Semisimple Lie Group

There are three equivalent characterization of semisimple Lie algebras. The first characterization is one which is isomorphic to a direct sum of simple Lie algebras. The second characterization is that complexification of the Lie algebra of a compact simply-connected group, for example, that sl(n;C) ≅ su(n)_C is semisimple. The third characterization is that a Lie algebra g is semisimple if and only if it has the complete reducibility property, that is, if and only if every finite-dimensional representation of g decomposes as a direct sum of irreducibles. 

Recall that a group or Lie algebra is said to have the complete reducibility property if every finite-dimensional representation of it decomposes as a direct sum of irreducible invariant subspaces. A connected compact matrix Lie group always has this property. It follows that the Lie algebra of compact simply-connected matrix Lie group also has the complete reducibility property, since there is a one-to-one correspondence between the representations of the compact group and its Lie algebra. Because there is a one-to-one correspondence between the representations of a real Lie algebra and the complex-linear representations of its complexification, we see also that if a complex Lie algebra g is isomorphic to the complexification of the Lie algebra of a compact simply-connected group, then g has the complete reducibility property. We have seen this reasoning to sl(2;C) (the complexification of the Lie algebra of SU(2) and to sl(3;C) and the complexification of the Lie algebra SU(3)).

Complex semisimple Lie algebras are complex Lie algebras that are isomorphic to the complexification of the Lie algebra of a compact simply-connected matrix Lie group.

Definition

If g is a complex Lie algebra, the an ideal in g is a complex subalgebra h of g with the property that for all X in g and H in h, we have [X, H] in h. 

A complex Lie algebra g is called indecomposable if the only ideals in g are g and {0}. A complex Lie algebra g is called simple if g is indecomposable and dim ≥ 2.

A complex Lie algebra is called reductive if it is isomorphic to a direct sum of indecomposable Lie algebras. A complex Lie algebra is called semisimple if it is isomorphic to a direct sum of simple Lie algebras. Note that a reductive Lie algebra is a direct sum of indecomposable algebras, which are either simple or one-dimensional commutative. Thus, a reductive Lie algebra is one that decomposes as a direct sum of a semisimple algebra and a commutative algebra. 

The following table lists the complex Lie algebras that are either reductive (not semisimple) or semisimple.

sl(n;C) (n≥2) semisimple
so(n;C) (n≥3) semisimple 
so(2;C) reductive 
gl(n;C) (n≥1) reductive 
sp(n;C) (n≥1) semisimple

All of the above listed semisimple algebras are actually simple, except for so(4;C), which is isomorphic to sl(2;C) ⊕ sl(2;C). Every complex simple Lie algebra is isomorphic to one of sl(n;C), so(n;C) (n≠4), sp(n;C), or to one of the five "exceptional" Lie algebras conventionally called G₂, F₄, E₆, F₇, and E₈.

For real Lie algebra,

su(n) (n≥2) semisimple 
so(n) (n≥3)  semisimple 
so(2) reductive 
sp(n) (n≥1) semisimple 
so(n,k) (n+k ≥3) semisimple 
so(1,1) reductive 
sp(n;R) (n≥1) semisimple 
sl(n;R) (n≥2) semisimple
gl(n;R) (n≥1) reductive

In each case, the complexification of the listed Lie algebra is isomorphic to one of the complex Lie algebras in the above table. Note that the Heisenberg group, the Euclidean group, and the Poincare group are neither reductive nor semisimple. 

Highest Weight

If we have a representation with a weight μ = (m₁, m₂), then by applying the root vectors X_α X₂, X₃, Y₁, Y₂, Y₃, we can get some new weights of the form μ + α, where α is the root (recall that π(H₁)π(Z_α)v = (m₁+ a₁)π(Z_α)v). If π(Z_α)v = 0, then μ + α is not necessarily a weight. In analogy to the classification of the representation sl(2;C) : In each irreducible representation of sl(2;C), π(H) is diagonalizable, and there is a largest eigenvalue of π(H). Two irreducible representations of sl(2;C) with the same largest eigenvalue are equivalent. The highest eigenvalue is always a non-negative integer, and, conversely, for every non-negative integer m, there is an irreducible representation with highest eigenvalue m. 

Let α₁ = (2, -1) and  α₂ = (-1, 2) be the roots introduced in "Weights & Roots". Let μ₁  and μ₂ be two weights. Then, μ₁is higher than μ₂ if μ₁- μ₂ can be written in the form  μ₁ - μ₂ = aα₁ + bα₂ with a ≥ 0 and b ≥ 0. If π is a representation of sl(3;C), then a weight μ₀ for π is said to be a highest weight if for all weights μ of π, μ ≤ μ₀.

Note that the relation of "higher" is only a partial ordering because μ₁ is neither higher nor lower than μ₂. For example, {0, α₁ - α₂} has no highest element. Moreover, the coefficients a and b do not have to be integers, even if both μ₁ and μ₂ have integer entries. For example, (1,0) is higher than (0,0) since (1,0) = 2/3α₁ + 1/3α₂.

Theorem of Highest Weight

The theorem of highest weight is a main theorem regarding the irreducible representation of sl(3;C).

1. Every irreducible representation π of sl(3;C) is the direct sum of its weight spaces; that is π(H₁) and  π(H₂) are simultaneously diagonalizable in every irreducible representation.

2. Every irreducible representation of sl(3;C) has a unique highest weight μ₀, and two equivalent irreducible representations have the same highest weight.

3. two irreducible representations of sl(3;C) with the same highest weight are equivalent.

4. If π is an irreducible representation of sl(3;C), then the highest weight μ₀ of π is of the form μ₀ = (m₁, m₂) with m₁ and m₂ being non-negative integers. 

An ordered pair (m₁, m₂) with m₁ and m₂ being non-negative integers is called a dominant integral element. The theorem says that the highest weight of each irreducible representation of sl(3;C) is a dominant integral element and, conversely, that every dominant integral element occurs as the highest weight of some irreducible representation. 

However, if μ has integer coefficients and is higher than zero, this does not necessarily mean that μ is dominant integral. For example, α₁ = (2, -1) is higher than zero but is not dominant integral. Note that the condition on which weights can be highest weights is m₁ and m₂ being non-negative integer.

The dimension of the irreducible representation with highest weight (m₁, m₂) is

1/2(m₁ + 1)(m₂ + 1)(m₁ + m₂ + 2).

Weyl Group in SU(3)

The representations of sl(3;C) are invariant under the adjoint action of SU(3). Let π be a finite-dimensional representation of sl(3;C) acting on a vector space V and let Π be the associated representation of SU(3) acting on the same space. For any A ∈ SU(3), we can define a new representation π_A of sl(3;C), acting on the same vector space V, by setting π_A(X) = π(AXA^-1). Since the adjoint action of A on sl(3;C) is a Lie algebra automorphism, this is a representation of sl(3;C). Π(A) is an intertwining map between (π, V) and (π_A, V). We say that adjoint action of SU(3) is a symmetry of the set of equivalence classes of representations of sl(3;C). 

The two-dimensional subspace h of sl(3;C) spanned by H₁ and H₂ is called a Cartan subalgebra. In general, the adjoint action of A  ∈ SU(3) will not preserve the space h and so the equivalence of π and π_A does not tell us anything about the weights of π. However, there are elements A in SU(3) for which Ad_A does preserve h. These elements make up the Weyl group for SU(3) and give rise to a symmetry of the set of weights of any representation π. 

Let N be the subgroup of SU(3) consisting of those A ∈ SU(3) such that Ad_A(H) is an element of h for all H in h. And let Z be the subgroup of SU(3) consisting of those  A  ∈ SU(3) such that Ad_A(H) = H for all H ∈ h. The Weyl group of SU(3), denoted by W, is the quotient group N/Z. 

The group Z consists precisely of the diagonal matrices inside SU(3), namely the matrices of the form A = (e^iθ,0,0 : 0,e^iΦ,0 : 0,0,e^-i(θ+Φ)) for θ and Φ in R. The group N consists of precisely those matrices A ∈ SU(3) such that for each k = 1,2,3, there exist l ∈ {1,2,3} and θ ∈ R such that Ae_k = e^iθe_l. Here e₁, e₂,e₃ is the standard basis for C³. The Weyl group W = N/Z is isomorphic to the permutation group on three elements. 

In order to show that Weyl group is a symmetry of the weights of any finite-dimensional representation of sl(3;C), we need to adopt a less basis-dependent view of weights. A vector v is an eigenvector of π(H₁) and π(H₂) then it is also an eigenvector for π(H) for any element H of the space h spanned by H₁ and H₂. Furthermore, the eigenvalues must depend linearly on H. For π(J) = λ₂v, then π(aH + bJ)v = (aπ(H) + bπ(J))v = (aλ₁ + bλ₂)v. We have the following basis-independent notion of weight :  a linear functional μ ∈ h^* is called a weight for π if there exists a nonzero vector v in V such that π(H)v = μ(H)v for all H in h. Such a vector v is called a weight vector with weight μ. So a weight is just a collection of simultaneously eigenvalues of all the elements H of h, which depends linearly on H and, therefore, define a linear functional on h. The reason for adopting this basis-independent approach is that the action of the Weyl group does not preserve the basis {H₁, H₂} for h.

In another words, the Weyl group is a group of linear transformation of h. This means that W acts on h, and we denote this action as w⋅H. We can define an associated action on the dual space h^*. Thus, for μ ∈ h^* and w ∈ W, we define w⋅μ to be the element of  h^* given by (w⋅μ)(H) = μ(w^-1⋅H). 

Weights & Roots

There is a one-to-one correspondence between the finite-dimensional complex representations Π of SU(3) and the finite-dimensional complex-linear representation π of sl(3;C). This correspondence is determined by the property that Π(e^X) = e^π(X) for all X ∈ su(3) ⊂ sl(3;C). The representation Π is irreducible if and only if the representation π is irreducible. 

Simultaneous Diagonalization

Suppose that V is a vector space and A is some collection of linear operators on V. Then a simultaneous eigenvector for A is a nonzero vector v ∈ V such that for all A ∈ A, there exists a constant λ_A with Av = λ_Av. The numbers λ_A are the simultaneous eigenvalues associated to v. For example, consider the space D of all diagonal nxn matrices. For each k = 1,...,n, the standard basis element e_k is a simultaneous eigenvector for D. For each diagonal matrix A, the simultaneous eigenvalue associated to e_k is the k-th diagonal entry of A.

If A is a simultaneously diagonalisable family of linear operators on a finite-dimensional vector space V, then the elements of A commute.

If A is commuting collection of linear operators on a finite-dimensional vector space V and each A ∈ A is diagonalizable, then the elements of A are simultaneously diagonalizable. 

Basis for sl(3;C)

Every finite-dimensional represetation of sl(2;C) or sl(3;C) decomposes as a direct sum of irreducible invariant subspaces. Consider the following basis for sl(3;C):

H₁ = (1,0,0 : 0,-1,0 : 0,0,0), H₁ = (0,0,0 : 0,1,0 : 0,0,-1),

X₁ = (0,0,0 : 1,0,0 : 0,0,0), X₂ = (0,0,0 : 0,0,0 : 0,1,0), X₃ = (0,0,0 : 0,0,0 : 1,0,0),

Y₁ = (0,1,0 : 0,0,0 : 0,0,0), Y₂ = (0,0,0 : 0,0,1 : 0,0,0), Y₃ = (0,0,1 : 0,0,0 : 0,0,0).

The span of {H₁, X₁, Y₁} is a subalgebra of sl(3;C) which is isomorphic to sl(2;C) (can be seen by ignoring the third row and the third column in each matrix). Similarly for  {H₂, X₂, Y₂}. Thus, the following commutation relations exists:

[X₁, Y₁] = H₁, [X₂, Y₂] = H₂,

[H₁, X₁] = 2X₁, [H₂, X₂] = 2X₂,

[H₁, Y₁] = -2Y₁, [H₂, Y₂] = -2Y₂.

Other commute relations among the basis elements which involve at least one H₁ and H₂:

[H₁, H₂] = 0;

[H₁, X₁] = 2X₁, [H₁, Y₁] = -2Y₁,

[H₂, X₁] = -X₁, [H₂, Y₁] = Y₁;

[H₁, X₂] = -X₂, [H₁, Y₂] = Y₂,

[H₂, X₂] = 2X₂, [H₂, Y₂] = -2Y₂;

[H₁, X₃] = X₃, [H₁, Y₃] = -Y₃,

[H₂, X₃] = X₃, [H₂, Y₃] = -Y₃;

Adding all of the remaining commutation relations:

[X₁, Y₁] = H₁,

[X₂, Y₂] = H₂,

[X₃, Y₃] = H₁ + H₂;

[X₁, X₂] = X₃, [Y₁, Y₂] = -Y₃,

[X₁, Y₂] = 0, [X₂, Y₁] = 0;

[X₁, X₃] = 0, [Y₁, Y₃] = 0,

[X₂, X₃] = 0, [Y₂, Y₃] = 0;

[X₂, Y₃] = Y₁, [X₃, Y₂] = X₁,

[X₁, Y₃] = -Y₂, [X₃, Y₁] = -X₂.

Weights of sl(3;C)

A strategy to classify the representation sl(3;C) is to simultaneously diagonalize π(H₁) and π(H₂). Since H₁ and H₂ commute, π(H₁) and π(H₂) will also commute and so there is at least a chance that π(H₁) and π(H₂) can be simultaneously diagonalized. 

If (π, V) is a representation of sl(3;C), then an ordered pair μ = (m₁, m₂) ∈ C² is called a weight of π if there exists v ≠ 0 in V such that π(H₁)v = m₁v, π(H₂)v = m₂v. A nonzero vector v satisfying this is called a weight vector corresponding to the weight μ. If μ = (m₁, m₂) is a weight, then the space of all vectors v satisfying π(H₁)v = m₁v, π(H₂)v = m₂v is the weight space corresponding to the weight μ. The multiplicity of a weight is the dimension of the corresponding weight space. Equivalent representations have the same weights and multiplicities. 

If π is a representation of sl(3;C), then all of the weights of π are of the form μ = (m₁, m₂) with m₁ and m₂ being integers.

Roots of sl(3;C)

An ordered pair α = (a₁, a₂) ∈ C² is called a root if 

1. a₁ and a₂ are not both zero, and

2. there exists a nonzero Z ∈ sl(3;C) such that [H₁, Z] = a₁Z, [H₂, Z] = a₁Z. The element Z is called a root vector corresponding to the root α. 

Recall that ad_X(Y) = [X, Y], e^adx = Ad(e^X), and the adjoint mapping Ad_A(X) = AXA^-1. Condition 2 of above says that Z is a simultaneous eigenvector for ad_H1 and ad_H2.  This means that Z is a weight vector for the adjoint representation and weight (a₁, a₂). By condition 1 the roots are precisely the nonzero weights of the adjoint representation. 

There are six roots of sl(3;C). They form a "root system", called A₂.

α            Z

(2, -1)    X₁

(-1, 2)    X₂

(1, 1)     X₃

(-2, 1)    Y₁

(1, -2)    Y₂

(-1, -1)   Y₃

It is convenient to single out the two roots corresponding to X₁ and X₂ and given them special names: α₁ = (2, -1), α₂ = (-1,2). α₁ and α₂ are called the positive simple roots. They have the property that all of the roots can be expressed as linear combinations of α₁ and α₂ with integer coefficients :

(2, -1) = α₁

(-1, 2) = α₂

(1, 1) = α₁ + α₂

(-2, 1) = -α₁

(1, -2) = -α₂

(-1, -1) = -α₁ - α₂ .

Let α = (a₁, a₂) be a root and Z_α a corresponding root vector in sl(3;C). Let π be the representation of sl(3;C), μ = (m₁, m₂) a weight for π, and v ≠ 0 a corresponding weight vector. Then

π(H₁)π(Z_α)v = (m₁+ a₁)π(Z_α)v,

π(H₂)π(Z_α)v = (m₂+ a₂)π(Z_α)v.

Thus, either π(Z_α)v = 0 or π(Z_α)v is a new weight vector with weight

μ + α = (m₁+ a_1, m₂+ a₂) .

Chateau Paradis Casseuil 2006

Château Paradis Casseuil gets its name from the combination of the registered name of the main parcel of vineyards called “Vines of Paradise” and Casseuil county. Taken under Domaines Barons de Rothschild (Lafite)’s wing in 1984, Château Paradis Casseuil then included 14 hectares of vines. In 1989, the estate grew by 9 hectares, and chais were included in the heart of the Sainte Foy la Longue vineyard. The Château Paradis Casseuil chais, located in the heart of the Sainte Foy la Longue vineyard are used for producing red wines. White wines are made at Château Rieussec, benefiting from the technical capacities of that great estate.

Lively and intense ruby color. Fine nose, red fruit and slight licorice aromas. The attack is supple (slight sensation of sweetness) with silky tannins.

2007
After a wet winter, the high temperatures in March and April helped give a good start to the vegetation. The following months were moderate until August. The fine weather settled early September encouraging the ripening of the grapes.

Beautiful crimson colour. A fresh nose with a touch of redcurrant and mint. The first impression on the palate is pleasant, frank with intense fruit. This wine can be appreciated now or kept for a few years when it will have reached its peak.

2006
The winter of 2005/2006 was dry and cold and the spring months were mild with little rain. August was disconcerting – cool and wet – and at the beginning of September ripeness levels were very low. However, the weather then became summery allowing the grapes to finish ripening well.

Beautiful straw yellow colour with hints of green. Discreet on the nose, but when swirled, the aromas are revealed with a fine lime bouquet. The attack is supple and the finish is fresh.

2005
The end of 2004 and the first few months of 2005 were dry. Moreover, maturation took place in perfect conditions.

Pale yellow colour. Notes of citrus fruit, mainly grapefruit, on the nose.

First impressions on the palate are full, lively and well rounded, leading to a silky finish marked by hints of hazelnuts.

2004
The year was marked by stormy weather until July, with no effects on the vines.
The beginning of the year was warmer than in 2003, but from March onwards the trend was reversed and an average drop of 2°C was noted. Rainfall was about the average for the past three vintages, with a dry June. July and August were very damp. Maturation was therefore slow but at harvest time the grapes for the dry whites were ripe.

Pale yellow colour. Very open and fresh on the nose: aromas of white flowers and violets.

Delicate first impressions on the palate. Notes of fresh fruit and Granny Smith apples.

Region : Sainte Foy la Longue, Médoc
Grape Varietals : Cabernet Sauvignon 50%, Merlot 45% and Cabernet franc 5%
Average wine production : 12 000 cases per year.

Use of SU(2) & SO(3)

Every Lie group homomorphism gives rise to a Lie algebra homomorphism. In the case of a simply-connected matrix Lie group G, a Lie algebra homomorphism also gives rise to a Lie group homomorphism. In fact, for a simply-connected matrix Lie group, there is a natural one-to-one correspondence between the representations of G and the representations of the Lie algebra g. Each of the representations π_m of su(2) was constructed from the corresponding representation Π_m of the group SU(2).

SU(2) is simply connected but SO(3) is not (SU(2) can be thought of (topologically) as the three-dimensional sphere S³ sitting inside R⁴. It is well known that S³ is simply connected). There exists a Lie group homomorphism Φ which maps SU(2) onto SO(3) and which is two-to-one. Therefore, SU(2) and SO(3) are almost isomorphic.

Consider the space V of all 2x2 complex matrices which are self-adjoint (i.e., A^* = A) and have trace zero. This is a three-dimensional real vector space with the following basis: A₁ = (0,1: 1,0) ; A₂ = (0,-i : i,0) ; A₃ = (1,0 : 0,-1) . Define the inner product on V by <A, B> = 1/2 trace(AB). {A₁, A₂, A₃} is an orthonormal basis for V. Next we are going to identify V with R³. Suppose U is an element of SU(2) and A is an element of V. Consider UAU^-1, trace(UAU^-1) = trace(A) = 0 and (UAU^-1)^* = UAU^-1. So UAU^-1is again in V. Because the map A --> UAU^-1 is linear. Therefore, we can define a linear map Φ_U of V to itself by Φ_U = UAU^-1. Given A, B ∈ V, <Φ_U(A), Φ_U(B)> = <A, B>. Thus, Φ_U is an orthogonal transformation of V.

Once we identify V with R³ using the above orthonormal basis, we may think of Φ_U as an element of O(3). Since Φ_U1U2 = Φ_U1Φ_U2, we see that Φ (the map U --> Φ_U) is a homomorphism of SU(2) into O(3). SU(2) is connected, Φ is continuous, and Φ_I is equal to I, which has determinant one. It follows that Φ must map SU(2) into the identity component of O(3), namely SO(3). However, the map Φ_U is not one-to-one, since for any U ∈ SU(2), Φ_U = Φ_-U. Actually Φ_U is a two-to-one map of SU(2) onto SO(3) (recall that every element of O(3) has determinant ± 1).

If we have the basis E₁ = 1/2(i,0 : 0,-i) ; E₂ = 1/2(0,-1: 1,0) ; E₃ = (0,i : i,0) for su(2) and the basis F₁ = (0,0,0 : 0,0,1 : 0,-1,0) ; F₂ = (0,0,-1 : 0,0,0 : 1,0,0) ; F₃ = (0,1,0 : -1,0,0 : 0,0,0) for so(3). Then, we have [E₁ , E₂] = E₃ , [E₂ , E₃] = E₁ and [E₃ , E₁] = E₂ , and similarly with the E's replaced by the F's. Thus the linear map Φ : su(2) --> so(3) which takes E_i to F_i will be a linear algebra isomorphism.

Let σ_m = π_m º Φ^-1 be the irreducible complex representations of the Lie algebra so(3) (m ≥ 0). If m is even, then there is a representation ∑_m of the group SO(3) such that ∑_m(exp X) = exp(σ_m(X)) for all X in so(3). If m is odd, then there is no such representation of SO(3).

Representation of su(2) ≅ so(3) in Physics

Representation of su(2) ≅ so(3) in Physics are labeled by the parameter l = m/2. In terms of this notation, a representation of so(3) comes from a representation of SO(3) if and only if l is an integer. The representations with l an integer are called "integer spin"; the others are called the "half-integer spin."Consider the path in SO(3) consisting of rotations by angle 2πt in the (x, y)-plane, which comes back to the identity when t = 1. However, this path is not homotopic to the constant path.

If one defines ∑_m along the constant path, then one gets the value ∑_m(I) = I, as expected. If m is odd and one defines ∑_m along the path of rotations in the (x, y)-plane, then one gets the value ∑_m(I) = -I. There is no way to define ∑_m(m odd) as a "single-valued" representations of SO(3).

An electron is a "spin-1/2" particle, which means that it is described in quantum machines in a way that involves the representation σ₁ of so(3). In the quantum machines, one finds statements to the effect that performing a 360º rotation on the wave function of the electron gives back the negative of the original wave function.This reflects that if one attempts to construct the nonexistent representation ∑₁ of SO(3), then when defining ∑₁ along a path of rotations in some plane, one gets that ∑₁(I) = -I.

A Unitary Representations of SO(3)

Consider the unit sphere S² ⊂ R³, with the usual surface measure Ω. Any R ∈ SO(3) maps S² into S² . For each R, we can define Π₂(R) acting on L²(S², dΩ) by [ Π₂(R) f](x) = f(R^-1x). Then, Π₂ is a unitary representation of SO(3). Here, L²(S², dΩ) has a very nice decomposition as the orthogonal direct sum of finite-dimensional invariant subspaces. This decomposition is the theory of "spherical harmonics" in physics.

Representations of SU(2) & su(2)

su(2) ≅ so(3) and the representation of so(3) is important in the computation of angular momentum. By studying the representation theory of su(2) we can know:

(1) how to commutation relations to determine the representations of a Lie algebra.

(2) how to determine the representations of semisimple Lie algebras, e.g., su(3).

Some Representations of SU(2)

By definition, an element of U of SU(2) is a linear transformation of C². Let z denotes the (z₁, z₂) pair in C². Then, we may define a linear transformation Π_m(U) on the space V_m by the formula [Π_m(U)f] (z) = f(U^-1z). The inverse is necessary in order to make Π_m a representation.  And V_m is the space of functions of the form f(z₁, z₂) = a₀z₁^m + a₁z₁^m-1z₂  + a₂z₁^m-2z₂²  + ... + a_mz₂^m .
 

Therefore, [Π_m(U)f] (z₁, z₂) =  ∑ a_k(U₁₁^-1z₁ + U₁₂^-1z₂)^m-k (U₂₁^-1z₁ + U₂₂^-1z₂)^k for k = 0 .. m.  Π_m(U)f is a homogeneous polynomial of degree m. Thus, Π_m(U) maps V_m into V_m. Moreover, Π_m(U₁) [Π_m(U₂)f] (z) = Π_m(U₁U₂) f(z). So Π_m is a (finite-dimensional complex) representation of SU(2).

The Lie algebra representation of Π_m can be computed as π_m(X) = d/dt Π_m(e^tX) |_t=0. So (π_m(X)f) (z) = d/dt f(e^-tXz) |_t=0. Let z(t) be the curve e^-tXz. We have z(0) = z and (dz/dt) |_t=0 = -Xz. Since z(t) can also be written as z(t) = (z₁(t), z₂(t)), with z_i(t) ∈ C. By chain rule, π_m(X)f = ∂f / ∂z₁(dz₁/dt) |_t=0 + ∂f / ∂z₂(dz₂/dt) |_t=0 . We have

π_m(X)f = -∂f / ∂z₁(X₁₁z₁ + X₁₂z₂)  - ∂f / ∂z₂(X₂₁z₁ + X₂₂z₂). 

Because every finite-dimensional complex representation of the Lie algebra su(2) extends uniquely to a complex-linear representation of the complexification of su(2). And the complexification of su(2) is sl(2;C). Therefore, the representation π_m of su(2) given by above extends to a representation of sl(2;C). 

Consider H = (1,0 : 0,-1)

(π_m(H)f) (z) = -(∂f / ∂z₁)z₁ + (∂f / ∂z₂)z₂ .

 π_m(H) = -z₁(∂f / ∂z) + z₂(∂f / ∂z₂). Apply this to a basis element z₁^kz₂^m-k , we have

π_m(H) z₁^kz₂^m-k = -k z₁^kz₂^m-k  + (m-k)z₁^kz₂^m-k  = (m-2k) z₁^kz₂^m-k .

Thus, z₁^kz₂^m-k is an eigenvector for π_m(H) with eigenvalue (m-2k). In particular, π_m(H) is diagonalizable. 

Let X and Y be the elements, X = (0,0 : 1,0) , Y = (0,1 : 0,0) in sl(2;C). We have π_m(X) = -z₂(∂f / ∂z₁) and π_m(X) = -z₁(∂f / ∂z₂). Apply these to the basis element 

π_m(X) z₁^kz₂^m-k = -k z₁^k-1z₂^m-k+1  

π_m(Y) z₁^kz₂^m-k = -(m-k) z₁^k+1z₂^m-k-1  

It suffices to show that every nonzero invariant subspace of V_m is equal to V_m. Let W be such a space. Since W is assumed nonzero, there is at least one nonzero element w in W. Then w can be written uniquely in the form 

w =  a₀z₁^m + a₁z₁^m-1z₂  + a₂z₁^m-2z₂²  + ... + a_mz₂^m

with at least one of the a_k's nonzero. Let k₀ be the smallest value of k for which  a_k ≠ 0 and consider π_m(X)^m-k0 w. Since π_m(X) lowers the power of z₁ by 1, it will kill all the terms in w except a_k0 z₁^m-k0 z₂^k0 . So we have  π_m(X)^m-k0 z₁^m-k0 z₂^k0 = (-1)^m-k0(m-k₀)! z₂^m . Since W is assumed invariant, W must contain z₂^m . Furthermore, π_m(Y)^k z₂^m is a nonzero multiple of z₁^kz₂^m-k for all 0 ≤ k ≤ m. Because these elements form a basis for V_m . In fact W = V_m. Therefore, representation π_m is an irreducible representation of sl(2;C). 

Irreducible Representations of su(2)

Every finite-dimensional complex representation π of su(2) extends to a complex-linear representation of the complexification of su(2), namely sl(2;C). Studying the irreducible representations of su(2) is equivalent to studying the irreducible representation of sl(2;C). Passing to the complexified Lie algebra makes computations easier, in that there is a nice basis for sl(2;C) that has no counterpart among the bases of su(2).

We can use commutation relations to determine the representation of a Lie algebra. Consider the following basis for sl(2;C) and commutation relations:

H = (1,0 : 0,-1);  X = (0,0 : 1,0);  Y = (0,1 : 0,0)

[H, X] = 2X,  [H, Y] = -2Y,  [X, Y] = H.

If V is a (finite-dimensional complex) vector space and A, B, and C are operators on V satisfying

[A, B] = 2B,  [A, C] = -2C,  [B, C] = A, then

because of the skew symmetry and bilinearity of brackets, the linear map π : sl(2;C) --> gl(V) satisfying π(H) = A, π(X) = B, π(Y) = C will be a representation of sl(2;C).

We call π(X) the "raising operator", because it has the effect of raising the eigenvalue of π(H) by 2, and call π(Y) the "lowering operator". Since [π(H), π(X)] = π([H, X]) = 2π(X). Let u be an eigenvector of π(H) with eigenvalue α ∈ C. Thus,

π(H)π(X)u = π(X)π(H)u + 2π(X)u

= π(X)(αu) + 2π(X)u = (α + 2)π(X)u.

Either π(X)u = 0 or π(X)u is an eigenvector for π(H) with eigenvalue α+2. More general, π(H)π(X)ⁿu = (α + 2n)π(X)ⁿu. Similarly, for [π(H), π(Y)] = -2π(Y), we have π(H)π(Y)u = (α - 2)π(Y)u.

An operator on a finite-dimensional space can have only finitely many distinct eigenvalues. Therefore, there is some N ≥ 0 such that π(X)^N+1u = 0.

Define u₀ = π(X)^Nu and λ = α + 2N. Then,

π(H)u₀ = λu₀, π(X)u₀ = 0. 

Define u_k = π(Y)^ku₀ , for k ≥ 0. Thus, we have π(H)u_k = (λ - 2k)u_k . Since π(H) can have only finitely many eigenvalues, the u_k's cannot be all be nonzero.

For k = 1,

π(H)u₁ = π(H)π(Y)u₀ = (α - 2)π(Y)u₀ = (α - 2)u₁.

π(X)u₁ = π(X)π(Y)u₀ = (π(Y)π(X) + π(H))u₀ = π(H)u₀ = λu₀ (as π(X)u₀ = 0)

π(Y)u₁ = π(Y)π(Y)u₀  = π(Y)²u₀ = u₂  

If π(X)u_k = [kλ - k(k-1)]u_k-1. By induction, π(X)u_k+1= π(X)π(Y)u_k

= (π(Y)π(X) + π(H))u_k = π(Y)π(X)u_k + (λ - 2k)u_k

= π(Y)[kλ - k(k-1)]u_k-1 + (λ - 2k)u_k  = [kλ - k(k-1) + (λ - 2k)]u_k

= [(k+1)λ - (k+1)k]u_k  

Because π(H) can have only finitely many eigenvalues, the u_k's cannot all be nonzero. For all k ≤ m, u_m+1 = π(Y)^m+1u₀  = 0. If u_m+1 = 0. Then π(X)u_m+1  = (m+1)(λ - m)u_m  = 0. Since m ≠ 0 and m + 1 ≠ 0. So we have λ = m, where m is a non-negative integer. 

In summary, given a finite-dimensional irreducible representation π of sl(2;C) acting on a space V and putting λ = m, there exists an integer m ≥ 0 and nonzero vectors u₀, ..., u_m such that

π(H)u_k = (m - 2k)u_k, 

π(Y)u_k = u_k+1 (k < m),

π(Y)u_m = 0, 

π(X)u_k = [km - k(k-1)]u_k-1 (k > 0),

π(X)u₀ = 0.

The vectors u₀, ..., u_m must be linearly independent, since they are eigenvectors of π(H) with distinct eigenvalues. Moreover, the (m+1)-dimensional span u₀, ..., u_m is explicitly invariant under π(H), π(X), and π(Y). Hence under π(Z) for all Z ∈ sl(2;C). Since π is irreducible, this space must be all of V. 

The (m+1)-dimensional representation Π_m described above must be equivalent to π. This can be seen explicitly by introducing the following basis for V_m :

u_k = [π_m(Y)]^k (z₂)^m = (z₂)^m (m! / (m-k)!) z₁^kz₂^m-k  (k ≤ m).

In other words, π_m have a basis of the form π(H), π(X), and π(Y). π's have the right commutation relations to forma representation of sl(2;C) and that this representation is irreducible.