8 SL3

We now discuss the finite-dimensional complex-linear representation theory of 𝔰⁢𝔩3,ℂ. This then tells us everything about the holomorphic representations of SL3⁡(ℂ) and, via the machinery of real forms, tells us about representations of SU⁢(3) and SL3⁡(ℝ).

Much of what we say, particularly at first, will generalise in a fairly obvious way to 𝔰⁢𝔩n,ℂ.

In this section, all representations of 𝔰⁢𝔩3,ℂ will be assumed to be complex-linear.

8.1 The Lie algebra sl3,C

We study the Lie algebra

𝔤=𝔰⁢𝔩3,ℂ={X∈𝔤⁢𝔩3,ℂ:tr⁡(X)=0}

of traceless 3×3 matrices. It has dimension 8. We first need to find the analog of the standard basis H,X,Y of 𝔰⁢𝔩2,ℂ

First, notation: we write Ei⁢j for the matrix with a ’1’ in row i and column j , and ’0’ elsewhere. Then Ei⁢j∈𝔰⁢𝔩3,ℂ if and only if i≠j.

The analogue of H will be the entire subalgebra of diagonal matrices.

Definition 8.1.

The (standard) Cartan subalgebra of 𝔤 is 𝔥, given by

𝔥={(h1h2h3):h1+h2+h3=0,}.

Note that 𝔥 is an abelian subalgebra, because diagonal matrices commute with each other.

We pick as a basis of 𝔥 the elements

H12=E11−E22 =(1−10)
and
H23=E22−E33 =(01−1),

and also define H13=E11−E33=H12+H23.

Next we consider the adjoint action of 𝔥 on 𝔤, seeking eigenvectors and eigenvalues. The key calculation is:

[(h1h2h3),Ei⁢j]=(hi−hj)⁢Ei⁢j.
Exercise 8.2.

Check this!

Thus {Ei⁢j:i≠j}∪{H12,H23} is a basis of simultaneous eigenvectors in 𝔰⁢𝔩3,ℂ for the adjoint action of 𝔥.

8.2 Weights

Suppose that (ρ,V) is a finite-dimensional complex-linear representation of 𝔰⁢𝔩3,ℂ. Suppose that v∈V is a simultaneous eigenvector for all ρ⁢(H), H∈𝔥. Then, for each H∈𝔥, there is an α⁢(H)∈ℂ such that ρ⁢(H)⁢v=α⁢(H)⁢v. Since ρ is complex-linear, α is a complex-linear map 𝔥→ℂ. In other words, α is an element of the dual space 𝔥∗. This motivates the following definition:

Definition 8.3.

Suppose that (ρ,V) is a representation of 𝔰⁢𝔩3,ℂ. Then a weight vector in V is a nonzero vector v∈V such that there is α∈𝔥∗ (the weight) with:

ρ⁢(H)⁢v=α⁢(H)⁢v

for all H∈𝔥.

The weight space of α is

Vα={v∈V:ρ⁢(H)⁢v=α⁢(H)⁢v⁢ for all H∈𝔥}.

On 𝔥 we have some ‘obvious’ functionals Li∈𝔥∗ given by

Li⁢(h1h2h3)=hi.

These span 𝔥∗, subject to the relation66 6 More precisely, 𝔥∗ is isomorphic to the quotient of the three dimensional vector space with basis {L1,L2,L3} by the subspace spanned by L1+L2+L3.

L1+L2+L3=0.

We compute the weights of some particular representations.

Example 8.4.

If V=ℂ3 is the standard representation of 𝔰⁢𝔩3,ℂ (for which ρ⁢(A)=A for all A∈𝔰⁢𝔩3,ℂ), then the standard basis vectors e1,e2,e3 are all weight vectors:

(h1h2h3)⁢ei=hi⁢ei

from which we see that H⁢ei=Li⁢(H)⁢ei for all H∈𝔥. See table 5.

Weight L1 L2 L3
Weight vector e1 e2 e3
Table 5: Weights of the standard representation
Example 8.5.

If V=(ℂ3)∗ is the dual of the standard representation then it has a basis e1∗,e2∗,e3∗ defined by

ei∗⁢(ej)=δi⁢j.

One can show that ei∗ is a weight vector of weight −Li, so the weights are −L1,−L2,−L3. See problem 85.

Example 8.6.

Let V=Sym2⁡(ℂ3) be the symmetric square of the standard representation. The rules for calculating the weights of V are the same as for 𝔰⁢𝔩2,ℂ — so, for the symmetric square, we add all unordered pairs of weights of ℂ3. For details see section 8.5 The weights of ℂ3 are {L1,L2,L3} and so the weights of Sym2⁡(ℂ3) are

{2⁢L1,2⁢L2,2⁢L3,L1+L2,L2+L3,L1+L3}.

Note that, if we wanted, we could also write Li+L2=−L3 etc.

Example 8.7.

Let V=𝔤 with the adjoint representation defined by ρ⁢(X)⁢Y=[X,Y]. As already observed, we have

[H,Ei⁢j]=Li−Lj

for H∈𝔥 and i≠j, while [H,H′]=0 for H,H′∈𝔥. Thus the weights of the adjoint representation are Li−Lj (i≠j) and 0. The weight space for 0 is 𝔥, which has dimension two with basis H12 and H23; we say that the weight 0 has multiplicity two in V. We obtain table 6.

WeightWeight space basis\ldelim⁢.20⁢m⁢m⁢\ldelim⁢.30⁢m⁢m⁢L1−L2E12\rdelim}3∗Positive roots\rdelim}2∗Simple rootsL2−L3E23L1−L3E13\ldelim⁢.30⁢m⁢m⁢L2−L1E21\rdelim}3∗Negative rootsL3−L2E32L3−L1E310H12,H13
Table 6: Weights of the adjoint representation
Definition 8.8.

A root of 𝔤=𝔰⁢𝔩3,ℂ is a nonzero weight of the adjoint representation. A root vector is a weight vector of a root, and a root space is the weight space of a root.

In other words, a root α with root vector 0≠E∈𝔤 is a nonzero element α∈𝔥∗ such that

[H,E]=α⁢(H)⁢E.

We write

Φ={±(L1−L2),±(L2−L3),±(L1−L3)}

for the set of roots of 𝔰⁢𝔩3,ℂ. Out of these, we call Φ+={L1−L2,L2−L3,L1−L3} the positive roots and Φ−={L2−L1,L3−L2,L3−L1} the negative roots. We write Δ={L1−L2,L2−L3}; these are the simple roots. Note that L1−L3 is the sum of the two simple roots. We will sometimes write αi⁢j for the root Li−Lj.

Finally, we have the root space or Cartan decomposition

𝔤=𝔥⊕⨁α∈Φ𝔤α,

where the 𝔤α are the root spaces, which are all one-dimensional.

Exercise 8.9.

Work through all the above theory in the case of 𝔰⁢𝔩2. What are the roots and root spaces? What is the relation between the weights (as linear functionals on 𝔥) and between the weights defined in section 7?

8.3 Visualising weights

Shortly we will prove that, if (ρ,V) is a finite dimensional representation of 𝔰⁢𝔩3,ℂ, then its weights are integer linear combinations of the Li. In other words, they lie in the weight lattice

ΛW={a1⁢L1+a2⁢L2+a3⁢L3:a1,a2,a3∈ℤ}.
Exercise 8.10.

Show that a1⁢L1+a2⁢L2+a3⁢L3∈ΛW if and only if a1−a2,a2−a3∈ℤ. Must the ai be integers?

We want to visualise this in a way that treats L1,L2,L3 symmetrically. Noticing that they sum to zero, we regard them them as the position vectors of the vertices an equilateral triangle with unit side length, centred on the origin. The weight lattice ΛW is then the set of vertices of equilateral triangles tiling the plane. For any representation (ρ,V), its weight diagram is then obtained by circling the weights that occur in that representation.

Example 8.11.

We draw the weight diagram for the standard representation, in Figure 7.

Refer to caption
Figure 7: Weights for ℂ3.
Example 8.12.

We draw the weight diagram for the adjoint representation in Figure 8. Note that in this case the dimension of the weight space for the weight 0 is two. We say the weight has multiplicity two, and indicate this on the weight diagram by circling the weight twice. If the multiplicity was much higher, we would need another method (like writing the multiplicity next to the circle as a number).

Refer to caption
Figure 8: Weights for the adjoint representation.

8.4 Representations and weights

Firstly, we recall from above that any finite-dimensional representation of 𝔰⁢𝔩3,ℂ is completely reducible (Theorem 7.27).

Theorem 8.13.

Let (ρ,V) be a finite-dimensional representation of 𝔰⁢𝔩3,ℂ. Then:

  • •

    There is a basis of V consisting of weight vectors.

  • •

    Every weight of V is in the weight lattice ΛW.

We may combine these two into the single equality

V=⨁α∈ΛWVα.
Proof.

Consider the embedding ι12:𝔰⁢𝔩2,ℂ→𝔰⁢𝔩3,ℂ embedding a 2×2 matrix into the ‘top left’ of a 3×3 matrix:

ι12:(abc−a)↦(ab0c−a0000).

This is a Lie algebra homomorphism, and ρ∘ι12 is a representation of 𝔰⁢𝔩2,ℂ. We know from the 𝔰⁢𝔩2,ℂ theory that

(ρ∘ι12)⁢(H)=ρ⁢(1−10)=ρ⁢(H12)

is diagonalizable with integer eigenvalues. If v∈V is a weight vector with weight a1⁢L1+a2⁢L2+a3⁢L3, then it is an eigenvector for ρ⁢(ι12⁢(H)) with eigenvalue a1−a2. Thus a1−a2 is an integer.

Now, there is another embedding ι23 putting a 2×2 matrix in the ‘bottom right’ corner. The same argument then shows that ρ⁢(H13) is diagonalisable with integer eigenvalues, which shows that a2−a3 is an integer for every weight.

Thus every weight is in the weight lattice. Moreover, ρ⁢(H12) and ρ⁢(H23) are diagonalizable, and they commute with each other since H12 and H23 commute and ρ is a Lie algebra homomorphism. A theorem from linear algebra states that commuting, diagonalizable matrices are simultaneously diagonalizable. It follows that there is a basis of V consisting of simultaneous eigenvectors for ρ⁢(H12) and ρ⁢(H23). Since H12 and H23 span 𝔥, this is a basis of weight vectors. ∎

Remark 8.14.

There is a third homomorphism

ι23:(abc−a)↦(a0b000c0−a).

We have ι23⁢(H)=H23.

Note that, for i<j, ιi⁢j⁢(X)=Ei⁢j and ιi⁢j⁢(Y)=Ej⁢i, so Ei⁢j and Ej⁢i will play the role of raising and lowering operators.

Remark 8.15.

We could also prove Theorem 8.13 by exponentiating ρ to a representation of SL3⁡(ℂ) and considering the action of the subgroup of diagonal matrices with entries in U⁢(1), which is compact (isomorphic to U⁢(1)). Compare the proof of the statement that the weights of representations of 𝔰⁢𝔩2,ℂ are integers.

8.5 Tensor constructions

We record how the various linear algebra constructions we know about interact with the theory of weights. If (ρ,V) is a representation of 𝔤 then we consider its weights as a multiset

{α1,…,αn}

where n=dimV and each α∈𝔥∗ is written in this list dimVα — the multiplicity of α — times.

Suppose that (σ,W) is another representation of 𝔤 with multiset of weights

{β1,…,βm}.
Theorem 8.16.

Suppose that V,W,αi,βj are as above. Then:

  1. 1.

    The weights of V∗ are {−α1,…,−αn}.

  2. 2.

    The weights of V⊗W are

    {αi+βj:1≤i≤n,1≤j≤m}.
  3. 3.

    The weights of Symk⁡(V) are

    {αi1+…+αik:1≤i1≤i2≤…≤ik≤n}.
  4. 4.

    The weights of Λk⁢(V) are

    {αi1+…+αik:1≤i1<i2<…<ik≤n}.
Proof.

Let v1,…,vn be a basis of weight vectors of V such that vi has weight αi, and let w1,…,wm be similar for W with weights βi.

  1. 1.

    The dual basis v1∗,…,vn∗ is a basis of weight vectors in V∗ with vi∗ having weight −αi.

  2. 2.

    If v has weight α and w has weight β, then for all H∈𝔥,

    H⁢(v⊗w) =(H⁢v⊗w)+v⊗(H⁢w)
    =α⁢(H)⁢v⊗w+v⊗β⁢(H)⁢w
    =(α+β)⁢(H)⁢(v⊗w).

    So {vi⊗wj} is a basis of V⊗W with the given weights.

  3. 3.

    Similarly to the previous part,

    {vi1⁢vi2⁢…⁢vik:1≤i1≤i2≤…≤ik≤n}

    is a basis of Symk⁡V with the given weights.

  4. 4.

    Similar.

∎

Remark 8.17.

Similar considerations apply to representations of 𝔰⁢𝔩2,ℂ (or 𝔰⁢𝔩n,ℂ).

8.6 Highest weights

We now develop the theory of highest weights, analogous to that for 𝔰⁢𝔩2,ℂ. We first carry out the fundamental weight calculation (the analogue of Lemma 7.7).

Lemma 8.18.

(Fundamental Weight Calculation). Let V be a representation of 𝔰⁢𝔩3,ℂ and let v∈Vβ be a weight vector with weight β∈𝔥∗. Let α∈𝔥∗ be a root and let Xα∈𝔤α be a root vector. Then

Xα⁢(v)∈Vα+β.

Thus we obtain a map

Xα:Vβ⟶Vα+β.
Proof.

Let H∈𝔥. Then

H⁢(X⁢(v))=([H,X]+X⁢H)⁢(v)=α⁢(H)⁢X⁢(v)+X⁢(β⁢(H)⁢v)=(α+β)⁢(H)⁢X⁢(v).∎
Example 8.19.

We work this out for the adjoint representation. Recall that, for i≠j, we have the root αi⁢j=Li−Lj with root vector Ei⁢j. The above calculation shows that, if α and β are roots, then [𝔤α,𝔤β]⊂𝔤α+β. Here are some examples:

  • •

    If α=α12, β=α13, then α+β is not a root so 𝔤α+β=0. Thus [E12,E23]=0 (which could also be checked directly).

  • •

    If α=−β then we get

    [𝔤α,𝔤−α]⊂𝔤0=𝔥.
  • •

    If α=α12, β=α23, then α+β=α13 and we get

    [E12,E23]∈𝔤α13=⟨E13⟩.

    In fact, you can check that [E12,E23]=E13.

Exercise 8.20.

Let V be a finite-dimensional irreducible representation of 𝔰⁢𝔩3,ℂ. Then the weights occurring in V all differ by integral linear combinations of the roots of 𝔰⁢𝔩3,ℂ, that is, by integral linear combinations of Li−Lj.

With regard to the weight diagram, we observe that the ’positive’ root vectors E12, E23, E13 move in the ’northeast’ direction while the ’negative’ root vectors move in the ’southwest’ direction (roughly speaking). See Figure 9.

Refer to caption
Figure 9: Effect of roots
Definition 8.21.

Let (ρ,V) be a representation of 𝔰⁢𝔩3,ℂ. A highest weight vector in V is a vector v∈V such that:

  1. 1.

    v is a weight vector; and

  2. 2.

    ρ⁢(E12)⁢v=ρ⁢(E23)⁢v=0.

The weight of v is then a highest weight for V.

Remark 8.22.

Since [E12,E23]=E13, it follows that a highest weight vector is also killed by E13. So it is killed by all the positive root vectors.

Example 8.23.
  1. 1.

    The standard representation V has highest weight L1 with highest weight vector e1.

  2. 2.

    The dual V∗ has highest weight −L3 with highest weight vector e3∗.

  3. 3.

    The adjoint representation has highest weight L1−L3 with highest weight vector E13.

  4. 4.

    The symmetric square Sym2⁡(ℂ3) has highest weight 2⁢L1 with highest weight vector e12.

Lemma 8.24.

Let V be a finite-dimensional representation of 𝔰⁢𝔩3,ℂ. Then V has a highest weight vector.

Proof.

For a weight α=α1⁢L1+α1⁢L2+α3⁢L3, define l⁢(α)=α1−α3. Of all the finitely many weights of V, choose a weight α such that l⁢(α) is maximal.

Let v be a weight vector with this weight. Then E12⁢v, if nonzero, has weight

α+L1−L2

and

l⁢(α+L1−L2)=l⁢(α)+l⁢(L1−L2)=l⁢(α)+1>l⁢(α).

This is not a weight of V by maximality of l⁢(α). Thus E12⁢v=0. Similarly E23⁢v, if nonzero, has weight

α+L2−L3

and l⁢(α+L2−L3)=l⁢(α)+1, so E23⁢v=0. ∎

8.7 Dominant weights and Weyl symmetry

Let s1, s2, and s3 be, respectively, reflections in the lines through L1, L2, and L3. Then any two of these generate the Weyl group W, which is the group of symmetries of the triangle with vertices L1,L2,L3, so W≅D3≅S3. The dominant Weyl chamber is the region

{a⁢L1−b⁢L3:a,b∈ℝ≥0},

and weights inside it (including the boundary) are called dominant weights. See Figure 10.

Refer to caption
Figure 10: Dominant Weyl chamber

The main result of this section is that every highest weight is dominant.

Proposition 8.25.

Every highest weight of a finite-dimensional representation of 𝔰⁢𝔩3,ℂ is of the form a⁢L1−b⁢L3 for non-negative integers a,b≥0.

Proof.

Let v be a highest weight vector with weight α=a1⁢L1+a2⁢L2+a3⁢L3.

Since v is killed by E12=ι12⁢(X), it is a highest weight vector for the restriction of V to 𝔰⁢𝔩2,ℂ via ι12. Its 𝔰⁢𝔩2,ℂ-weight is

ρ⁢(ι12⁢(H))⁢v=ρ⁢(H12)⁢v=(a1−a2)⁢v,

so by the 𝔰⁢𝔩2,ℂ-theory, a1−a2 is a non-negative integer.

Likewise, since v is killed by E23=ι23⁢(X), the weight a2−a3=α⁢(H23) is also a non-negative integer.

Since L2=−L1−L3, we can write

α=(a1−a2)⁢L1−(a2−a3)⁢L3,

which is of the required form with a=a1−a2≥0 and b=a2−a3≥0. ∎

The following result shows that the weight diagram of any finite-dimensional representation has the six-fold symmetry of the Weyl group. Its proof is nonexaminable.

Theorem 8.26 (Weyl symmetry).

Let (ρ,V) be a finite-dimensional representation of 𝔰⁢𝔩3,ℂ. Then the weights of V are symmetric with respect to the action of the Weyl group.

Refer to caption
Figure 11: Simple reflections
Proof (nonexaminable).

We will prove they are symmetric with respect to s3 by using the inclusion

ι=ι12:𝔰⁢𝔩2,ℂ→𝔰⁢𝔩3,ℂ

that puts a 2×2 matrix in the top left corner of a 3×3 matrix. We consider the restriction of V to 𝔰⁢𝔩2,ℂ.

The kernel of E12 on V is preserved by ℌ (check it!) and so has a basis made up of 𝔰⁢𝔩3,ℂ-weight vectors v1,…,vr. These are a maximal set of linearly independent highest weight vectors for 𝔰⁢𝔩2,ℂ, so as an 𝔰⁢𝔩2,ℂ-representation,

V=⨁i=1rVi,

where Vi is generated by vi. Fix i; it suffices to show that Vi has a basis of 𝔰⁢𝔩3,ℂ-weight vectors whose weights are preserved by s3. Let vi have weight a⁢L1−b⁢L3. By Proposition 8.25, a≥0, and — recalling ι⁢(Y)=E21 — the 𝔰⁢𝔩2,ℂ-theory gives a basis

vi,E21⁢vi,…,E21a⁢vi.

By the FWC (Lemma 8.18) these are weight vectors with weights

a⁢L1−b⁢L3,(a−1)⁢L1+L2−b⁢L3,…,L1+(a−1)⁢L2−b⁢L3,a⁢L2−b⁢L3,

which are symmetrical under s3 (this reflection swaps L1 and L2), as required. This argument is illustrated in Figure 12.

Refer to caption
Figure 12: Proof of Weyl symmetry

Invariance with respect to the other reflections is proved similarly using the other inclusions ιi⁢j of 𝔰⁢𝔩2,ℂ in 𝔰⁢𝔩3,ℂ. ∎

Exercise 8.27.

We can give another proof of Weyl symmetry using the Lie group SL3⁡(ℂ). Let (ρ,V) be a representation of 𝔰⁢𝔩3,ℂ. As SL3⁡(ℂ) is simply-connected ρ exponentiates to a representation, ρ~, of SL3⁡(ℂ). Let

σ3=(010−100001)∈SL3⁡(ℂ).

Show that, for every weight α, ρ~⁢(σ3) is an isomorphism

Vα→Vs3⁢α.

Give another proof of Theorem 8.26.

8.8 Irreducible representations of sl3,C.

We are now in a position to state the main theorem of 𝔰⁢𝔩3,ℂ-theory.

Theorem 8.28.

For every pair a,b of nonnegative integers, there is a unique (up to isomorphism) irreducible finite-dimensional representation V(a,b) of 𝔰⁢𝔩3,ℂ with a highest weight vector of weight a⁢L1−b⁢L3.

Note that the highest weights occurring in the theorem are exactly the dominant elements of the weight lattice.

Since every irreducible finite-dimensional representation does have a highest weight, necessarily dominant, every irreducible representation is isomorphic to V(a,b) for some integers a,b≥0.

Example 8.29.

We have already seen some examples:

  1. 1.

    The standard representation ℂ3 is irreducible with highest weight L1, therefore

    V(1,0)=ℂ3.
  2. 2.

    The dual to the standard representation is irreducible with highest weight L3, and so

    V(0,1)=(ℂ3)∗.
  3. 3.

    The adjoint representation 𝔤 is irreducible with highest weight L1−L3, and so

    V(1,1)=𝔤.
  4. 4.

    The symmetric square Sym2⁡(ℂ3) has highest weight 2⁢L1 with highest weight vector e12.

Exercise 8.30.

(Problem 92.) Use complete reducibility to show that, if V is a finite-dimensional representation of 𝔰⁢𝔩3,ℂ with a unique highest weight vector (up to scalar), then V is necessarily irreducible.

Deduce that the three representations listed above are indeed irreducible.

For a more general example, we have:

Exercise 8.31.

(Problem 94). Show that the representation Symn⁡(ℂ3) has a unique highest weight vector with weight n⁢L1. Deduce that

V(n,0)=Symn⁡(ℂ3).

Convince yourself that the weights in this case are as shown in Figure 13 (which illustrates the case n=5).

Refer to caption
Figure 13: Weights for Symn⁡(ℂ3).

8.9 Proof of theorem 8.28 — nonexaminable

Lemma 8.32.

Let (ρ,V) be a finite-dimensional representation of 𝔰⁢𝔩3,C. Let v∈V be a highest weight vector of weight α, and let

W={ρ(Yn)ρ(Yn−1)…ρ(Y1)v:n≥0,Yi∈{E21,E32}.

Then

  1. 1.

    W is a subrepresentation of V

  2. 2.

    Wα=⟨v⟩ i.e. v is the unique weight vector in W of weight α, up to scaling.

  3. 3.

    W is irreducible.

Proof.
  1. 1.

    Let

    Wn={ρ(Ym)ρ(Ym−1)…ρ(Y1)v:n≥m≥0,Yi∈{E21,E32}.

    Then

    W=⋃n=0∞Wn.

    Firstly, it is clear that ρ⁢(E21) and ρ⁢(E32) take Wn to Wn+1, and so preserve W. Since

    ρ⁢(E31)=ρ⁢([E32,E21])=[ρ⁢(E32),ρ⁢(E21)]

    we see that ρ⁢(E31) also preserves W.

    Secondly, every ρ⁢(Ym)⁢ρ⁢(Ym−1)⁢…⁢ρ⁢(Y1)⁢v is a weight vector (by the fundamental weight calculation) and so an eigenvector for all ρ⁢(H), H∈𝔥. Thus ρ⁢(H) preserves each Wn (and hence also W).

    Finally, we show that ρ⁢(E12) preserves Wn. A similar proof then applies for ρ⁢(E23), and then ρ⁢(E13) preserves W by the same argument as for ρ⁢(E31). We prove the statement for ρ⁢(E12) by induction on n.

    For n=0, W0=⟨v⟩. Since v is a highest weight vector, ρ⁢(E12)⁢v=0 and so ρ⁢(E12)⁢(W0)⊂W0.

    Suppose that the claim is true for n. Consider w=ρ⁢(Yn+1)⁢…⁢ρ⁢(Y1)⁢v∈Wn+1 with Yi∈{E21,E32}. We must show that ρ⁢(E12)⁢w∈Wn+1. Suppose first that Yn+1=E21. Then, as [E12,E21]=H12∈ℍ, we have

    ρ⁢(E12)⁢w =ρ⁢(E12)⁢ρ⁢(E21)⁢ρ⁢(Yn)⁢…⁢ρ⁢(Y1)⁢v
    =ρ⁢(E21)⁢ρ⁢(E12)⁢ρ⁢(Yn)⁢…⁢ρ⁢(Y1)⁢v+ρ⁢(H12)⁢ρ⁢(Yn)⁢…⁢ρ⁢(Y1)⁢v
    ∈ρ⁢(E21)⁢ρ⁢(E12)⁢Wn+ρ⁢(H12)⁢Wn
    ⊂ρ⁢(E21)⁢Wn+Wn
    by the induction hypothesis and the fact that Wn is preserved by 𝔥
    ⊂Wn+1+Wn
    =Wn+1,

    as required. The proof in the case Yn+1=E32 is similar, using that [E12,E32]=0.

  2. 2.

    Note that if β the weight of ρ⁢(Yn)⁢ρ⁢(Ym−1)⁢…⁢ρ⁢(Y1)⁢v, with n and Yi as in the lemma, then a calculation using the fundamental weight calculation, as in the proof of lemma [[ref:lem-hwv-exists]], shows that

    l⁢(β)=l⁢(α)−n

    and so β≠α if n>0. Since these vectors span W, v is the unique (up to scalar) weight vector in W of weight α.

  3. 3.

    Suppose that W is reducible. By complete reducibility (Theorem [[ref:thm-complete-reducible-sl]]) we have

    W=U⊕U′

    for U,U′ nonzero proper subrepresentations of W. We must have v=u+u′ for unique u∈U,u′∈U′. The unicity implies that u and u′ are both weight vectors of weight α; by part 2, either u=0 or u′=0. So without loss of generality v=u∈U. But then all ρ⁢(Yn)⁢…⁢ρ⁢(Y1)⁢v∈U as U is a subrepresentation, so W=U contradicting that U is a proper subrepresentation.

∎

Remark 8.33.

It follows that W as in Lemma 8.32 is actually the subrepresentation generated by v, that is, the span of all vectors obtained by applying arbitrary elements of 𝔰⁢𝔩3,ℂ some number of times. The content of the lemma is then that it suffices to apply only E21 and E32.

We are now ready to prove Theorem 8.28:

Proof.

First we show the existence. Let a,b∈ℤ≥0. Consider

V=Syma⁡(ℂ3)⊗Symb⁡((ℂ3)∗).

This has a highest weight vector v=e1a⊗(e3∗)b of weight a⁢L1−b⁢L3. Let W be the representation generated by v. Then W is irreducible by 8.32 part 3, and has a highest weight vector v of weight a⁢L1−b⁢L3. Thus we can take V(a,b)=W.

Next we show the uniqueness. Suppose that V,W are two irreducible representations with highest weight vectors v and w, respectively, of weight a⁢L1−b⁢L3. Let U⊂V⊕W be the representation generated by u=(v,w). Then U is irreducible by 8.32 part 3. The projection V⊕W→V sending (v′,w′) to v′ restricts to a homomorphism U→V which sends u to v. This is therefore a nonzero homomorphism between irreducible representations, and so must be an isomorphism. Thus U≅V. Similarly U≅W, and so V≅W as required. ∎

In fact, it is possible to give an explicit description of the irreducible representations; see problem 95.

Theorem 8.34.

Let a,b≥0. Let V=ℂ3

ϕ:Syma⁡(V)⊗Symb⁡(V∗)→Syma−1⁡(V)⊗Symb−1⁡(V∗)

be the map

(v1⁢…⁢va)⊗(λ1⁢…⁢λb)↦∑i=1a∑j=1bλj⁢(vi)⁢(v1⁢…⁢v^i⁢…⁢va)⊗(λ1⁢…⁢λj^⁢…⁢λb).

Then ϕ is a surjective 𝔰⁢𝔩3,ℂ-homomorphism, and its kernel is the irreducible representation with highest weight a⁢L1−b⁢L3.