A question on Lie groups

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  • #1
asm
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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
 

Answers and Replies

  • #2
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 [tex]\gamma(t)\colon R\rightarrow GL(n,H)[/tex] such that [tex]\gamma(0)=\tilde1[/tex] and then take the derivative of that path at the identity.

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

ZM

PS: typically the requirement for [tex]\hbox{det} g=1[/tex] for [tex]g\in G[/tex] means that the element in the Lie Algebra has zero trace.
 
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  • #3
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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
 
  • #4
matt grime
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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
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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
matt grime
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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).
 
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  • #7
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Thanks, how about sl(n,H) ?
 
  • #8
matt grime
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Do the same thing: it must be satisfy two conditions.
 

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