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For Lie algebra of 9 elements $e1\dots e9$, the commutation relations are defined as:

$[{\it e1},{\it e4}]=-{\it e4},[{\it e1},{\it e6}]=-{\it e6},[{\it e1} ,{\it e7}]=-{\it e7},[{\it e1},{\it e8}]=-{\it e8},[{\it e2},{\it e3}] ={\it e5},[{\it e2},{\it e4}]={\it e6},[{\it e2},{\it e5}]=-{\it e3},[ {\it e2},{\it e6}]=-{\it e4},[{\it e3},{\it e4}]={\it e7},[{\it e3},{ \it e5}]={\it e2},[{\it e3},{\it e7}]=-{\it e4},[{\it e5},{\it e6}]={ \it e7},[{\it e5},{\it e7}]=-{\it e6} $

For $S = \langle a_{3}e2+\frac{a_{3}^{2}}{a_{2}}e3+e7,a_{2}e2+a_{3}e3+e6,e8,e9\rangle$

$N = \langle a_{2}e2+a_{3}e3+e6+a_{8}e8+a_{9}e9\rangle$.

How I can find factor algebra $S/N$ ?

IgotiT
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1 Answers1

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Your notation is a little confusing, as you've used square brackets for the Lie product and in the definition of S and N. In the latter I assume the square brackets stand for the subspace spanned by the elements within the brackets. In which case, S is 4-dimensional and N is 1-dimensional, so S/N is 3-dimensional. If we denote the basis vectors of S by x1, x2, x3 and x4 for simplicity, then S/N is spanned by x1+N, x2+N, x3+N and x4+N. Note that x2+N, x3+N and x_4+N are linearly dependent, so a basis for S/N is, for example, x1+N, e8+N and e9+N. The products are easy to compute

David Towers
  • 566
  • Can you please give further details as I am very immature in linear algebra. Please provide complete basis of $S/N$ if you can. – IgotiT Aug 19 '16 at 11:47
  • Have you taken $x_{1} = a_{3}e2+\frac{a_{3}^{2}}{a_{2}}e3+e7,x_{2} = a_{2}e2+a_{3}e3+e6, x_{3} = e8, x_{4} = e9$ ? – IgotiT Aug 19 '16 at 12:03
  • Yes I have. So the basis for S/N is a_{3}e2+\frac{a_{3}^{2}}{a_{2}}e3+e7 + N, e8 + N, e9 + N. The products are all zero, so S/N is the abelian Lie algebra of dimension 3. – David Towers Aug 19 '16 at 12:25