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

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Discussion Overview

The discussion revolves around the isomorphism between the Lie algebras su(n) and sl(n,C), particularly focusing on the process of tensoring su(n) with complex numbers and the implications of this operation. Participants explore the concepts of complexification and the relationships between the generators of these algebras.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification, Debate/contested

Main Points Raised

  • One participant asserts that tensoring su(n) with complex numbers results in sl(n,C), suggesting a structural relationship between the two algebras.
  • Another participant introduces the concept of complexification, explaining that su(2) is a Lie algebra over the reals, and by allowing complex coefficients, one can generate all traceless 2x2 matrices, which correspond to sl(2,C).
  • A different viewpoint emphasizes that any traceless matrix can be expressed as a sum of a traceless Hermitian and a traceless anti-Hermitian matrix, linking this to the elements of su(n).
  • Some participants express confusion regarding the nature of the isomorphism and seek clarification on how it is established.

Areas of Agreement / Disagreement

Participants exhibit a mix of understanding and confusion regarding the isomorphism between su(n) and sl(n,C). While some concepts are agreed upon, such as the role of complexification, the exact nature of the isomorphism remains contested and not fully resolved.

Contextual Notes

There are unresolved questions about the specific definitions and properties that underpin the isomorphism, as well as the assumptions regarding the bases of the algebras involved.

koolmodee
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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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