Editing Representation theory of SU(2) (section)

===Weights and the structure of the representation===
In this setting, the eigenvalues for <math>H</math> are referred to as the '''weights''' of the representation. The following elementary result<ref>{{harvnb|Hall|2015}} Lemma 4.33</ref> is a key step in the analysis. Suppose that <math>v</math> is an [[eigenvector]] for <math>H</math> with eigenvalue <math>\alpha</math>; that is, that <math>H v = \alpha v.</math> Then
:<math>\begin{alignat}{5}
H (X v) &= (X H + [H,X]) v &&= (\alpha + 2) X v,\\[3pt]
H (Y v) &= (Y H + [H,Y]) v &&= (\alpha - 2) Y v.
\end{alignat}</math>

In other words, <math>Xv</math> is either the zero vector or an eigenvector for <math>H</math> with eigenvalue <math>\alpha + 2</math> and <math>Y v</math> is either zero or an eigenvector for <math>H</math> with eigenvalue <math>\alpha - 2.</math> Thus, the operator <math>X</math> acts as a '''raising operator''', increasing the weight by 2, while <math>Y</math> acts as a '''lowering operator'''.

Suppose now that <math>V</math> is an irreducible, finite-dimensional representation of the [[complexification|complexified]] Lie algebra. Then <math>H</math> can have only finitely many eigenvalues. In particular, there must be some final eigenvalue <math>\lambda \in \mathbb{C}</math> with the property that <math>\lambda + 2</math> is ''not'' an eigenvalue. Let <math>v_0</math> be an eigenvector for <math>H</math> with that eigenvalue <math>\lambda:</math> 

:<math>H v_0 = \lambda v_0,</math>

then we must have

:<math>X v_0 = 0,</math>

or else the above identity would tell us that <math>X v_0</math> is an eigenvector with eigenvalue <math>\lambda + 2  .</math>

Now define a "chain" of vectors <math>v_0, v_1, \ldots</math> by
:<math>v_k = Y^k  v_0</math>.

A simple argument by [[mathematical induction|induction]]<ref>{{harvnb|Hall|2015}}, Equation (4.15)</ref> then shows that
:<math>X  v_k = k(\lambda - (k - 1))v_{k-1}</math> 
for all <math>k = 1, 2, \ldots .</math> Now, if <math> v_k </math> is not the zero vector, it is an eigenvector for <math>H</math> with eigenvalue <math> \lambda - 2k .</math> Since, again, <math> H </math> has only finitely many eigenvectors, we conclude that <math> v_\ell </math> must be zero for some <math> \ell </math> (and then <math>v_k = 0</math> for all <math> k > \ell </math>).

Let <math>v_m</math> be the last nonzero vector in the chain; that is, <math> v_m \neq 0 </math> but <math> v_{m+1} = 0 .</math> Then of course <math> X  v_{m+1} = 0 </math> and by the above identity with <math>k = m + 1 ,</math> we have
:<math> 0 = X  v_{m+1} = (m + 1)(\lambda - m)v_m .</math>

Since <math> m + 1 </math> is at least one and <math> v_m \neq 0  ,</math> we conclude that <math> \lambda </math> ''must be equal to the non-negative integer'' <math> m  .</math>

We thus obtain a chain of <math> m + 1 </math> vectors, <math> v_0, v_1, \ldots, v_m  ,</math> such that <math> Y </math> acts as
:<math> Y  v_m = 0, \quad Y  v_k = v_{k+1} \quad (k < m) </math>
and <math> X </math> acts as
:<math> X  v_0 = 0, \quad X  v_k = k  (m - (k - 1))  v_{k-1} \quad (k \ge 1)</math>
and <math> H </math> acts as
:<math>H  v_k = (m - 2k)  v_k .</math>
(We have replaced <math>\lambda</math> with its currently known value of <math> m </math> in the formulas above.)

Since the vectors <math> v_k </math> are eigenvectors for <math>H</math> with distinct eigenvalues, they must be linearly independent. Furthermore, the span of <math> v_0, \ldots , v_m </math> is clearly invariant under the action of the complexified Lie algebra. Since <math>V</math> is assumed irreducible, this span must be all of <math> V .</math> We thus obtain a complete description of what an irreducible representation must look like; that is, a basis for the space and a complete description of how the generators of the Lie algebra act. Conversely, for any <math> m \geq 0 </math> we can construct a representation by simply using the above formulas and checking that the commutation relations hold. This representation can then be shown to be irreducible.<ref>{{harvnb|Hall|2015}}, proof of Proposition&nbsp;4.11</ref>

'''Conclusion''': For each non-negative integer <math> m ,</math> there is a unique irreducible representation with highest weight <math> m  .</math> Each irreducible representation is equivalent to one of these. The representation with highest weight <math> m </math> has dimension <math> m + 1 </math> with weights <math> m, m - 2, \ldots, -(m - 2), -m  ,</math> each having multiplicity one.