Complexifying su(2) to get sl(2,C)---group thread footnote


by marcus
Tags: cgroup, complexifying, footnote
marcus
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#1
Aug10-03, 10:07 AM
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On the group thread midterm exam (which we never had to take!) it says what is the LA of the matrix group SL(2, C)
and the answer is the TRACE ZERO 2x2 matrices.
So that is what sl(2,C) is.
When you exponentiate one of the little critters, det = exp trace,
so the determinant is one which is what SL means.

Any X in sl(2,C) has a unique decomposition into skew hermitians that goes like this

X = (X - X*)/2 + i(X + X*)/2i

and these two skew hermitians
(X - X*)/2 and (X + X*)/2i
are trace zero, because trace is linear

check the skew hermitiandom of them:
(X - X*)* = (X* - X) = - (X - X*)

the other one checks because (1/2i)* = - (1/2i)
since conjugation does not change (X + X*)* = (X + X*)

so the upshot is that any X in sl(2,C) is composed
X = A + iB
of two matrices A and B in su(2)

Also on the midterm was the fact that su(2) is the skew hermitian ones: A* = - A.

There was this footnote on complexification of LAs and the above suffices to show, without much further ado, that su(2)C the complexification of su(2) is isomorphic to sl(2, C)
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Tyger
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#2
Aug10-03, 11:40 AM
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SL(2,C) is a representation of the group of boosts and turns, so why doesn't it show up in our descriptions instead of the 4×4 Dirac spinors?
r637h
#3
Aug12-03, 01:00 AM
P: n/a
Well, there you go: Topology/Non-Euclidian Geomerty, like poverty and ignorance: We will always have them with us.

Rudy

"Go Figure." - Archimedes


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