The irreducible representations of of su(2): Highest weight method

[Hydaspex]

TL;DR Summary

I need to understand this passage from "An Elementary Introduction to Groups and Representations" Brian C. Hall

Hi all I need to understand the following passage from Hall link page 78 :

Some notation first:

Basis for $sl(2;C)$:

$H=\begin{pmatrix} 1&0\\0&−1\end{pmatrix} ;X=\begin{pmatrix} 0&1\\0&0\end{pmatrix} ;Y=\begin{pmatrix} 0&0\\1&0\end{pmatrix}$

which have the commutation relations

$[H,X] = 2X ~ ~, [H,Y] =−2Y ~ , [X,Y] =H$$π(X)$ acts as the raising operator such that:

$π(H)π(X)u= (α+ 2)π(X)u$

$π(Y)$ acts as the lowering operator such that:

$π(H)π(Y)u= (α−2)π(Y)u$

There is some N≥0 such that $π(X)^Nu \neq 0$

but $π(X)^{N+1}u= 0$

We define $u_0=π(X)^Nu$ then

$(H)u_0=λu_0$

$π(X)u_0= 0$
Now, by definition

$u_{k+1}=π(Y)u_k$

Using $π(H)u_k= (λ−2k)u_k$ and induction we have$π(X)u_{k+1}=π(X)π(Y)u_{k} \\= (π(Y)π(X) +π(H))u_k \\=π(Y) [kλ−k(k−1)]u_{k−1}+ (λ−2k)u_k \\= [kλ−k(k−1) + (λ−2k)]u_k$I don't understand how to get $kλ−k(k−1)]$ at the third passage and why $(λ−2k)u_k$ should be zero to get

$π(X)u_{k+1}= [kλ−k(k−1)]u_k$.

 

[fresh_42]

I have difficulties to understand how the representation is finally defined. There seem to be some mistakes in what you have written. E.g. the lines where you explain the action of $\pi(X)$ is actually the description of $\pi(H)$, which leaves $\pi(X).u$ invariant. And I do not see where $k^2$ comes from, and $(\lambda - 2k)u_k$ isn't zero, why should it be?

I hope I gave a better description of the theorem here:
https://www.physicsforums.com/insights/lie-algebras-a-walkthrough-the-representations/

 

[Hydaspex]

please read page 77 of the pdf at the link...the full proof is laid out... the theorem is clear and I don't see mistakes yet that passage to prove the lemma is obscure. Wikipedia refers to that book...https://en.wikipedia.org/wiki/Representation_theory_of_SU(2) ...
"Weights and the structure of the representation"

 

[fresh_42]

So the key is Lemma 5.11. I will omit the $\pi$ and write shortly $\pi(X)(v)=X.v$.

As always, let us list what we have:
$u_0:=X^N.u$ such that $H.u_0=\lambda u_0=(2N+\alpha)u_0$ and $X.u_0=0\,.$
$u_k:=Y^k.u_0$ and thus $H.u_k=(\lambda-2k)u_k\,.$

The only equation which is used repeatedly is $A.B.v=[A,B].v+B.A.v$ where $[A,B]=C$ is known, so $A.B.v=C.v+B.A.v$ This is the general procedure. In Lemma 5.11. we want to know, what $X.u_k$ is.

$X.u_0 = 0$ per definition of $u_0\,.$
$X.u_1=X.Y^1.u_0= [X,Y].u_0+Y.X.u_0= H.u_0+Y.0=\lambda u_0 \,.$
Now per induction we have
\begin{align*}
X.u_{k+1}&=X.Y^{k+1}.u_0 =X.Y.(Y^k.u_0)=[X,Y].(Y^k.u_0) +Y.X.(Y^k.u_0)\\
&=H.Y^k.u_0+Y.(X.Y^k.u_0)=H.u_k+Y.(X.u_k)\\&=(\lambda-2k)u_k+[k\lambda-k(k-1)]Y.u_{k-1}\\
&=\left(\lambda - 2k + k \cdot \lambda - k^2 +k\right)u_k\\&=\left[(k+1)\lambda - k^2 - k\right]u_k\\
&=\left[(k+1)\lambda - (k+1)k\right]u_k
\end{align*}

 

[Hydaspex]

It was so obvious...Thank you for your help and I will read your insight.