Schur Functors in This Lecture We Will Examine the Schur Polynomials and Young Tableaux to Obtain Irreducible Representations of Sl(N +1, C)

MATH 223A NOTES 2011 LIE ALGEBRAS 91

23. Schur functors In this lecture we will examine the Schur polynomials and Young tableaux to obtain irreducible representations of sl(n +1, C). The construction is based on Fulton’s book “Young Tableaux” which is highly recommended for beginning students because it is a beautifully written book with minimal prerequisites. This will be only in the case of L = sl(n +1,F). Other cases and the proof of the Weyl character formula will be next. First, we construct the irreducible representations V (λi) corresponding to the fundamental dominant weights λi = 1 + + i. Then we use Young tableaux to ﬁnd the modules V (λ) inside tensor products··· of these fundamental representations.

23.1. Fundamental representations.

Theorem 23.1.1. The fundamental representation V (λj) is equal to the ith exterior power of V = F n+1: V (λ )= jV j ∧ L = sl(n +1,F) acts by the isomorphism sl(n +1,F) ∼= sl(V ). Deﬁnition 23.1.2. Recall that the jth exterior power of a vector space V is the quotient of the jth tensor power by all elements of the form w1 wj where the wi are not j ⊗···⊗ distinct. The image of w1 wj in V is denoted w1 wj. When the characteristic of F is not equal to 2, this⊗··· is equivalent⊗ ∧ to saying that∧ the··· wedge∧ is skew symmetric: w w = sgn(σ)w w σ(1) ∧···∧ σ(j) 1 ∧···∧ j If v , ,v is a basis for V then 1 ··· n+1 v v v i1 ∧ i2 ∧···∧ ij for 1 i

110 1 2

2 1 011 101 3 3

! = ! 2 92 MATH 223A NOTES 2011 LIE ALGEBRAS

Example 23.1.3. Take n = 2 and λ = λ2 = 1 + 2. The Schur polynomial is

s11 = x1x2 + x2x3 + x1x3 The Young diagram is ﬁlled in with numbers 1, ,n+1 so that the numbers are increasing as we go down and nondecreasing as we go across.··· Let V = F n+1 = F 3 in this case with basis v ,v ,v . Then the corresponding representation is 2V with basis 1 2 3 ∧ v v ,v v ,v v 1 ∧ 2 2 ∧ 3 1 ∧ 3 The action of the Cartan subalgebra H is given on basis elements by h(v v )=h(v ) v + v h(v )= (h)v v + v (h)v i ∧ j i ∧ j i ∧ j i i ∧ j i ∧ j j Therefore, v v lies in the + weight space. Since + + =0weget: i ∧ j i j 1 2 3 vector v v v v v v 1 ∧ 2 2 ∧ 3 1 ∧ 3 1 + 2 2 + 3 1 + 3 weight λ2 λ1 λ1 λ2 maximal − − Proof of Theorem. In the general case, the maximal vector in jV is v v v with ∧ 1 ∧ 2 ∧···∧ j highest weight λj. One of the key points is that the wedge V jV is a functor (i.e., natural). In particular, any endomorphism g : V V induces→ an ∧ endomorphism jg : jV jV making jV into a module over sl(V )→= sl(n +1,F). ∧ ∧ →∧ ∧ ∼ 23.2. Symmetric powers. As I pointed out before, the irreducible module V (λ)for λ = ciλi is a direct summand of a tensor product of V (λi), taking ci copies of V (λi). For example, take n = 2 and λ = λ1 + λ2 =21 + 2. But, the tensor product of 2 1 V (λ2)= V and V (λ1)= V = V is 3 3 = 9 dimensional, whereas, V (λ1 + λ2) is 8 dimensional∧ (see worksheet).∧ The Schur-Weyl× construction tells us which 1-dimensional subspace to mod out.

Example 23.2.1. Take the example λ =2λ1. This Young diagram and Schur polynomial 2 2 2 s20 = x1 + x2 + x3 + x1x2 + x2x3 + x1x3

This is the sum of all monomials of degree 2 in the variables x1,x2,x3. Therefore, it corresponds to the second symmetric power of V : V (2λ )= 2(V ) 1 S 2 A basis for (V ) is given by vi vj where 1 i j n + 1. These basis elements are representedS by ﬁlling in the boxes:⊗ ≤ ≤ ≤ i j More generally, for any n and k, we have V (kλ )= k(V ) 1 S MATH 223A NOTES 2011 LIE ALGEBRAS 93

23.3. Schur functors. Let λ = λ , where b b , be given by a Young diagram bj 1 ≥ 2 ≥··· D (whose jth column has bj boxes). Then the Schur functor λ(V ) is given as a quotient of S b1 V b2 V b3 V ∧ ⊗∧ ⊗∧ ⊗··· by exchange relation which I will explain in class using the two examples on the worksheet.

1 2 1 1

120 210 2 2

1 2 111 021 201 2 2 3 1 1 3 3 1 3 012 102 2

2 3 1 3 3 3

! = ! 1 + !2

Example 23.3.1. Take the example λ = λ1 +λ2 =21 +2. The irreducible module V (λ), which is 8 dimensional by the calculation in the diagrams above, is a direct summand of 2 1 2 V (λ2) V (λ1)= V V = V V which is 3 3 = 9 dimensional. What is the missing⊗ basis vector?∧ ⊗∧ ∧ ⊗ × The missing vector is (v v ) v which corresponds to the Young tableaux 2 ∧ 3 ⊗ 1 2 1 3 This is not admissible since 2 > 1 but it is a linear combination of admissible Young tableaux by the exchange relation which says that the contents of any box can be exchanged with the contents of all boxed in any other column: A D D A A B A C B = B + D + B C C C D In this case we have: 2 1 1 2 2 3 = + 3 3 1 The second term is equal to 2 3 1 3 = 1 − 2 94 MATH 223A NOTES 2011 LIE ALGEBRAS since the ﬁrst column represents v v = v v . Thus: 2 ∧ 1 − 1 ∧ 2 (v v ) v =(v v ) v (v v ) v 2 ∧ 3 ⊗ 1 1 ∧ 3 ⊗ 2 − 1 ∧ 2 ⊗ 3 where indicates tensor product symmetrised by the exchange relations. ⊗ More generally, the exchange relation says that any set of squares in any column can be exchanged with squares in one other column as long as one takes the sum of all ways to do that and, also, the second column is to the left of the ﬁrst column. For example, if two columns have the same number of squares then they can be switched. Another example is: A D D A A B D A B E = E B + D C + B C C C E E

Theorem 23.3.2. If λ is the sum of fundamental dominant weights λji then V (λ) is the quotient of V (λji ) by the exchange relations. This quotient is the Schur functor (V ). Sλ Proof. The quotient λ(V ) is a representation of L = sl(V ). By a combinatorial argu- ment, we can see thatS it has a basis given by the admissible Young tableaux. The Schur polynomial s is the sum of the corresponding monomials. So, the character of (V ) is λ Sλ equal to chλ as given by the Weyl character formula. But we know that representations are uniquely determined by their formal characters. So, we conclude that (V )=V (λ). Sλ