What Is the Adjoint Representation of SU(2)?

[nigelscott]

Homework Statement

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I am looking at this document. http://www.math.columbia.edu/~woit/notes3.pdf

Homework Equations

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ad(x)y = [x,y]

Ad(X) = gXg^-1

The Attempt at a Solution

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I understand how ad(S₁) and X is found but I don't understand what g and g^-1 to use to find Ad(X). Also I need to show that ad and Ad are connected via matrix exponentiation.

 

[fresh_42]

To be honest, I don't really want to read four pages to answer a possibly simple question, the more as I've already written two insight articles, which deal with $SU(2)$ representations. I worked out quite a few explicit formulas for expressions in certain bases. So maybe you'll find there what you are looking for (you can skip the first two sections).

https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/
https://www.physicsforums.com/insights/journey-manifold-su2-part-ii/

 

[nigelscott]

Thanks. I worked through your paper and understand how you got to part 2 (12) and (13). The part 1 am stuck on now is solving Ad(expX) = exp(ad(X)) for the matrices you have calculated. When I try to calculate the matrix exponential I get strange results. Any pointers you could give would be of great help. Thanks again.

 

[fresh_42]

When I try to calculate the matrix exponential I get strange results. Any pointers you could give would be of great help.

It's not really funny to calculate the exponentials of $(3\times 3)-$matrices and easy to make mistakes. I guess, that's why I haven't really seen it anywhere. I would start with $(2\times 2)-$matrices and the Lie group of matrices of the form $\begin{bmatrix}a&b\\0&0\end{bmatrix}$ with $a \neq 0$ and the Lie algebra $[X,Y]=Y$ which is easier than with simple Lie groups which have entries on both sides of the diagonal. It's also the reason why I used the differentiation of curves in the Lie group to show the connection to its Lie algebra. Exponentiation is basically the way back, an integration.