The e-core, the e-quotient, and the e-weight

Let $e\ge 2$, let $\lambda ⊢n$ and let $X=\{h_1>h_2>\cdots >h_m\}$ be a $\beta$-set for $\lambda$, with first-column hook lengths $b_i(\lambda )$ and $e$-abacus configuration $A_X$ as in Beta-sets, the shift, and the e-runner abacus of a partition; hooks and their lengths are those of Hook, arm, leg, and hook length of a box, and the hook/bead-gap dictionary is Hooks are bead-gap pairs on the abacus.

$e$-core partition. The partition $\lambda$ is an $e$-core when $e\mathrm{\notin }H(\lambda )$, that is, when no hook of $\lambda$ has length $e$; by Hooks are bead-gap pairs on the abacus(iii) this is the case exactly when no bead of $A_X$ has a gap in the position $e$ labels above it on its own runner. The empty partition is an $e$-core, and so is every $\lambda$ with $\mathrm{\mid }\lambda \mathrm{\mid }<e$.

Runner label sets. For $i\in \{0,1,\dots ,e-1\}$ let $X_i^{(e)}:=\{a\in {\mathbb{N}}_0:ae+i\in X\}$ be the set of row labels of the beads of $A_X$ on runner $i$. The $X_i^{(e)}$ are finite subsets of ${\mathbb{N}}_0$ and may overlap. The disjoint union ${⨆}_i\{i\}\times X_i^{(e)}$ records each bead with its runner and row; the map $(i,a)\mapsto ae+i$ bijects it with $X$.

$e$-core. The top-packed $\beta$-set of $X$ is $X^{(e)}:={\bigcup }_{i=0}^{e-1}\{ae+i:0\le a\le \mathrm{\mid }{X}_i^{(e)}\mathrm{\mid }-1\},$ the configuration obtained from $A_X$ by sliding every bead up as far as it goes on its own runner, i.e. into the topmost $\mathrm{\mid }{X}_i^{(e)}\mathrm{\mid }$ positions of runner $i$. The $e$-core of $\lambda$ is $C_e(\lambda ):=P(X^{(e)})\mathrm{.}$

$e$-quotient and $e$-weight. Choose the $\beta$-set $X$ of $\lambda$ to be normalised, meaning $\mathrm{\mid }X\mathrm{\mid }\equiv 0(\mathrm{m}\mathrm{o}\mathrm{d}e)$. Then define the $e$-quotient of $\lambda$ to be the $e$-tuple of partitions $Q_e(\lambda ):=({\lambda }^{(0)},{\lambda }^{(1)},\dots ,{\lambda }^{(e-1)}),{\lambda }^{(i)}:=P(X_i^{(e)}),$ each runner read as a $1$-abacus, and the $e$-weight of $\lambda$ to be $w_e(\lambda ):=\mathrm{\mid }{Q}_e(\lambda )\mathrm{\mid }:={\sum }_{i=0}^{e-1}\mathrm{\mid }{\lambda }^{(i)}\mathrm{\mid }\mathrm{.}$ A normalised $\beta$-set exists: by Beta-sets, the shift, and the e-runner abacus of a partition the $\beta$-sets of $\lambda$ are the sets $X_{\lambda }+l$, $l\in {\mathbb{N}}_0$, and $\mathrm{\mid }{X}_{\lambda }+l\mathrm{\mid }=\mathrm{\mid }{X}_{\lambda }\mathrm{\mid }+l$, so among the $e$ consecutive shifts $l,\dots ,l+e-1$ exactly one has size divisible by $e$.

No choice principle is used.

• Well-definedness. Under the shift $X\mapsto X+1$, the runner sets satisfy $(X+1{)}_0^{(e)}=(X_{e-1}^{(e)})+1$ and $(X+1{)}_i^{(e)}=X_{i-1}^{(e)}$ for $1\le i<e$, where $+1$ is the beta-set shift, including the new bead at row $0$. Thus the new counts are $(c_{e-1}+1,c_0,\dots ,c_{e-2})$, not just a permutation of the old counts. Top-packing commutes with this shift: $(X+1{)}^{(e)}=(X^{(e)})+1$, so $P$ is unchanged. A shift by $e$ applies a beta-set shift by $1$ to every runner, preserving each component partition. Hence any two normalised beta-sets give the same ordered quotient and weight. These identities and the identification of the core with terminal rim $e$-hook removal are proved in The e-core is independent of the rim hook removal order ↗.

• $e$-cores are $e$-regular and $e$-restricted. If a positive part $a$ occurs at least $e$ times, put $r={\lambda }_a^{\mathrm{\prime }}$. The bottom $e$ rows of length $a$ are $r-e+1,\dots ,r$, and the node $(r-e+1,a)$ has hook length $a-a+r-(r-e+1)+1=e$. Thus an $e$-core is $e$-regular. Conjugation preserves hook lengths, since it exchanges arm and leg; applying this argument to ${\lambda }^{\mathrm{\prime }}$ and using e-regular and e-restricted partitions gives $e$-restrictedness.

• Normalisation and the quotient's ordering. Shifting by $1$ rotates the component partitions cyclically. Shifting by $e$ increases every runner count by $1$ and replaces each row-label set $Y$ by $Y+1=\{a+1:a\in Y\}\cup \{0\}$; it preserves $P(Y)$, rather than $Y$ itself. The condition $e\mid \mathrm{\mid }X\mathrm{\mid }$ therefore fixes the ordered quotient, as in Lemma 4.9 and its normalising remark in the lecture notes.