references I have found so far are in characteristic 0.

I don't need the exact bound, just something close will do. (I have a Lie algebra L with an abelian subalgebra of dimension, say, 43, and I want to know that L cannot be embedded in E7, and in particular is not E7. But I'm in characteristic 7, for example, or worse, 2 or 3.)