Symmetry, Integrability and Geometry: Methods and Applications

SIGMA 13 (2017), 070, 10 pages

Restriction of Odd Degree Characters of Sn

arXiv:1705.08655v2 [math.RT] 5 Sep 2017

Christine BESSENRODT † , Eugenio GIANNELLI

‡

and Jørn B. OLSSON

§

†

Institute for Algebra, Number Theory and Discrete Mathematics,
Leibniz Universität Hannover, Welfengarten 1, D-30167 Hannover, Germany
E-mail: bessen@math.uni-hannover.de

‡

Department of Pure Mathematics and Mathematical Statistics, University of Cambridge,
Cambridge CB3 0WA, United Kingdom
E-mail: eg513@cam.ac.uk

§

Department of Mathematical Sciences, University of Copenhagen,
DK-2100 Copenhagen Ø, Denmark
E-mail: olsson@math.ku.dk

Received May 25, 2017, in final form August 30, 2017; Published online September 05, 2017
https://doi.org/10.3842/SIGMA.2017.070
Abstract. Let n and k be natural numbers such that 2k < n. We study the restriction
to Sn−2k of odd-degree irreducible characters of the symmetric group Sn . This analysis
completes the study begun in [Ayyer A., Prasad A., Spallone S., Sém. Lothar. Combin. 75
(2015), Art. B75g, 13 pages] and recently developed in [Isaacs I.M., Navarro G., Olsson J.B.,
Tiep P.H., J. Algebra 478 (2017), 271–282].
Key words: characters of symmetric groups; hooks in partitions
2010 Mathematics Subject Classification: 20C30; 05A17

1

Introduction

Let n be a natural number, and let χ be an irreducible character of odd degree of the symmetric
group Sn . Then there exists a unique odd-degree irreducible constituent of the restriction χSn−1 .
This interesting fact was discovered recently in [1]. The result had immediate applications in
the study of natural correspondences of characters of finite groups (see for example [2]). In [3,
Theorem A] the result mentioned above was generalized, by showing that given any k ∈ N such
that 2k < n, there exists a unique odd-degree irreducible constituent fkn (χ) of χSn−2k appearing
with odd multiplicity. The main goal of this article is to study for all n, k ∈ N the map
fkn : Irr20 (Sn ) −→ Irr20 (Sn−2k ),
naturally defined by Theorem A of [3]. All our results are proved using a description of fkn in
terms of the natural partition labels of the involved irreducible characters.
Before describing the main results of this paper, we introduce some vocabulary. If 2k appears
in the binary expansion of n we say that 2k is a binary digit of n. Similarly we say that two
natural numbers m and n are 2-disjoint if they do not have any common binary digit. On the
other hand, if m ≤ n and all the binary digits of m appear in the binary expansion of n, then we
say that m is a binary subsum of n. This will be denoted by m ⊆2 n. Let ν2 (n) be the exponent
of the highest power of 2 dividing the integer n.
This paper is a contribution to the Special Issue on the Representation Theory of the Symmetric Groups
and Related Topics. The full collection is available at https://www.emis.de/journals/SIGMA/symmetric-groups2018.html

2

C. Bessenrodt, E. Giannelli and J.B. Olsson

A question raised in [3] may be phrased as: For which n and k is fkn surjective? The
authors showed that fkn is surjective whenever 2k is a binary digit of n, and they observed that
otherwise fkn could be both surjective or not (see [3, Proposition 4.5 and Remark 4.6]). In this
paper we answer the question of surjectivity completely with the following result.
 
Theorem A. Let n ∈ N, k ∈ N0 be such that 2k < n. Let d(n, k) = ν2 2nk .
• If k = 0 then fkn is surjective if and only d(n, k) ≤ 2.
• If k > 0 then fkn is surjective if and only d(n, k) ≤ 1.
Theorem A is a consequence of Theorem 3.5 below, which describes the images of the maps fkn .
For all n ∈ N, k ∈ N0 with 2k < n and any ψ ∈ Irr20 (Sn−2k ) we define the set
 
E ψ, 2k = χ ∈ Irr20 (Sn ) | fkn (χ) = ψ ,


and set e ψ, 2k = E ψ, 2k . We show in Corollary 3.8 that the maps fkn are regular on their
images. This means that for any ψ in the image of fkn , the number e(ψ, 2k ) depends only on n
and k and not on the specific ψ. We also give a complete description of those ψ ∈ Irr20 (Sn−2k )
such that e(ψ, 2k ) = 0, in Theorem 3.5.
In the final part of the paper we study commutativity. For convenience, we sometimes
denote fkn just by fk , when the natural number n is clear from the context. Then, for k, ` ∈ N0 ,
k < `, such that 2k + 2` ≤ n, we may ask: when is fk f` = f` fk ? or more specifically:
k
`
when is fkn−2 f`n = f`n−2 fkn ? In [3, Proposition 4.3] it was proved that fk f` = f` fk whenever
2` < n < 2`+1 . This is the case ` = t in our second main result, which answers the question
completely.
Theorem B. Let n = 2t + m where 0 ≤ m < 2t . Suppose that k, ` satisfy 0 ≤ k < ` ≤ t and
2k + 2` ≤ n. Then, with the exception of the case n = 6, k = 0, ` = 1,
fk f` = f` fk if and only if 2k > m or ` = t.

2

Notation and background

Let n be a natural number. We let Irr(Sn ) denote the set of irreducible characters of Sn
and P(n) the set of partitions of n. The notation λ ∈ P(n) is sometimes replaced by λ ` n and
we write |λ| = n. There is a natural correspondence λ ↔ χλ between P(n) and Irr(Sn ). We
say then that λ labels χλ . We denote by Irr20 (Sn ) the set of irreducible characters of Sn of odd
degree. If χλ ∈ Irr20 (Sn ) we say that χλ is an odd character, we call λ an odd partition of n and
write λ `o n. Also the empty partition will be considered as an odd partition.
Remark 2.1. Let n, k be such that 2k < n. In [3, Theorem A and Proposition 4.2] it is shown
that the map fkn : Irr20 (Sn ) → Irr20 (Sn−2k ) may be described in terms of the odd partitions
labelling the odd characters as follows:
fkn (χλ ) = χµ ⇔ µ `o n − 2k can be obtained from λ `o n by removing a 2k -hook.
Correspondingly we write (by abuse of notation) fkn (λ) = µ. In fact when λ is odd, there is only
one 2k -hook of λ whose removal leads again to an odd partition; we will refer to such a hook
as an odd hook of λ. This combinatorial description of fkn will be used throughout this paper,
and we will regard fkn alsoas a map between
the corresponding sets of odd partitions. Also, for

µ `o n − 2k we set e µ, 2k = e χµ , 2k .

Restriction of Odd Degree Characters of Sn

3

We need some concepts and basic facts concerning hooks in partitions. For any integer
e ∈ N we denote by Ce (λ) and Qe (λ) the e-core and the e-quotient of λ, respectively. Then
e−1
P
Qe (λ) = (λ0 , . . . , λe−1 ) is an e-tuple of partitions satisfying n = |Ce (λ)| + e
|λi |. It is welli=0

known that a partition is uniquely determined by its e-core and e-quotient (we refer the reader
to [6] or [4, Chapter 2.7] for a detailed discussion on this topic).
Let He (λ) be the set of hooks of λ having length divisible by e, and let H(Qe (λ)) = ∪e−1
i=0 H(λi ).
As explained in [6, Theorem 3.3], there is a bijection between He (λ) and H(Qe (λ)) mapping
hooks in λ of length ex to hooks in the quotient of length x. Moreover, the bijection respects
the process of hook removal. Namely, the partition µ obtained by removing a ex-hook from λ
is such that Ce (µ) = Ce (λ) and the e-quotient of µ is obtained by removing an x-hook from one
of the partitions involved in Qe (λ).
For e = 2 we want to repeat the process of taking 2-cores and 2-quotients to obtain the
2-quotient tower Q2 (λ) and the 2-core tower C2 (λ) of λ. They have rows numbered by k ≥ 0.
(k)
(k)
(k)
The kth row Q2 (λ) of Q2 (λ) contains 2k partitions λi , 0 ≤ i ≤ 2k −1, and the kth row C2 (λ)

(k)
of C2 (λ) contains the 2-cores of these partitions in the same order, i.e., C2 λi , 0 ≤ i ≤ 2k − 1.
(0)

(1)

(1)

The 0th row of Q2 (λ) contains λ = λ0 itself, row 1 contains the partitions λ0 , λ1 occurring in the 2-quotient Q2 (λ), row 2 contains the partitions occurring in the 2-quotients of
(k) 
(k+1) (k+1) 
partitions occurring in row 1, and so on. Specifically we have Q2 λi
= λ2i , λ2i+1 for

(k)
i ∈ 0, 1, . . . , 2k − 1 . We remark that the 2k partitions in Q2 (λ) are the same as those in the
2k -quotient Q2k (λ) of λ, but in a different order for k ≥ 2.
(k)
We also introduce the k-data D2 (λ) of λ. This is a table containing the following k + 1 rows:
(j)
(k)
the k rows C2 (λ), j = 0, . . . , k − 1, and in addition the row Q2 (λ).
Remark 2.2. A partition λ may be recovered from its 2-core tower. For k > 0, it may also be
(k)
(l)
recovered from the knowledge of the k-data D2 (λ) of λ, because the rows C2 (λ) with l ≥ k
(k)
of C2 (λ) consist of the 2-core towers of the partitions in Q2 (λ).
Lemma 2.3. Suppose that λ ` n − 2k and µ ` n. The following are equivalent.
(i) λ is obtained from µ by removing a 2k -hook.

(k)
(k)
(k)
(ii) The k-data D2 (µ) and D2 (λ) coincide, except that for one i ∈ 0, . . . , 2k − 1 λi is
(k)

obtained from µi

by removing a 1-hook.

Proof . A 2k -hook H0 in µ corresponds in a canonical way to a 2k−1 -hook H1 in a partition
(1)
in Q2 (µ), i.e., in row 1 of the 2-quotient tower Q2 (µ). Continuing we see that H0 corresponds
(k)
(k)
in a canonical way to a 1-hook Hk in a partition µi in Q2 (µ), row k of Q2 (µ). If λ is obtained
(k)
by removing H0 from µ, this corresponds to λi being obtained by removing the 1-hook Hk
(k)
(k)
from µi (by repeated applications of [6, Theorem 3.3]). Apart from this the rows Q2 (µ)
(k)
(j)
(j)
and Q2 (λ) coincide. Note also that the rows C2 (µ) and C2 (λ) coincide for j = 0, . . . , k − 1,
since the removal of the hooks Hj of even length do not change the 2-cores.

Odd-degree characters of Sn and thus odd partitions were completely described in [5]. We
(k)
restate this result in a language which is convenient for our purposes. We let c2 (λ) be the sum
(k)
of the cardinalities of the partitions in the kth row C2 (λ) of C2 (λ).
(k)

Lemma 2.4 ([5]). Let λ be a partition. Then λ is odd if and only if c2 (λ) ≤ 1 for all k ≥ 0.
(k)

It may be decided from the k-data D2 (λ) whether λ is odd. The case k = 1 of the following
result appeared in [3, Lemma 4.1] and also in [1, Lemma 6].

4

C. Bessenrodt, E. Giannelli and J.B. Olsson
(k)

(k) 

Theorem 2.5. Let λ ` n, and let k ≥ 0 be fixed. Consider Q2 (λ) = λi
and only if the following conditions are all fulfilled:

. Then λ is odd if

(j)

(i) c2 (λ) ≤ 1 for all j < k.
(k)

(ii) The partitions λi , 0 ≤ i ≤ 2k − 1, are all odd.
(k)

(iii) The numbers λi , 0 ≤ i ≤ 2k − 1, are pairwise 2-disjoint.
 
P (k)
In this case
λi = 2nk .
i≥0

Proof . This is proved by induction on k ≥ 0, using Remark 2.2 and Lemma 2.4.



We illustrate the result above by giving an example.

Example 2.6. Let n = 15 and take λ = 5, 4, 22 , 12 ` 15. To decide whether λ is odd, we
(2)

(0)

choose k = 2 and compute the 2-data D2 (λ). The
2-core
is C2 (λ) = (1), giving C2 (λ) = ((1)).



2
2
Furthermore, the 2-quotient is Q2 (λ) = 2 , 1 , (1) , and computing the 2-cores C2 22 , 12 =

(1)
2 , 12
(0), C2 ((1))
=
(1),
we
obtain
the
next
row:
C
(λ)
=
((0),
(1)).
The
2-quotients
are
Q
2
2
2


= 12 , (1) , Q2 ((1)) = ((0), (0)); hence the final row of the 2-data table is obtained as


(2)
Q2 (λ) = 12 , (1), (0), (0) .
(2)

We visualize D2 (λ) like this:
(0)

C2 (λ) :

(1)

(1)

C2 (λ) :
(2)

Q2 (λ) :

(0)
(12 )

(1)
(1)

(0)

(0)

Theorem 2.5 shows that λ is odd and thus it contains a unique odd 4-hook. Again using the
(2)
theorem, it is clear that removing this 4-hook corresponds to the second partition (1) in Q2
being replaced by (0). Thus, removing the corresponding 4-hook of λ we obtain the odd partition

(2)
(2)
µ = 3, 23 , 12 ` 11 with the property that D2 (λ) and D2 (µ) differ only in their final row.
Remark 2.7. Using the construction of partitions from their 2-cores and 2-quotients already
mentioned, the criterion above can be applied to construct all odd partitions of n with a specific
kth row in the 2-quotient tower. For this, let n, k ∈ N, and take any sequencePof odd parti 
|νi | = 2nk .
tions νi , 0 ≤ i ≤ 2k − 1, such that the numbers |νi | are pairwise 2-disjoint, and
i≥0

Then there are exactly

Q

2m

odd partitions λ of n with

m<k
2m ⊆2 n

(k)
Q2 (λ)

= (νi ), obtained by choosing

one 2-core in row m of the k-data table to be (1), for each m < k such that 2m ⊆2 n.
The following easy consequence of Theorem 2.5 will be used repeatedly.
Lemma 2.8. Let 2t be the largest binary digit of n. A partition λ of n is odd if and only if λ
contains a unique 2t -hook and the partition obtained from λ by removing this 2t -hook is an odd
partition of n − 2t .

3

Surjectivity and regularity

The aim of this section is to study the images of the maps fkn for all n, k such that 2k ≤ n. For
this purpose we introduce the concept of d-good partitions (see Definition 3.1 below). This will
allow us to prove Theorem 3.5 (describing the images) and thus Theorem A (describing exactly
when fkn is surjective) and to show that the maps fkn are always regular on their image (see
Corollary 3.8).

Restriction of Odd Degree Characters of Sn

5

Definition 3.1. Let d ≥ 0. We call an odd partition λ d-good, if
(i) |λ| ≡ 2d − 1 mod 2d+1 .
(ii) C2d (λ) is a hook partition.
Let us remark that condition (i) may be reformulated as
(i∗ ) ν2 (|λ| + 1) = d.
In particular, if λ is d-good, then |λ| is odd if and only if d > 0.
The relevance of d-good partitions in our context is illuminated by the following reformulation
of [1, Theorem 2]:
Lemma 3.2. Let λ `o n. Let d = ν2 (n + 1). Then e(λ, 1) 6= 0 if and only if λ is d-good. In this
case, e(λ, 1) = 1 if d = 0, and e(λ, 1) = 2 if d > 0.
Lemma 3.3. Let λ be an odd partition, and let d ≥ 0. Then the following hold.
(1) For d ≤ 2, λ is d-good if and only if |λ| ≡ 2d − 1 mod 2d+1 .
(2) If λ is d-good, then C2d (λ) is a partition of 2d − 1.

Proof . If the odd partition λ is d-good, then |λ| = 2d − 1 + m where the binary digits of m
are at least 2d+1 . The hooks of λ corresponding to the binary digits of m may be decomposed
into 2d -hooks and thus do not contribute to C2d (λ). Thus |C2d (λ)| = 2d − 1. This shows (2).
For d = 0, 1, 2 we have |C2d (λ)| = 0, 1 and 3, respectively. Since all partitions of 0, 1 and 3 are
hook partitions, (1) follows.


Definition 3.4. If 2k ≤ n, we define d(n, k) = ν2 b 2nk c . Thus d(n, k) is the smallest integer
k+d
d ≥ 0 satisfying the condition
d(n, k) = 0 if and only if 2k ⊆2 n.
 n  2 d(n,k)⊆2 n. In particular,d(n,k)+1
Moreover, we may write 2k = 2
+ m(n, k) where 2
| m(n, k).
As mentioned in the introduction, the results in [3] show that fkn is a surjective (2k -to-1)-map
whenever 2k ⊆2 n, i.e., d(n, k) = 0. In the spirit of [1, Theorem 2], we now give a characterization
of the image of the map fkn for all n, k such that 2k < n.

Theorem 3.5. Let n ∈ N, k ∈ N0 be such that 2k < n. Let λ `o n−2k . Then e λ, 2k 6= 0if and
only if there exists a d(n, k)-good partition in the kth row of Q2 (λ). In this case, e λ, 2k = 2k
if d(n, k) = 0, and e λ, 2k = 2 if d(n, k) > 0.
Proof . If k = 0 then the statement
follows from Lemma 3.2. Hence assume that k ≥ 1. Let
n
d
d = d(n, k). By assumption 2k = 2 + m, where the binary digits of m are at least 2d+1 . Thus
 n−2k 

= 2d − 1 + m.
2k

Suppose first that e λ, 2k 6= 0 and that µ `o n satisfies fk (µ) = λ. From Remark 2.1 and
(k) 
(k)
(k)
Lemma 2.3 we get that there exists an i ∈ {0, 1, . . . , 2k − 1} such that f0 µi
= λi . Since µi
(k)
(k) 
(k)
(k)
and λi are odd, we get e λi , 1 6= 0. We have that λi and µi are both 2-disjoint with
 n−2k 

 
P (k)
P (k)
k
m1 :=
λj =
µj ⊆2 2k , by Theorem 2.5. Since m1 ⊆2 n−2
and m1 ⊆2 2nk ,
2k
j6=i

j6=i

(k)

(k)

we get m1 ⊆2 m. Thus λi = (2d − 1) + m2 and µi = 2d + m2 , where m2 = m − m1 ⊆2 m.

(k)
(k) 
(k)
In particular ν2 λi + 1 = ν2 µi
= d. Then Lemma 3.2 shows that λi is d-good.

(k)
Conversely, if λi is a d-good partition for some i ∈ 0, 1, . . . , 2k − 1 , then there exists
(k)

a µ∗ `o λi

(k)

(k)

+ 1 such that f0 (µ∗ ) = λi , by Lemma 3.2. We let µ be the partition where the
(k)

(k)

k-data D2 (µ) and D2 (λ) coincide, except that µi

(k)

= µ∗ . Since λ is odd and λi

is d-good,

6

C. Bessenrodt, E. Giannelli and J.B. Olsson
(k)

we know that λi


(k)
= 2d − 1 + m0 where m0 ⊆2 m, and λj ⊆2 m − m0 for all j 6= i. Hence
(k)

(k)

|µ∗ | = λi + 1 = 2d + m0 is 2-disjoint from all λj , j 6= i. Thus µ is an odd partition of n by
Theorem 2.5, and fk (µ) = λ by Lemma 2.3 and Remark 2.1.


P
k
(k) 
We conclude that e λ, 2k =
e λi , 1 . If d = 0 then n−2
is even. This implies
k
2
(k)

λi d−good
(k)
(k) 
that all λi are of even cardinality and thus d-good. Thus e λi , 1 = 1 for all i, and we get

(k)
(k)
(k)
e λ, 2k = 2k . If d > 0 there is exactly one λi in Q2 (λ) of odd cardinality. Only this λi



(k)
may be d-good and then e λ, 2k = e λi , 1 = 2. Otherwise e λ, 2k = 0.


 
Corollary 3.6. Let n ∈ N, k ∈ N0 be such that 2k < n, and let d = ν2 2nk . Let λ `o n − 2k .

(k)
Then e λ, 2k 6= 0 if and only if there exists a partition λi in the kth row of Q2 (λ) such that


(k)
(k)
λi ≡ 2d −
1 mod 2d+1 , and C2d λi
is a hook partition. In this case, e λ, 2k = 2k if d = 0,

and e λ, 2k = 2 if d > 0.
We are now ready to prove Theorem A. In fact, this is a consequence of Theorem 3.5 and it
is stated here as the following corollary.
Corollary 3.7 (Theorem A). Let n ∈ N, k ∈ N0 be such that 2k < n.
• If k = 0 then fkn is surjective if and only if d(n, k) ≤ 2.
• If k > 0 then fkn is surjective if and only if d(n, k) ≤ 1.
Proof . By Theorem 3.5, fkn is surjective if and only if for all λ `o n − 2k we have that the kth
(k)
row of Q2 (λ) contains a d(n, k)-good partition λj . By Theorem 2.5 and Definition 3.4, for any



P
k
(k)
d(n,k) − 1 + m(n, k).
=
2
λj = n−2
λ `o n − 2k we have
k
2
j≥0
(0)
Q2 (λ)

(0)

If k = 0 then
contains only λ = λ0 . Hence f0n is surjective if and only all odd
partitions of n − 1 are d(n, 0)-good. By Lemma 3.3(1), the latter condition holds when d =
d(n, 0) ≤ 2. On the other hand, if d = ν2 (n) > 2, then λ = (n − 5, 2, 2) is an odd partition of
n − 1 by Theorem 2.5, but C8 (λ) = (3, 2, 2) is not a hook, and hence C2d (λ) is not a hook. So λ
is not d-good, and thus f0n is not surjective.
(k)
Now assume k ≥ 1. Then Q2 (λ) contains at least two odd partitions. If d(n, k) ≥ 2 then

k
any d(n, k)-good partition µ satisfies 3 ⊆2 2d(n,k) − 1 ⊆2 |µ|. Write n−2
= 1 + m1 where m1 is
2k
(k)

even. Applying Remark 2.7, take any λ `o n−2k such that λ0
(k)

(k)

= 1 and λ1 is an odd partition

(k)

with λ1 = m1 . Then no partition in Q2 (λ) is d(n, k)-good. Thus fkn is not surjective. On
the other hand, if d(n, k) = 0 then 2k ⊆2 n and fkn is surjective [3, Proposition 4.5]. If d(n, k) = 1

k
(k)
then n−2
= 1 + m(n, k), where 4 | m(n, k). Thus any Q2 (λ) contains a partition with odd
2k
cardinality; this partition is 1-good, by Lemma 3.3. Again fkn is surjective.

It is an immediate consequence of Theorem 3.5 that fkn is regular on its image for all relevant
choices of n, k such that 2k < n. We have:
 
Corollary 3.8. Let n ∈ N, k ∈ N0 be such that 2k < n; set d = ν2 2nk . Let λ `o n − 2k .
Then

k

2 if d = 0;

e λ, 2k = 2 if d > 0, and the kth row of Q2 (λ) contains a d-good partition;


0 otherwise.

Restriction of Odd Degree Characters of Sn

7

Example 3.9. For an illustration, we consider odd extensions of odd partitions
by a 4-hook,

i.e., we take k = 2 above. For n > 22 we first compute d(n, k) = ν2 2nk , and then consider
odd partitions of n − 4 and their 4-extensions.

 For 4n = 6, d(6, 2) = 0. Thus e((2), 4) = 4. The
2
2
2
odd 4-extensions of (2) are (6), 3 , 2 , 1 , 2, 1 . For n = 10, d(10, 2) = 1. In this case,
e(λ, 4) = 2 for all odd partitions λ of 6. For instance, the odd 4-extensions of (6)
 are (10) and
2
2
(6, 3, 1). For n = 19, d(19, 2) = 2. Example 2.6 shows that for λ = 5, 4, 2 , 1 `o 15 there is
(2)

no 2-good partition in Q2 (λ), hence e(λ, 4) = 0.

4

Deciding commutativity of the maps fk and f`

Let n ∈ N, and suppose that 0 ≤ k < ` satisfy 2k + 2` ≤ n. As stated in the introduction,
we want to complete the discussion of the commutativity of the maps fk and f` . Since the
relevant n will always be apparent for the maps fkn in this section, we just write fk .
We write (n; k, `) ∈ T if for all λ `o n we have fk f` (λ) = f` fk (λ). Otherwise we write
(n; k, `) ∈ F.
In this section we will prove Theorem B, which may be reformulated as follows.
Theorem 4.1. Let n = 2t + m where 0 ≤ m < 2t . Suppose that k, ` satisfy 0 ≤ k < ` and
2k + 2` ≤ n. Then with the exception of (6; 0, 1)
(n; k, `) ∈ F if and only if ` < t and 2k ≤ m.
The proof of Theorem 4.1 is based on a series of lemmas. The first lemmas concern two
extreme cases, where fk and f` commute.
In the case ` = t we have the following result as a reformulation of [3, Proposition 4.3].
Lemma 4.2. Let n = 2t + m with 0 ≤ m < 2t . If 2k ≤ m, then (n; k, t) ∈ T .
It is also known that in the case where n is a power of 2, the maps fk and f` commute [3,
Remark 4.4], and we include a short proof here.
Lemma 4.3. If n = 2t then (n; k, `) ∈ T for all k, `.

Proof . If 0 ≤ b ≤ a are integers then the binomial coefficient ab is odd if and onlyif b ⊆2 a,
by Lucas’ theorem. The odd partitions of 2t are exactly the hook partitions 2t − b, 1b , 0 ≤ b ≤

t
2t − 1, of degree 2 b−1 . Hence for k ∈ {0, 1, . . . , t − 1} we have
(

2t − b − 2k , 1b
if 2k 6⊆2 b,

fk (λ) =
k
2t − b, 1b−2
if 2k ⊆2 b.
It follows that for any k, ` < t and odd partition λ of 2t , we have f` fk (λ) = fk f` (λ).



Lemma 4.4. Let n = 2t + m with 0 ≤ m < 2t . Suppose that k, ` satisfy 0 ≤ k < ` and
2k + 2` ≤ n. If m < 2k then (n; k, `) ∈ T .
Proof . We use induction on k ≥ 0. For k = 0 we have m = 0 and the claim follows from
Lemma 4.3. Suppose that k ≥ 1 and that the claim has been proved up to k − 1. Let λ `o n.
Odd hooks of length 2k and 2` in λ correspond to odd hooks of length 2k−1 and 2`−1 in the
2-quotient Q2 (λ) =(λ0 , λ1 ) of λ. From Theorem 2.5 we deduce that |λ0 | and
2-disjoint
 m|λ
 1 | are
n
t−1
k−1
binary subsums of 2 , so one of them contains 2 , say |λ0 |; then |λ1 | ≤ 2 < 2
< 2`−1 .
k−1
Thus the odd 2 -hook in Q2 (λ) has to be in λ0 . Therefore
Q2 (fk (λ)) = (fk−1 (λ0 ), λ1 ).

8

C. Bessenrodt, E. Giannelli and J.B. Olsson

Applying f` , the odd 2`−1 -hook cannot be in λ1 , hence
Q2 (f` fk (λ)) = (f`−1 fk−1 (λ0 ), λ1 )).
 
 
In particular, we know that |λ0 | ≥ 2`−1 + 2k−1 . Also |λ0 | + |λ1 | = n2 = 2t−1 + m
2 . We have
t−1
t−1
already
that 2
is the largest binary digit of |λ0 |; furthermore |λ0 |−2
is a binary subsum
  seenk−1
of m
<
2
.
We
may
therefore
apply
the
inductive
hypothesis
to
λ
to
get
f`−1 fk−1 (λ0 ) =
0
2
fk−1 f`−1 (λ0 ). This implies that Q2 (fk f` (λ)) = Q2 (f` fk (λ)) and thus fk f` (λ) = f` fk (λ).

Lemmas 4.2 and 4.4 show that the only if part of the theorem is true. We now turn to the if
part. We start by proving the statement for k = 0 and use this as part of an inductive argument.
Lemma 4.5. Let n = 2t + m with 0 < m < 2t . If 0 < ` < t then (n; 0, `) ∈ F, with the exception
of (6; 0, 1).
Proof . The result is easily checked for n ≤ 8, which includes the exception (6; 0, 1). So we
assume that t ≥ 3.

Case 1: 2` < m. Then m ≥ 3, since ` > 0. Consider the partition λ = m, m, 1a ` n where
a = n − 2m = 2t − m. The (1,1)-hook length of λ is 2t + 1. The (2,1)-hook length of λ is 2t .
Removing the (2,1)-hook hook we get the odd partition (m), so λ is odd, by Lemma 2.8. We
claim that

f0 (λ) = m, m, 1a−1 .

Indeed we cannot have f0 (λ) = m, m − 1, 1a because this partition does not have a hook of
length 2t , and thus it is not odd. Now


f` (f0 (λ)) = f` m, m, 1a−1 = m, m − 2` , 1a−1

`
since m, m, 1a−1−2 and m − 1, m − 2` + 1, 1a−1 both do not have a hook of length 2t and
thus are not odd (again by Lemma 2.8).
On the other hand,
 
f` (λ) = m − 1, m − 2` − 1 , 1a .

`
Indeed, the other candidates for f` (λ), which are m, m − 2` , 1a and m, m, 1a−2 , do not have
hooks of length 2t . Then
 

f0 (f` (λ)) = f0 m − 1, m − 2` − 1 , 1a = m − 1, m − 2` , 1a .
This follows (again) by observing that all the other partitions of n − 2` − 1 obtained from
m − 1, m − 2` − 1 , 1a by removing a node do not have hooks of length 2t . Thus f0 (f` (λ)) 6=
f` (f0 (λ)).

Case 2: m < 2` . Consider the partition λ = n − 2` , m + 1, 1a , where a = 2` − (m + 1).
Note that n − 2` ≥ m + 1 since ` < t by assumption, and that a ≥ 0. The (1,1)-hook length of
λ is n − m = 2t . Removing this hook we get the odd partition (m), so λ is odd. The (2,1)-hook
length of λ is 2` . Now

f0 (λ) = n − 2` , m, 1a
since the other candidates do not have hooks of length 2t . Then

f` (f0 (λ)) = f` n − 2` , m, 1a = µ,

Restriction of Odd Degree Characters of Sn

9

where µ is obtained from f0 (λ) by removing a 2` -hook in the first
 row. (There are only hooks
of length < 2` in the other rows.) In fact, µ = n − 2`+1 , m, 1a since n − 2`+1 ≥ n − 2t = m.
Thus f` (f0 (λ)) has at least 2 parts. On the other hand

f` (λ) = n − 2`
since this odd partition is obtained from the odd partition λ by removing a 2` -hook (the one
in (2, 1)). It follows that

f0 (f` (λ)) = n − 2` − 1
and again f0 (f` (λ)) 6= f` (f0 (λ)).

Case 3: m = 2` . Then n = 2t + 2` . If ` ≥ 2 then choose λ = 2t , 2` − 1, 1 . The (1, 2)-hook
length of λ is 2t ; thus λ is an odd partition since removing this 2t -hook gives an odd partition
2` − 2, 1, 1 of 2` . We have f0 (λ) = 2t , 2` − 2, 1 since the other candidates are not odd. Then

f` (f0 (λ)) = 2t − 2` , 2` − 2, 1 .

The (2, 1)-hook length of λ is 2` , so f` (λ) = 2t and

f0 (f` (λ)) = 2t − 1 ,
showing f0 (f` (λ)) 6= f` (f0 (λ)).

On the other hand, if ` = 1 then chooseλ = 2t − 2, 2, 2 `o 2t + 2 = n. Since t ≥ 3, it is now
easy to show that f1 (f0 (λ)) = 2t − 4, 2, 1 . On the other hand we see that f0 (f1 (λ)) is a hook
partition of 2t − 1 = n − 3 and therefore is not equal to f1 (f0 (λ)).

Lemma 4.6. If (n; k, `) ∈ F then also (2n; k + 1, ` + 1) ∈ F and (2n + 1; k + 1, ` + 1) ∈ F.
Proof . Let the odd partition µ of n satisfy fk f` (µ) 6= f` fk (µ). Let λ be a partition of 2n or
2n + 1 having 2-quotient Q2 (λ) = (µ, (0)). Then λ is odd, by Theorem 2.5. We have
Q2 (fk+1 f`+1 (λ)) = (fk f` (µ), (0)) 6= (f` fk (µ), (0)) = Q2 (f`+1 fk+1 (λ)),
so that fk+1 f`+1 (λ) 6= f`+1 fk+1 (λ).



We are now ready to conclude this section with the proof of Theorem B.
Proof of Theorem 4.1. The only if part follows from Lemmas 4.2 and 4.4. To prove the if
part we use induction on k ≥ 0. If k = 0, then the statement follows from Lemma 4.5. Let k > 1
and suppose that the assertion is true up to and including k − 1. To show that (n; k, `) ∈ F
t
it suffices to prove (b n2 c; k − 1, ` − 1) ∈ F, by Lemma 4.6.
We are

 massuming

 mn = 2t−1+ m,
n
t
k
`
t−1
0 ≤ m < 2 , 0 ≤ k < ` ≤ t and 2 + 2 ≤ n. This implies 2 = 2  +
and
 2 , 0 ≤ 2 < 2
n
2k−1 +2`−1 ≤ n2 . We mayapply
the
inductive
hypothesis
to
get
;
k−1,
`−1
∈
F,
and
then
2


(n; k, `) ∈ F except when n2 ; k − 1, ` − 1 = (6; 0, 1). In that case we are considering (12;1,2)
or (13;1,2) which are both in F, by direct computation (consider for example (6, 4, 2) `o 12 and
(6, 4, 3) `o 13, respectively).


Acknowledgements
We thank Gabriel Navarro for providing some useful tables which helped us find the patterns
explained by the results of this paper. The second author is grateful to Trinity Hall, University
of Cambridge, for funding his research. Thanks also go to the referee for helpful comments on
a previous version of the article.

10

C. Bessenrodt, E. Giannelli and J.B. Olsson

References
[1] Ayyer A., Prasad A., Spallone S., Odd partitions in Young’s lattice, Sém. Lothar. Combin. 75 (2015),
Art. B75g, 13 pages, arXiv:1601.01776.
[2] Giannelli E., Kleshchev A., Navarro G., Tiep P.H., Restriction of odd degree characters and natural correspondences, Int. Math. Res. Not., to appear, arXiv:1601.04423.
[3] Isaacs I.M., Navarro G., Olsson J.B., Tiep P.H., Character restrictions and multiplicities in symmetric
groups, J. Algebra 478 (2017), 271–282.
[4] James G., Kerber A., The representation theory of the symmetric group, Encyclopedia of Mathematics and
its Applications, Vol. 16, Addison-Wesley Publishing Co., Reading, Mass., 1981.
[5] Macdonald I.G., On the degrees of the irreducible representations of symmetric groups, Bull. London Math.
Soc. 3 (1971), 189–192.
[6] Olsson J.B., Combinatorics and representations of finite groups, Vorlesungen aus dem Fachbereich Mathematik der Universität Essen, Heft 20, 1994, available at http://www.math.ku.dk/~olsson/manus/comb_
rep_all.pdf.

