Restriction of Odd Degree Characters of Sn
We study the restriction of odd-degree irreducible characters of the symmetric group Sn.
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irk-123456789-1487572019-02-19T01:23:33Z Restriction of Odd Degree Characters of Sn Bessenrodt, C. Giannelli, E. Olsson, J.B. We study the restriction of odd-degree irreducible characters of the symmetric group Sn. 2017 Article Restriction of Odd Degree Characters of Sn / C. Bessenrodt, E. Giannelli, J.B. Olsson // Symmetry, Integrability and Geometry: Methods and Applications. — 2017. — Т. 13. — Бібліогр.: 6 назв. — англ. 1815-0659 2010 Mathematics Subject Classification: 20C30; 05A17 DOI:10.3842/SIGMA.2017.070 http://dspace.nbuv.gov.ua/handle/123456789/148757 en Symmetry, Integrability and Geometry: Methods and Applications Інститут математики НАН України |
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We study the restriction of odd-degree irreducible characters of the symmetric group Sn. |
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Restriction of Odd Degree Characters of Sn |
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Restriction of Odd Degree Characters of Sn |
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Restriction of Odd Degree Characters of Sn |
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Restriction of Odd Degree Characters of Sn |
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Restriction of Odd Degree Characters of Sn / C. Bessenrodt, E. Giannelli, J.B. Olsson // Symmetry, Integrability and Geometry: Methods and Applications. — 2017. — Т. 13. — Бібліогр.: 6 назв. — англ. |
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Symmetry, Integrability and Geometry: Methods and Applications SIGMA 13 (2017), 070, 10 pages
Restriction of Odd Degree Characters of Sn
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
groupSn. 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 fnk (χ) of χS
n−2k
appearing
with odd multiplicity. The main goal of this article is to study for all n, k ∈ N the map
fnk : Irr2′(Sn) −→ Irr2′(Sn−2k),
naturally defined by Theorem A of [3]. All our results are proved using a description of fnk 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-groups-
2018.html
mailto:bessen@math.uni-hannover.de
mailto:eg513@cam.ac.uk
mailto:olsson@math.ku.dk
https://doi.org/10.3842/SIGMA.2017.070
https://www.emis.de/journals/SIGMA/symmetric-groups-2018.html
https://www.emis.de/journals/SIGMA/symmetric-groups-2018.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 fnk surjective? The
authors showed that fnk is surjective whenever 2k is a binary digit of n, and they observed that
otherwise fnk 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
(⌊
n
2k
⌋)
.
• If k = 0 then fnk is surjective if and only d(n, k) ≤ 2.
• If k > 0 then fnk 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 fnk .
For all n ∈ N, k ∈ N0 with 2k < n and any ψ ∈ Irr2′(Sn−2k) we define the set
E
(
ψ, 2k
)
=
{
χ ∈ Irr2′(Sn) | fnk (χ) = ψ
}
,
and set e
(
ψ, 2k
)
=
∣∣E(ψ, 2k)∣∣. We show in Corollary 3.8 that the maps fnk are regular on their
images. This means that for any ψ in the image of fnk , 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 ∈ Irr2′(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 fnk 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 fkf` = f`fk? or more specifically:
when is fn−2
`
k fn` = fn−2
k
` fnk ? In [3, Proposition 4.3] it was proved that fkf` = 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,
fkf` = 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 Irr2′(Sn) the set of irreducible characters of Sn of odd
degree. If χλ ∈ Irr2′(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 fnk : Irr2′(Sn) → Irr2′(Sn−2k) may be described in terms of the odd partitions
labelling the odd characters as follows:
fnk (χ
λ) = χµ ⇔ µ `o n− 2k can be obtained from λ `o n by removing a 2k-hook.
Correspondingly we write (by abuse of notation) fnk (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 fnk will be used throughout this paper,
and we will regard fnk also as 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
Qe(λ) = (λ0, . . . , λe−1) is an e-tuple of partitions satisfying n = |Ce(λ)| + e
e−1∑
i=0
|λi|. It is well-
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).
LetHe(λ) be the set of hooks of λ having length divisible by e, and letH(Qe(λ)) = ∪e−1i=0H(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.
The kth row Q(k)
2 (λ) of Q2(λ) contains 2
k partitions λ
(k)
i , 0 ≤ i ≤ 2k−1, and the kth row C(k)2 (λ)
of C2(λ) contains the 2-cores of these partitions in the same order, i.e., C2
(
λ
(k)
i
)
, 0 ≤ i ≤ 2k − 1.
The 0th row of Q2(λ) contains λ = λ
(0)
0 itself, row 1 contains the partitions λ
(1)
0 , λ
(1)
1 oc-
curring in the 2-quotient Q2(λ), row 2 contains the partitions occurring in the 2-quotients of
partitions occurring in row 1, and so on. Specifically we have Q2
(
λ
(k)
i
)
=
(
λ
(k+1)
2i , λ
(k+1)
2i+1
)
for
i ∈
{
0, 1, . . . , 2k − 1
}
. We remark that the 2k partitions in Q(k)
2 (λ) are the same as those in the
2k-quotient Q2k(λ) of λ, but in a different order for k ≥ 2.
We also introduce the k-data D(k)
2 (λ) of λ. This is a table containing the following k+1 rows:
the k rows C(j)2 (λ), j = 0, . . . , k − 1, and in addition the row Q(k)
2 (λ).
Remark 2.2. A partition λ may be recovered from its 2-core tower. For k > 0, it may also be
recovered from the knowledge of the k-data D(k)
2 (λ) of λ, because the rows C(l)2 (λ) with l ≥ k
of C2(λ) consist of the 2-core towers of the partitions in Q(k)
2 (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.
(ii) The k-data D(k)
2 (µ) and D(k)
2 (λ) coincide, except that for one i ∈
{
0, . . . , 2k − 1
}
λ
(k)
i is
obtained from µ
(k)
i 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
in Q(1)
2 (µ), i.e., in row 1 of the 2-quotient tower Q2(µ). Continuing we see that H0 corresponds
in a canonical way to a 1-hook Hk in a partition µ
(k)
i in Q(k)
2 (µ), row k of Q2(µ). If λ is obtained
by removing H0 from µ, this corresponds to λ
(k)
i being obtained by removing the 1-hook Hk
from µ
(k)
i (by repeated applications of [6, Theorem 3.3]). Apart from this the rows Q(k)
2 (µ)
and Q(k)
2 (λ) coincide. Note also that the rows C(j)2 (µ) and C(j)2 (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
restate this result in a language which is convenient for our purposes. We let c
(k)
2 (λ) be the sum
of the cardinalities of the partitions in the kth row C(k)2 (λ) of C2(λ).
Lemma 2.4 ([5]). Let λ be a partition. Then λ is odd if and only if c
(k)
2 (λ) ≤ 1 for all k ≥ 0.
It may be decided from the k-data D(k)
2 (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
Theorem 2.5. Let λ ` n, and let k ≥ 0 be fixed. Consider Q(k)
2 (λ) =
(
λ
(k)
i
)
. Then λ is odd if
and only if the following conditions are all fulfilled:
(i) c
(j)
2 (λ) ≤ 1 for all j < k.
(ii) The partitions λ
(k)
i , 0 ≤ i ≤ 2k − 1, are all odd.
(iii) The numbers
∣∣λ(k)i ∣∣, 0 ≤ i ≤ 2k − 1, are pairwise 2-disjoint.
In this case
∑
i≥0
∣∣λ(k)i ∣∣ = ⌊ n2k ⌋.
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
choose k = 2 and compute the 2-data D(2)
2 (λ). The 2-core is C2(λ) = (1), giving C(0)2 (λ) = ((1)).
Furthermore, the 2-quotient is Q2(λ) =
((
22, 12
)
, (1)
)
, and computing the 2-cores C2
((
22, 12
))
=
(0), C2((1)) = (1), we obtain the next row: C(1)2 (λ) = ((0), (1)). The 2-quotients are Q2
((
22, 12
))
=
((
12
)
, (1)
)
, Q2((1)) = ((0), (0)); hence the final row of the 2-data table is obtained as
Q(2)
2 (λ) =
((
12
)
, (1), (0), (0)
)
.
We visualize D(2)
2 (λ) like this:
C(0)2 (λ) : (1)
C(1)2 (λ) : (0) (1)
Q(2)
2 (λ) : (12) (1) (0) (0)
Theorem 2.5 shows that λ is odd and thus it contains a unique odd 4-hook. Again using the
theorem, it is clear that removing this 4-hook corresponds to the second partition (1) in Q(2)
2
being replaced by (0). Thus, removing the corresponding 4-hook of λ we obtain the odd partition
µ =
(
3, 23, 12
)
` 11 with the property that D(2)
2 (λ) and D(2)
2 (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 sequence of odd parti-
tions νi, 0 ≤ i ≤ 2k − 1, such that the numbers |νi| are pairwise 2-disjoint, and
∑
i≥0
|νi| =
⌊
n
2k
⌋
.
Then there are exactly
∏
m<k
2m⊆2n
2m odd partitions λ of n with Q(k)
2 (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 fnk 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 fnk is surjective) and to show that the maps fnk 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 n
2k
c
)
. Thus d(n, k) is the smallest integer
d ≥ 0 satisfying the condition 2k+d ⊆2 n. In particular, d(n, k) = 0 if and only if 2k ⊆2 n.
Moreover, we may write
⌊
n
2k
⌋
= 2d(n,k) +m(n, k) where 2d(n,k)+1 |m(n, k).
As mentioned in the introduction, the results in [3] show that fnk 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 fnk 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= 0 if 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
d = d(n, k). By assumption
⌊
n
2k
⌋
= 2d+m, where the binary digits of m are at least 2d+1. Thus⌊
n−2k
2k
⌋
=
(
2d − 1
)
+m.
Suppose first that e
(
λ, 2k
)
6= 0 and that µ `o n satisfies fk(µ) = λ. From Remark 2.1 and
Lemma 2.3 we get that there exists an i ∈ {0, 1, . . . , 2k−1} such that f0
(
µ
(k)
i
)
= λ
(k)
i . Since µ
(k)
i
and λ
(k)
i are odd, we get e
(
λ
(k)
i , 1
)
6= 0. We have that
∣∣λ(k)i ∣∣ and ∣∣µ(k)i ∣∣ are both 2-disjoint with
m1 :=
∑
j 6=i
∣∣λ(k)j ∣∣ = ∑
j 6=i
∣∣µ(k)j ∣∣ ⊆2
⌊
n−2k
2k
⌋
, by Theorem 2.5. Since m1 ⊆2
⌊
n−2k
2k
⌋
and m1 ⊆2
⌊
n
2k
⌋
,
we get m1 ⊆2 m. Thus
∣∣λ(k)i ∣∣ = (2d − 1) +m2 and
∣∣µ(k)i ∣∣ = 2d +m2, where m2 = m−m1 ⊆2 m.
In particular ν2
(∣∣λ(k)i ∣∣+ 1
)
= ν2
(∣∣µ(k)i ∣∣) = d. Then Lemma 3.2 shows that λ
(k)
i is d-good.
Conversely, if λ
(k)
i is a d-good partition for some i ∈
{
0, 1, . . . , 2k − 1
}
, then there exists
a µ∗ `o
∣∣λ(k)i ∣∣+ 1 such that f0(µ
∗) = λ
(k)
i , by Lemma 3.2. We let µ be the partition where the
k-data D(k)
2 (µ) and D(k)
2 (λ) coincide, except that µ
(k)
i = µ∗. Since λ is odd and λ
(k)
i is d-good,
6 C. Bessenrodt, E. Giannelli and J.B. Olsson
we know that
∣∣λ(k)i ∣∣ = (2d − 1
)
+m′ where m′ ⊆2 m, and
∣∣λ(k)j ∣∣ ⊆2 m−m′ for all j 6= i. Hence
|µ∗| =
∣∣λ(k)i ∣∣+1 = 2d+m′ is 2-disjoint from all
∣∣λ(k)j ∣∣, 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.
We conclude that e
(
λ, 2k
)
=
∑
λ
(k)
i d−good
e
(
λ
(k)
i , 1
)
. If d = 0 then
⌊
n−2k
2k
⌋
is even. This implies
that all λ
(k)
i are of even cardinality and thus d-good. Thus e
(
λ
(k)
i , 1
)
= 1 for all i, and we get
e
(
λ, 2k
)
= 2k. If d > 0 there is exactly one λ
(k)
i in Q(k)
2 (λ) of odd cardinality. Only this λ
(k)
i
may be d-good and then e
(
λ, 2k
)
= e
(
λ
(k)
i , 1
)
= 2. Otherwise e
(
λ, 2k
)
= 0. �
Corollary 3.6. Let n ∈ N, k ∈ N0 be such that 2k < n, and let d = ν2
(⌊
n
2k
⌋)
. Let λ `o n− 2k.
Then e
(
λ, 2k
)
6= 0 if and only if there exists a partition λ
(k)
i in the kth row of Q2(λ) such that∣∣λ(k)i ∣∣ ≡ 2d−1 mod 2d+1, and C2d
(
λ
(k)
i
)
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 fnk is surjective if and only if d(n, k) ≤ 2.
• If k > 0 then fnk is surjective if and only if d(n, k) ≤ 1.
Proof. By Theorem 3.5, fnk is surjective if and only if for all λ `o n− 2k we have that the kth
row of Q2(λ) contains a d(n, k)-good partition λ
(k)
j . By Theorem 2.5 and Definition 3.4, for any
λ `o n− 2k we have
∑
j≥0
∣∣λ(k)j ∣∣ = ⌊n−2k2k
⌋
=
(
2d(n,k) − 1
)
+m(n, k).
If k = 0 then Q(0)
2 (λ) contains only λ = λ
(0)
0 . Hence fn0 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 fn0 is not surjective.
Now assume k ≥ 1. Then Q(k)
2 (λ) contains at least two odd partitions. If d(n, k) ≥ 2 then
any d(n, k)-good partition µ satisfies 3 ⊆2 2
d(n,k)−1 ⊆2 |µ|. Write
⌊
n−2k
2k
⌋
= 1+m1 where m1 is
even. Applying Remark 2.7, take any λ `o n−2k such that
∣∣λ(k)0
∣∣ = 1 and λ
(k)
1 is an odd partition
with
∣∣λ(k)1
∣∣ = m1. Then no partition in Q(k)
2 (λ) is d(n, k)-good. Thus fnk is not surjective. On
the other hand, if d(n, k) = 0 then 2k ⊆2 n and fnk is surjective [3, Proposition 4.5]. If d(n, k) = 1
then
⌊
n−2k
2k
⌋
= 1 +m(n, k), where 4 | m(n, k). Thus any Q(k)
2 (λ) contains a partition with odd
cardinality; this partition is 1-good, by Lemma 3.3. Again fnk is surjective. �
It is an immediate consequence of Theorem 3.5 that fnk 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
(⌊
n
2k
⌋)
. Let λ `o n − 2k.
Then
e
(
λ, 2k
)
=
2k if d = 0;
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
(⌊
n
2k
⌋)
, and then consider
odd partitions of n− 4 and their 4-extensions. For n = 6, d(6, 2) = 0. Thus e((2), 4) = 4. The
odd 4-extensions of (2) are (6),
(
32
)
,
(
22, 12
)
,
(
2, 14
)
. 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
(6, 3, 1). For n = 19, d(19, 2) = 2. Example 2.6 shows that for λ =
(
5, 4, 22, 12
)
`o 15 there is
no 2-good partition in Q(2)
2 (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 fnk in this section, we just write fk.
We write (n; k, `) ∈ T if for all λ `o n we have fkf`(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
(
a
b
)
is odd if and only if b ⊆2 a,
by Lucas’ theorem. The odd partitions of 2t are exactly the hook partitions
(
2t− b, 1b
)
, 0 ≤ b ≤
2t − 1, of degree
(
2t−1
b
)
. Hence for k ∈ {0, 1, . . . , t− 1} we have
fk(λ) =
{(
2t − b− 2k, 1b
)
if 2k 6⊆2 b,(
2t − b, 1b−2k
)
if 2k ⊆2 b.
It follows that for any k, ` < t and odd partition λ of 2t, we have f`fk(λ) = fkf`(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 |λ1| are 2-disjoint
binary subsums of
⌊
n
2
⌋
, so one of them contains 2t−1, say |λ0|; then |λ1| ≤
⌊
m
2
⌋
< 2k−1 < 2`−1.
Thus the odd 2k−1-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`−1fk−1(λ0), λ1)).
In particular, we know that |λ0| ≥ 2`−1 + 2k−1. Also |λ0|+ |λ1| =
⌊
n
2
⌋
= 2t−1 +
⌊
m
2
⌋
. We have
already seen that 2t−1 is the largest binary digit of |λ0|; furthermore |λ0|−2t−1 is a binary subsum
of
⌊
m
2
⌋
< 2k−1. We may therefore apply the inductive hypothesis to λ0 to get f`−1fk−1(λ0) =
fk−1f`−1(λ0). This implies that Q2(fkf`(λ)) = Q2(f`fk(λ)) and thus fkf`(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 choose λ =
(
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 fkf`(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+1f`+1(λ)) = (fkf`(µ), (0)) 6= (f`fk(µ), (0)) = Q2(f`+1fk+1(λ)),
so that fk+1f`+1(λ) 6= f`+1fk+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
it suffices to prove (bn2 c; k − 1, ` − 1) ∈ F , by Lemma 4.6. We are assuming n = 2t + m,
0 ≤ m < 2t, 0 ≤ k < ` ≤ t and 2k+2` ≤ n. This implies
⌊
n
2
⌋
= 2t−1+
⌊
m
2
⌋
, 0 ≤
⌊
m
2
⌋
< 2t−1 and
2k−1+2`−1 ≤
⌊
n
2
⌋
. We may apply the inductive hypothesis to get
(⌊
n
2
⌋
; k−1, `−1
)
∈ F , and then
(n; k, `) ∈ F except when
(⌊
n
2
⌋
; 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
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https://doi.org/10.1093/imrn/rnw174
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http://www.math.ku.dk/~olsson/manus/comb_rep_all.pdf
http://www.math.ku.dk/~olsson/manus/comb_rep_all.pdf
1 Introduction
2 Notation and background
3 Surjectivity and regularity
4 Deciding commutativity of the maps fk and f
References
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