Let gl(n, R) be the Lie algebra of matrices with real entries and GL(n, R) its associated Lie group. Recall that a linear subgroup G ⊂ GL(n,R) acts by conjugation on gl(n, R), that is, for g ∈ G its action on A ∈ gl(n, R), is defined by
g(A) = g^{-1}Ag.
Definition: Let G ⊂ GL(n, R) be a subgroup. A polynomial f ∈ R[(Xij)1≤i,j≤n] is called invariant on gl(n, R) with respect to conjugation by elements in G iff
f(g^{-1}Ag) = f(A) (4) ∀g ∈ G, ∀A ∈ gl(n,R).
We will denote by G' the set of all invariant polynomials.
Question: What is known about G'?
@Mihail, G' is the invariant ring. If the G is finite (or solvable of particular type) the G' is finitely generated and you can find a basis using Reynolds operator. Tak e a look at the following book (page 16)
http://jones.math.unibas.ch/~kraft/Papers/KP-Primer.pdf
and
https://www.youtube.com/watch?v=s1wH6GOWANc
@Mohamed Thank you so much Mohamed. I will most definitely take a look at the book you suggested.
If G=SL(n,R) then G' consisnts only one function namely the deteminant. For G=GL(n,R) the G' is generated by traces of powers of the matrix X_{i,j}. If G is finite group then to get G' you have use the Reinolds operators to all monomials (of variables X_{i,l}) up to degrees |G|.
Mihail This MO conversation contains some information on invariant polynomials
http://mathoverflow.net/questions/182871/generators-of-invariant-polynomials-of-semisimple-lie-algebra
Best regards
Ali
- Badges
- Science topic
- Similar topics
- Mathematics
- Statistics