Finding the Real Lie Algebra of SL(n,H) in GL(n,H)

[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

 

[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.

 

[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

 

[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}

 

[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?

 

[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).

 

[arz2000]

Thanks, how about sl(n,H) ?

 

[matt grime]

Do the same thing: it must be satisfy two conditions.