# Representations of SU(2)

1. Jan 27, 2008

### jdstokes

I'm taking a course on Lie groups and am reading alongisde Cahn's semi-simple lie algebras and their representations.

On page 4 he starts to construct a representation T of the Lie group corresponding to SU(2) acting on a linear space V, by defining the action of $T_z$ and $T_+$ on a vector $v_j$ by

$T_z v_j = jv_j, \quad T_+ v_j = 0$

and then constructs a $(2j+1)$-dimensional representation.

I don't understand what allows him to assume that there exist vectors in V with this property.

Any help would be appreciated.

2. Jan 28, 2008

### George Jones

Staff Emeritus
First, Cahn has complexified the real Lie algebra $su\left(2\right)$, i.e., he looks for representations of $\mathbb{C} \otimes su\left(2\right)$. If this is not done, the definitions of $t_+$ and $t_-$ make no sense. Physicists usually do this without explicitly saying so. Since $sl\left(2 , \mathbb{C}\right)$ and $\mathbb{C} \otimes su\left(2\right)$ are isomorphic as complex Lie algebras, physics books' treatments of what they call $su\left(2\right)$ and $so\left(3\right)$ look like math books' treatments of $sl\left(2,\mathbb{C}\right)$.

Consider a representation of $su\left(2\right)$ on an n-dimensional complex vector space $V$. $T_z$ is a non-zero linear operator on $V$, so its eigenvalue equation is a complex $n^{th}$ degree polynomial that has at most $n$ distinct roots. Consequently, $T_z$ has at most $n$ distinct eigenvalues and at least one eigenvalue, with corresponding non-zero eigenvectors. Therefore, the set of eigenvalues has a member with maximal value $j$, say. Call the corresponding eigenvector $v_j$, so $T_zv_j = jv_j$.

From this its follows that $T_+v_j = 0$. Why?

Last edited: Jan 28, 2008
3. Jan 28, 2008

### jdstokes

Hi George,

Thanks for replying. When I wrote SU(2) I actually meant SO(3). In this case do we still need to complexify the Lie algebra so(3)?

In any case I don't understand why this complexification is necessary. Could you please explain that to me?

Ok, but $j$ can be any complex number here right? Not just integers or half-integers. I don't have time to check why $T_+v_j = 0$ right now but I'm guessing it follows from the commutation relations.

4. Jan 28, 2008

### jdstokes

The commutation relation $[T_z,T_+] = T_+$ implies that $T_+v_j$ is an eigenvector of $T_z$ with eigenvalues j + 1. But since j is the largest eigenvalues of T_z, this implies $T_+v_j$ is the zero vector.

I'm not too happy with this line of reasoning, however, since the eigenvalues can be complex numbers, and how exactly do we define larger than in this case?

5. Feb 2, 2008

### dextercioby

$T_{z}$ is self adjoint, its spectrum is real.

6. Feb 2, 2008

### George Jones

Staff Emeritus

A unitary representation of the Lie group SO(3) on a complex vector space V gives rise to a skew-Hermitian representation of the real Lie algebra so(3) on V. Consequently, i times a representative of so(3) is a self-adjoint operator on V.

More later.

7. Feb 2, 2008

### jdstokes

That makes sense since the matrix logarithm of a unitary matrix is skew-Hermitian so if $iT_z$ is skew-Hermtian then $T_z$ is certainly Hermitian, hence real eigenvalues.

Thanks for your help George and bigubau.

What exactly is meant by the statement: the weight space corresponding to the weight $\alpha$ will be the sum of the one-dimensional subspaces of the representation space that are irreducible representations of T, with weight $\alpha$''?
The weights are defined to be the irreps of the maximal torus T. These can be thought of as linear functionals in $\mathfrak{t}^\ast$ giving integers on the integer lattice.