Let $G\subset GL_n(\mathbb{C})$ be an algebraic group of dimension n, and let $\mathfrak{g}$ its Lie algebra.Is there a relations between the maximal number of independent semisimple elements of $G$ and the maximal number of independent semisimple elements $\mathfrak{g}$?
$\begingroup$
$\endgroup$
3
-
1$\begingroup$ What is "independent" in either case? $\endgroup$– user1688Commented Mar 10, 2011 at 12:39
-
1$\begingroup$ independent = linearly independent. sorry. $\endgroup$– Michele TorielliCommented Mar 10, 2011 at 14:21
-
3$\begingroup$ Linear independence doesn't make sense for elements of a group. Perhaps you mean to ask what the relation is between the dimension of a maximal subalgebra consisting of semisimple elements (a maximal toral subalgebra) and the dimension of a maximal algebraic subgroup consisting of semisimple elements (a maximal torus). $\endgroup$– mephistoCommented Mar 10, 2011 at 18:52
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
Assuming you mean "linearly independent" here, it's important to understand that the maximal numbers in question depend on the chosen faithful linear representation of the group rather than on its intrinsic properties as an algebraic group. For example, there are many conjugate maximal tori (of the same dimension) living inside the full upper triangular matrix group, but such a torus may or may not consist of diagonal matrices depending on the choice you make.
Also, the maximal number of independent matrices in the Lie algebra may well differ, as seen in the case of the trivial algebraic group whose Lie algebra is zero.
I'm not at all sure what the question has to do with Lie theory (?)
answered Mar 10, 2011 at 13:18