Question 1: Using only the language of high school linear algebra, how does one find the generators of SU(2)?

Question

My final objective is to try to understand the business of roots and weights in group theory.

I want to focus on SU(2). I'm doing this because abstractions are a bit hard for me right this minute.

I'm going to think of SU(2) very simply as the group of 2 x 2 matrices that are unitary and have determinant 1. This is to keep things as simple as possible.

The Lie Algebra is generated by

$\begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}$

$\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$

$\begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}$

I'm not clear on how these generators come about.

This could be good to know.

One can then deduce a root from this somehow, but I am getting ahead of myself. I am not yet sure how this all works.

I'm inclined to believe it should be simple.

Using only the language of high school linear algebra, how does one find the generators of SU(2)?

** The below is just some stuff I have tried, very likely wrong but might be worth commenting on at least it will guide me a but.

I have now performed some numerical experiments on this, can someone let me know if I am anywhere close?

Someone copy modify and try this code and let me know.. . .

(* Step 1: Define g(t) as a 2x2 matrix *)

g[t_] := {{a[t], b[t]}, {-Conjugate[b[t]], Conjugate[a[t]]}}

(* Step 2: Compute the anti-Hermitian matrix X *)

X = -I * MatrixLog[g[t]]

(* Step 3: Diagonalize X to obtain its eigenvectors *)

eig = Eigensystem[X];
eigV = eig[[2]];

(* Step 4: Construct the generators of su(2) *)

(* Step 4a: Construct the first generator T1 *)
T1 = (1/2) * (eigV[[1]] * ConjugateTranspose[eigV[[1]]] - eigV[[2]] * 
ConjugateTranspose[eigV[[2]]]);

(* Step 4b: Construct the second generator T2 *)
 Y = I * X; (* Define Y = iX *)
T2 = (1/(2*I)) * (eigV[[1]] * ConjugateTranspose[eigV[[2]]] - eigV[[2]] * 
ConjugateTranspose[eigV[[1]]]);

(* Step 4c: Construct the third generator T3 *)
T3 = (1/2) * (eigV[[1]] * ConjugateTranspose[eigV[[1]]] + eigV[[2]] * 
ConjugateTranspose[eigV[[2]]]) - (1/2) * IdentityMatrix[2];

(* Step 5: Verify that the generators satisfy the commutation relations *)

(* Compute the commutators of the generators *)
Comm[T1, T2]
Comm[T2, T3]
Comm[T3, T1]

(* Step 6: Verify that the generators form a basis for the Lie algebra *)

(* Compute the determinant of the matrix formed by the generators *)
Det[Transpose[{Flatten[T1], Flatten[T2], Flatten[T3]}]]

(* The determinant should be non-zero for the generators to form a basis *)

Below is some other embryonic Mathematica experimental code where I try to delineate between SU(2) and su(2)

Answer 1

Someone, please edit this.

My previous answer was not even adequate. This is my second attempt.

Here goes.

SU(2) the group and su(2) the algebra are different.

$G \in SU(2)$ satisfies

\begin{pmatrix} \alpha & - \bar{\beta} \\ \beta & \bar{\alpha} \end{pmatrix}

such that $\alpha^2 + \beta^2 = 1$

If the group elements in SU(2) are parameterized,

There is such a thing as Lie group- Lie algebra correspondence

The correspondence says there is a relationship between the tangent space of the Lie group at the identity element and the lie algebra of the group.

It would then seem that for some parameterization, oh God help me

\begin{pmatrix} \cos{\frac{\theta}{2}} & - e^{i \phi}\sin{\frac{\theta}{2}} \\ e^{-i \phi}sin{\frac{\theta}{2}} & \cos{\frac{\theta}{2}} \end{pmatrix}

So

$X = i \frac{d}{d \theta} G|_{\theta -> 0, \phi -> 0}$

$Y = i \frac{d}{d \theta} G|_{\phi -> 0, \phi -> 0}$

$Z = \frac{1}{2 i}(X.Y - Y.X)$

I want to also believe that there is a correspondence between $G \in SU(2)$ and $g \in su(2)$

$g = Log(G)/i$

$G = Exp(t g)$

t is a parameter.