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Su(n) to sl(n,C)

  1. Dec 6, 2008 #1
    su(n) is isomorphic to sl(n,C), when we tensor su(n) with the complex numbers we get sl(n,C).

    Say we have su(2) with E_1= 1/2 [i, 0;0, -i], E_2=1/2[0,1;-1,0], E_3=1/2[0, i; i,0]

    sl(2,C) with F_1=[1, 0; 0, -1], F_2=[0, 1; 0, 0], F_3=[0, 0; 1, 0]

    so that [E_1, E_2]=E_3, [E_2, E_3]= E_1, [E_3, E_1]=E_2

    and [F_1, F_2]=2F_3, [F_1, F_3]=-2F_3, [F_2,F_3]=F_1

    Now I could write F_1=-2E_1, F_2=E_2-iE_3, F_3=E_2+iE_3, which i guess means tensoring su(2) with complex numbers and by what I get from the su(2) bracket relations to the sl(2,C) bracket relations.

    But where is the isomorphism?
  2. jcsd
  3. Dec 7, 2008 #2
    Shorter version of my question above:

    What does ' tensoring su(n) with the complex numbers we get sl(n,C), which shows that su(n) and sl(n,C) are isomorphic' mean?

    thank you
  4. Dec 8, 2008 #3
    I think you are talking about what's called the "complexification" of the Lie algebra su(2).

    Normally, su(2) is a Lie algebra over the real numbers. But if you allow yourself to multiply the generators by complex numbers as well as real numbers you will find that you can make any traceless 2x2 matrix (i.e. sl(2,C)).

    I.e. you can form any matrix in sl(2,C) by taking combinations of su(2) matrices with complex coefficients (but it's not possible using only real coefficients).
  5. Dec 9, 2008 #4
    Thanks for answering!

    Right, complexifaction is it also called by others.

    But why and how are su(n) and sl(n,C) isomorphic?
  6. Dec 10, 2008 #5
    Write down a basis for su(n) then you will be able to form a basis for sl(n,c) by taking linear combinations of the su(n) basis with complex coefficients.

    Or maybe a more elegant way to see it is to say that any traceless matrix can be written as a sum of a traceless hermitian and a traceless anti-hermitian matrix (M = H + A). Since su(n) ARE the traceless hermitian matrices then any matrix in sl(n,c) can be written as H + i(-iA) and both H and -iA are hermitian so they are in su(n).
  7. Dec 10, 2008 #6
    I know, this is what I did in post 1. But why and how makes that su(n) and sl(n,c) isomorphic?
  8. Dec 10, 2008 #7
    What exactly do you mean by isomorphic?
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