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03 - Lecture 03 Example 1 A 2-dimensional non abelian Lie...

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Lecture 03 Example 1. A 2-dimensional, non abelian Lie algebra. Consider: g = { b 0 a 0 : a, b F } spanned by x = 0 0 1 0 , y = 1 0 0 0 Notice then that [ x, y ] = x Z ( g ) , g gl 2 ( F ) , dim g = 2 , Z ( g ) 6 = { 0 } . Example 2. A 3-dimensional, non abelian Lie algebra, with [ g , g ] Z ( g ) and dim [ g , g ] = 1 . Consider: g = { 0 a b 0 0 c 0 0 0 : a, b, c F } spanned by x = 0 1 0 0 0 0 0 0 0 , y = 0 0 0 0 0 1 0 0 0 , z = 0 0 1 0 0 0 0 0 0 This is a subalgebra of gl 3 ( F ) with the properties [ x, y ] = z Z ( g ) , [ z, x ] = 0 , [ z, y ] = 0 , called the Heisenberg Algebra. 1

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Example 3. A 3-dimensional, non abelian Lie algebra, with [ g , g ] * Z ( g ) and dim [ g , g ] = 1 . Consider: g = { b 0 0 a 0 0 0 0 c a, b, c F } spanned by x = 0 0 0 1 0 0 0 0 0 , y = 1 0 0 0 0 0 0 0 0 , z = 0 0 0 0 0 0 0 0 1 This is a subalgebra of gl 3 ( F ) satisfying the properties [ x, y ] = x, [ x, z ] = 0 , [ y, z ] = 0 . It is also isomorphic to the “direct sum” of the algebra presented at the first example and the 1-dimensional algebra F (look at the splitting of the matrix!). Back to the classification. Dim g = 3 dim [ g , g ] = 2 Claim: [ g , g ] is abelian. Proof. Let { x, y } be a basis for [ g , g ] and pick z g - [ g , g ]. Notice that [ x, y ] [ g , g ] thus [ x, y ] = ax + by for some a, b F Then define X := [ x, · ] : g g This is a linear map denoted ad x . Notice that Im ( X ) [ g , g ]. Thus X ( x ) = 0 , X ( y ) = ax + by, X ( z ) = cx + dy . Clearly Trace ( X ) = tr ( X ) = 0+ b +0 = b . Consider the corresponding map Y := [ y, · ] : g g Likewise tr ( Y ) = - a . On the other hand notice that if u, v g then [[ u, v ] , · ] = [[ u, · ] , [ v, · ]] = [ u, · ] [ v, · ] - [ v, · ] [ u, · ] where is composition of linear maps. Thus: tr [[ u, v ] , · ] = tr ([ u, · ] [ v, · ] - [ v, · ] [ u, · ]) = 0 2
Then remember that x belongs to [ g , g ]. So x = n i =1 a i · [ u i , v i ] for some a i F , u i , v i g . Thus since Trace is a linear map we get: tr ( X ) = n X i =1 a i · tr ([ u i , v i ]) = n X i =1 a i · 0 = 0 Similarly tr ( Y ) = 0. So a = b = 0 yielding [ x, y ] = 0.

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