Isomorphism of su(2) and sl(2,C): Tensor w/ Complex Numbers

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The discussion centers on the isomorphism between the Lie algebras su(2) and sl(2,C) through the process of complexification. It is established that tensoring su(2) with complex numbers results in sl(2,C), allowing for the formation of any traceless 2x2 matrix. The relationship is further clarified by demonstrating that any traceless matrix can be expressed as a combination of traceless Hermitian and anti-Hermitian matrices, with su(2) representing the traceless Hermitian matrices. This confirms the isomorphic nature of su(n) and sl(n,C) when complex coefficients are employed.

PREREQUISITES
  • Understanding of Lie algebras, specifically su(n) and sl(n,C)
  • Familiarity with complex numbers and their application in linear algebra
  • Knowledge of matrix theory, particularly traceless matrices
  • Concept of complexification in the context of algebraic structures
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  • Study the concept of complexification in Lie algebras
  • Explore the properties of traceless Hermitian and anti-Hermitian matrices
  • Learn about the structure and applications of sl(n,C) in representation theory
  • Investigate the implications of isomorphism in algebraic structures
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Mathematicians, physicists, and students of algebra interested in the properties and applications of Lie algebras, particularly those working with su(n) and sl(n,C).

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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?
 
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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
 
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).
 
Thanks for answering!

Right, complexifaction is it also called by others.

But why and how are su(n) and sl(n,C) isomorphic?
 
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).
 
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.

I know, this is what I did in post 1. But why and how makes that su(n) and sl(n,c) isomorphic?
 
What exactly do you mean by isomorphic?
 

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