Séminaire Lotharingien de Combinatoire 71 (2014), Article B71d

IRREDUCIBLE REPRESENTATIONS OF SU(n) WITH PRIME POWER DEGREE

Sarah Peluse^{1}

Department of Mathematics, University of Chicago 5734 S. University Ave., Chicago, Illinois 60637, USA

Abstract. The correspondence between irreducible representations of the symmetric group S_{n} and partitions of n is well-known. Less well-known is the connection between irreducible representations of the special unitary group SU(n) and partitions with less than n parts. This paper uses this correspondence to classify the irreducible representations of SU(n) with prime power degree.

1. Introduction and Statement of Results

It is well-known that there is a natural one-to-one correspondence between irreducible representations of the symmetric group S_{n} and partitions of n (see [3].) In fact, these representations are characterized by the action of S_{n} on the set of tableaux for each partition of n. The famous Frame–Robinson–Thrall hook-length formula computes the degree of an irreducible representation of S_{n} using only properties of the partition associated to it.

Let λ = (a_{k}, …, a_{1}) be a partition of n, so a_{k} ≥ a_{k−1} ≥ ⋯ ≥ a_{1} ≥ 1 and a_{k} + a_{k−1} + ⋯ + a_{1} = n. The Young diagram for λ is an array of k left-justified rows of boxes such that the i^{th} row contains a_{i} boxes.

Figure 1. Young diagram of λ = (5, 4, 1).

We attach a coordinate system to a Young diagram as follows. The upper left-hand box is labeled by (1, 1), and the x and y coordinates increase moving down and right, respectively (as one labels entries in a matrix.) For a given coordinate (i, j) in a Young diagram, we define the hook-length, h_{i,j}, to be the number of boxes directly to the right or below the box with coordinates (i, j) plus 1. For more on the beautiful subject of partition hook-lengths, see Han’s excellent article [2]. Figure 2 shows a Young diagram where each box is labeled with its hook-length.

1The author is grateful for the NSF’s support of the REU at Emory University.

Current address: Department of Mathematics, Stanford University, 450 Serra Mall, Building 380, Stanford, California 94305, USA..

7 5 4 3 1
5 3 2 1

1

Figure 2. Hook lengths for λ = (5, 4, 1).

Denote the degree of a group representation T by d_{T}. If T is an irreducible representation of S_{n} and λ is its associated partition of n, then the Frame–Robinson–Thrall formula is

d_{T} = \frac{n!}{∏_{(i,j)} h_{i,j}} . (1.1)

It is less well-known that the irreducible representations of SU(n) are naturally indexed by partitions with less than n parts. Further, similar to the case of S_{n}, SU(n) too possesses a degree formula depending only on n and properties of an irreducible representation’s associated partition. For a detailed exposition of this topic, see pages 246–254 and 327–352 of [5].

Let λ = (a_{k}, …, a_{1}) be a partition and n > k. Suppose that T is the irreducible representation of SU(n) corresponding to λ. Draw the Young diagram of λ and fill in the (1, 1) box with n. Then fill in the remaining boxes in the diagram by increasing by 1 moving right and decreasing by 1 moving down. The result of this is that the (i, j) box is filled in with m_{i,j} = n − i + j. Then, we have (see page 253 of [5])

d_{T} = \frac{∏_{(i,j)} m_{i,j}}{∏_{(i,j)} h_{i,j}} . (1.2)

Remark. We shall reformulate (1.2) in Section 3.

We define an admissible diagram to be a Young diagram filled in according to the rules in the preceding paragraph. The diagram below is an example of an admissible diagram.

8 9 10 11 12
7 8 6 5

Figure 3. An admissible diagram for λ = (5, 2, 1, 1)

The degree of the representation associated to the above diagram is

\frac{8 · 9 · 10 · 11 · 12 · 7 · 8 · 6 · 5}{8 · 5 · 3 · 2 · 1 · 4 · 1 · 2 · 1} = 83160.

Balog, Bessenrodt, Olsson, and Ono [1] classified all of the irreducible representations of S_{n} with prime power degree by combining combinatorial observations related to the Frame– Robinson–Thrall formula (1.1) with results on the distribution of prime numbers. The aim of this paper is to classify the irreducible representations of SU(n) of prime power degree. We have the following theorem.

IRREDUCIBLE REPRESENTATIONS OF SU(n) WITH PRIME POWER DEGREE 3

Theorem 1.1. The only prime power degree irreducible representations of SU(n) are those corresponding to the following families of admissible diagrams:

(1) The first family is the collection of admissible diagrams where λ = (1) and n = p^{k} for some prime p and k ∈ ℕ.

(2) The second family is the collection of admissible diagrams where

λ = ((n − 1)(p^{ℓ} − 1), (n − 2)(p^{ℓ} − 1), …, p^{ℓ} − 1),

p is a prime, and ℓ ∈ ℕ.

(3) The third family is the collection of admissible diagrams where

λ = (n(p^{ℓ} − 1) + 1, (n − 2)(p^{ℓ} − 1), …, (p^{ℓ} − 1))

or

λ = (n(p^{ℓ} − 1) + 1, (n − 1)(p^{ℓ} − 1) + 1, …, 2(p^{ℓ} − 1) + 1)

where p is a prime, ℓ ∈ ℕ, and n is a power of p.

It is not difficult to check that the three infinite families correspond to prime power degree irreducible representations. The bulk of the proof of Theorem 1.1 consists of a reformulation of (1.2) and an inductive argument. The proof then boils down to a characterization of what we call the “hands” of a partition and our ability to detect inadmissible prime factors.

This paper is organized as follows. In Section 2, we write down the degree formula for SU(n) in a tangible form and deal with the families listed in Theorem 1.1, proving that they do in fact correspond to prime power degree representations. In Section 3, we rewrite the degree formula in terms of certain entries in an admissible diagram. This reformulation helps to illuminate how the divisibility properties of a representation’s degree depend on the shape of its associated partition. In Section 4 we prove Theorem 1.1 using this formula.

Acknowledgments

The author would like to thank Ken Ono for his encouragement throughout this project, the referee for many helpful comments, and Christian Krattenthaler for pointing out references for the derivation of formula (2.1).

2. The Three Families

In this section we prove that the families described in Theorem 1.1 do indeed correspond to prime power degree representations of SU(n). First, we will translate the degree formula for SU(n) into a closed-form formula depending on n and the parts of a partition. Let λ = (a_{k}, a_{k−1}, …, a_{1}) be a partition and n > k. Then

d_{T} = ∏_{i=1}^{k} \frac{(n − (k − i + 1) + a_{i})! (a_{i} − a_{i−1} + 1) ⋯ (a_{i} − a_{1} + i − 1)}{(n − (k − i + 1))! (a_{i} + i − 1)!}. (2.1)

This follows from well-known manipulations involving specializations of Schur functions. See pages 44 and 45 in [4].

We will now use this formula for the degree to show that all of the families described in Theorem 1.1 correspond to prime power degree representations.

Proposition 2.1. Let p be a prime and k, ℓ, m ∈ ℕ, and let λ = (a_{k}, a_{k−1}, …, a_{1}).

(1) The partition

λ = (1)

corresponds to a prime power degree irreducible representation of SU(p^{m}).

(2) The partition

λ = (k(p^{ℓ} − 1), (k − 1)(p^{ℓ} − 1), …, p^{ℓ} − 1)

corresponds to a prime power degree irreducible representation of SU(k + 1).

(3) The partition

λ = (p^{m}(p^{ℓ} − 1) + 1, (p^{m} − 2)(p^{ℓ} − 1), …, p^{ℓ} − 1)

corresponds to a prime power degree irreducible representation of SU(p^{m}).

(4) The partition

λ = (p^{m}(p^{ℓ} − 1) + 1, (p^{m} − 1)(p^{ℓ} − 1) + 1, …, 2(p^{ℓ} − 1) + 1)

corresponds to a prime power degree irreducible representation of SU(p^{m}).

Proof. For each family, we will call the representation specified by the family T.

(1) Clearly, in this case

d_{T} = \frac{p^{m}}{1} = p^{m}.

(2) Note that, in this case, a_{i} = i(p^{ℓ} − 1). So, using Formula (2.1), we have

d_{T} = ∏_{i=1}^{k} \frac{(ip^{ℓ})! p^{ℓ} ⋯ (i − 1)p^{ℓ}}{i! (ip^{ℓ} − 1)!}

= ∏_{i=1}^{k} \frac{ip^{ℓ}p^{ℓ(i−1)}(i − 1)!}{i!}

= ∏_{i=1}^{k} p^{ℓ}p^{ℓ(i−1)}.

(3) Note that, in this case, a_{p^{m}−1} = p^{m}(p^{ℓ} − 1) + 1 and a_{i} = i(p^{ℓ} − 1) for i ≠ p^{m} − 1. So, using (2.1), we have

d_{T} = \frac{p^{m+ℓ}! (2p^{ℓ}) ⋯ (p^{m} − 1)(p^{ℓ})}{(p^{m} − 1)! (p^{m+ℓ} − 1)!} ∏_{i=1}^{p^{m}−2} \frac{(ip^{ℓ})! p^{ℓ} ⋯ (i − 1)p^{ℓ}}{i! (ip^{ℓ} − 1)!}

= \frac{p^{m+ℓ}p^{ℓ(p^{m}−2)}(p^{m} − 1)!}{(p^{m} − 1)!} ∏_{i=1}^{p^{m}−2} \frac{(ip^{ℓ})p^{ℓ(i−1)}(i − 1)!}{i!}

= p^{m+ℓ}p^{ℓ(p^{m}−2)} ∏_{i=1}^{p^{m}−2} p^{ℓ}p^{ℓ(i−1)}.

(4) Note that, in this case, a_{p^{m}−1} = p^{m}(p^{ℓ} − 1) + 1 and a_{i} = (i + 1)(p^{ℓ} − 1) + 1 for i ≠ p^{m} − 1. So, using (2.1), we have

d_{T} = \frac{p^{m+ℓ}! (2p^{ℓ}) ⋯ (p^{m} − 1)(p^{ℓ})}{(p^{m} − 1)! (p^{m+ℓ} − 1)!} ∏_{i=1}^{p^{m}−2} \frac{(i(p^{ℓ}))! p^{ℓ} ⋯ (i − 1)p^{ℓ}}{i! (i(p^{ℓ}) − 1)!}

= \frac{p^{m+ℓ}p^{ℓ(p^{m}−2)}(p^{m} − 1)!}{(p^{m} − 1)!} ∏_{i=1}^{p^{m}−2} \frac{i(p^{ℓ})p^{ℓ(i−1)}(i − 1)!}{i!}

= p^{m+ℓ}p^{ℓ(p^{m}−2)} ∏_{i=1}^{p^{m}−2} p^{ℓ}p^{ℓ(i−1)}.

□

3. A Reformulation of the Degree Formula

In this section, we derive a version of the degree formula depending on the “hands” of a partition which is much more compact than (2.1). Let λ = (a_{k}, a_{k−1}, …, a_{i}) be a partition and define the i^{th} hand to be q_{i} = a_{i} + i. In an admissible diagram for λ when n = k + 1, i.e., when n is minimal, q_{i} is also the number in the last box in the i^{th} row.

5 6 7 8 9
4 5 3 2

Figure 4. The hands of the partition (5, 2, 1, 1) are 2, 3, 5, and 9

Substitution of the definition of the hands into (2.1) yields the formula in the following lemma.

Lemma 3.1. Let λ = (a_{k}, a_{k−1}, …, a_{1}) be a partition, q_{i} = a_{i} + i, and j ≥ 0. If T is the irreducible representation of SU(k + 1 + j) corresponding to λ, then we have

d_{T} = ∏_{i=1}^{k} \frac{(q_{i} + j)! (q_{i} − q_{i−1}) ⋯ (q_{i} − q_{1})}{(i + j)! (q_{i} − 1)!}. (3.1)

Proof. By (2.1), we have

d_{T} = ∏_{i=1}^{k} \frac{(j + a_{i} + i)! (a_{i} − a_{i−1} + i − (i − 1)) ⋯ (a_{i} − a_{1} + i − 1)}{(i + j)! (a_{i} + i − 1)!}

= ∏_{i=1}^{k} \frac{(q_{i} + j)! (q_{i} − q_{i−1}) ⋯ (q_{i} − q_{1})}{(i + j)! (q_{i} − 1)!}.

□

In the case that n is as small as possible, i.e., when j = 0, expression (3.1) becomes

d_{T} = ∏_{i=1}^{k} \frac{q_{i}(q_{i} − q_{i−1}) ⋯ (q_{i} − q_{1})}{i!}. (3.2)

This suggests the key idea used in the proof of Theorem 1.1: the prime factors of d_{T} depend highly on the distribution of the hands modulo i = 1, 2, …, k.

4. Proof of Theorem 1.1

In this section we first use (3.2) to prove that the only λ that correspond to prime power degree representations for minimal n are those listed in Theorem 1.1. We then use a similar technique and (3.1) to prove that, if λ ≠ (1), then λ has k parts, and n > k + 1, then λ does not correspond to a prime power degree representation of SU(n). These two results will complete the proof of Theorem 1.1.

We begin with a simple combinatorial lemma.

Lemma 4.1. If a, k ∈ ℕ and μ is a strictly decreasing sequence q_{k} > q_{k−1} > ⋯ > q_{1} > 0, for each b = 0, …, a − 1 let k_{b}^{μ} denote the number of q_{i} in μ that are congruent to b modulo a. Define

m_{a,μ} := #{i : q_{i} ≡ 0 (mod a)} + #{(i_{1}, i_{2}) : q_{i_{1}} ≡ q_{i_{2}} (mod a), i_{1} > i_{2}}.

Let m_{a,k} be the minimum over all such sequences μ of the value of m_{a,μ}. Then m_{a,k} = m_{a,μ} if and only if k_{b}^{μ} − 2 ≤ k_{0}^{μ} ≤ k_{b}^{μ} and |k_{b}^{μ} − k_{b′}^{μ}| ≤ 1 for all b, b′ ∈ {1, …, a − 1}. Further,

m_{a,k} = ∑_{i=1}^{k} #{j : 1 ≤ j ≤ i, j ≡ 0 (mod a)}.

Proof. Let μ be a strictly decreasing sequence q_{k} > q_{k−1} > ⋯ > q_{1} > 0. Then

m_{a,μ} = k_{0}^{μ} + ∑_{b=0}^{a−1} \binom{k_{b}^{μ}}{2}.

Note that, if k_{b}^{μ} ≥ k_{b′}^{μ} + 1, then since \binom{n+1}{2} − \binom{n}{2} = n for all n ∈ ℕ, we have

\binom{k_{b}^{μ}}{2} + \binom{k_{b′}^{μ}}{2} ≥ \binom{k_{b}^{μ} − 1}{2} + \binom{k_{b′}^{μ} + 1}{2},

where equality holds exactly when k_{b}^{μ} = k_{b+1}^{μ} + 1. Hence, m_{a,μ} = m_{a,k} exactly when k_{b}^{μ} − 2 ≤ k_{0}^{μ} ≤ k_{b}^{μ} for all b = 0, 1, …, a − 1 and the distance between k_{b}^{μ} and k_{b′}^{μ} is at most 1 for all b and b′ in {1, …, a − 1}. Clearly, q_{i} = i is one sequence with this property. Set m = ∑_{i=1}^{k} #{j : 1 ≤ j ≤ i, j ≡ 0 (mod a)}. So, letting U = {1, …, k} and U_{i} = {u ∈ U : u ≡ 0 (mod a), i ≥ u}, we have

m_{a,k} = #U_{k} + ∑_{i=2}^{k} #{j : q_{i} − q_{j} ≡ 0 (mod a), i > j}

= #U_{k} + ∑_{i=2}^{k} #{j : i − j ≡ 0 (mod a), i > j}

= #U_{k} + ∑_{i=2}^{k} #U_{i−1} = ∑_{i=1}^{k} #U_{i} = m.

□

This lemma says that m_{a,k} is equal to the number of multiples of a in ∏_{i=1}^{k} i!, which is the denominator in (3.2). There are two important consequences of the lemma. The first is that it shows that d_{T} is an integer without any representation theory. The second is that it shows what is required for a representation to have prime power degree: q_{k} > ⋯ > q_{1} > 1 must achieve the minimum in the definition of m_{a,k} for all a not equal to a multiple of some fixed prime. This motivates the following definition.

Definition 4.2. We say a strictly decreasing sequence of k integers μ is good with respect to p if m_{a,k} = m_{a,μ} for all a ≢ 0 (mod p).

The following lemma makes the observation from the previous paragraph precise.

Lemma 4.3. Suppose that T is an irreducible representation of SU(k + 1) corresponding to the partition λ with hands μ = q_{k} > ⋯ > q_{1}. The degree d_{T} is a power of some prime p if and only if μ is good with respect to p.

Proof. To see this, simply note that μ is good with respect to p if and only if any power of a prime q ≠ p appearing in the numerator of d_{T} is canceled by something in the denominator. This is because, if q is a prime distinct from p, then Lemma 4.1 says that μ is good with respect to p if and only if, for all ℓ ∈ ℕ, there are the same number of multiples of q^{ℓ} in the numerator as in the denominator, and hence q ∤ d_{T}. □

Suppose that μ = q_{k} > ⋯ > q_{1} > 0 is good with respect to p. The following three facts follow easily from Lemma 4.1, and will be used repeatedly throughout the proofs of the next several lemmas:

(1) m_{k,k} = 1.

(2) If p ∤ q_{i}, then q_{i} ≤ k.

(3) If p ∤ q_{i} − q_{j} for i > j, then q_{i} − q_{j} ≤ k. Fact 1 is immediate from the Lemma 4.1. Note also that m_{a,k} > 0 only when k ≥ a. If p ∤ q_{i}, then m_{q_{i},k} = m_{q_{i},μ} ≥ 1 since μ is good with respect to p and q_{i} contributes to m_{q_{i},μ}. This implies that q_{i} ≤ k. The proof of Fact 3 is similar.

Lemma 4.4. Suppose that k ≥ 2 and μ = q_{k} > ⋯ > q_{1} > 0 is good with respect to some prime p. Then either q_{k} and q_{k−1} are both divisible by p, q_{i} = i for all i = 1, …, k, q_{i} = i + 1 for all i = 1, …, k, or q_{k} = k + 1 and q_{i} = i for all i = 1, …, k − 1.

Proof. Suppose that p ∤ q_{k}. Then q_{k} ≤ k by Fact 2. This forces q_{i} = i for all i = 1, …, k since k ≥ q_{k} > q_{k−1} > ⋯ > q_{1} ≥ 1.

Now suppose that p | q_{k} and p ∤ q_{k−1}. By Fact 2, q_{k−1} ≤ k. Since the q_{i} are strictly increasing, q_{k−1} ≥ q_{1} + k − 2, which implies that q_{1} = 1 or q_{1} = 2. If q_{k} = k + 1 and q_{k−1} = k, then, if q_{i} ≠ i + 1 for some i (i.e., we are not in the third case in the statement of the lemma), we must have q_{1} = 1 because the q_{i} are strictly decreasing. In this case, since p ∤ q_{k−1}, p ∤ k as well. However, k = q_{k−1} = q_{k} − q_{1}, so m_{k,μ} ≥ 2. Since μ is good with respect to p, m_{k,k} = m_{k,μ} ≥ 2, which contradicts Fact 1. If q_{k} = k + 1 and q_{k−1} = k − 1, this forces q_{i} = i for all i = 1, 2, …, k − 1, the final case in the statement of the lemma. If q_{k} ≥ k + 2 and q_{1} = 1, then p ∤ q_{k} − 1 and q_{k} − 1 > k, contradicting Fact 3. Finally, if q_{k} ≥ k + 2 and q_{1} = 2, then q_{k−1} = k and p ∤ k. Either q_{k} > k + 2 or q_{k} = k + 2. In the first case, q_{k} − q_{1} > k, contradicting Fact 3. In the second case, m_{k,k} = m_{k,μ} ≥ 2 since k = q_{k−1} = q_{k} − q_{1}, contradicting Fact 1. □

The goal of the next collection of lemmas is to show that, if k ≥ 3, then μ = q_{k} > q_{k−1} > ⋯ > q_{1} > 0 is good with respect to a prime p, and p divides both q_{k} and q_{k−1}, then p must in fact divide q_{i} for every i = 1, …, k. The next lemma gives an upper bound for the smallest q_{i} in μ which is not divisible by p based on how many consecutive large terms in μ are known to be multiples of p.

Lemma 4.5. Suppose that μ = q_{k} > ⋯ > q_{1} > 0 is good with respect to some prime p and that p | q_{k}, q_{k−1}, …, q_{k−m+1}with m ≥ 2, but not every term in μ is divisible by p. Then there exists an i = 1, …, k such that p ∤ q_{i} and q_{i} ≤ \frac{p}{p−1}(q_{k−m} + m + 2 − k − p) + p − 1.

Proof. To construct the sequence μ′ = q′_{k} > q′_{k−1} > ⋯ > q′_{1} > 0 with q′_{i} = q_{i} for i = k − m, …, k with maximum possible min{q′_{i} : p ∤ q′_{i}}, we must set q′_{i} = pi for as many of the first k −m i’s as possible, and then increase q′_{i} by p−1 once and then only by 1 as i increases by 1 up to k − m − 1 in order to have a sequence of the required length. So let q′_{i} = pi for i = 1, …, ℓ_{p}, q′_{ℓ_{p}+1} = pℓ_{p} + p − 1, and q′_{ℓ_{p}+1+j} = pℓ_{p} + p − 1 + j for j = 1, …, k − m − ℓ_{p} − 1. Then, we have k − m − ℓ_{p} − 1 + p − 1 + pℓ_{p} = q_{k−m}, i.e., ℓ_{p} = \frac{q_{k−m}+m+2−k−p}{p−1}. Thus, for some i ≤ k − m, p ∤ q_{i} with q_{i} ≤ pℓ_{p} + p − 1 = \frac{p}{p−1}(q_{k−m} + m + 2 − k − p) + p − 1.

□

Using the previous lemma in conjunction with Fact 3, we are able to show that, when p > 2, if μ is good with respect to p and p divides q_{k} and q_{k−1}, then p must divide every term in μ. When p = 2, the following lemma also gives the same result in all but two cases.

Lemma 4.6. Suppose that μ = q_{k} > ⋯ > q_{1} > 0 is good with respect to some prime p, p | q_{k}, q_{k−1}, …, q_{k−m+1}, p ∤ q_{k−m}, and k > m ≥ 2. Then p = 2, k is odd, and either q_{k−m} = k or q_{k−m} = k − 2.

Proof. Let q_{i} be the smallest term in μ which is not divisible by p. We have, by Lemma 4.5,

q_{k} − q_{i} ≥ q_{k−m} + p(m − 1) + 1 − \frac{p}{p − 1}(q_{k−m} + m + 2 − k − p) − p + 1

= k + \frac{1}{p − 1}(k − q_{k−m}) + m (\frac{p^{2} − 2p}{p − 1}) − \frac{p^{2}}{p − 1} + 2

≥ k + \frac{1}{p − 1}(k − q_{k−m}) + \frac{p^{2} − 4p}{p − 1} + 2,

since m ≥ 2. When p ≥ 3, since q_{k−m} ≤ k, we get that q_{k} − q_{i} ≥ k − \frac{3}{2} + 2 = k + \frac{1}{2}, which contradicts Fact 3 since p ∤ q_{k} − q_{i}.

When p = 2, the inequality above yields q_{k} − q_{i} ≥ k − 2 + (k − q_{k−m}). Because p ∤ q_{k} − q_{i}, by Fact 3 we have k ≥ q_{k} − q_{i} ≥ k − 2 + (k − q_{k−m}), and hence q_{k−m} ≥ k − 2. Since q_{k−m} ≤ k by Fact 2, we see that either q_{k−m} = k or q_{k−m} = k − 2, and, in addition, k must be odd.

□

In the next lemma, we show that, if μ = q_{k} > … q_{1} > 0 is good with respect to 2 but not every term in μ is even, then μ must be of a specific form depending on whether q_{k} = k or q_{k} = k − 2.

Lemma 4.7. Suppose that μ = q_{k} > ⋯ > q_{1} > 0 is good with respect to 2 and that 2 | q_{k}, q_{k−1}, …, q_{k−m+1}, 2 ∤ q_{k−m}, and m ≥ 2. If q_{k−m} = k, then

q_{j} = \begin{cases}2j, & j = 1, …, m, \\ m + j, & j = m + 1, …, m + 2ℓ + 1, \\ 2(j − ℓ − 1), & j = m + 2ℓ + 2, …, 2m + 2ℓ + 1, \end{cases}

for some ℓ ≥ 0, and, if q_{k−m} = k − 2, then

q_{j} = \begin{cases}2j, & j = 1, …, m − 2, \\ m + j − 2, & j = m − 1, …, m + 2ℓ − 1, \\ 2(j − ℓ − 1), & j = m + 2ℓ, …, 2m + 2ℓ − 1, \end{cases}

for some ℓ ≥ 0.

Proof. Let i be the smallest integer for which 2 ∤ q_{i}. Assume that q_{k−m} = k. Since m_{k,μ} = m_{k,k} because μ is good with respect to 2 and k is odd, by Fact 1 m_{k,μ} = 1. Hence, we must have q_{k} −q_{i} = k−2. It is also clear that q_{k} ≥ q_{k−m}+2(m−1)+1 = k+2m−1. Subtracting q_{i} from both sides, we see that k−2 ≥ k+2m−1−q_{i}, which implies that q_{i} ≥ 2m+1. However, by the definition of q_{i} and Lemma 4.5, we also have that q_{i} ≤ 2(k + m + 2 − k − 2) + 2 − 1 = 2m + 1, and hence q_{i} = 2m + 1.

Since k = q_{k−m} ≥ q_{i} + (k − m − i) because the q_{i} are strictly increasing, we have k ≥ 2m + 1 + k − m − i, i.e., i ≥ m + 1. By the definition of q_{i}, for all j < i, we have 2 | q_{j}. Hence, q_{i} ≥ 2(i − 1) + 1, i.e., 2m + 2 ≥ 2i, and m + 1 ≥ i. This implies that i = m + 1 and forces q_{j} = 2j for all j = 1, …, m.

By Fact 2, since 2 ∤ q_{i}, 2m + 1 ≤ k. Since k is odd, k = 2m + 2ℓ + 1 for some ℓ ≥ 0. Note that, if k ≠ 2m + 1 and if we do not increase q_{j} by 1 as j increases by 1 from q_{i} up to q_{k−m}, then we must have k > 2m + 1 + (k − m − (m + 1)) = k, which is impossible. Hence, q_{m+1+j} = 2m + 1 + j for j = 1, …, (k − m − (m + 1)), and a choice of ℓ uniquely determines μ: q_{j} = 2j, j = 1, …, m, q_{j} = m + j for j = m + 1, … k − m, and q_{j} = 2j + 2m − k − 1 for j = k − m + 1, …, k. Replacing k with 2m + 2ℓ + 1, we get the result.

The q_{k−m} = k − 2 case is proved analogously. □

The following lemma and proposition comprise one half of Theorem 1.1.

Lemma 4.8. Suppose that μ = q_{k} > ⋯ > q_{1} > 0 is good with respect to some prime p. Then either the q_{1}, …, q_{k} are all divisible by p, q_{i} = i for all i = 1, …, k, q_{i} = i + 1 for all i = 1, …, k, or q_{k} = k + 1 and q_{i} = i for all i = 1, …, k − 1.

Proof. By Lemmas 4.4 and 4.6, it suffices to show that both of the two cases described in Lemma 4.7 lead to a contradiction. Our main tool will be Lemma 4.1.

Assume that q_{k−m} = k. When q_{k−m} = k − 2 and m ≥ 3 (so that we are in not in the last case described in the statement of the lemma), a contradiction is reached in an almost identical manner. Suppose that ℓ = 0. Then k = 2m + 1. Since μ is good with respect to 2, 3 = m_{k−2,k} = m_{k−2,μ}. However, q_{k}−q_{k−m} = k−2, q_{k−m}−q_{1} = k−2, q_{k−2} = 4m−2 = 2(k−2), and q_{k} − q_{1} = 4m − 2 = 2(k − 2). So, m_{k−2,μ} ≥ 4, a contradiction. Suppose that ℓ = 1, so then k = 2m + 2ℓ + 1 ≥ 7, and since μ is good with respect to 2, 5 = m_{k−4,k} = m_{k−2,μ}. However, q_{k} − q_{2} = 2(k − 4), q_{k} − q_{k−m} = k − 4, q_{k−1} − q_{1} = 2(k − 4), q_{k−1} − q_{k−m−2} = k − 4, q_{k−m−2} − q_{1} = k − 4, and q_{k−m} − q_{2} = k − 4. So, m_{μ,k−2} ≥ 6, a contradiction.

Now suppose that ℓ ≥ 2. There are two cases: either m+ℓ is odd or m+ℓ+1 is odd. In the first case, it suffices to show that there are three q_{j}’s which are congruent to each of 2 and 4 modulo m+ℓ, and that, if there is more than one q_{j} congruent to 1 modulo m+ℓ, then there are two q_{j}’s divisible by m+ℓ. Then, in either case, there must be a nonzero congruence class modulo m + ℓ which contains at most one element, while the congruence class of 2 contains three elements, contradicting Lemma 4.1. When m + ℓ is odd, q_{j} ≡ q_{k−m+j} (mod m + ℓ) for all j = 1, …, m, so we have at least two q_{j}’s equivalent to each of 2 and 4, and two q_{j}’s from this range congruent to 1 if and only if 2m ≥ m + ℓ. For j = m + 1, …, k − m, q_{j} ranges from 2m + 1 to 2m + 2ℓ + 1, and 2m + 2ℓ + 1 ≥ m + ℓ + 4 since m, ℓ ≥ 2, and hence there are q_{j}’s in this range that are equivalent to each of 1, 2, and 4 modulo m + ℓ, giving us a total of at least three in the classes of 2 and 4. If 2m ≥ m + ℓ, then the class of 1 contains at least three elements, and so there must be a nonzero equivalence class with at most one element. If 2m < m + ℓ, then m + ℓ < 2ℓ, which implies that, since q_{k−m} ≡ 1 (mod m + ℓ), that at least two of the q_{j} in the range of j = m + 1, …, k − m are divisible by m + ℓ, and so, again, there must be a nonzero equivalence class with at most one element. The argument when m + ℓ + 1 is odd is similar. □ Proposition 4.9. Let T be an irreducible representation of SU(k + 1) corresponding to the partition with hands q_{k} > ⋯ > q_{1} > 1. If d_{T} = p^{ℓ} for ℓ ∈ ℕ, then T is one of the representations in Theorem 1.1. Proof. Let m be the highest power of p that divides all of the q_{i}, so setting q′_{i} = q_{i}/p^{m}, define d′_{T} to be the result of substituting q′_{k}, …, q′_{1} into (3.2). Then not all of the q′_{i} are divisible by p, so either q′_{i} = i or q′_{i} = i + 1 for all i = 1, …, k − 1 by Lemma 4.8, which implies that T is one of the representations in Theorem 1.1. □

We now finish the proof of Theorem 1.1 by proving that, if n > k + 1 where k is the number of parts of the partition λ and T is the irreducible representation corresponding to λ of SU(n), then d_{T} is not a prime power unless λ = (1). In order to do this, we modify the strategy used to prove the previous proposition. First, we have a result analogous to Lemma 4.1, which has essentially the same proof. Lemma 4.10. If a, k, j ∈ ℕ and μ is a strictly decreasing sequence q_{k} > q_{k−1} > ⋯ > q_{1} > 0, define

m_{a,μ,j} := #{(i, s) : q_{i} + s ≡ 0 (mod a), s = 0, …, j} + #{(i_{1}, i_{2}) : q_{i_{1}} ≡ q_{i_{2}} (mod a), i_{1} > i_{2}}.

Let m_{a,k,j} be the minimum over all such sequences μ of the value m_{a,μ,j}. Then

m_{a,k,j} = ∑_{i=1}^{k} #{m : 1 ≤ m ≤ i + j, m ≡ 0 (mod a)}.

In analogy with the n = k + 1 case, we make the following definition. Definition 4.11. We say a decreasing sequence of k integers is j-good with respect to p if m_{a,k,j} = m_{a,μ,j} for all a ≠ 0 (mod p).

Note that the definition of 0-good is the same as the original definition of good. Corresponding to Lemma 4.3, we have the obvious result. Lemma 4.12. Suppose that T is an irreducible representation of SU(k+1+j) corresponding to the partition λ with hands q_{k} > ⋯ > q_{1}. The degree d_{T} is a power of some prime p if and only if q_{k}, …, q_{1} is j-good with respect to p.

Finally, we complete the proof of Theorem 1.1 with the following proposition. Proposition 4.13. Let T be an irreducible representation of SU(k + 1 + j) with j > 0 corresponding to the partition with hands q_{k} > ⋯ > q_{1} > 1. The degree d_{T} is a prime power if and only if k = 1 and q_{1} = 2. Proof. Suppose that q_{k} > ⋯ > q_{1} > 1 is j-good with respect to p. Note that either q_{k} + j is not a multiple of p or q_{k} +j −1 is not a multiple of p. In the first case, we have m_{q_{k}+j,k,j} = 1, and so by Lemma 4.10,

1 = ∑_{i=1}^{k} #{m : 1 ≤ m ≤ i + j, m ≡ 0 (mod q_{k} + j)},

which means that there exists some 1 ≤ m ≤ j + k such that q_{k} + j divides m, which means that q_{k} + j ≤ m ≤ j + k. In the second case, we have m_{q_{k}+j−1,k,j} = 1 or 2, and so by a similar argument, q_{k} + j − 1 ≤ j + k. So, we certainly have q_{k} + j − 1 ≤ k + j, which means that q_{k} ≤ k + 1. This implies that q_{i} = i + 1 for all i = 1, …, k since q_{1} ≥ 2. Now, suppose that k > 1. We will show that in this case d_{T} cannot be a prime power.

We have

d_{T} = (k + 1) ∏_{i=1}^{k} \frac{(q_{i} + 1) ⋯ (q_{i} + j)}{(i + 1) ⋯ (i + j)}

= (k + 1) ∏_{i=1}^{k} \frac{(i + 2) ⋯ (i + j + 1)}{(i + 1) ⋯ (i + j)}

= (k + 1) ∏_{i=1}^{k} \frac{(i + j + 1)}{(i + 1)}

= \frac{1}{k!} ∏_{i=1}^{k}(i + j + 1)

= \frac{(j + 2) ⋯ (j + k + 1)}{k!}

= \binom{j + k + 1}{j + 1}.

But \binom{j+k+1}{j+1} is a prime power only when j + 1 = 1 or j + 1 = j + k + 1 − 1. However, j > 0 and k > 1, so d_{T} is not a prime power. □

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