What are weight spaces and how do they relate to the representation of SU(2)?

[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.

 

[George Jones]

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.

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?

 

[jdstokes]

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).

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?

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?

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.

 

[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?

 

[dextercioby]

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

 

[George Jones]

I haven't forgotten about this thread, but I've been too busy to answer in the detail that I wanted. I will, however, add a little to what bigubau posted.

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.

 

[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.


I had another question which I was hoping you could answer.

I'm reading about roots and weights using the lecture notes

http://www.math.columbia.edu/~woit/notes7.pdf

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.
The weight space is defined as the subspace of the representation space which transform under the action of T according to the given weight. I'm also not clear as to why the nontrivial weight spaces are necessarily 2-dimensional in the case of the adjoint representation.