A question on Lie groups

1. Mar 15, 2007

asm

Hi all,

Anybody knowes how to find, or at least knows the reference that shows, the real lie algebra of sl(n,H)?
By sl(n,H), I mean the elements in Gl(n,H) [i.e. the invertible quaternionic n by n matrices] whose real determinant is one.

Many Thanks
Asi

2. Mar 15, 2007

zenmaster99

Given that the Lie Algebra of a Lie Group is identified with the tangent space of the Lie Group at the identity, I would create an arbitrary path $$\gamma(t)\colon R\rightarrow GL(n,H)$$ such that $$\gamma(0)=\tilde1$$ and then take the derivative of that path at the identity.

I haven't seen a representation of $$GL(n,H)$$ so, if you find one, post it and we'll see if we can take it's derivative.

ZM

PS: typically the requirement for $$\hbox{det} g=1$$ for $$g\in G$$ means that the element in the Lie Algebra has zero trace.

Last edited: Mar 15, 2007
3. Mar 16, 2007

arz2000

Actually, GL(n,H) is the set of all matrices A in GL(2n,C) such that AJ=Ja where a is complex conjugate of A and J is 2n by 2n matrice with the rows
(0 -I) and (I 0), I is the n by n identity matrix.

Thanks

4. Mar 16, 2007

matt grime

Differentiate the condition. The lie algebra is the set of matrices D satisfing Id+eD is in GL(n,H) mod e^2.

Thus (id+eD)J=J(Id+eD)^*

Thus eDJ=eJD* or DJ-JD^*=0. So the lie algebra is

gl(n,H):={ D : DJ-JD^*=0}

5. Mar 16, 2007

arz2000

Many thanks, But how did you get the lie algebra is the set of matrices D satisfing Id+eD is in GL(n,H) mod e^2 by differentiation?

6. Mar 16, 2007

matt grime

Because that is the definition of the lie algebra. I just wrote down the conditions necessary to be a tangent vector (and ignored convergence issues - everytihing is defined in terms of polynomials so there is no concern here).

Last edited: Mar 16, 2007
7. Mar 16, 2007