07.05.E1 · representation-theory / symmetric

Lie-group and Lie-algebra representation exercise pack (Fulton-Harris supplement)

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Formal definition of the pack Intermediate

Fulton-Harris develops the representation theory of semisimple Lie algebras through a long chain of examples: the irreducibles of $sl_{2} C$ indexed by a single non-negative integer, those of $sl_{3} C$ laid out on the weight lattice, the general highest-weight theory organised by root systems and the Weyl group, and the bridge to the symmetric group through Schur-Weyl duality and Young symmetrizers. Many of its exercises cut across several of these threads at once: a Clebsch-Gordan computation uses the $sl_{2}$ classification and the character ring together; a dimension count uses the Weyl dimension formula and the hook-length formula together.

This pack collects ten such exercises — three easy, four medium, three hard — each with a hint and a full solution. It is meant to be read alongside its prerequisite units rather than as a standalone development. The exercises are loosely grouped: $sl_{2}$ and $sl_{3}$ irreducibles and tensor decompositions (easy to medium), weights, roots, and the Weyl character/dimension formulae (medium), and the Schur-Weyl/Young-symmetrizer combinatorics (hard).

The conventions throughout are Fulton-Harris's: $sl_{2}$ has standard basis $H, X, Y$ with $[H, X] = 2 X$ , $[H, Y] = - 2 Y$ , $[X, Y] = H$ ; the irreducible of highest weight $n$ is written $V_{n} = Sym^{n} (C^{2})$ with $dim V_{n} = n + 1$ ; weights are recorded as integer eigenvalues of $H$ (or of the diagonal Cartan for $sl_{3}$ ).

Key theorem with full solution Intermediate

Before the pack proper, we work one exercise in full as an exemplar of the format. The remaining nine follow the same structure (problem, hint, full answer in <details> blocks).

Lead exercise. Classify the finite-dimensional irreducible representations of $sl_{2} C$ and read off the Clebsch-Gordan rule $V_{a} \otimes V_{b}$ .

Solution. Let $V$ be a finite-dimensional irreducible $sl_{2} C$ -representation. Since $H$ acts on the finite-dimensional space $V$ , it has an eigenvector; from the commutation relations, $H (X v) = (λ + 2) X v$ and $H (Y v) = (λ - 2) Y v$ whenever $H v = λ v$ , so $X$ raises and $Y$ lowers the $H$ -eigenvalue by $2$ . Finite-dimensionality forces a highest-weight vector $v_{0}$ with $X v_{0} = 0$ and $H v_{0} = n v_{0}$ for some scalar $n$ .

Set $v_{k} = Y^{k} v_{0}$ . Then $H v_{k} = (n - 2 k) v_{k}$ and one computes by induction $X v_{k} = k (n - k + 1) v_{k - 1}$ . The chain $v_{0}, v_{1}, \dots$ must terminate: there is a least $m$ with $v_{m + 1} = 0$ , and then $0 = X v_{m + 1} = (m + 1) (n - m) v_{m}$ forces $n = m$ . So $n$ is a non-negative integer, $dim V = n + 1$ , and the weights are $n, n - 2, \dots, - n$ each with multiplicity one. This is $V_{n} = Sym^{n} (C^{2})$ .

For the tensor product, the character (the formal sum of weights) of $V_{n}$ is $χ_{n} = q^{n} + q^{n - 2} + \dots + q^{- n} = \frac{q ^{n + 1} - q ^{- (n + 1)}}{q - q ^{- 1}}$ . Characters multiply under $\otimes$ , and the ring of such symmetric Laurent polynomials has ${χ_{n}}$ as a basis. Multiplying $χ_{a} χ_{b}$ and re-expanding gives, for $a \geq b$ ,

V_{a} \otimes V_{b} ≅ V_{a + b} \oplus V_{a + b - 2} \oplus \dots \oplus V_{a - b} .

This is the Clebsch-Gordan rule. The summands run from $∣ a - b ∣$ to $a + b$ in steps of $2$ , exactly $b + 1$ of them, and a dimension check confirms $(a + 1) (b + 1) = \sum_{j = 0}^{b} (a + b - 2 j + 1)$ . $□$

Exercises Intermediate

Exercise 2 (easy, numeric). Dimension of an $sl_{3}$ irreducible by the Weyl dimension formula.

The irreducible of $sl_{3} C$ with highest weight $(a, b)$ in the basis of fundamental weights $ω_{1}, ω_{2}$ has dimension $$ \dim V_{(a,b)} = \tfrac12(a + 1)(b + 1)(a + b + 2). $$ Compute the dimensions for $(a, b) = (1, 0)$ , $(0, 1)$ , $(1, 1)$ , and $(2, 0)$ , and identify each representation.

Hint

$(1, 0)$ is the standard $C^{3}$ , $(0, 1)$ its dual, $(1, 1)$ the adjoint, $(2, 0)$ the symmetric square $Sym^{2} C^{3}$ .

Answer

Plugging into $\frac{1}{2} (a + 1) (b + 1) (a + b + 2)$ :

$(1, 0)$ : $\frac{1}{2} \cdot 2 \cdot 1 \cdot 3 = 3$ — the standard representation $C^{3}$ .
$(0, 1)$ : $\frac{1}{2} \cdot 1 \cdot 2 \cdot 3 = 3$ — the dual $(C^{3})^{*}$ .
$(1, 1)$ : $\frac{1}{2} \cdot 2 \cdot 2 \cdot 4 = 8$ — the adjoint representation $sl_{3}$ itself.
$(2, 0)$ : $\frac{1}{2} \cdot 3 \cdot 1 \cdot 4 = 6$ — the symmetric square $Sym^{2} C^{3}$ .

The two fundamental representations $C^{3}$ and $(C^{3})^{*}$ are distinct (unlike $sl_{2}$ , where every representation is self-dual), reflecting the order-two diagram automorphism of $A_{2}$ .

Exercise 3 (easy, symbolic). The Casimir eigenvalue on $V_{n}$ .

The quadratic Casimir of $sl_{2}$ is $C = \frac{1}{2} H^{2} + X Y + Y X$ . Show it acts on $V_{n}$ as the scalar $\frac{1}{2} n (n + 2)$ .

Hint

$C$ is central, so by Schur's lemma it is a scalar on the irreducible $V_{n}$ . Evaluate it on the highest-weight vector $v_{0}$ , using $X v_{0} = 0$ .

Answer

$C = \frac{1}{2} H^{2} + X Y + Y X$ is central (it commutes with $H, X, Y$ ), so by Schur's lemma it acts as a single scalar on the irreducible $V_{n}$ . Evaluate on the highest-weight vector $v_{0}$ with $H v_{0} = n v_{0}$ and $X v_{0} = 0$ . Rewrite $X Y + Y X = 2 Y X + [X, Y] = 2 Y X + H$ . Then

C v_{0} = \frac{1}{2} H^{2} v_{0} + (2 Y X + H) v_{0} = \frac{1}{2} n^{2} v_{0} + (0 + n) v_{0} = \frac{1}{2} (n^{2} + 2 n) v_{0} = \frac{1}{2} n (n + 2) v_{0},

using $X v_{0} = 0$ . So $C = \frac{1}{2} n (n + 2) Id$ on $V_{n}$ . The eigenvalue $\frac{1}{2} n (n + 2) = \frac{1}{2} ⟨ λ, λ + 2 ρ ⟩$ matches the general Casimir eigenvalue $⟨ λ, λ + 2 ρ ⟩$ up to the normalisation of the Killing form.

Exercise 4 (medium, proof). Weights of the standard $sl_{3}$ -representation and its dual.

Let $h \subset sl_{3}$ be the diagonal Cartan with linear functionals $L_{1}, L_{2}, L_{3}$ (the diagonal entries, subject to $L_{1} + L_{2} + L_{3} = 0$ ). List the weights of the standard representation $C^{3}$ and of its dual $(C^{3})^{*}$ , and show the two have no weight in common.

Hint

The standard representation has weights $L_{1}, L_{2}, L_{3}$ on the coordinate basis. Dualising negates weights.

Answer

On the coordinate basis $e_{1}, e_{2}, e_{3}$ of $C^{3}$ , a diagonal $H = diag (h_{1}, h_{2}, h_{3})$ acts by $H e_{i} = h_{i} e_{i}$ , so $e_{i}$ has weight $L_{i}$ . The weights of $C^{3}$ are ${L_{1}, L_{2}, L_{3}}$ , the three vertices of a triangle in the plane $L_{1} + L_{2} + L_{3} = 0$ .

The dual representation has weights negated: ${- L_{1}, - L_{2}, - L_{3}}$ , the vertices of the inverted triangle.

These two weight sets are disjoint: $L_{i} = - L_{j}$ would force $L_{i} + L_{j} = 0$ , hence $L_{k} = 0$ (since the three sum to zero), but $L_{k}$ is a nonzero functional on $h$ . So no weight is shared. The two inequivalent fundamental representations sit as opposite triangles, and their non-isomorphism is the failure of self-duality that distinguishes $A_{2}$ from $A_{1}$ .

Exercise 5 (medium, proof). The Weyl character formula recovers Clebsch-Gordan for $sl_{2}$ .

Specialise the Weyl character formula to $sl_{2}$ and use it to re-derive the Clebsch-Gordan decomposition of $V_{a} \otimes V_{b}$ .

Hint

For $sl_{2}$ the Weyl denominator is $q - q^{- 1}$ and the character of $V_{n}$ is $χ_{n} = (q^{n + 1} - q^{- (n + 1)}) / (q - q^{- 1})$ . Multiply $χ_{a} χ_{b}$ and divide.

Answer

The Weyl character formula for $sl_{2}$ gives $χ_{n} = \frac{q ^{n + 1} - q ^{- (n + 1)}}{q - q ^{- 1}}$ (numerator the antisymmetrised highest weight $ρ = 1$ , denominator the Weyl denominator). Take $a \geq b$ and compute

χ_{a} χ_{b} = \frac{q ^{a + 1} - q ^{- (a + 1)}}{q - q ^{- 1}} \cdot (q^{b} + q^{b - 2} + \dots + q^{- b}),

using the polynomial form of $χ_{b}$ . Distributing, $χ_{a} χ_{b} = \sum_{j = 0}^{b} \frac{q ^{a + 1 + b - 2 j} - q ^{- (a + 1 + b - 2 j)}}{q - q ^{- 1}} = \sum_{j = 0}^{b} χ_{a + b - 2 j}$ .

Reading off the indices, the irreducible characters appearing are $χ_{a + b}, χ_{a + b - 2}, \dots, χ_{a - b}$ , each with multiplicity one. Since characters of irreducibles are linearly independent, this is the decomposition:

V_{a} \otimes V_{b} ≅ V_{a + b} \oplus V_{a + b - 2} \oplus \dots \oplus V_{a - b} .

The character ring isomorphism turns the tensor decomposition into ordinary multiplication of Laurent polynomials.

Exercise 6 (medium, numeric). Hook-length formula for the dimension of a Specht module.

Compute $dim S^{λ}$ , the dimension of the irreducible $S_{n}$ -representation indexed by $λ = (3, 2)$ (so $n = 5$ ), via the hook-length formula $dim S^{λ} = n! / \prod_{c} h (c)$ .

Hint

Draw the Young diagram of $(3, 2)$ and compute the hook length $h (c) = (arm) + (leg) + 1$ at each of the five boxes.

Answer

The Young diagram of $λ = (3, 2)$ has boxes in rows of lengths $3$ and $2$ . Hook lengths (arm to the right plus leg below plus one), box by box:

Row 1: $(1, 1)$ has arm $2$ , leg $1$ , hook $4$ ; $(1, 2)$ has arm $1$ , leg $1$ , hook $3$ ; $(1, 3)$ has arm $0$ , leg $0$ , hook $1$ .
Row 2: $(2, 1)$ has arm $1$ , leg $0$ , hook $2$ ; $(2, 2)$ has arm $0$ , leg $0$ , hook $1$ .

Product of hooks: $4 \cdot 3 \cdot 1 \cdot 2 \cdot 1 = 24$ . With $n! = 5! = 120$ ,

dim S^{(3, 2)} = \frac{120}{24} = 5.

Check against the standard-Young-tableau count: there are exactly $5$ standard fillings of $(3, 2)$ , matching $dim S^{(3, 2)} = 5$ .

Exercise 7 (medium, proof). Roots and the Weyl group of $sl_{3}$ .

Describe the root system of $sl_{3}$ as a subset of the plane $L_{1} + L_{2} + L_{3} = 0$ , and identify its Weyl group.

Hint

The roots are $L_{i} - L_{j}$ for $i \neq = j$ . Reflections in the root hyperplanes permute the $L_{i}$ .

Answer

The roots of $sl_{3}$ are $α_{ij} = L_{i} - L_{j}$ for $i \neq = j$ , six in total, forming the hexagonal root system $A_{2}$ . A choice of simple roots is $α_{1} = L_{1} - L_{2}$ and $α_{2} = L_{2} - L_{3}$ ; the positive roots are $α_{1}, α_{2}, α_{1} + α_{2} = L_{1} - L_{3}$ . The angle between $α_{1}$ and $α_{2}$ is $12 0^{\circ}$ , and all roots have equal length (the system is simply laced).

The Weyl group is generated by the reflections $s_{α_{ij}}$ in the hyperplanes $L_{i} = L_{j}$ . Such a reflection swaps $L_{i}$ and $L_{j}$ , so the Weyl group acts on ${L_{1}, L_{2}, L_{3}}$ by all permutations:

W (A_{2}) ≅ S_{3},

the symmetric group on three letters, of order $6$ . The six chambers of the hexagon correspond to the six elements of $S_{3}$ , and $ρ = α_{1} + α_{2}$ (half the sum of positive roots, times two) sits at the center of the dominant chamber.

Exercise 8 (hard, proof). Schur-Weyl duality on $(C^{n})^{\otimes d}$ .

State and prove the double-centralizer half of Schur-Weyl duality: on $V^{\otimes d}$ with $V = C^{n}$ , the image of $C [S_{d}]$ (permuting factors) and the image of $GL (V)$ (diagonal action) are each other's commutants. Deduce the decomposition $V^{\otimes d} = ⨁_{λ} S^{λ} \otimes S_{λ} (V)$ .

Hint

One inclusion (the $S_{d}$ -image lands in the $GL$ -commutant) is immediate from commuting actions. For the reverse, use that the $GL (V)$ -image is spanned by $g^{\otimes d}$ and these span the symmetric tensors in $End (V)^{\otimes d}$ . Then apply the double-commutant theorem.

Answer

$S_{d}$ acts on $V^{\otimes d}$ by permuting tensor factors and $GL (V)$ acts diagonally by $g \cdot (v_{1} \otimes \dots \otimes v_{d}) = g v_{1} \otimes \dots \otimes g v_{d}$ . These commute, so each image lands in the other's commutant.

Let $A =$ image of $C [S_{d}]$ and $B =$ image of $C [GL (V)]$ in $End (V^{\otimes d})$ . The span of ${g^{\otimes d} : g \in GL (V)}$ equals the span of all $g^{\otimes d}$ for $g \in End (V)$ (a Zariski-density argument), which is exactly the space of $S_{d}$ -symmetric tensors $(End V)^{\otimes d, S_{d}}$ . A linear operator commutes with the $S_{d}$ -action iff it is $S_{d}$ -symmetric, so $B = A^{'}$ (the commutant of $A$ ). Since $A = C [S_{d}] / ann$ is semisimple, the double-commutant theorem gives $A = B^{'}$ as well.

By the double-centralizer theorem for the commuting actions of the two semisimple algebras $A$ and $B$ , the bimodule $V^{\otimes d}$ decomposes multiplicity-free as

V^{\otimes d} ≅ λ ⊢ d, ℓ (λ) \leq n ⨁ S^{λ} \otimes S_{λ} (V),

where $S^{λ}$ is the irreducible $S_{d}$ -module (Specht module) and $S_{λ} (V)$ is the irreducible $GL (V)$ -module (the Schur functor applied to $V$ ), nonzero exactly when $ℓ (λ) \leq n = dim V$ . The Schur functors $S_{λ}$ thus arise as the $λ$ -isotypic pieces under $S_{d}$ .

Exercise 9 (hard, proof). Young symmetrizer is idempotent up to a scalar.

For a Young tableau of shape $λ ⊢ d$ , with row-symmetrizer $a_{λ} = \sum_{p \in P} p$ and column-antisymmetrizer $b_{λ} = \sum_{q \in Q} sgn (q) q$ , the Young symmetrizer is $c_{λ} = a_{λ} b_{λ}$ . Show $c_{λ}^{2} = n_{λ} c_{λ}$ for a nonzero scalar $n_{λ}$ , and identify $n_{λ} = d! / dim S^{λ}$ .

Hint

Show $a_{λ} x b_{λ}$ is a scalar multiple of $c_{λ}$ for every $x \in C [S_{d}]$ (a "key lemma"). Apply with $x = b_{λ} a_{λ}$ to get $c_{λ}^{2} = n_{λ} c_{λ}$ . The image of $c_{λ}$ is the Specht module, so the trace of the projection $c_{λ} / n_{λ}$ equals $dim S^{λ}$ .

Answer

Key lemma. For any $x \in C [S_{d}]$ , $a_{λ} x b_{λ} = k (x) c_{λ}$ for a scalar $k (x)$ . This follows because $a_{λ} (σ) b_{λ} = a_{λ} b_{λ} = c_{λ}$ scaled by $\pm 1$ for permutations $σ$ that lie in the row-then-column structure, and any $σ$ outside that structure contains a transposition fixed by some row group and some column group, which makes $a_{λ} σ b_{λ} = 0$ by the sign mismatch.

Applying the lemma with $x = b_{λ} a_{λ}$ ,

c_{λ}^{2} = a_{λ} b_{λ} a_{λ} b_{λ} = a_{λ} (b_{λ} a_{λ}) b_{λ} = k c_{λ}

for a scalar $k =: n_{λ}$ . To evaluate $n_{λ}$ , note $e_{λ} = c_{λ} / n_{λ}$ is idempotent and its image is the Specht module $S^{λ}$ , so $e_{λ}$ is a projection onto a subspace isomorphic to $dim S^{λ}$ copies of $S^{λ}$ inside the regular representation. The trace of right-multiplication by $e_{λ}$ on $C [S_{d}]$ equals $dim S^{λ} \cdot (that count)$ ; a direct trace computation gives $tr (c_{λ}) = d! \cdot (coefficient of identity in c_{λ}) = d!$ , while $tr (c_{λ}) = n_{λ} tr (e_{λ}) = n_{λ} dim S^{λ}$ . Hence

n_{λ} = \frac{d !}{dim S ^{λ}} .

For $λ = (d)$ (one row) $c_{λ} = a_{λ}$ projects onto the unit-character line, $dim S^{λ} = 1$ , $n_{λ} = d!$ ; for $λ = (1^{d})$ (one column) it projects onto the sign line, same scalar.

Exercise 10 (hard, proof). Frobenius character formula consistency: dimension of $S_{λ} (C^{n})$ .

The dimension of the Schur functor $S_{λ} (C^{n})$ is given by the Weyl dimension formula for $GL_{n}$ : $$ \dim\mathbb{S}\lambda(\mathbb{C}^n) = \prod{1\leq i < j\leq n}\frac{\lambda_i - \lambda_j + j - i}{j - i}. $$ Verify, for $λ = (2, 1)$ and $n = 3$ , that this equals the count from the standard tableaux of shape $λ$ with entries in ${1, \dots, n}$ (semistandard tableaux), and reconcile with Schur-Weyl.

Hint

Compute the product over the three pairs $(i, j) \in {(1, 2), (1, 3), (2, 3)}$ with $λ = (2, 1, 0)$ . Separately count semistandard Young tableaux of shape $(2, 1)$ filled from ${1, 2, 3}$ .

Answer

Take $λ = (2, 1, 0)$ for $n = 3$ . The three factors:

$(i, j) = (1, 2)$ : $\frac{λ _{1} - λ _{2} + 1}{1} = \frac{2 - 1 + 1}{1} = 2$ .
$(i, j) = (1, 3)$ : $\frac{λ _{1} - λ _{3} + 2}{2} = \frac{2 - 0 + 2}{2} = 2$ .
$(i, j) = (2, 3)$ : $\frac{λ _{2} - λ _{3} + 1}{1} = \frac{1 - 0 + 1}{1} = 2$ .

Product: $2 \cdot 2 \cdot 2 = 8$ . So $dim S_{(2, 1)} (C^{3}) = 8$ , the adjoint representation of $GL_{3}$ (modulo the determinant twist, the $sl_{3}$ adjoint of Exercise 2).

Semistandard count: tableaux of shape $(2, 1)$ with entries in ${1, 2, 3}$ , weakly increasing along rows and strictly increasing down columns. Enumerating by the column pair (top-left $<$ bottom-left) and the top-right entry $\geq$ top-left: the count is exactly $8$ . This is the Kostka-number total $\sum_{μ} K_{λ μ}$ over all content $μ$ , equivalently $s_{λ} (1, 1, 1)$ .

Reconciliation with Schur-Weyl: in $V^{\otimes 3}$ with $V = C^{3}$ , the $λ = (2, 1)$ isotypic piece is $S^{(2, 1)} \otimes S_{(2, 1)} (C^{3})$ with $dim S^{(2, 1)} = 2$ and $dim S_{(2, 1)} = 8$ , contributing $16$ . Summing over $λ \in {(3), (2, 1), (1^{3})}$ gives $1 \cdot 10 + 2 \cdot 8 + 1 \cdot 1 = 27 = 3^{3} = dim V^{\otimes 3}$ , confirming the decomposition.

Exercise pack EP1 for Chapter 07.05. Fulton-Harris supplement: $sl_{2}$ / $sl_{3}$ representations, weights and roots, the Weyl character and dimension formulae, Schur-Weyl duality, and Young symmetrizers.

Prerequisites

07.06.11 pending
07.06.12 pending
07.06.03 pending
07.06.07 pending
07.05.04 pending
07.05.02 pending
07.05.01 pending

Tier anchors

beginner
intermediate: Fulton-Harris *Representation Theory: A First Course* exercises (§11-§12 for $\mathfrak{sl}_2$/$\mathfrak{sl}_3$, §14 weights and roots, §24-§25 Weyl character formula, §6 and §15.3 Schur-Weyl and Young symmetrizers)
master

References

Fulton-Harris — Representation Theory: A First Course · Exercises across §11-§12 ($\mathfrak{sl}_2\mathbb{C}$, $\mathfrak{sl}_3\mathbb{C}$ irreducibles and the Clebsch-Gordan decomposition), §14 (weight lattices and root systems), §24-§25 (Weyl character and dimension formulae), §6 and §15.3 (Young symmetrizers, Schur-Weyl duality, the hook-length formula)
Humphreys — Introduction to Lie Algebras and Representation Theory · §7 (representations of $\mathfrak{sl}_2$), §13 (weights), §24 (Weyl character formula)
Sagan — The Symmetric Group · Ch. 2-4 — Young tableaux, Specht modules, the hook-length formula

Estimated time

intermediate: 6h