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Direct Sum of Lie Algebras

  1. Feb 3, 2008 #1
    We are asked to show that L(SO(4)) = L(SU(2)) (+) L(SU(2))

    where L is the Lie algebra and (+) is the direct product.

    We are given the hint to consider the antisymmetric 4 by 4 matrices where each row and column has a single 1/2 or -1/2 in it.

    By doing this we generate four matrices, call them S1, S2, T1, T2
    we can show that the commutator of S1, S2 generates another matrix S3 and that the same occurs for the T's. We can show that the 6 matrices form a general basis for L(SO(4))

    we then get the following commutation relations:

    [Sa, Sb] = epsilon(a,b,c) Sc
    [Ta, Tb] = epsilon(a,b,c) Tc
    and [Ta, Sb] = 0

    We can then see that the S and T matrices form a basis for L(SU(2)) as they obey the correct commutation relations, and therefore the direct sum of these two Lie algebras forms the lie algebra for L(SO(4)).

    My question is if we require the two L(SU(2))'s to commute (as they in this case do) in order use their direct sum, or is it okay to take a direct sum of two non-commuting lie algebras. thanks
  2. jcsd
  3. Feb 3, 2008 #2

    George Jones

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    Let [itex]L_1[/itex] and [itex]L_2[/itex] be any Lie algebras. Since [itex]L_1[/itex] and [itex]L_2[/itex] are vector spaces, the external vector space direct sum [itex]L_1 \oplus L_2[/itex] can be formed. The underlying set for the vector space direct sum is

    [tex]L_1 \times L_2 = \left\{ \left( x,y \right) | x \in L_1, y \in L_2 \right\}.[/tex]

    In order for this set to have vector space structure, scalar multiplication and addition of ordered pairs must be defined. This is done in a natural way:

    [tex]c \left( x,y \right) := \left( cx,cy \right);[/tex]

    [tex]\left( x_1,y_1 \right) + \left( x_2,y_2 \right) := \left( x_1 + x_2,y_1 + y_2 \right).[/tex]

    Now in order to get a Lie algebra, a commutator must be defined on ordered pairs:

    [tex]\left[\left( x_1,y_1 \right) , \left( x_2,y_2 \right)\right] := \left( \left[x_1,x_2\right] , \left[y_1,y_2\right]\right).[/tex]

    Note that

    [tex]\left[\left( x , 0 \right) , \left( 0,y \right)\right] = \left( \left[x , 0\right] , \left[0,y\right \right)] = 0.[/tex]
    Last edited: Feb 3, 2008
  4. Feb 5, 2008 #3
    thanks, i've got it now
  5. Feb 6, 2008 #4


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    How would you show those 6 matrices forma a basis in so(4) ?
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