A question on Lie groups

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

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.

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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
Homework Helper
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}

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