1 Introduction
1.1 Quantum orthosymplectic superalgebras and q-oscillator modules
The crystal base of the negative half of a quantized enveloping algebra associated with a symmetrizable Kac–Moody algebra is an object with a fundamental combinatorial structure, into which the crystal bases of integrable highest weight modules are naturally incorporated [Reference Kashiwara14]. It can be obtained by taking the limit of crystal bases of integrable highest weight modules.
Let
${\mathcal U}(\mathfrak {g})$
be a quantized enveloping algebra over
$\Bbbk =\mathbb {Q}(v)$
(
$v=q$
or
$q^{\frac {1}{2}}$
) associated to an orthosymplectic superalgebra
$\mathfrak {g}=\mathfrak {b}_{m|n}, \mathfrak {c}_{m|n}, \mathfrak {d}_{m|n}$
(
$m\ge 2$
) with a standard fundamental system
$\Pi =\{\,\alpha _i\,|\,i\in I\,\}$
. In this paper we construct a crystal base of the negative half
${\mathcal U}(\mathfrak {g})^-$
of
${\mathcal U}(\mathfrak {g})$
as a continuation of our previous works [Reference Jang, Kwon and Uruno12, Reference Jang, Kwon and Uruno13].
In the case of an orthosymplectic superalgebra
$\mathfrak {g}$
and even a general linear Lie superalgebra
$\mathfrak {gl}_{m|n}$
, it is not known in general whether a finite-dimensional irreducible
${\mathcal U}(\mathfrak {g})$
-module has a crystal base. On the other hand, a family of irreducible highest weight
${\mathcal U}(\mathfrak {g})$
-modules was studied in [Reference Kwon20, Reference Kwon21], which are q-analogues of oscillator representations of
$\mathfrak {g}$
[Reference Cheng and Zhang6] naturally arising from the theory of Howe dual pairs [Reference Howe8]. It is shown that this module, which we denote by
$V(\lambda ,\ell )$
for
$(\lambda ,\ell )\in \mathscr {P}(\mathfrak {g})$
, has a unique crystal base
$(\mathscr {L}(V(\lambda ,\ell )),\mathscr {B}(V(\lambda ,\ell )))$
together with a combinatorial model of the crystal
$\mathscr {B}(V(\lambda ,\ell ))$
called the spinor model. The classical limit of
$V(\lambda ,\ell )$
(after taking a limit of the rank n) corresponds to an integrable highest weight module over the Lie algebras of type
$BCD$
under the theory of super duality [Reference Cheng, Lam and Wang5]. From this point of view, it would be natural to construct a crystal base of
${\mathcal U}(\mathfrak {g})$
by taking a limit of
$(\mathscr {L}(V(\lambda ,\ell )),\mathscr {B}(V(\lambda ,\ell )))$
.
Another major difference from the case of symmetrizable Kac-Moody algebras is that a q-oscillator representation does not have an explicit presentation as a quotient of
${\mathcal U}(\mathfrak {g})^-$
. This makes it more complicated to take the limit of crystals of
$V(\lambda ,\ell )$
and describe the associated crystal embeddings. So in order to overcome these difficulties, we consider a sequence of parabolic Verma modules following the ideas in [Reference Jang, Kwon and Uruno12] in the case of
$\mathfrak {gl}_{m|n}$
and also use combinatorial objects and algorithms.
1.2 Parabolic Verma modules
Let
$\mathfrak {l}$
be the subalgebra of
$\mathfrak {g}$
corresponding to
$\Pi \setminus \{\alpha _0\}$
which is isomorphic to
$\mathfrak {gl}_{m|n}$
, and let
${\mathcal U}(\mathfrak {l})$
be the associated subalgebra of
${\mathcal U}(\mathfrak {g})$
. Let
$V_{\mathfrak l}(\lambda )$
be the irreducible polynomial
${\mathcal U}(\mathfrak {l})$
-module corresponding to an
$(m|n)$
-hook partition
$\lambda $
, which has a crystal base with its crystal
$\mathscr {B}(V_{\mathfrak l}(\lambda ))$
being the set of
$(m|n)$
-hook semistandard tableaux of shape
$\lambda $
[Reference Benkart, Kang and Kashiwara1]. Let
$P(\lambda )={\mathcal U}(\mathfrak {g})\otimes _{{\mathcal U}(\mathfrak {p})}V_{\mathfrak {l}}(\lambda )$
, where
${\mathcal U}(\mathfrak {p})$
is the subalgebra associated to a parabolic subalgebra
$\mathfrak {p}\subset \mathfrak {g}$
generated by
$\mathfrak {l}$
and the Borel subalgebra with respect to
$\Pi $
. As a
$\Bbbk $
-space,
$P(\lambda )=\mathcal {N}\otimes _{\Bbbk } V_{\mathfrak l}(\lambda )$
, where
$\mathcal {N}$
is the subalgebra of
${\mathcal U}(\mathfrak {g})^-$
generated by the root vectors associated to
$\mathfrak {g}/\mathfrak {p}$
. In [Reference Jang, Kwon and Uruno13], it is shown that
$P(\lambda )$
has a unique crystal base
$(\mathscr {L}(P(\lambda )),\mathscr {B}(P(\lambda )))$
. The construction of the crystal base of
$\mathcal {N}=P(0)$
is the main part in the proof of the existence.
Let
$Q(\lambda )$
and
$R(\lambda )$
be the parabolic Verma modules induced from polynomial representations of
${\mathcal U}(\mathfrak {gl}_{m|0}\oplus \mathfrak {gl}_{0|n})$
and
${\mathcal U}(\mathfrak {gl}_{m|0})$
, respectively. Then we have the commuting diagram as follows:

where the horizontal maps in the top row are canonical
${\mathcal U}(\mathfrak {g})^-$
-linear projections, the vertical maps are
$\Bbbk $
-linear isomorphisms, and
$K(\lambda )$
and
$X(\lambda )$
are the corresponding parabolic Verma modules over
${\mathcal U}(\mathfrak {l})$
.
One can show by adopting arguments similar to those in [Reference Jang, Kwon and Uruno12] that
${\mathcal U}(\mathfrak {g})^-$
,
$R(\lambda )$
, and
$Q(\lambda )$
have crystal bases (which are unique in the case of the last two) such that they are isomorphic to the tensor product of crystal bases of
$\mathcal {N}$
and
${\mathcal U}(\mathfrak {l})$
-modules in the bottom line of (1.1) (as modules over
${\mathcal U}(\mathfrak {l})$
or
${\mathcal U}(\mathfrak {l})^-$
), and compatible with the canonical projections (or their duals). Then the
${\mathcal U}(\mathfrak {g})$
-crystal structures of
${\mathcal U}(\mathfrak {g})^-$
,
$R(\lambda )$
, and
$Q(\lambda )$
can also be described explicitly from that of
$\mathcal {N}=P(0)$
.
The crystal
$\mathscr {B}({\mathcal U}(\mathfrak {g})^-)$
of
${\mathcal U}(\mathfrak {g})^-$
can be given in terms of a PBW type basis [Reference Clark, Hill and Wang3, Reference Yamane26] together with an explicit description of crystal operators, where the roots for
$\mathcal {N}$
precede the negative roots of
$\mathfrak {l}$
. A description of the associated embeddings from
$\mathscr {B}(P(\lambda ))$
to
$\mathscr {B}({\mathcal U}(\mathfrak {g})^-)$
follows directly from the ones for
${\mathcal U}(\mathfrak {l})$
-modules in (1.1) (see [Reference Jang, Kwon and Uruno12]) by tensoring with
$\mathrm {id}_{\mathscr {B}(\mathcal {N})}$
.
1.3 Irreducible quotient, crystal embedding and Burge correspondence
In order to complete a connection between crystal bases of
$V(\lambda ,\ell )$
and
${\mathcal U}(\mathfrak {g})^-$
, it remains to consider
$\pi _{V(\lambda ,\ell )}$
in (1.1).
We first show that
$\pi _{V(\lambda ,\ell )}$
preserves the crystal bases. It is done by showing that
$(\mathscr {L}(P(\lambda )),\mathscr {B}(P(\lambda )))$
is a limit of
$(\mathscr {L}(V(\lambda ,\ell )),\mathscr {B}(V(\lambda ,\ell )))$
when
$\ell \rightarrow \infty $
. Next, we describe the crystal embedding
which is indeed the main result in this paper. It consists of two steps: As a first step, we define an embedding

The
${\mathcal U}(\mathfrak {g})$
-crystal
$\lim _{\ell \rightarrow \infty }\mathscr {B}(V(0,\ell ))$
can be realized as the set of
$(m|n)$
-hook semistandard tableaux of arbitrary shape
$\delta ^\pi $
(rotation of
$\delta $
by
$180^\circ $
) with an even parity condition on its rows or columns of an
$(m|n)$
-hook partition
$\delta $
depending on the type of
$\mathfrak {g}$
. The map (1.3) is given by applying the combinatorial algorithms in [Reference Jang and Kwon11, Reference Kwon22] (called separation algorithm) to the spinor model of
$\mathscr {B}(V(\lambda ,\ell ))$
, which were introduced to compute Lusztig data of Kashiwara-Nakashima tableaux of type
$BCD$
. As a second step, we establish a
${\mathcal U}(\mathfrak {g})$
-crystal isomorphism
It is realized by the (super-analogue of) RSK map on symmetric matrices when
$\mathfrak {g}=\mathfrak {b}_{m|n}, \mathfrak {c}_{m|n}$
, and Burge correspondence of type D [Reference Burge2] when
$\mathfrak {g}=\mathfrak {d}_{m|n}$
. Hence we get (1.2) by combining (1.3) and (1.4) since
$\mathscr {B}(P(\lambda ))=\mathscr {B}(\mathcal {N})\times \mathscr {B}(V_{\mathfrak l}(\lambda ))$
. We remark that (1.2) is given in [Reference Jang and Kwon11, Reference Kwon22] when
$n=0$
, while its generalization to the case of
$\mathfrak {d}_{m|n}$
with
$n>0$
is nontrivial and indeed the most technical part in this paper.
1.4 Organization
The paper is organized as follows. In Section 2, we recall necessary notations and definitions. In Section 3, we recall the crystal bases of
$V(\lambda ,\ell )$
and
$P(\lambda )$
, and prove their compatibility under
$\pi _{V(\lambda ,\ell )}$
. In Section 4, we recall the crystal base of
${\mathcal U}(\mathfrak {l})^-$
[Reference Jang, Kwon and Uruno12]. In Section 5, we construct the crystal base of
${\mathcal U}(\mathfrak {g})^-$
. In Section 6, we describe (1.2). The proof for the crystal theoretic interpretation of Burge correspondence providing (1.4) for
$\mathfrak {g}=\mathfrak {d}_{m|n}$
is presented in Section 7.
2 Preliminary
2.1 Notations
Throughout the paper, we assume that m and n are positive integers with
$m\ge 2$
. We keep the same notations as in [Reference Jang, Kwon and Uruno13, Section 2], which we briefly recall.
-
• $\mathbb {I} = \{\, 1, \dots ,m , m+1, \dots , m+n \,\}$
with a usual linear order, -
• $\mathbb {I} = \mathbb {I}_{\overline 0} \cup \mathbb {I}_{\overline 1}$
, where
$\mathbb {I}_{\overline 0} = \{\, 1, \dots , m \,\}$
and
$\mathbb {I}_{\overline 1} = \{ \, m+1,\dots , m+n \,\}$
, -
• $\epsilon =(\epsilon _a)_{a\in \mathbb {I}}$
: a sequence with
$\epsilon _a=\varepsilon $
for
$a\in \mathbb {I}_{\varepsilon }$
(
$\varepsilon ={\overline 0},{\overline 1}$
), -
• $P = \bigoplus _{a\in \mathbb {I}}\mathbb {Z} \delta _a \oplus \mathbb {Z}\Lambda _0$
: a free abelian group, -
• $I=\{\,0,1, \ldots ,m,m+1,\dots , m+n-1\}$
, -
• $\mathfrak {g}=\mathfrak {b}_{m|n}, \mathfrak {c}_{m|n}, \mathfrak {d}_{m|n}$
: an orthosymplectic Lie superalgebra associated to the Dynkin diagram with the set of simple roots
$\Pi =\{\,\alpha _i\,|\,i\in I\,\}\subset P$
as follows:
$$ \begin{align*} \alpha_i= \begin{cases} -\delta_{1} & \text{if }i=0,\\ \delta_{i}-\delta_{i+1} & \text{if }1 \le i \le m+n-1, \end{cases} \end{align*} $$
$$ \begin{align*} \alpha_i= \begin{cases} -2\delta_{1} & \text{if }i=0,\\ \delta_i-\delta_{i+1} & \text{if }1\le i \le m+n-1, \end{cases} \end{align*} $$
$$ \begin{align*} \alpha_i= \begin{cases} -\delta_{1}-\delta_{2} & \text{if }i=0,\\ \delta_i-\delta_{i+1} & \text{if }1\le i \le m+n-1. \end{cases} \end{align*} $$
-
• $(\, \cdot \, |\, \cdot \, )$
: a symmetric bilinear form on P satisfying
$(\delta _a|\delta _b)=(-1)^{\epsilon _a}r_{\mathfrak {g}}\delta _{ab}$
,
$(\delta _a|\Lambda _0)=-\frac {r_{\mathfrak {g}}}{2}\ (a,b\in \mathbb {I})$
with
$r_{\mathfrak {b}_{m|n}}=2$
,
$r_{\mathfrak {c}_{m|n}}=r_{\mathfrak {d}_{m|n}}=1$
(note that
$(\Lambda _0|\alpha _i)=\delta _{0i}$
for
$i\in I$
), -
• $\Phi ^+$
: the set of reduced positive roots, -
• $\Phi ^+_{\overline {0}}$
(resp.
$\Phi ^+_{\overline {1}}$
): the set of even (resp. odd) roots in
$\Phi ^+$
, -
• $\mathfrak {l}=\mathfrak {l}_{m|n}$
: the subalgebra of
$\mathfrak {g}$
corresponding to
$\Pi \setminus \{\alpha _0\}$
isomorphic to
$\mathfrak {gl}_{m|n}$
, -
• $\Phi ^+(\mathfrak {l})$
: the set of positive roots of
$\mathfrak {l}$
, -
• $\mathfrak {l}_{\overline 0}$
: the subalgebra of
$\mathfrak {l}$
corresponding to even roots in
$\Phi ^+(\mathfrak {l})$
, -
• $\mathfrak {l}_{m|0}$
,
$\mathfrak {l}_{0|n}$
: the subalgebras of
$\mathfrak {l}_{\overline 0}$
corresponding to
$\mathfrak {gl}_m$
and
$\mathfrak {gl}_n$
, respectively, -
• $\mathfrak {u}$
: the subalgebra of
$\mathfrak {g}$
generated by the root vectors corresponding to
$\Phi ^+\setminus \Phi ^+(\mathfrak {l})$
, -
• $\Phi ^+(\mathfrak {u})= \Phi ^+\setminus \Phi ^+(\mathfrak {l})$
.
2.2 Quantum orthosymplectic superalgebra
We also recall the following [Reference Jang, Kwon and Uruno13]:
-
• $\Bbbk =\mathbb {Q}(v)$
and
$\Bbbk ^{\times }=\Bbbk \setminus \left \{0\right \}$
where
$v=q$
if
$\mathfrak {g}=\mathfrak {b}, \mathfrak {c} $
and
$v=q^{\frac {1}{2}}$
if
$\mathfrak {g}=\mathfrak {d} $
with
$ q^{\frac {1}{2}}$
an indeterminate, -
• ${\texttt {q}}_a=(-1)^{\epsilon _a} q^{(-1)^{\epsilon _a} r_{\mathfrak {g}}}\ (a\in \mathbb {I})$
, -
• $q_i = q^{d_i}\ (i\in I\setminus \{m\})$
, where
$d_i=|(\alpha _i|\alpha _i)|/2$
, and
$q_m = q^{r_{\mathfrak g}}$
, -
• $\mathbf {q}(\mu ,\nu ) = v^{\sum _{a\in \mathbb {I}} \ell ' \mu _a + \ell \nu _a } \prod _{a\in \mathbb {I}}\mathtt {q}_a^{\mu _a\nu _a}$
for
$\mu =\sum _{a\in \mathbb {I}}\mu _a\delta _a +\ell \Lambda _0 $
and
$\nu =\sum _{a\in \mathbb {I}}\nu _a\delta _a + \ell ' \Lambda _0$
(note that
$\mathbf { q}(\Lambda _0,\alpha _i)=q_0^{\delta _{0 i}}$
), -
• For $s\in \mathbb {Z}_{\ge 0}$
and
$z\in q^{\mathbb {Z}_{>0}}$
,
$[s]_z=\frac {z^s-z^{-s}}{z-z^{-1}}$
, and
$[s]_z!=[s]_z [s-1]_z \cdots [1]_z (s\geq 1)$
with
$[0]_z!=1$
, where we simply write
$[s]$
when
$z=q$
, and
$[s]_i=[s]_{q_i}$
when
$z=q_i$
(
$i\in I$
), -
• ${{\mathcal U}}(\mathfrak {g})$
: the associative
$\Bbbk $
-algebra with
$1$
generated by
$k_\mu , e_i, f_i$
for
$\mu \in P$
and
$i\in I$
satisfying $$ \begin{align*} \begin{aligned} k_{\mu}=1 \quad (\mu=0), \quad k_{\nu +\nu'}=k_{\nu}k_{\nu'} \quad (\nu, \nu' \in P), \qquad\qquad\\k_\mu e_i k_{-\mu}=\mathbf{q}(\mu,\alpha_i)e_i,\quad k_\mu f_i k_{-\mu}=\mathbf{q}(\mu,\alpha_i)^{-1}f_i\quad (i\in I, \mu\in P), \\e_if_j - f_je_i =\delta_{ij}\frac{k_{i} - k^{-1}_{i}}{q_i-q_i^{-1}}\quad (i,j\in I),\qquad\qquad\qquad\\e_m^2= f_m^2 =0,\qquad\qquad\qquad\qquad\qquad\qquad \end{aligned} \end{align*} $$where $k_i=k_{\alpha _i}$
for
$i\in I$
, and $$ \begin{align*}\begin{aligned} x_i x_j - x_j x_i =0 \quad (i,j \in I\text{ and }(\alpha_i|\alpha_j) = 0), \qquad\qquad\qquad\qquad\quad\\x_i^2 x_j- (-1)^{\epsilon_i}[2]_i x_i x_j x_i + x_j x_i^2= 0 \quad (i\in I\setminus\{0, m\}, j\in I\setminus\{0\}, i\neq j\text{ and }(\alpha_i|\alpha_j)\neq 0), \\\begin{array}{ll} x_0^3 x_1- [3]_0 x_0^2 x_1 x_0 + [3]_0 x_0 x_1 x_0^2 - x_1 x_0^3 = 0\\ x_1^2 x_0- [2]_1 x_1 x_0 x_1 + x_0 x_1^2= 0 \end{array} \quad (\mathfrak{g} = \mathfrak{b}_{m|n}), \qquad\qquad\qquad\\\begin{array}{ll} x_0^2 x_1- [2]_0 x_0 x_1 x_0 + x_1 x_0^2= 0 \\ x_1^3 x_0- [3]_1 x_1^2 x_0 x_1 + [3]_1 x_1 x_0 x_1^2 - x_0 x_1^3 = 0 \end{array} \quad (\mathfrak{g} = \mathfrak{c}_{m|n}), \qquad\qquad\qquad\\\begin{array}{ll} x_0^2 x_2- [2]_0 x_0 x_2 x_0 + x_2 x_0^2= 0 \\ x_2^2 x_0- [2]_2 x_2 x_0 x_2 + x_0 x_2^2= 0 \end{array} \quad (\mathfrak{g} = \mathfrak{d}_{m|n}\ \text{with }m\ge 3),\qquad\qquad\qquad\\\begin{array}{ll} x_{m}x_{m'}x_{m}x_{m+1} - x_{m}x_{m+1}x_{m}x_{m'} + x_{m+1}x_{m}x_{m'}x_{m} \\ \qquad \qquad \qquad \qquad \qquad - x_{m'}x_{m}x_{m+1}x_{m} + [2]_{m} x_{m}x_{m'}x_{m+1}x_{m} =0 \\\hskip 6cm (m'=0,1 \text{ if }\mathfrak{g}={\mathfrak d}_{2|n}, \text{ and } m'=m-1 \text{ otherwise} ), \end{array} \end{aligned} \end{align*} $$where $x=e,f$
,
-
• $\Delta $
: the comultiplication on
${\mathcal U}(\mathfrak {g})$
given by
$\Delta (k_\mu )=k_\mu \otimes k_\mu $
,
$\Delta (e_i)= 1\otimes e_i + e_i\otimes k_i^{-1}$
,
$\Delta (f_i)= f_i\otimes 1 + k_i\otimes f_i$
for
$\mu \in P$
and
$i\in I$
, -
• $e^{\prime }_i, e^{\prime \prime }_i$
:
$\Bbbk $
-linear maps on
${\mathcal U}(\mathfrak {g})^-$
defined by (2.1) $$ \begin{align} \begin{aligned} e^{\prime}_i(f_j)&=\delta_{ij},\quad e^{\prime}_i(uv)=e^{\prime}_i(u)v + \mathbf{q}{(\alpha_i,|u|)}ue^{\prime}_i(v), \\ e^{\prime\prime}_i(f_j)&=\delta_{ij},\quad e^{\prime\prime}_i(uv)=e^{\prime\prime}_i(u)v + \mathbf{q}{(\alpha_i,|u|)}^{-1}ue^{\prime\prime}_i(v), \end{aligned} \end{align} $$for $i,j\in I$
and
$u,v\in {\mathcal U}(\mathfrak {g})^-$
with u homogeneous.
We also consider the following subalgebras of
${\mathcal U}(\mathfrak {g})$
:
-
• ${\mathcal U}(\mathfrak {g})^+= \langle \, e_i \,|\,i\in I\,\rangle $
,
${\mathcal U}(\mathfrak {g})^- = \langle \, f_i \,|\,i\in I\,\rangle $
,
${\mathcal U}(\mathfrak {g})^0 = \langle \, k_\mu \,|\,\mu \in P\,\rangle $
, -
• ${\mathcal U}(\mathfrak {l})=\langle \, k_\mu , e_i, f_i \,|\,\mu \in P,\ i\in I\setminus \{0\}\,\rangle $
, -
• ${\mathcal U}(\mathfrak {l}_{\overline 0})=\langle \, k_\mu , e_i, f_i \,|\,\mu \in P,\ i\in I\setminus \{0,m\}\,\rangle $
, -
• ${\mathcal U}(\mathfrak {l}_{m|0})=\langle \, k_\mu , e_i, f_i \,|\,\mu \in P,\ i=1,\dots ,m-1 \,\rangle $
, -
• ${\mathcal U}(\mathfrak {l}_{0|n})=\langle \, k_\mu , e_i, f_i \,|\,\mu \in P,\ i=m+1,\dots ,m+n-1 \,\rangle $
, -
• ${\mathcal U}(\mathfrak {l}_\ast )^\pm = {\mathcal U}(\mathfrak {l}_\ast )\cap {\mathcal U}(\mathfrak {g})^\pm $
, where
$\mathfrak {l}_\ast = \mathfrak {l}, \mathfrak {l}_{\overline 0}, \mathfrak {l}_{m|0}, \mathfrak {l}_{0|n}$
.
Remark 2.1. As noted in [Reference Jang, Kwon and Uruno13, Section 2.2], the algebra
$\mathcal {U}(\mathfrak {g})$
defined above is the generalized quantum group associated with
$\mathfrak {g}$
introduced in [Reference Kuniba, Okado and Sergeev15], and it is isomorphic, up to a mild extension, to the quantum superalgebra associated with
$\mathfrak {g}$
defined in [Reference Yamane26]. This implies that there exists a natural correspondence between representations of these two algebras (see [Reference Jang, Kwon and Uruno13, Remark 2.2]).
Note that
${\mathcal U}(\mathfrak {l}_{\overline 0})$
is isomorphic to
$U_q(\mathfrak {gl}_m)\otimes U_{-q^{-1}}(\mathfrak {gl}_n)$
as a
$\Bbbk $
-algebra up to the Cartan part, where
$U_q(\mathfrak {gl}_m)$
is the usual quantized enveloping algebra associated to
$\mathfrak {gl}_m$
. Recall that
${\mathcal U}(\mathfrak {g})^\pm $
is naturally graded by
$Q^+:=\sum _{i\in I}\mathbb {Z}_{\ge 0}\alpha _i$
and
$Q^-:=-Q^+$
.
Let
$\mathfrak {a}$
denote either
$\mathfrak {g}$
or
$\mathfrak {l}_\ast $
above. For a
${\mathcal U}(\mathfrak a)$
-module V and
$\mu \in P$
, let
$V_\mu = \{\,u\in V\,|\,k_{\nu } u= \mathbf {q}(\mu ,\nu ) u \ \ (\nu \in P) \,\}$
be the
$\mu $
-weight space of V. For
$u\in V_\mu \setminus \{0\}$
, we call u a weight vector with weight
$\mu $
and put
$\mathrm {wt}(u)=\mu $
. Denote by
$\mathrm {wt}(V) =\{ \mu \in P \, | \, V_\mu \ne 0 \}$
the set of weights of V. For
$\Lambda \in P$
, let
$V_{\mathfrak a}(\Lambda )$
be the irreducible highest weight
${\mathcal U}(\mathfrak {a})$
-module generated by a highest weight vector
$v^{\mathfrak a}_\Lambda $
or simply
$v_\Lambda $
of weight
$\Lambda $
.
We also remark that the representation theory of
${\mathcal U}(\mathfrak {g})$
is equivalent to that of the quantum superalgebra associated to
$\mathfrak {g}$
[Reference Yamane26] (see [Reference Jang, Kwon and Uruno13, Section 2.2] for more details).
2.3 Root vectors
Let
$\mathbf {f}_\beta $
(
$\beta \in \Phi ^+$
) be a root vector in
${\mathcal U}(\mathfrak {g})^-$
, which is defined in [Reference Jang, Kwon and Uruno13, Section 3.1] following [Reference Clark, Hill and Wang3, Reference Leclerc24]. Let
$\prec $
be the linear order on
$\Phi ^+$
induced from that on the set of good Lyndon words
$\mathcal {GL}(\mathfrak {g})$
under the bijection with
$\Phi ^+$
. Then we have a PBW type basis B of
${\mathcal U}(\mathfrak {g})^-$
with respect to
$\prec $
. Note that
$\alpha \prec \beta $
for
$\alpha \in \Phi ^+(\mathfrak {u})$
and
$\beta \in \Phi ^+(\mathfrak {l})$
. For
$\beta \in \Phi ^+(\mathfrak {u})$
, we simply write
Then
and for
$(i_1,j_1), (i_2,j_2)\in \Phi ^+(\mathfrak {u})$
,
Let
${\mathcal U}(\mathfrak {u}^-)$
be the
$\Bbbk $
-space spanned by the vectors in B where the product is over
$\beta \in \Phi ^+(\mathfrak {u})$
. It is a
$\Bbbk $
-subalgebra of
${\mathcal U}(\mathfrak {g})^-$
generated by
$\mathbf {f}_\beta $
(
$\beta \in \Phi ^+(\mathfrak {u})$
), which we denote by
$\mathcal {N}$
. As a
$\Bbbk $
-space
where
$\otimes =\otimes _\Bbbk $
.
2.4 Irreducible oscillator modules
$V(\lambda ,\ell )$
Let us recall a semisimple tensor category of
${\mathcal U}(\mathfrak {g})$
-modules studied in [Reference Kwon20]. Let
-
• $\mathtt {P}=\sum _{a\in \mathbb {I}}\mathbb {Z}\delta _a \subset P$
, -
• $\mathtt {P}^+=\{\,\lambda =\sum _{i\in \mathbb {I}}\lambda _i\delta _i \in \mathtt {P}\ |\ \lambda _{1}\geq \ldots \geq \lambda _{m},\ \ \lambda _{m+1}\geq \ldots \geq \lambda _{m+n}\,\}$
, -
• $\lambda _{+}=\sum _{i\in \mathbb {I}_0}\lambda _{i}\delta _i$
,
$\lambda _{-}=\sum _{j\in \mathbb {I}_1}\lambda _{j}\delta _j$
for
$\lambda \in \mathtt { P}$
, -
• $\mathtt {P}_{\geq 0}=\sum _{a\in \mathbb {I}}\mathbb {Z}_{\ge 0}\delta _a$
.
Let
$\mathcal {O}(\mathfrak l)_{\geq 0}$
be the category of
${\mathcal U}(\mathfrak {l})$
-modules V such that
$V=\bigoplus _{\mu \in \mathtt {P}_{\ge 0}}V_\mu $
with
$\dim V_\mu < \infty $
. Let
$\mathscr {P}$
be the set of all partitions
$\lambda =(\lambda _i)_{i\ge 1}$
, and let
$\mathscr {P}_{m|n}$
be the set of
$\lambda \in \mathscr {P}$
such that
$\lambda _{m+1}\leq n$
. For
$\lambda \in \mathscr {P}_{m|n}$
, let
where
$(\mu _1,\ldots ,\mu _n)$
is the conjugate of the partition
$(\lambda _{m+1},\lambda _{m+2},\ldots )$
. We often identify
$\lambda \in \mathscr {P}_{m|n}$
with
$\Lambda _\lambda $
if there is no confusion. We simply write
$V_{\mathfrak {l}}(\lambda )=V_{\mathfrak {l}}(\Lambda _{\lambda })$
and
$v_\lambda =v_{\Lambda _\lambda }$
. Then any irreducible
${\mathcal U}(\mathfrak {l})$
-module in
$\mathcal {O}(\mathfrak l)_{\geq 0}$
is isomorphic to
$V_{\mathfrak {l}}(\lambda )$
for some
$\lambda \in \mathscr {P}_{m|n}$
[Reference Benkart, Kang and Kashiwara1, Propositions 3.4 and 4.5] (see also [Reference Jang, Kwon and Uruno13, Remark 4.2]). Moreover,
$\mathcal {O}(\mathfrak l)_{\geq 0}$
is semisimple [Reference Jang, Kwon and Uruno13, Proposition 4.3].
Definition 2.2. Let
$\mathcal {O}(\mathfrak {g})_{\ge 0}$
be the category of
${\mathcal U}(\mathfrak {g})$
-modules V satisfying
-
(1) $V=\bigoplus _{\gamma \in P}V_\gamma $
and
$\dim V_\gamma <\infty $
for
$\gamma \in P$
, -
(2) $\mathrm {wt}(V)\subset \bigcup _{i=1}^r\left (\ell _i\Lambda _{0}+ \mathtt {P}_{\ge 0} \right )$
for some
$r\geq 1$
and
$\ell _i\in \mathbb {Z}_{\geq 0}$
, -
(3) $f_{0}$
acts locally nilpotently on V.
We may understand that
$V\in \mathcal {O}(\mathfrak {g})_{\ge 0}$
belongs to
$\mathcal {O}(\mathfrak l)_{\ge 0}$
as a
${\mathcal U}(\mathfrak {l})$
-module if we ignore
$\Lambda _0$
, and hence is a direct sum of
$V_{\mathfrak l}(\mu )$
for
$\mu \in \mathscr {P}_{m|n}$
with finite multiplicity. Let
$\mathscr {P}(\mathfrak {g})$
be given by
and for
$(\lambda ,\ell )\in \mathscr {P}(\mathfrak {g})$
, let
Theorem 2.3 [Reference Kwon20, Theorem 3.8].
$\mathcal {O}(\mathfrak {g})_{\ge 0}$
is a semisimple tensor category with nontrivial irreducible objects
$V(\Lambda (\lambda ,\ell ))$
for
$(\lambda ,\ell )\in \mathscr {P}(\mathfrak {g})$
.
For simplicity, let us write
$V(\lambda ,\ell )=V(\Lambda (\lambda ,\ell ))$
and
$v_{(\lambda ,\ell )}=v_{\Lambda (\lambda ,\ell )}$
. For
$V\in \mathcal {O}(\mathfrak {g})_{\ge 0}$
, we have
for some
$m(\lambda ,\ell )\in \mathbb {Z}_{\ge 0}$
.
2.5 Parabolic Verma modules
Let
${\mathcal U}(\mathfrak {p})$
be the subalgebra of
${\mathcal U}(\mathfrak {g})$
generated by
${\mathcal U}(\mathfrak {l})$
and
${\mathcal U}(\mathfrak {g})^{+}$
. For
$\lambda \in \mathscr {P}_{m|n}$
, let
where we regard
$V_{\mathfrak {l}}(\lambda )$
as a
${\mathcal U}(\mathfrak {p})$
-module with
$e_0 v_\lambda =0$
. We have an isomorphism of
$\Bbbk $
-spaces
sending
$u\otimes v$
to itself since
${\mathcal U}(\mathfrak {g})\cong {\mathcal U}(\mathfrak {u}^-)\otimes {\mathcal U}(\mathfrak {l})^-\otimes {\mathcal U}(\mathfrak {g})^{\ge 0}\cong \mathcal {N} \otimes {\mathcal U}(\mathfrak {p})$
, where
${\mathcal U}(\mathfrak {g})^{\ge 0} = {\mathcal U}(\mathfrak {g})^+{\mathcal U}(\mathfrak {g})^0$
. It is a highest weight
${\mathcal U}(\mathfrak {g})$
-module with highest weight
$\Lambda _\lambda $
(2.3).
Let
${\mathcal U}(\mathfrak {q})$
be the subalgebra of
${\mathcal U}(\mathfrak {g})$
generated by
${\mathcal U}(\mathfrak {l}_{\overline 0})$
and
${\mathcal U}(\mathfrak {g})^{+}$
. For
$\lambda \in \mathtt {P}^+$
, let
where we regard
$V_{\mathfrak {l}_{\overline 0}}(\lambda )$
as a
${\mathcal U}(\mathfrak {q})$
-module with
$e_0 v_{\lambda }=e_m v_{\lambda }=0$
.
Finally, let
${\mathcal U}(\mathfrak {r})$
be the subalgebra of
${\mathcal U}(\mathfrak {l})$
generated by
${\mathcal U}(\mathfrak {l}_{0|n})$
and
${\mathcal U}(\mathfrak {g})^{+}$
. For
$\lambda \in \mathtt {P}$
with
$\lambda _-\in \mathtt {P}^+$
, let
where we regard
$V_{\mathfrak {l}_{0|n}}(\lambda )$
as a
${\mathcal U}(\mathfrak {r})$
-module with
$e_i v_{\lambda }=0$
for
$i=0,\dots ,m$
.
For
$(\lambda ,\ell )\in \mathscr {P}(\mathfrak {g})$
, we have natural surjective
${\mathcal U}(\mathfrak {g})^-$
-linear maps
where each map sends a highest weight vector to a highest weight vector.
3 Crystal bases of
$V(\lambda ,\ell )$
and
$P(\lambda )$
3.1 Crystal bases of
$V(\lambda ,\ell )$
and
$P(\lambda )$
Let us briefly recall the results on crystal bases of
$V(\lambda ,\ell )$
[Reference Kwon20, Reference Kwon21] and
$P(\lambda )$
[Reference Jang, Kwon and Uruno13].
3.1.1 Crystal base of
$V(\lambda ,\ell )$
Let V be a
${\mathcal U}(\mathfrak {g})$
-module in
$\mathcal {O}(\mathfrak {g})_{\ge 0}$
. For a weight vector
$u\in V$
and
$i\in I$
, we define
$\tilde {e}_i u$
and
$\tilde {f}_i u$
as follows:
-
(1) Suppose that $i\neq m$
. We have
$u=\sum _{k \geq 0} f_i^{(k)}u_k$
, where
$f_i^{(k)}=f_i^k/[k]_i!$
and
$e_iu_k=0$
for
$k\geq 0$
. We define $$ \begin{align*}\begin{aligned} \tilde{e}_iu&=\sum_{k\geq1}f_i^{(k-1)}u_k,\quad \tilde{f}_iu=\sum_{k\geq0}f_i^{(k+1)}u_k \quad (i<m),\\\tilde{e}_iu&=\sum_{k\geq1}q_i^{-l_k+2k-1}f_i^{(k-1)}u_k,\quad \tilde{f}_iu=\sum_{k\geq 0}q_i^{l_k-2k-1}f_i^{(k+1)}u_k\quad (i>m), \end{aligned}\end{align*} $$where $l_k= - (\mathrm {wt}(u_k)|\alpha _i)/r_{\mathfrak {g}}$
. Here we follow the convention in [Reference Jang, Kwon and Uruno12, Section 4.2].
-
(2) Suppose that $i=m$
. We define $$ \begin{align*} \tilde{e}_m u =\eta(f_m) u =q_m^{-1}k_me_m u,\quad \tilde{f}_m u=f_m u, \end{align*} $$where $\eta $
is the anti-involution on
${\mathcal U}(\mathfrak {g})$
defined by
$\eta (k_\mu )=k_\mu $
,
$\eta (e_i)=q_if_ik^{-1}_i$
, and
$\eta (f_i)=q^{-1}_ik_ie_i$
for
$\mu \in P$
and
$i\in I$
.
Let
$A_0$
be the subring of rational functions
$f(q)\in \Bbbk $
regular at
$q=0$
. We define a crystal base of V to be a pair
$(L,B)$
in the usual way with respect to
$\tilde {e}_i$
and
$\tilde {f}_i$
(
$i\in I$
), where L is an
$A_0$
-lattice of V, and B is a signed basis of
$L/qL$
, that is,
$B=\mathbf {B}\cup -\mathbf {B}$
, where
$\mathbf {B}$
is a
$\mathbb {Q}$
-basis of
$L/qL$
. We call the set
$B/\{\pm 1\}$
, which has a colored oriented graph structure, a
${\mathcal U}(\mathfrak {g})$
-crystal of V or simply a crystal of V. We identify B with
$B/\{\pm 1\}$
when we consider B as a crystal, if there is no confusion. The following is obtained in [Reference Kwon20, Theorem 8.8] for
$\mathfrak {g}=\mathfrak {b}_{m|n}, \mathfrak {c}_{m|n}$
and [Reference Kwon21, Theorem 5.7] for
$\mathfrak {g}=\mathfrak {d}_{m|n}$
.
Theorem 3.1. For
$(\lambda ,\ell )\in \mathscr {P}(\mathfrak {g})$
,
$V(\lambda ,\ell )$
has a crystal base
$(\mathscr {L}(V(\lambda ,\ell )), \mathscr {B}(V(\lambda ,\ell )))$
, where the crystal
$\mathscr {B}(V(\lambda ,\ell ))$
is connected.
Since
$\mathscr {B}(V(\lambda ,\ell ))$
is connected, we have a unique crystal base of
$V(\lambda ,\ell )$
up to scalar multiplication, that is,
where
$x = e,f$
for each
$i_k$
. Note that (3.1) holds in general for a highest weight module that has a crystal base (with respect to
$\tilde {e}_i$
and
$\tilde {f}_i$
defined suitably) such that the associated crystal is connected. A combinatorial model for
$\mathscr {B}(V(\lambda ,\ell ))$
is also given in [Reference Kwon20, Reference Kwon21], which will be reviewed in Section 6.2.
Regarding
$V\in \mathcal {O}(\mathfrak {g})_{\ge 0}$
as a
${\mathcal U}(\mathfrak {l})$
-module, we may consider a crystal base of
$V\in \mathcal {O}(\mathfrak l)_{\ge 0}$
with respect to
$\tilde {e}_i$
and
$\tilde {f}_i$
(
$i\in I\setminus \{0\}$
). For
$\lambda \in \mathscr {P}_{m|n}$
,
$V_{\mathfrak {l}}(\lambda )$
has a crystal base
$(\mathscr {L}(V_{\mathfrak l}(\lambda )), \mathscr {B}(V_{\mathfrak l}(\lambda )))$
[Reference Benkart, Kang and Kashiwara1], where the connected crystal
$\mathscr {B}(V_{\mathfrak l}(\lambda ))$
can be realized as the set
$SST_{m|n}(\lambda )$
of
$(m|n)$
-hook semistandard tableaux of shape
$\lambda $
. Hence
$(\mathscr {L}(V(\lambda ,\ell )), \mathscr {B}(V(\lambda ,\ell )))$
is a direct sum of crystal bases of
$V_{\mathfrak {l}}(\mu )$
’s for
$\mu \in \mathscr {P}_{m|n}$
as a
${\mathcal U}(\mathfrak {l})$
-module.
3.1.2 Crystal base of
$P(\lambda )$
Let
$\lambda \in \mathscr {P}_{m|n}$
be given. For a weight vector
$u\in P(\lambda )$
, we define
where
$u=\sum _{k\ge 0}f_0^{(k)}u_k$
for
$e^{\prime }_0 ( u_k ) = 0$
. Then we define a crystal base of
$P(\lambda )$
in the same way as in Section 3.1.1, where
$\tilde {e}_0$
and
$\tilde {f}_0$
are replaced by (3.2). We may define
$\tilde {e}_0$
and
$\tilde {f}_0$
on
${\mathcal U}(\mathfrak {g})^-$
in the same way, and we have
$(\widetilde {x}_0P)(1\otimes v_\lambda ) = \widetilde {x}_0(P(1\otimes v_\lambda ))$
for
$P\in {\mathcal U}(\mathfrak {g})^-$
(cf. [Reference Jang, Kwon and Uruno13, Section 4.4]).
Theorem 3.2 [Reference Jang, Kwon and Uruno13, Section 4.4].
For
$\lambda \in \mathscr {P}_{m|n}$
,
$P(\lambda )$
has a crystal base
$(\mathscr {L}(P(\lambda )), \mathscr {B}(P(\lambda )))$
, where the crystal
$\mathscr {B}(P(\lambda ))$
is connected.
Let
$\mathrm {ad}_q: {\mathcal U}(\mathfrak {g}) \longrightarrow \mathrm {End}_{\Bbbk }({\mathcal U}(\mathfrak {g}))$
be the adjoint representation of
${\mathcal U}(\mathfrak {g})$
with respect to
$\Delta $
, where for
$\mu \in P$
,
$i \in I$
, and
$u \in {\mathcal U}(\mathfrak {g})$
,
Let us write
$\mathrm {ad}_q(x)(u)=x\cdot u$
for simplicity. Then the subalgebra
$\mathcal {N}$
is a
${\mathcal U}(\mathfrak {l})$
-submodule of
${\mathcal U}(\mathfrak {g})$
with respect to
$\mathrm {ad}_q$
such that
$x\cdot m_{\mathcal N}(u_1\otimes u_2) = m_{\mathcal N}(\Delta (x)\cdot (u_1\otimes u_2))$
for
$x\in {\mathcal U}(\mathfrak {l})$
and
$u_1\otimes u_2\in \mathcal {N}\otimes \mathcal {N}$
, where
$m_{\mathcal N}$
denotes the multiplication on
$\mathcal {N}$
[Reference Jang, Kwon and Uruno13, Proposition 3.16]. We have
$\mathcal {N}\in \mathcal {O}(\mathfrak l)_{\ge 0}$
as a
${\mathcal U}(\mathfrak {l})$
-module, and the map (2.4) is an isomorphism of
${\mathcal U}(\mathfrak {l})$
-modules.
For
$\beta \in \Phi ^+(\mathfrak {u})$
and
$k\in \mathbb {Z}_{\ge 0}$
, let
(see [Reference Jang, Kwon and Uruno13, Section 3.1] for the definition of
$\mathbf {f}_\beta ^{(k)}$
). Let
where
$N=|\Phi ^+(\mathfrak {u})|$
and
$\Phi ^+_{\text {n-iso}}$
is the set of nonisotropic roots in
$\Phi ^+_{\overline 1}$
. For
$\mathbf {c}=(c_\beta )_{\beta \in \Phi ^+(\mathfrak {u})}\in M(\mathfrak {u})$
, let
where the product is taken in increasing order with respect to
$\prec $
.
Theorem 3.3 [Reference Jang, Kwon and Uruno13, Theorem 4.15].
Let
Then
$(\mathscr {L}\left (\mathcal {N}\right ),\mathscr {B}\left (\mathcal {N}\right ))$
is a crystal base of
$\mathcal {N}$
as a
${\mathcal U}(\mathfrak {l})$
-module.
By [Reference Jang, Kwon and Uruno13, Theorems 4.22 and 4.24], we have
The crystal operators
$\tilde {e}_i, \tilde {f}_i$
for
$i\in I\setminus \{0\}$
act on
$\mathscr {B}(\mathcal {N})\otimes \mathscr {B}(V_{\mathfrak {l}}(\lambda ))$
by the tensor product rule [Reference Benkart, Kang and Kashiwara1, Proposition 2.8], while
$\tilde {e}_0, \tilde {f}_0$
act by decreasing and increasing the multiplicity
$c_{\alpha _0}$
in
$\mathbf {F}^{(\mathbf c)}$
by
$1$
, respectively.
3.2 Compatibility of crystal bases of
$V(\lambda ,\ell )$
and
$P(\lambda )$
Let
$(\lambda ,\ell )\in \mathscr {P}(\mathfrak {g})$
be given. We show that the crystal base of
$P(\lambda )$
is compatible with that of
$V(\lambda ,\ell )$
under
$\pi _{V(\lambda ,\ell )}$
. This can be done by following the same idea as in [Reference Kashiwara14, Section 4].
Let
$\ell '>\ell \ge 0$
be given. Note that
$(\lambda ,\ell ')\in \mathscr {P}(\mathfrak {g})$
. For
$t=\ell '-\ell $
, we have a
${\mathcal U}(\mathfrak {g})$
-linear embedding

and a
${\mathcal U}(\mathfrak {g})^-$
-linear map

for
$v'$
such that
$\mathrm {wt}(v') \ne \Lambda (0,t)=t\Lambda _0$
. Since
$\mathcal {O}(\mathfrak {g})_{\ge 0}$
is a semisimple tensor category,
$\Phi $
is a well-defined
${\mathcal U}(\mathfrak {g})$
-linear map, and in particular it commutes with
$\tilde {e}_i$
and
$\tilde {f}_i$
for
$i\in I$
.
Lemma 3.4. Under the above hypothesis,
-
(1) $S\circ \Phi (\mathscr {L}(V(\lambda ,\ell '))) =\mathscr {L}(V(\lambda ,\ell ))$
, -
(2) the map induced from $S\circ \Phi $
at
$q=0$
gives a bijection $$ \begin{align*} \overline{S\circ \Phi} : \{ \, b \in \mathscr{B}(V(\lambda,\ell')) \, | \, \overline{S\circ \Phi} (b) \ne 0 \, \} \longrightarrow \mathscr{B}(V(\lambda,\ell)), \end{align*} $$which commutes with $\tilde {e}_i$
and
$\tilde {f}_i$
for
$i\in I$
.
Proof. Since
$\Phi $
commutes with
$\tilde {e}_i$
and
$\tilde {f}_i$
for
$i\in I$
, it follows from the connectedness of
$\mathscr {B}(V(\lambda ,\ell '))$
that
$\Phi (\mathscr {L}(V(\lambda ,\ell ')))\subset \mathscr {L}(V(\lambda ,\ell ))\otimes \mathscr {L}(V(0,t))$
. Hence
$S\circ \Phi (\mathscr {L}(V(\lambda ,\ell ')))\subset \mathscr {L}(V(\lambda ,\ell ))$
since
$S(\mathscr {L}(V(\lambda ,\ell ))\otimes \mathscr {L}(V(0,t)))=\mathscr {L}(V(\lambda ,\ell ))$
by definition of S.
Let
$b \in \mathscr {B}(V(\lambda ,\ell ))$
be given, where
$b=\widetilde {x}_{i_k}\dots \widetilde {x}_{i_1} v_{(\lambda ,\ell )}$
for some
$i_1, \dots , i_k \in I $
, where
$x=e,f$
for each
$i_j$
. By the tensor product rule (cf. [Reference Benkart, Kang and Kashiwara1, Proposition 2.8]), we have
since
$\tilde {f}_i v_{(0,t)}=0$
for all
$i>m$
. Hence
$\overline {S\circ \Phi }(\tilde {x}_{i_k}\dots \tilde {x}_{i_1} v_{(\lambda ,\ell ')})=b$
. In particular,
$\overline {S\circ \Phi }$
is surjective and hence
$S\circ \Phi (\mathscr {L}(V(\lambda ,\ell '))) =\mathscr {L}(V(\lambda ,\ell ))$
. This proves (1).
To prove the injectivity of
$\overline {S\circ \Phi }$
, suppose
$b, b' \in \mathscr {B}(V(\lambda ,\ell '))$
satisfy
$\overline {S\circ \Phi }(b) = \overline {S\circ \Phi }(b') \neq 0$
. Write
$b=\widetilde {x}_{i_k}\dots \widetilde {x}_{i_1} v_{(\lambda ,\ell ')}$
,
$b'=\widetilde {x}_{j_l}\dots \widetilde {x}_{j_1} v_{(\lambda ,\ell ')}$
for some
$i_1, \dots , i_k, j_1, \dots j_l \in I $
, where each x stands for e or f. Then by assumption, we have
$\widetilde {x}_{i_k}\dots \widetilde {x}_{i_1} v_{(\lambda ,\ell )} = \widetilde {x}_{j_l}\dots \widetilde {x}_{j_1} v_{(\lambda ,\ell )}$
. On the other hand,
Since
$\overline {\Phi }$
is an embedding of crystals, it follows that
$b=b'$
.
Let
$\Phi _{\ell ', \ell }=S\circ \Phi $
, which is
${\mathcal U}(\mathfrak {g})^-$
-linear, and let
$\Psi _{\ell }:{\mathcal U}(\mathfrak {g})^- \longrightarrow V(\lambda ,\ell )$
be the
${\mathcal U}(\mathfrak {g})^-$
-linear projection sending
$1$
to
$v_{(\lambda ,\ell )}$
. Since
$\Phi _{\ell ',\ell } \circ \Psi _{\ell '} = \Psi _{\ell }$
, we have
$\mathrm {Ker} \, \Psi _{\ell '} \subset \mathrm {Ker} \, \Psi _{\ell }$
, especially
$(\mathrm {Ker} \, \Psi _{\ell '})_{\xi } \subset (\mathrm {Ker} \, \Psi _{\ell })_{\xi } \subset {\mathcal U}(\mathfrak {g})^-_\xi $
for
$\xi \in Q^-$
. Hence for a given
$\xi \in Q^-$
there is
$N>0$
such that
$(\mathrm {Ker} \, \Psi _{\ell +t})_{\xi } = (\mathrm {Ker} \, \Psi _{\ell +N})_{\xi }$
for all
$t\ge N$
.
This implies that the restriction
$\Phi _{\ell +t,\ell +N} : V(\lambda ,\ell +t)_{\Lambda (\lambda ,\ell +t)+\xi } \longrightarrow V(\lambda ,\ell +N)_{\Lambda (\lambda ,\ell +N)+\xi }$
is a
$\Bbbk $
-linear isomorphism, and
is an isomorphism of
$A_0$
-modules for all
$t\ge N$
.
Lemma 3.5 (cf. [Reference Kashiwara14, Lemma 4.4.1]).
Let
$\xi \in Q^-$
be given. For
$P \in {\mathcal U}(\mathfrak {g})^-_{\xi }$
, there exists
$N>0$
such that for all
$t>N$
,
where
$\widetilde {x}_0$
on
${\mathcal U}(\mathfrak {g})^-$
is defined in the same way as in (3.2).
Proof. The argument is essentially the same as in [Reference Kashiwara14, Lemma 4.4.1] after a minor modification. Assume that
$P=f_0^{(k)}Q$
with
$e^{\prime }_0 Q=0$
. Then
$\tilde {e}_0 P = f_0^{(k-1)}Q$
and
$\tilde {f}_0 P = f_0^{(k+1)}Q$
. We will show that
for all sufficiently large t.
We remark that [Reference Kashiwara14, Corollary 3.4.6, Lemma 4.3.9] still hold for
${\mathcal U}(\mathfrak {g})^-$
and
$V(\lambda ,\ell )$
, where
$(L(\lambda ),B(\lambda ))$
and u are replaced by
$(\mathscr {L}(V(\lambda ,\ell )),\mathscr {B}(V(\lambda ,\ell )))$
and
$Qv_{(\lambda ,\ell +t)}$
respectively. Thus by [Reference Kashiwara14, Lemma 4.3.9], it is enough to show that
where we have by [Reference Kashiwara14, Corollary 3.4.6],
Let N be as in (3.5). Then we have
$P v_{(\lambda ,\ell +t)} {\kern-1pt}\in{\kern-1pt} \mathscr {L}(V(\lambda ,\ell {\kern-1pt}+{\kern-1pt}t))$
if and only if
${P v_{(\lambda ,\ell {\kern-1pt}+{\kern-1pt}N)} {\kern-1pt}\in{\kern-1pt} \mathscr {L}(V(\lambda ,\ell +N))}$
for all
$t\ge N$
. On the other hand, there exists
$M_\nu \in \mathbb {Z}$
such that
$q^{M_\nu } ({e^{\prime \prime }_0}^{\, \nu } Q) v_{(\lambda ,\ell )} \in \mathscr {L}(V(\lambda ,\ell +t))$
for all
$t \geq N$
. Then one can choose
$N'\geq N$
such that
for all
$t\geq N'$
and
$1 \leq \nu \leq k+1$
. This completes the proof.
The following lemma shows that
$P(\lambda )$
can be viewed as the limit of
$V(\lambda ,\ell )$
as
$\ell \to \infty $
.
Lemma 3.6. Let
$(\lambda ,\ell )\in \mathscr {P}(\mathfrak {g})$
and
$\xi \in Q^-$
be given. Then, for sufficiently large integers t, we have
$P(\lambda )_{\Lambda (\lambda ,0)+\xi } \cong V(\lambda ,\ell )_{\Lambda (\lambda ,\ell +t)+\xi }$
as vector spaces.
Theorem 3.7. Let
$(\lambda ,\ell )\in \mathscr {P}(\mathfrak {g})$
be given. Then
-
(1) $\pi _{V(\lambda ,\ell )}(\mathscr {L}(P(\lambda )))=\mathscr {L}(V(\lambda ,\ell ))$
, -
(2) $\overline {\pi }_{V(\lambda ,\ell )}(\mathscr {B}(P(\lambda )))= \mathscr {B}(V(\lambda ,\ell )) \cup \{0\}$
, where $$ \begin{align*}\overline{\pi}_{V(\lambda,\ell)} : \mathscr{L}(P(\lambda))/q\mathscr{L}(P(\lambda)) \rightarrow \mathscr{L}(V(\lambda,\ell))/q\mathscr{L}(V(\lambda,\ell))\end{align*} $$is the induced $\mathbb {Q}$
-linear map at
$q=0$
,
-
(3) $\overline {\pi }_{V(\lambda ,\ell )}$
restricts to a bijection $$ \begin{align*}\overline{\pi}_{V(\lambda,\ell)} : \{\, b\in \mathscr{B}(P(\lambda))\,|\,\overline{\pi}_{V(\lambda,\ell)}(b)\neq 0 \,\} \longrightarrow \mathscr{B}(V(\lambda,\ell)),\end{align*} $$which commutes with $\tilde {e}_i$
and
$\tilde {f}_i$
for
$i\in I$
.
Proof. Since
$P(\lambda )$
and
$V(\lambda ,\ell )$
are semisimple
${\mathcal U}(\mathfrak {l})$
-modules, every
${\mathcal U}(\mathfrak {l})$
-irreducible component L of
$P(\lambda )$
is isomorphic to its image under
$\pi _{V(\lambda ,\ell )}$
if it is not zero. Thus
$\pi _{V(\lambda ,\ell )}$
commutes with
$\tilde {e}_i$
and
$\tilde {f}_i$
if
$i\in I\setminus \{ 0 \}$
.
Now we claim that
for all
$(\lambda ,\ell ) \in \mathscr {P}(\mathfrak {g})$
and
$i_i,\dots ,i_k\in I$
. We use induction on k. It is clearly true when
$k=0$
. Suppose that (3.6) is true for
$k-1$
.
If
$i_k \ne 0$
, then (3.6) is true since
$\tilde {x}_i$
commutes with
$\pi _{V(\lambda ,\ell )}$
. So we assume
$i_k=0$
. Let
$P\in {\mathcal U}(\mathfrak {g})^-$
such that
$\tilde {x}_{i_{k-1}}\dots \tilde {x}_{i_1}(1 \otimes v_\lambda )=P(1\otimes v_\lambda )$
. By the induction hypothesis,
for all
$(\lambda ,\ell ) \in \mathscr {P}(\mathfrak {g})$
. By Lemma 3.5 and (3.7), there exists N such that for all
$t \geq N$
Applying
$\Phi _{\ell +t,\ell }$
to (3.8) and using Lemma 3.4 together with the induction hypothesis, we obtain
Finally, since
$\tilde {x}_0 (P(1\otimes v_\lambda )) = (\tilde {x}_0 P)(1 \otimes v_\lambda )$
by definition, we have
where the second-to-last equivalence follows from (3.9). This completes the proof of (3.6). Then (1) and (2) follow from (3.6) and the connectedness of
$\mathscr {B}(P(\lambda ))$
and
$\mathscr {B}(V(\lambda ,\ell ))$
. We now prove (3). Suppose there exist
$b,b' \in \mathscr {B}(P(\lambda ))_{\Lambda (\lambda ,0)+\xi }$
for some
$\xi \in Q^-$
with
$b \neq b'$
and
${\overline {\pi }_{V(\lambda ,\ell )}(b)=\overline {\pi }_{V(\lambda ,\ell )}(b') \neq 0}$
. By Lemma 3.6, for a sufficiently large integer t we have an isomorphism
$\mathscr {B}(P(\lambda ))_{\Lambda (\lambda ,0)+\xi } \cong \mathscr {B}(V(\lambda ,\ell +t))_{\Lambda (\lambda ,t)+\xi }$
, and therefore
$\overline {\pi }_{V(\lambda ,\ell +t)}(b) \neq \overline {\pi }_{V(\lambda ,\ell +t)}(b')$
. Combining this with Lemma 3.4 (2) and
$\pi _{V(\lambda ,\ell )} = \Phi _{\ell +t,\ell +N} \circ \pi _{V(\lambda ,\ell +t)}$
yields
$\overline {\pi }_{V(\lambda ,\ell )}(b) \neq \overline {\pi }_{V(\lambda ,\ell )}(b')$
, a contradiction. Together with (3.6), the proof of (3) is complete.
4 Crystal base of
${\mathcal U}(\mathfrak {l})^-$
Let us briefly recall the results on the crystal base of
${\mathcal U}(\mathfrak {l})^-$
in [Reference Jang, Kwon and Uruno12].
4.1 Crystal base of a Kac module
Let us first recall the crystal bases of q-deformed Kac modules [Reference Kwon19, Section 4]. Let
$\Phi ^+(\mathfrak {l})_{\kappa }=\Phi ^+(\mathfrak {l})\cap \Phi ^+_{\kappa }$
(
$\kappa ={\overline 0}, {\overline 1}$
). Let
$\textbf {f}^{\prime }_\beta $
be the root vector for
$\beta \in \Phi ^+(\mathfrak {l})$
with a partial order
$\prec '$
on
$\Phi ^+(\mathfrak {l})$
defined as in [Reference Jang, Kwon and Uruno12, Section 2.5], where
$\alpha \prec '\beta $
for
$\alpha \in \Phi ^+(\mathfrak {l})_{\overline 1}$
and
$\beta \in \Phi ^+(\mathfrak {l})_{\overline 0}$
. Note that
$\textbf {f}^{\prime }_\beta $
is different from
$\textbf {f}_\beta $
in Section 2.3.
Let
$\mathcal {K}$
be the
$\Bbbk $
-span of
$\mathbf {f}^{\prime }_S$
for
$S\subset \Phi ^+(\mathfrak {l})_{\overline 1}$
, where
$\mathbf {f}^{\prime }_S$
is the product of
$\textbf {f}^{\prime }_\alpha $
(
$\alpha \in S$
) with respect to
$\prec '$
. With respect to the associated PBW type basis, we have
Moreover
$\mathcal {K}$
is equal to the subalgebra generated by
$\textbf {f}^{\prime }_\alpha $
for
$\alpha \in \Phi ^+(\mathfrak {l})_{\overline 1}$
, and it is a
${\mathcal U}(\mathfrak {l}_{\overline 0})$
-module with respect to (3.3), which is a direct sum of
$V_{\mathfrak {l}_{\overline 0}}(\lambda )$
for some
$\mathfrak {l}_{\overline 0}$
-dominant weights
$\lambda \in \texttt {P}^+$
. The pair
$(\mathscr {L}({\mathcal K}),\mathscr {B}({\mathcal K}))$
with
is a crystal base of
$\mathcal {K}$
as a
${\mathcal U}(\mathfrak {l}_{\overline 0})$
-module.
For
$\lambda \in \mathtt {P}^+$
, let
where
$\mathcal {Q}$
is the subalgebra of
${\mathcal U}(\mathfrak {l})$
generated by
${\mathcal U}(\mathfrak {l}_{\overline 0})$
and
${\mathcal U}(\mathfrak {l})^+$
, and
$V_{\mathfrak {l}_{\overline 0}}(\lambda )$
is extended to a
$\mathcal {Q}$
-module trivially. It is a highest weight
${\mathcal U}(\mathfrak {l})$
-module generated by
$1\otimes v_{\lambda }$
with highest weight
$\lambda $
. As a
${\mathcal U}(\mathfrak {l}_{\overline 0})$
-module, we have
We define a crystal base of
$K(\lambda )$
as in Section 3.1.1, where
$\tilde {e}_m, \tilde {f}_m$
are replaced by
Then
$K(\lambda )$
has a unique crystal base
$\left (\mathscr {L}({K(\lambda )}),\mathscr {B}(K(\lambda ))\right )$
, where
$\mathscr {B}(K(\lambda ))$
is connected. Let
$(\mathscr {L}(V_{\mathfrak {l}_{\overline 0}}(\lambda )), \mathscr {B}(V_{\mathfrak {l}_{\overline 0}}(\lambda )))$
be the crystal base of
$V_{\mathfrak {l}_{\overline 0}}(\lambda )$
generated by
$v_{\lambda }$
. Then we have
under (4.2), where
$\cong $
means the isomorphism of crystal bases.
For
$\lambda \in \mathscr {P}_{m|n}$
, let
be the
${\mathcal U}(\mathfrak {l})$
-module homomorphism such that
$\pi _{V_{\mathfrak {l}}(\lambda )}(1\otimes v_\lambda )=v_\lambda $
. Then
$\pi _{V_{\mathfrak {l}}(\lambda )}$
preserves the crystal bases in the sense of Theorem 3.7.
For
$\lambda ,\mu \in \mathtt {P}^+$
, we define
$\lambda <\mu $
if and only if
$\mu -\lambda \in \mathtt {P}^+$
. It is shown in [Reference Jang, Kwon and Uruno12] that
$\{\,\mathscr {B}(K(\lambda ))\,|\,\lambda \in \mathtt {P}^+\,\}$
forms an inductive system of (abstract)
${\mathcal U}({\mathfrak l}_{\overline 0})$
-crystals (not
${\mathcal U}(\mathfrak l)$
-crystal), with respect to the injective morphism of
${\mathcal U}({\mathfrak l}_{\overline 0})$
-crystals;
for
$\lambda < \mu $
. Then its limit, say
$\mathscr {B}(K(\infty ))$
, becomes an (abstract)
${\mathcal U}(\mathfrak l)$
-crystal (see [Reference Jang, Kwon and Uruno12, Section 5.3] for more details), where for
$b\in \mathscr {B}(K(\infty ))$
where
$\mathbf {0}$
denotes the formal zero (when we consider an abstract crystal).
4.2 Crystal base of
${\mathcal U}(\mathfrak {l})^-$
Let us recall a crystal base of
${\mathcal U}(\mathfrak {l})^-$
[Reference Jang, Kwon and Uruno12, Section 6]. It is given in such a way that the associated crystal is
$\mathscr {B}(K(\infty ))$
.
We first need a parabolic Verma module and its crystal base, which interpolates those of
$K(\lambda )$
and
${\mathcal U}(\mathfrak {l})^-$
[Reference Jang, Kwon and Uruno12, Section 6.2]. For
$\lambda \in \mathtt {P}$
with
$\lambda _-\in \mathtt {P}^+$
, let
where
$\mathcal {R}$
is the subalgebra of
${\mathcal U}(\mathfrak {l})$
generated by
${\mathcal U}(\mathfrak {l}_{0|n})$
and
${\mathcal U}(\mathfrak {l})^+$
, and
$V_{\mathfrak {l}_{0|n}}(\lambda )$
is extended to an
$\mathcal {R}$
-module trivially. It is also a highest weight
${\mathcal U}(\mathfrak {l})$
-module generated by
$1\otimes v_{\lambda }$
with highest weight
$\lambda $
. As a
$\Bbbk $
-space, we have
We define a crystal base of
$X(\lambda )$
as in
$K(\lambda )$
, where
$\tilde {e}_i, \tilde {f}_i$
for
$i<m$
are replaced by those defined as in (3.2) with respect to
$e^{\prime }_i$
. Then
$X(\lambda )$
also has a unique crystal base
$\left (\mathscr {L}({X(\lambda )}),\mathscr {B}(X(\lambda ))\right )$
, where
$\mathscr {B}(X(\lambda ))$
is connected.
Let
$(\mathscr {L}({\mathcal U}(\mathfrak {l}_{m|0})^-),\mathscr {B}({\mathcal U}(\mathfrak {l}_{m|0})^-))$
and
$(\mathscr {L}(V_{\mathfrak {l}_{0|n}}(\lambda )),\mathscr {B}(V_{\mathfrak {l}_{0|n}}(\lambda )))$
be the crystal bases of
${\mathcal U}(\mathfrak {l}_{m|0})^-$
and
$V_{\mathfrak {l}_{0|n}}(\lambda )$
, respectively. We have
under (4.6).
For
$\lambda \in \mathtt {P}^+$
, let
be the
${\mathcal U}(\mathfrak {l})$
-module homomorphism such that
$\pi _{K(\lambda )}(1\otimes v_{\lambda })=1\otimes v_{\lambda }$
. Then
$\pi _{K(\lambda )}$
preserves the crystal bases in the sense of Theorem 3.7.
Let
$u\in {\mathcal U}(\mathfrak {l})^-$
be given with
$u=u_1 u_2 u_3$
for
$u_1\in \mathcal {K}$
,
$u_2\in {\mathcal U}(\mathfrak {l}_{m|0})^-$
and
$u_3\in {\mathcal U}(\mathfrak {l}_{0|n})^-$
by (4.1). For
$i\in I$
with
$i>m$
, we define
where
$\tilde {e}_i$
and
$\tilde {f}_i$
are those on
${\mathcal U}(\mathfrak {l}_{0|n})^-$
. Then we define a crystal base of
${\mathcal U}(\mathfrak {l})^-$
as in
$X(\lambda )$
, where
$\tilde {e}_i, \tilde {f}_i$
for
$i>m$
are replaced by (4.7).
Theorem 4.1 [Reference Jang, Kwon and Uruno13, Theorem 6.1].
Let
where
$\cdot $
denotes the multiplication in
${\mathcal U}^-$
and the induced one in
$\mathscr {L}({\mathcal U}(\mathfrak {l})^-)/q\mathscr {L}({\mathcal U}(\mathfrak {l})^-)$
respectively. Then
$(\mathscr {L}({\mathcal U}(\mathfrak {l})^-),\mathscr {B}({\mathcal U}(\mathfrak {l})^-))$
is a crystal base of
${\mathcal U}(\mathfrak {l})^-$
, where
$\mathscr {B}({\mathcal U}(\mathfrak {l})^-)$
is isomorphic to
$\mathscr {B}(K(\infty ))$
.
Let
$\lambda \in \mathtt {P}$
with
$\lambda _-\in \mathtt {P}^+$
. Since
$\left (\mathscr {L}({X(\lambda )}),\mathscr {B}(X(\lambda ))\right )$
is an upper crystal base of
$X(\lambda )$
as a
${\mathcal U}(\mathfrak {l}_{0|n})$
-module, it is not compatible with that of
${\mathcal U}(\mathfrak l)^-$
under the canonical projection from
${\mathcal U}(\mathfrak {l})^-$
to
$X(\lambda )$
.
Let
$(\mathscr {L}({\mathcal U}(\mathfrak {l}_{0|n})^-),\mathscr {B}({\mathcal U}(\mathfrak {l}_{0|n})^-))$
be the crystal base of
${\mathcal U}(\mathfrak {l}_{0|n})^-$
(see [Reference Jang, Kwon and Uruno12, (3.8)] for more details). For
$\lambda \in \mathtt {P}$
with
$\lambda _-\in \mathtt {P}^+$
, let
$\pi _{\lambda }: {\mathcal U}(\mathfrak {l}_{0|n})^- \longrightarrow V_{\mathfrak {l}_{0|n}}(\lambda )$
be the canonical projection, and let
be the dual of
$\pi _{\lambda }$
induced from nondegenerate bilinear forms on
${\mathcal U}(\mathfrak {l}_{0|n})^-$
and
$V_{\mathfrak {l}_{0|n}}(\lambda )$
, which is injective. Then we have
where
$\overline {\pi }_\lambda ^\vee $
is the induced map at
$q=0$
and
$\overline {\pi }_\lambda ^\vee (\tilde {e}_i b)= \tilde {e}_i \overline {\pi }_\lambda ^\vee (b)$
for
$i>m$
and
$b\in \mathscr {B}(V_{\mathfrak {l}_{0|n}}(\lambda ))$
(cf. [Reference Jang, Kwon and Uruno12, Section 3.2.2] and references therein).
For
$\lambda \in \mathtt {P}$
with
$\lambda _-\in \mathtt {P}^+$
, let
$\pi _{X(\lambda )}: {\mathcal U}(\mathfrak {l})^- \longrightarrow X(\lambda )$
be the
${\mathcal U}(\mathfrak {l})^-$
-linear map such that
$\pi _{X(\lambda )}(1)=1\otimes v_{\lambda }$
. By (4.1), (4.6), and (4.8), we immediately have
given by
$\pi _{X(\lambda )}^\vee =1_{\mathcal {K}\otimes {\mathcal U}(\mathfrak {l}_{m|0})^-}\otimes \pi ^\vee _\lambda $
. Then we have
where
$ \overline {\pi }_{X(\lambda )}^\vee $
is the induced map at
$q=0$
. Moreover,
$\overline {\pi }^\vee _{X(\lambda )}$
restricts to a bijection
where
$b=b_1b_2b_3{\kern-1pt}\in{\kern-1pt} \mathscr {B}({\mathcal U}(\mathfrak {l})^-)$
is as in Theorem 4.1 and
$\{t_\lambda \}$
is the (abstract) crystal with
${\tilde {e}_i t_\lambda {\kern-1pt}={\kern-1pt}\tilde {f}_i t_\lambda =0}$
,
$\varepsilon _i(t_\lambda )=\varphi _i(t_\lambda )=-\infty $
for all
$i\in I$
, and
$\mathrm { wt}(t_\lambda )=\Lambda _{0,\lambda }$
. Here
$\overline {\pi }^\vee _{X(\lambda )}$
commutes with
$\tilde {e}_i$
,
$\tilde {f}_i$
for
$i<m$
, while it commutes with
$\tilde {e}_i$
,
$\tilde {f}_i$
for
$i\ge m$
and
$b'=\overline {\pi }^\vee _{X(\lambda )}(b)$
with
$\lambda $
sufficiently large (see [Reference Jang, Kwon and Uruno12, Remark 6.6]).
Let
$\lambda \in \mathscr {P}_{m|n}$
be given. Summarizing, the crystal bases mentioned in this subsection are compatible with the following maps:
The maps in (4.12) induce injective maps at
$q=0$
:
A combinatorial description of
$\iota _{V_{\mathfrak {l}}(\lambda )}$
can be seen in [Reference Kwon19], and
$\iota _{K(\lambda )}$
is essentially the one from
$\mathscr {B}(V_{\mathfrak {l}_{\overline 0}}(\lambda ^+))$
into
$\mathscr {B}({\mathcal U}(\mathfrak {l}_{m|0})^-)$
for
$\lambda =\lambda ^+\in \texttt {P}^+$
given by counting the number of entries in each row of a semistandard Young tableau to give multiplicities of root vectors (and hence Lusztig data for a PBW type crystal), and the description of
$\iota _{X(\lambda )}$
is similar.
We remark that the crystal
$\mathscr {B}({\mathcal U}(\mathfrak {l})^-)$
is not connected for
$n\ge 2$
unlike
$\mathscr {B}(X(\lambda ))$
,
$\mathscr {B}(K(\lambda ))$
, and
$\mathscr {B}(V_{\mathfrak l}(\lambda ))$
. More precisely, it is not connected if and only if
$n\ge 2$
, and
(see [Reference Jang, Kwon and Uruno12, Theorem 6.8] for more details).
5 Crystal base of
${\mathcal U}(\mathfrak {g})^-$
For
$\lambda \in \mathscr {P}_{m|n}$
, we have the following diagram:

where the vertical maps are
$\Bbbk $
-linear isomorphisms. We construct crystal bases of
$Q(\lambda )$
,
$R(\lambda )$
, and
${\mathcal U}(\mathfrak {g})^-$
using Theorem 3.3 and the results in Section 4.
5.1 Crystal base of
$Q(\lambda )$
Let
$B_q(\mathfrak {l}_{1|1})$
be the q-boson algebra associated with
$\mathfrak {l}_{1|1}$
, that is, an associative
$\Bbbk $
-algebra generated by
$e'$
and f subject to the relations
As these relations have no q-dependence,
$B_q(\mathfrak {l}_{1|1})$
is simply isomorphic to the rank 2 Clifford algebra with f (resp.
$e'$
) representing a single fermionic particle creation (resp. annihilation) operator. Note that
$\langle e^{\prime }_m, f_m \rangle \cong B_q(\mathfrak {l}_{1|1})$
, where
$e^{\prime }_m$
acts as in (2.1), and
$f_m$
acts by left multiplication on
${\mathcal U}(\mathfrak {g})^-$
.
It is easy to see that
${\mathcal U}(\mathfrak {l}_{1|1})^-=\Bbbk 1 \oplus \Bbbk f$
is the unique irreducible
$B_q(\mathfrak {l}_{1|1})$
-module up to isomorphism, where
${\mathcal U}(\mathfrak {l}_{1|1})=\langle e,f,k^{\pm 1} \rangle $
. For a
$B_q(\mathfrak {l}_{1|1})$
-module, we define crystal operators
$\widetilde {e}$
and
$\widetilde {f}$
as in (4.3). Then
${\mathcal U}(\mathfrak {l}_{1|1})^-$
has a unique crystal base
$(L,B)$
up to scalar multiplication, where
$L=A_0 1 \oplus A_0 f$
and
$B=\{1, f \ (\mathrm {mod} \ qL) \}$
with
$\tilde {f} 1= f (\mathrm {mod} \ qL)$
. Furthermore, the natural basis
$\{1, f\}$
of this unique simple module actually serves as a global crystal basis, which coincides with the canonical basis introduced in [Reference Clark4] for the
$m=1$
case.
For
$\lambda =a \delta _1 + b \delta _2$
with
$a, b\in \mathbb {Z}_{\ge 0}$
, let
$V_{\mathfrak {l}_{1|1}}(\lambda )$
be the irreducible highest weight
${\mathcal U}(\mathfrak {l}_{1|1})$
-module with highest weight
$\lambda $
. Then
$V_{\mathfrak {l}_{1|1}}(\lambda )\otimes {\mathcal U}(\mathfrak {l}_{1|1})^-$
is a
$B_q(\mathfrak {l}_{1|1})$
-module under the
$\Bbbk $
-algebra homomorphism
$\Delta _{B_q(\mathfrak {l}_{1|1})} : B_q(\mathfrak {l}_{1|1}) \longrightarrow {\mathcal U}(\mathfrak {l}_{1|1})\otimes B_q(\mathfrak {l}_{1|1})$
, which is defined by
Lemma 5.1. Let
$V_1=V_{\mathfrak {l}_{1|1}}(\lambda )$
and
$V_2={\mathcal U}(\mathfrak {l}_{1|1})^-$
with crystal bases
$(L_1,B_1)$
and
$({L}_2,{B}_2)$
respectively. Then
$(L_1 \otimes L_2,B_1 \otimes B_2)$
is a crystal base of
$V_1\otimes V_2$
, where
$\tilde {e}$
and
$\tilde {f}$
act on
$B_1 \otimes B_2$
by
Proof. The proof for the case when
$a=b=0$
is clear. Suppose that
$a+b>0$
. Then we have
$V_1=\Bbbk v_1 \oplus \Bbbk f v_1$
and it has a unique crystal base up to scalar multiplication. So we may assume that
$L_1= A_0 v_1 \oplus A_0 f v_1$
and
$L_2=A_0 v_2 \oplus A_0 fv_2$
with
$v_2=1\in {\mathcal U}(\mathfrak {l}_{1|1})^-$
. We have
where
$X=(1-q^{2(a+b)}) v_1 \otimes f v_2 - q^{a+b} f v_1 \otimes v_2 \equiv v_1 \otimes f v_2 \ (\mathrm {mod} \ qL_1\otimes L_2)$
.
For
$\lambda \in \mathtt {P}^+$
, consider
$Q(\lambda )$
in (2.5). As a
$\Bbbk $
-space, we have
where the second isomorphism is given by sending
$u_1\otimes u_2 \otimes v$
to
$u_1 \otimes u_2 v$
for
$u_1\in \mathcal {N}$
,
$u_2\in \mathcal {K}$
and
$v\in V_{\mathfrak {l}_{\overline 0}}(\lambda )$
from the right to the left.
Lemma 5.2. The
$\Bbbk $
-linear map in (5.1) is an isomorphism of
${\mathcal U}({\mathfrak l}_{\overline 0})$
-modules.
Proof. By the same arguments as in [Reference Jang, Kwon and Uruno13, Proposition 4.20], the first
$\Bbbk $
-linear map in (5.1) is
${\mathcal U}({\mathfrak l}_{\overline 0})$
-linear. Also by [Reference Kwon19, Proposition 4.1], the
$\Bbbk $
-linear isomorphism (4.2) is
${\mathcal U}({\mathfrak l}_{\overline 0})$
-linear, and hence the second
$\Bbbk $
-linear map in (5.1) is
${\mathcal U}({\mathfrak l}_{\overline 0})$
-linear.
We define a crystal base of
$Q(\lambda )$
as in Section 3.1.2 for
$P(\lambda )$
, where
$\tilde {e}_m, \tilde {f}_m$
are replaced by the ones in (4.3).
Theorem 5.3. For
$\lambda \in \mathtt {P}^+$
, let
where
$\cdot $
denotes the multiplication in
${\mathcal U}(\mathfrak {g})^-$
and the induced one at
$q=0$
, respectively. Then
$(\mathscr {L}(Q(\lambda )), \mathscr {B}(Q(\lambda )))$
is a crystal base of
$Q(\lambda )$
, and the crystal
$\mathscr {B}(Q(\lambda ))$
is connected.
Proof. Put
$\mathscr {L}=\mathscr {L}(Q(\lambda ))$
and
$\mathscr {B}=\mathscr {B}(Q(\lambda ))$
. First we show that
$(\mathscr {L},\mathscr {B})$
is a crystal base. It suffices to show that
$\widetilde {x}_i \mathscr {L}\subset \mathscr {L}$
and
$\widetilde {x}_i \mathscr {B}\subset \mathscr {B}\cup \{0\}$
for
$i\in I$
and
$x=e,f$
, which follows from arguments similar to those in [Reference Jang, Kwon and Uruno13, Theorem 4.22] when
$i=0$
, and Lemma 5.2 when
$i\in I\setminus \{m\}$
. If
$i=m$
, it follows from Lemma 5.1, since
$\mathcal {N}$
is a semisimple
${\mathcal U}(\mathfrak {l}_{1|1})$
-module and
$\mathcal {K}$
is a semisimple
$B_q(\mathfrak {l}_{1|1})$
-module.
Next, we show that
$\mathscr {B}$
is connected. As a set, we have
Note that the crystal graph
$\mathscr {B}(K(\lambda ))$
is equal to that of
$\mathscr {B}(K(\lambda +\ell \delta _+))$
(up to shifting weights by
$\ell \delta _+$
), where
$\delta _+=\delta _1+\cdots +\delta _m$
for all
$\ell \ge 0$
. Hence we may identify
$\mathscr {B}$
with
$\mathscr {B}(\mathcal {N})\times \mathscr {B}(K(\lambda +\ell \delta _+))$
(up to shifting weights by
$\ell \delta _+$
).
On the other hand, if
$\ell $
is large enough, then
$K(\lambda +\ell \delta _+)=V_{\mathfrak l}(\lambda +\ell \delta _+)$
, and
$\mathscr {B}(K(\lambda +\ell \delta _+))=\mathscr {B}(V_{\mathfrak l}(\lambda +\ell \delta _+))$
. Since
$\mathscr {B}(P(\lambda +\ell \delta _+))=\mathscr {B}(\mathcal {N})\times \mathscr {B}(V_{\mathfrak {l}}(\lambda +\ell \delta _+))$
is connected by Theorem 3.2,
$\mathscr {B}$
is connected (see [Reference Kwon19, Section 6.4] and references therein).
Corollary 5.4. For
$\lambda \in \mathscr {P}_{m|n}$
, the map
$\pi _{P(\lambda )}: Q(\lambda ) \longrightarrow P(\lambda )$
preserves the crystal bases in the sense of Theorem 3.7.
Proof. It follows from the fact that the map (4.4) preserves the crystal bases and
$\varepsilon _i, \varphi _i$
for
$ i \in I \setminus \{ 0, m \}$
since
$Q(\lambda )$
and
$P(\lambda )$
are semisimple
${\mathcal U}(\mathfrak {l}_{\overline 0})$
-modules.
The crystal structure on
$\mathscr {B}(Q(\lambda ))$
can be described explicitly from (5.2). As an
$\mathfrak {l}_{\overline 0}$
-crystal, we have
while
$\tilde {e}_0, \tilde {f}_0$
act only on
$\mathscr {B}(\mathcal {N})$
, and the actions of
$\tilde {e}_m, \tilde {f}_m$
are given by Lemma 5.1. A combinatorial description of
$\mathscr {B}(\mathcal {N})$
and
$\mathscr {B}(K(\lambda ))$
can be found in [Reference Jang, Kwon and Uruno13, Section 4.2] and [Reference Kwon19, Section 5.1], respectively.
5.2 Crystal base of
$R(\lambda )$
Let
$B_q=B_q({\mathfrak l}_{m|0})$
be the q-boson algebra associated with
$\mathfrak {l}_{m|0}$
generated by
$e^{\prime }_i$
and
$f_i$
(
$i=1,\dots ,m-1$
) [Reference Kashiwara14, Section 3.3]. Then
${\mathcal U}(\mathfrak {l}_{m|0})^-$
is the unique irreducible
$B_q$
-module, where
$e^{\prime }_i$
acts as in (2.1) and
$f_i$
acts as left multiplication.
For
$\lambda \in \mathtt {P}^+$
with
$\lambda =\lambda _+$
, let
$V_{\mathfrak {l}_{m|0}}(\lambda )$
be the irreducible highest weight
${\mathcal U}(\mathfrak {l}_{m|0})$
-module with highest weight
$\lambda $
. Then
$V_{\mathfrak {l}_{m|0}}(\lambda )\otimes {\mathcal U}(\mathfrak {l}_{m|0})^-$
is a
$B_q$
-module under the
$\Bbbk $
-algebra homomorphism
$\Delta _{B_q} : B_q \longrightarrow {\mathcal U}(\mathfrak {l}_{m|0})\otimes B_q$
given by
Let
$V_1=V_{\mathfrak {l}_{m|0}}(\lambda )$
and
$V_2={\mathcal U}(\mathfrak {l}_{0|n})^-$
with crystal bases
$(L_1,B_1)$
and
$({L}_2,{B}_2)$
, respectively. Then
$(L_1 \otimes L_2,B_1 \otimes B_2)$
is a crystal base of
$V_1\otimes V_2$
, where
$\tilde {e}_i$
and
$\tilde {f}_i\ (i \in I)$
act on
$B_1 \otimes B_2$
by the same formula as in the case of a tensor product of crystal bases of
${\mathcal U}(\mathfrak {l}_{m|0})$
-modules [Reference Kashiwara14, Remark 3.5.1].
For
$\lambda \in \mathtt {P}$
with
$\lambda _-\in \mathtt {P}^+$
, consider
$R(\lambda )$
in (2.6). As a
$\Bbbk $
-space, we have
where the isomorphism is given in a standard way (cf. (5.1)).
Lemma 5.5. The
$\Bbbk $
-linear map in (5.3) is an isomorphism of
${\mathcal U}({\mathfrak l}_{0|n})$
-modules, where we regard
${\mathcal U}(\mathfrak {l}_{m|0})^-$
as a trivial
${\mathcal U}({\mathfrak l}_{0|n})$
-module, and also it is an isomorphism of
$B_q$
-modules, where we regard
$V_{\mathfrak {l}_{0|n}}(\lambda )$
as a trivial
$B_q$
-module.
Proof. The proof is almost identical to that of Lemma 5.2. We leave it to the reader.
We define a crystal base of
$R(\lambda )$
as in
$Q(\lambda )$
, where
$\tilde {e}_i, \tilde {f}_i$
for
$i=1,\dots ,m-1$
are replaced by those defined as in (3.2) with respect to
$e^{\prime }_i$
.
Theorem 5.6. For
$\lambda \in \mathtt {P}$
with
$\lambda _-\in \mathtt {P}^+$
, let
Then
$(\mathscr {L}(R(\lambda )), \mathscr {B}(R(\lambda )))$
is a crystal base of
$R(\lambda )$
, where the crystal
$\mathscr {B}(R(\lambda ))$
is connected.
Proof. By the same arguments as in Theorem 5.3 using Lemma 5.5 in this case, we have that
$(\mathscr {L}(R(\lambda )), \mathscr {B}(R(\lambda )))$
is a crystal base of
$R(\lambda )$
. The proof of connectedness of
$\mathscr {B}(R(\lambda ))$
is almost identical to the proof of Theorem 6.2 in [Reference Jang, Kwon and Uruno12], since
$\mathscr {B}(R(\lambda ))$
is isomorphic to the limit of
$\mathscr {B}(P(\lambda ))$
when
$\lambda _+ \rightarrow \infty $
.
As a set, we have
Recall that
$\tilde {e}_0, \tilde {f}_0$
act only on
$\mathscr {B}(\mathcal {N})$
. The actions of
$\tilde {e}_i, \tilde {f}_i$
for
$i\in I\setminus \{0\}$
can be described by applying them to
Corollary 5.7. For
$\lambda \in \mathtt {P}^+$
, the map
$\pi _{Q(\lambda )}: R(\lambda ) \longrightarrow Q(\lambda )$
preserves the crystal bases in the sense of Theorem 3.7.
Proof. Note that
$\pi _{Q(\lambda )}=\mathrm {id}_{\mathcal {N}\cdot \mathcal {K}}\otimes \pi _{\lambda }\otimes \mathrm { id}_{V_{\mathfrak {l}_{0|n}}(\lambda )}$
, where
$\pi _{\lambda }: {\mathcal U}(\mathfrak {l}_{m|0})^- \longrightarrow V_{\mathfrak {l}_{m|0}}(\lambda )$
is the canonical projection. Since
$\pi _\lambda $
preserves the crystal bases and
$\varepsilon _i$
,
$\pi _{Q(\lambda )}$
also preserves crystal bases by Theorem 5.6.
5.3 Crystal base of
${\mathcal U}(\mathfrak {g})^-$
We define a crystal base of
${\mathcal U}(\mathfrak {g})^-$
as in
$R(\lambda )$
except for
$\tilde {e}_i$
and
$ \tilde {f}_i$
for
$i>m$
, which are defined by
for
$u=u_1u_2\in {\mathcal U}(\mathfrak {g})^-$
with
$u_1\in \mathcal {N}$
and
$u_2\in {\mathcal U}(\mathfrak l)^-$
, where
$\tilde {e}_iu_2$
and
$\tilde {f}_iu_2$
are given in (4.7).
As a set, let
We define a
${\mathcal U}(\mathfrak {g})$
-crystal structure on this set, which is intended to be the crystal of
${\mathcal U}(\mathfrak {g})^-$
, as follows: Let
$b=(S,b_0)$
be given. We let
$\mathrm {wt}(b) = \mathrm {wt}(S) + \mathrm {wt}(b_0)$
, and let
$\varepsilon _i(b)$
and
$\varphi _i(b)$
be as in (4.5). For
$i\in I$
and
$x=e,f$
, we define
where
$\tilde {x}_i (S\otimes b_0)$
is defined by using [Reference Kashiwara14, Remark 3.5.1] for
$1\le i<m$
and using Lemma 5.1 for
$i=m$
. Here we assume that
$\tilde {x}_i b=\mathbf {0}$
if any of its components on the right-hand side is
$\mathbf {0}$
.
Theorem 5.8. Let
Then
$(\mathscr {L}(\infty ),\mathscr {B}(\infty ))$
is a crystal base of
${\mathcal U}(\mathfrak {g})^-$
.
Proof. As in the proof of Theorem 5.3, it suffices to show that
$\widetilde {x}_i \mathscr {L}(\infty )\subset \mathscr {L}(\infty )$
and
$\widetilde {x}_i \mathscr {B}(\infty )\subset \mathscr {B}(\infty )\cup \{0\}$
for
$i\in I$
and
$x=e,f$
. For the cases
$i=0$
and
$1 \leq i \le m$
, the arguments are almost identical to those of Theorems 5.3 and 5.6, respectively. For
$i>m$
, the claim is directly obtained from Theorem 4.1 via (5.6). The crystal structure then immediately follows from these arguments.
Example 5.9. Let
$\mathfrak {g}=\mathfrak {d}_{2|2}$
. Then, we have
as a (
$\mathfrak {g}_{2|1}$
,
$\mathfrak {g}_{0|2}$
)-bicrystal. For
$\tilde {x}=\tilde {e}$
or
$\tilde {f}$
, it follows from (5.6) that
$\tilde {x}_3$
acts exclusively on
$\mathscr {B}({\mathcal U}(\mathfrak {l}_{0|2})^-)$
and commutes with
$\tilde {x}_i$
’s for
$0 \leq i \leq 2$
. Since
$\mathscr {B}({\mathcal U}(\mathfrak {l}_{0|2})^-) $
is isomorphic to
$ \mathscr {B}(U_q^-(\mathfrak {gl}_2))$
, the crystal graph of
$\mathscr {B}({\mathcal U}(\mathfrak {l}_{0|2})^-)$
is given as follows, where each arrow denotes
$\tilde {f}_3$
:
On the other hand, the operators
$\tilde {x}_i$
for
$0 \leq i \leq 2$
act on
$\mathscr {B}(\mathcal {N}) \otimes \mathscr {B}({\mathcal K}) \otimes \mathscr {B}({\mathcal U}(\mathfrak {l}_{2|0})^-)$
according to the two cases for
$i\leq m$
given in (5.5), and this crystal is disconnected (as is the case for general
$n \geq 2$
). The figure below illustrates the connected component containing
$1 \otimes 1 \otimes 1 \in \mathscr {B}(\mathcal {N}) \otimes \mathscr {B}({\mathcal K}) \otimes \mathscr {B}({\mathcal U}(\mathfrak {l}_{2|0})^-)$
(with
$\tilde {x}_3$
omitted). Here, solid, dashed, and dotted arrows represent
$\tilde {f}_0$
,
$\tilde {f}_1$
, and
$\tilde {f}_2$
, respectively:

For
$\lambda \in \mathtt {P}$
with
$\lambda _-\in \mathtt {P}^+$
, let
given by
$\mathrm {id}_{\mathcal {N}}\otimes \pi _{X(\lambda )}^\vee $
(see (4.9) for
$\pi _{X(\lambda )}^\vee $
).
Corollary 5.10. We have the following.
-
(1) $\pi ^\vee _{R(\lambda )}(\mathscr {L}(R(\lambda )))\subset \mathscr {L}(\infty )$
. -
(2) $\overline {\pi }^\vee _{R(\lambda )}(\mathscr {B}(R(\lambda )))\subset \mathscr {B}(\infty )$
, where
$\overline {\pi }^\vee _{R(\lambda )}$
is the
$\mathbb {Q}$
-linear map induced from
$\pi _{R(\lambda )}^\vee $
. -
(3) $\overline {\pi }^\vee _{R(\lambda )}$
restricts to a bijection $$ \begin{align*} \overline{\pi}^\vee_{R(\lambda)} : \mathscr{B}(R(\lambda)) \longrightarrow \{\,b=b_1b_2b_3b_4\in \mathscr{B}(\infty)\,|\ b_4 \otimes t_{\lambda_-} \in \mathscr{B}(V_{\mathfrak{l}_{0|n}}(\lambda)) \,\}, \end{align*} $$which commutes with $\tilde {e}_i$
,
$\tilde {f}_i$
for
$i<m$
, where
$b=b_1b_2b_3b_4 \in \mathscr {B}(\infty )$
is as in (5.4).
Remark 5.11. The map
$\overline {\pi }^\vee _{R(\lambda )}$
commutes with
$\tilde {e}_i, \tilde {f}_i$
for
$i\ge m$
only when
$\lambda _-$
is sufficiently large. More precisely, let
$b_0 \in \mathscr {B}(\infty )$
be given with
$b_0=\overline {\pi }^\vee _{R(\lambda )}(b)$
for some
$\lambda \in \mathtt {P}$
with
$\lambda _-\in \tt {P}^+$
and
$b \in \mathscr {B}(R(\lambda ))$
. Then
$\overline {\pi }^\vee _{R(\lambda )}(\tilde {x}_i b)= \tilde {x}_i \overline {\pi }^\vee _{R(\lambda )}(b)$
, for
$x=e, f$
if
$i>m$
and
$\lambda _- \gg 0$
, or if
$i=m$
and
$(\mathrm {wt}(b),\alpha _{m+1})\ll 0$
(cf. [Reference Jang, Kwon and Uruno12, Remark 6.6]).
Summarizing, the crystal bases constructed in Section 5 are compatible with the following maps (partly compatible in the case of
$\pi ^\vee _{R(\lambda )}$
):
This induces the following commuting diagram at
$q=0$
:

Since the maps in the bottom line are obtained from (4.13) by tensoring with
$\mathrm {id}_{\mathscr {B}(\mathcal {N})}$
, it is enough to describe
6 Combinatorial description of crystal embeddings
In this section, we give a combinatorial description of (5.7), which is a super-analogue of the results in [Reference Jang and Kwon11, Reference Kwon22]. So we may assume that
$n\ge 1$
.
The cases of
$\mathfrak {g}=\mathfrak {b}_{m|n}$
and
$\mathfrak {c}_{m|n}$
can be naturally generalized from the results for
$\mathfrak {b}_{m+n}$
and
$\mathfrak {c}_{m+n}$
[Reference Kwon22], while the case of
$\mathfrak {d}_{m|n}$
is not. In this case, we need a crystal theoretic super-analogue of the Burge correspondence of type D (Theorem 6.6). It is essentially the main result in this section and its proof will be given in Section 7.
6.1 Crystal
$\mathscr {B}(V_{\mathfrak l}(\lambda ))$
Let us recall the results on the crystal
$\mathscr {B}(V_{\mathfrak l}(\lambda ))$
for
$\lambda \in \mathscr {P}_{m|n}$
[Reference Benkart, Kang and Kashiwara1].
Let
$\mathcal {A}$
be a
$\mathbb {Z}_2$
-graded linearly ordered set. For a skew Young diagram
$\lambda /\mu $
, we denote by
$SST_{\mathcal {A}}(\lambda /\mu )$
the set of all
$\mathcal {A}$
-semistandard tableaux of shape
$\lambda /\mu $
with entries in
$\mathcal {A}$
. We write
$SST_{\mathcal {A}}(\lambda /\mu )$
as
$SST_{m|n}(\lambda /\mu )$
when
$\mathcal {A} = \mathbb {I}$
. For
$T \in SST_{\mathcal {A}}(\lambda /\mu )$
, let
$\mathrm {sh}(T)=\lambda /\mu $
denote the shape of T, and let
$w(T)$
be the word of T, which is the word with letters in
$\mathcal {A}$
obtained from T by reading its entries column by column from the right to the left, and from top to bottom in each column.
We may identify
$\mathbb {I} =SST_{m|n}((1))$
, a crystal of
$V_{\mathfrak l}((1))$
. Then
$SST_{m|n}(\lambda /\mu )$
has a structure of an (abstract)
${\mathcal U}(\mathfrak {l})$
-crystal (or simply
$\mathfrak l$
-crystal) under the identification of
$T\in SST_{m|n}(\lambda /\mu )$
with
$w(T)$
, where we regard
$w(T)$
as an element in
$SST_{m|n}((1))^{\otimes |\lambda /\mu |}$
(
$|\lambda /\mu |$
is the number of boxes in
$\lambda /\mu $
) and apply the tensor product rule. Then we have
$SST_{m|n}(\lambda )\cong \mathscr {B}(V_{\mathfrak l}(\lambda ))$
for
$\lambda \in \mathscr {P}_{m|n}$
.
For a skew Young diagram
$\lambda /\mu $
, let
$(\lambda /\mu )^\pi $
be the skew Young diagram obtained by rotating
$\lambda /\mu $
by
$180^\circ $
. We also write
$\lambda ^\pi /\mu ^\pi =(\lambda /\mu )^\pi $
. For
$T\in SST_{m|n}(\delta ^\pi )$
with
$\delta \in \mathscr {P}_{m|n}$
and
$i\in \mathbb {I}$
, we denote by
$(T\leftarrow i)$
the tableau obtained by applying Schensted’s column insertion of i into T in a reverse way starting from the rightmost column of T so that
$(T\leftarrow i)\in SST_{m|n}(\gamma ^\pi )$
for some
$\gamma \supset \delta $
, which is given by adding a box in a corner of
$\delta $
.
6.2 Crystal
$\mathscr {B}(V(\lambda ,\ell ))$
Let us recall the results on the realization of the crystal
$\mathscr {B}(V(\lambda ,\ell ))$
in [Reference Kwon20, Reference Kwon21]. Here we follow the convention in [Reference Kwon23, Section 3], which is slightly different from that in [Reference Kwon20, Reference Kwon21].
Remark 6.1. The definitions of relevant notions in [Reference Kwon20, Reference Kwon21, Reference Kwon23] are given only in terms of
$\mathbb {N}$
-semistandard tableaux with the usual linear order and
$\mathbb {Z}_2$
-grading
$\mathbb {N}=\mathbb {N}_{\overline 0}$
. But we may still obtain the definition for
$\mathscr {B}(V(\lambda ,\ell ))$
simply by replacing
$\mathbb {N}$
-semistandard tableaux with
$\mathbb {I}$
-semistandard ones. So we only recall necessary notations briefly and refer the reader to [Reference Kwon20, Reference Kwon21, Reference Kwon23] for full details.
First we assume the following notations:
-
• $\mathbf {T}^{\mathfrak {g}}(a)$
for
$a\in \mathbb {Z}_{\ge 0}$
, -
• $\mathbf {T}^{\texttt {sp}}$
for
$\mathfrak {g} = \mathfrak {b}_{m|n}$
and
$\mathfrak {d}_{m|n}$
, -
• $\overline {\mathbf {T}}(0)$
,
$\mathbf {T}^{\texttt {sp}+}$
,
$\mathbf {T}^{\texttt {sp}-}$
for
$\mathfrak {g}=\mathfrak {d}_{m|n}$
.
For precise definitions, we refer the reader to [Reference Kwon20, Definition 6.5] for the types
$\mathfrak {b}_{m|n}$
and
$\mathfrak {c}_{m|n}$
, and to [Reference Kwon21, Definition 3.1] for the type
$\mathfrak {d}_{m|n}$
, where the only difference is that
$\mathbb {N}$
-semistandardness is replaced by
$\mathbb {I}$
-semistandardness. For example,
$\mathbf {T}^{\mathfrak {c}_{m|n}}(a)=\bigsqcup _{c\ge 0}SST_{m|n}\left (\lambda (a,0,c)\right )$
, where
$\lambda (a, b, c) = (2^{b+c}, 1^a) / (1^b)$
, while
$\mathbf {T}^{\mathfrak {b}_{m|n}}(a)$
and
$\mathbf {T}^{\mathfrak {d}_{m|n}}(a)$
consist of
$T\in SST_{m|n}\left (\lambda (a,b,c)\right )$
(
$a,b\in \mathbb {Z}_{\ge 0}$
) satisfying certain mild constraints. When drawing tableaux, we often assume that a horizontal line L is given such that any box in a tableau T is either below or above L, and denote by
$T^{\texttt {body}}$
and
$T^{\texttt {tail}}$
the subtableaux of T placed above and below L, respectively (see Example 6.2).
Let
$\mathbf {B}$
be one of the sets above, and we regard it as an (abstract)
$\mathfrak {g}$
-crystal such that the
$\mathfrak {l}$
-crystal structure is given as in Section 6.1, and
$\tilde {e}_0$
,
$\tilde {f}_0$
are defined as in [Reference Kwon23, Section 3.1], where for
$T \in \mathbf {B}$
,
and
$\mathrm {wt}_{\mathfrak {l}}(T)$
denotes the
$\mathfrak {l}$
-weight of
$T \in \mathbf {B}$
. Indeed, they are crystals of
$V(\lambda ,\ell )$
as follows:
-
• when $\mathfrak {g} = \mathfrak {c}_{m|n}$
,
$ \mathbf {T}^{\mathfrak {c}_{m|n}}(a) \cong \mathscr {B}\left ( V((1^a),1) \right ), $
-
• when $\mathfrak {g} = \mathfrak {b}_{m|n}$
,
$ \mathbf {T}^{\mathfrak {b}_{m|n}}(a) \cong \mathscr {B}\left ( V((1^a),2) \right ) $
and
$ \mathbf {T}^{\texttt {sp}} \cong \mathscr {B}\left ( V((0), 1) \right ), $
-
• when $\mathfrak {g} = \mathfrak {d}_{m|n}$
, $$ \begin{align*} \begin{cases} \mathbf{T}^{\mathfrak{d}_{m|n}}(a) \cong \mathscr{B}\left( V((1^a),2) \right), \\ \overline{\mathbf{T}}(0) \cong \mathscr{B}\left( V((2), 2) \right), \end{cases} \quad \begin{cases} \mathbf{T}^{\texttt{sp}+} \cong \mathscr{B}\left( V((0),1) \right), \\ \mathbf{T}^{\texttt{sp}-} \cong \mathscr{B}\left( V((1),1) \right). \end{cases} \end{align*} $$
Here
$(0)$
denotes the empty partition (that is,
$(0,0,\dots )$
) and
$(1^0)$
is understood as
$(0)$
.
For a partition
$\lambda =(\lambda _i)_{i \ge 1}$
, let
$\lambda '=(\lambda ^{\prime }_i)_{i \ge 1}$
be its conjugate. For
$\mathfrak {g}=\mathfrak {d}_{m|n}$
and given
$(\lambda , \ell ) \in \mathscr {P}(\mathfrak {d}_{m|n})$
, let
$\overline {\lambda } = (\overline {\lambda }_i)_{i \ge 1}$
be given by
$\overline {\lambda }_1 = \ell - \lambda _1$
and
$\overline {\lambda }_i = \lambda _i$
for
$i \ge 2$
. Let
$\mu = \lambda '$
,
$\overline {\mu } = \left ( \overline {\lambda } \right )'$
,
$M_+ = \lambda _1$
,
$M_- = \overline {\lambda }_1$
, and let
$q_{\pm }$
and
$r_{\pm }\in \left \{ 0, 1 \right \}$
be such that
We define
$\widehat {\mathbf {T}}^{\mathfrak {g}}(\lambda , \ell )$
as follows:
-
• when $\mathfrak {g} = \mathfrak {b}_{m|n}$
or
$\mathfrak {c}_{m|n}$
, $$ \begin{align*} \widehat{\mathbf{T}}^{\mathfrak{g}}(\lambda, \ell) = \begin{cases} \mathbf{T}^{\mathfrak{g}}(\lambda^{\prime}_1) \times \dots \times \mathbf{T}^{\mathfrak{g}}(\lambda^{\prime}_\ell) & \text{if }\mathfrak{g} = \mathfrak{c}_{m|n}, \\ \mathbf{T}^{\mathfrak{g}}(\lambda^{\prime}_1) \times \dots \times \mathbf{T}^{\mathfrak{g}}(\lambda^{\prime}_{\ell/2}) & \text{if }\mathfrak{g} = \mathfrak{b}_{m|n}\text{ and }\ell - 2\lambda_1\text{ is even,} \\ \mathbf{T}^{\mathfrak{g}}(\lambda^{\prime}_1) \times \dots \times \mathbf{T}^{\mathfrak{g}}(\lambda^{\prime}_{(\ell-1)/2}) \times \mathbf{T}^{\texttt{sp}} & \text{if }\mathfrak{g} = \mathfrak{b}_{m|n}\text{ and }\ell - 2\lambda_1\text{ is odd,} \\ \end{cases} \end{align*} $$
-
• when $\mathfrak {g} = \mathfrak {d}_{m|n}$
, $$ \begin{align*} \widehat{\mathbf{T}}^{\mathfrak{g}}(\lambda, \ell) = \begin{cases} \mathbf{T}^{\mathfrak{g}}(\mu_1) \times \dots \times \mathbf{T}^{\mathfrak{g}}(\mu_{M_+}) \times \left( \mathbf{T}(0) \right)^{q_+} \times \left( \mathbf{ T}^{\texttt{sp}+} \right)^{r_+} & \text{if }\ell - 2\lambda_1 \ge 0, \\ \mathbf{T}^{\mathfrak{g}}(\overline{\mu}_1) \times \dots \times \mathbf{T}^{\mathfrak{g}}(\overline{\mu}_{M_-}) \times \left( \overline{\mathbf{T}}(0) \right)^{q_-} \times \left( \mathbf{T}^{\texttt{sp}-} \right)^{r_-} & \text{if }\ell - 2\lambda_1 < 0, \end{cases} \end{align*} $$
where one may define a
$\mathfrak {g}$
-crystal structure on
$\widehat {\mathbf {T}}^{\mathfrak {g}}(\lambda , \ell )$
by identifying
$(\dots , T_2, T_1)$
with
$T_1 \otimes T_2 \otimes \dots $
. Let
(see [Reference Kwon23, Definition 3.2] for the definition of
$T_{i+1} \prec T_i$
). Then
$\mathbf {T}^{\mathfrak {g}}(\lambda , \ell )$
is a
$\mathfrak {g}$
-subcrystal of
$\widehat {\mathbf {T}}^{\mathfrak {g}}(\lambda , \ell )$
, and it is shown in [Reference Kwon20, Reference Kwon21] that
Example 6.2. (1) Let
$\mathfrak {g}=\mathfrak {c}_{2|5}$
and take
$\lambda = (3, 2, 1, 1) \in \mathscr {P}_{2|5}$
and
$\ell = 3$
. Since
$\ell - \lambda _1= 0$
, we know that
$(\lambda , \ell ) \in \mathscr {P}(\mathfrak {c}_{2|5})$
. Let
$\mathbf {T}=(T_3,T_2,T_1) \in \mathbf {T}^{\mathfrak {g}}(4) \times \mathbf {T}^{\mathfrak {g}}(2) \times \mathbf {T}^{\mathfrak {g}}(1)$
be given by

where the tableaux
$T_1,\cdots ,T_3$
are placed on the plane with a horizontal line L drawn as the dotted line. In fact, the line L is determined by
$\lambda '$
, in the sense that the length of
$T_i^{\mathrm {tail}}$
is
$\lambda _i'$
. Let us check that
$T_3 \prec T_2$
. We have

where
$T_i^{\mathtt {L}}$
and
$T_i^{\mathtt {R}}$
denote the left and right columns of
$T_i$
, respectively, and
${}^{\mathtt {L}} T_i$
(resp.
${}^{\mathtt {R}} T_i$
) is the left (resp. right) column of the tableau obtained from
$T_i$
by applying
$\lambda _i'$
jeu de taquin slidings into the position below the bottom of
$T_i^{\mathtt {L}}$
(resp.
$T_i^{\mathtt {R}}$
) (see [Reference Kwon23, Section 3.2] for more details). Then
$({}^{\mathtt {R}}T_3,T^{\mathtt {L}}_2)$
(in blue) and
$(T^{\mathtt {R}}_3,{}^{\mathtt {L}}T_2)$
(in red) form semistandard tableaux as shown below:

This precisely corresponds to the definition of
$T_3 \prec T_2$
for type
$\mathfrak {c}_{m|n}$
(and similarly for type
$\mathfrak {b}_{m|n}$
). In a similar way, we can also check that
$T_2 \prec T_1$
, and hence
$\mathbf {T} \in \mathbf {T}^{\mathfrak {g}}(\lambda ,\ell )$
.
(2) Let
$\mathfrak {g} = \mathfrak {d}_{4|4}$
. Take
$\lambda = (4, 4, 2) \in \mathscr {P}_{4|4} $
and
$\ell = 8$
so that
$\ell - 2\lambda _1 = 0$
and
$\mu = \lambda ' = (3, 3, 2, 2)$
. Since
$\ell - \lambda _1 - \lambda _2 = 0$
and
$\ell - 2\lambda _1 = 0$
, we know that
$(\lambda , \ell ) \in \mathscr {P}(\mathfrak {d}_{4|4})$
and
$q_+ = r_+ = 0$
. Let
$\mathbf {T} = (T_4, T_3, T_2, T_1) \in \mathbf {T}^{\mathfrak {g}}(3) \times \mathbf {T}^{\mathfrak {g}}(3) \times \mathbf {T}^{\mathfrak {g}}(2) \times \mathbf {T}^{\mathfrak {g}}(2)$
be given by

One can check that
$T_4 \prec T_3 \prec T_2 \prec T_1$
, and hence
$\mathbf {T} \in \mathbf {T}^{\mathfrak {g}}(\lambda ,\ell )$
. We remark that the definition of admissibility for type
$\mathfrak {d}_{m|n}$
is more involved than for other types. See [Reference Kwon23, Section 3.2] for more details. Note that
$T_4 \ntriangleleft T_3 \triangleleft T_2 \triangleleft T_1$
, where the definition of
$\triangleleft $
is given in [Reference Jang and Kwon9, Definition 3.5].
6.3 Crystal
$\mathscr {B}(P(\lambda ))$
Let us give some comments on the combinatorial realization of
$\mathscr {B}(P(\lambda ))$
. By (3.4), it suffices to describe
$\mathscr {B}(\mathcal {N})$
. We may regard
$\mathscr {B}(\mathcal {N})= M(\mathfrak {u})$
by Theorem 3.3, and also identify it with the set
$\mathbf {M}^{\mathfrak {g}}$
of upper-triangular matrices
$\mathbf {c}=\left (c_{ij}\right )_{i, j \in \mathbb {I}, i \le j}$
such that
where
$c_{ij}$
is equal to the multiplicity
$c_\beta =c_{(i,j)}$
of
$\mathbf {F}_\beta $
with
$\beta $
corresponding to
$(i,j)$
following (2.2). For example, when
$\mathfrak {g} = \mathfrak {b}_{2|3}$
,

and when
$\mathfrak {g} = \mathfrak {d}_{2|4}$
,

Here the notation
means the k-th root
$\beta $
with respect to
$\prec $
and
denotes the type of
$\beta $
, where
(even),
$\otimes $
(isotropic) or
(nonisotropic odd). An explicit description of
$\tilde {e}_i$
and
$\tilde {f}_i$
on
$\mathbf {M}^{\mathfrak {g}}$
can be found in [Reference Jang, Kwon and Uruno13, Section 4.2].
For
$\lambda \in \mathscr {P}_{m|n}$
, we set
which we may identify with
$\mathscr {B}(P(\lambda ))$
.
6.4 RSK of BCD types
Let us give an alternative description of
$\mathbf {M}^{\mathfrak {g}}$
in terms of
$\mathbb {I}$
-semistandard tableaux. Let
where
Then
$\mathbf {V}^{\mathfrak {g}}$
is a
$\mathfrak {g}$
-crystal, where the
$\mathfrak {l}$
-crystal structure is given as in Section 6.1, and the actions of
$\tilde {e}_0$
and
$\tilde {f}_0$
are defined as in [Reference Kwon22, Remark 4.1] for
$\mathfrak {g}=\mathfrak {b}_{m|n}$
,
$\mathfrak {c}_{m|n}$
, and as in [Reference Kwon18, Section 5.2] for
$\mathfrak {g}=\mathfrak {d}_{m|n}$
. For example,
$\tilde {f}_0$
acts on
$T\in \mathbf {V}^{\mathfrak {g}}$
by adding

on T at a certain corner according to a signature rule. In this subsection, we give a combinatorial description of an isomorphism of
$\mathfrak {g}$
-crystals
6.4.1 The case when
$\mathfrak {g}=\mathfrak {b}_{m|n}$
,
$\mathfrak {c}_{m|n}$
In this case, the isomorphism (6.2) is obtained from the RSK correspondence on symmetric matrices as an
$\mathfrak {l}$
-crystal isomorphism [Reference Kwon16] and then extended to an isomorphism of
$\mathfrak {g}$
-crystals. The proof is essentially the same as in [Reference Kwon22, Section 4.3]. So we briefly explain how to construct an isomorphism without proof.
Let
Then it has an
$(\mathfrak {l},\mathfrak {l})$
-bicrystal structure, where we denote by
$\widetilde {\texttt e}_i$
and
$\widetilde {\texttt f}_i$
(resp.
$\widetilde {\texttt e}^*_i$
and
$\widetilde {\texttt f}^*_i$
) the crystal operators for the first (resp. second)
$\mathfrak {l}$
[Reference Kwon16].
Let
$\mathbf {m}\in \mathbf {M}$
be given. Let
$\mathbf {m}^j$
denote the j-th column of
$\mathbf {m}$
and
$m_j=\sum _{i}m_{ij}$
. We identify
$\mathbf {m}^j$
with an element in
$SST_{m|n}((m_j))$
(resp.
$SST_{m|n}((1^{m_j}))$
) when
$\epsilon _j=0$
(resp.
$\epsilon _j=1$
), and regard
$\mathbf {m}=\mathbf {m}^1\otimes \dots \otimes \mathbf {m}^{m+n}$
as an element of an
$\mathfrak {l}$
-crystal with respect to
$\widetilde {\texttt e}_i$
and
$\widetilde {\texttt f}_i$
(
$i\in I$
). Then
$\widetilde {\texttt e}^*_i$
and
$\widetilde {\texttt f}^*_i$
are given by
$\widetilde {\texttt e}^*_i\mathbf {m} =\left (\widetilde {\texttt e}_i \mathbf { m}^t\right )^t$
and
$\widetilde {\texttt f}^*_i\mathbf {m} =\left (\widetilde {\texttt f}_i \mathbf {m}^t\right )^t$
, where
$\mathbf {m}^t$
denotes the transpose of
$\mathbf {m}$
.
Let
$w_1\dots w_{\ell }=w^1 \cdots w^{m+n}$
be the word with letters
$w_k\in \mathbb {I}$
(
$1\le k\le \ell :=\sum _{i,j}m_{ij})$
, where
$w^j=w(\mathbf {m}^j)$
is the word of
$\mathbf {m}^j$
regarding
$\mathbf {m}^j$
as an element in
$SST_{m|n}((m_j))$
or
$SST_{m|n}((1^{m_j}))$
. Let
Then the following RSK correspondence

is an isomorphism of
$(\mathfrak {l},\mathfrak {l})$
-bicrystals [Reference Kwon16, Theorem 3.11], where
$\widetilde {\texttt e}_i$
and
$\widetilde {\texttt f}_i$
(resp.
$\widetilde {\texttt e}^*_i$
and
$\widetilde {\texttt f}^*_i$
) act on
$P(\mathbf {m})$
(resp.
$Q(\mathbf {m})$
) on the right-hand side.
Now, we regard as a set
by identifying
$\mathbf {c}=(c_{ij})$
with
$\mathbf {m}=(m_{ij})$
where
$m_{ii}=2c_{ii}/r_{\mathfrak {g}}$
and
$m_{ij}=c_{ij}$
(
$i\neq j$
). Under (6.4), we can check by the
$\mathfrak {l}$
-crystal structure on
$\mathbf {M}^{\mathfrak {g}}=\mathscr {B}(\mathcal {N})$
[Reference Jang, Kwon and Uruno13, Section 4.2] that
$\tilde {e}_i=\widetilde {\texttt e}_i\widetilde {\texttt e}_i^*$
and
$\tilde {f}_i=\widetilde {\texttt f}_i\widetilde {\texttt f}_i^*\ (i\in I\setminus \{0\})$
.
Then we have the following which proves (6.2).
Theorem 6.3. For
$\mathfrak {g}=\mathfrak {b}_{m|n}, \mathfrak {c}_{m|n}$
, the restriction of (6.3) to
$\mathbf {M}^{\mathfrak {g}}$
is an isomorphism of
$\mathfrak {g}$
-crystals

where
$\mathbf {m}\in \mathbf {M}^{\mathrm {sym}}$
corresponds to
$\mathbf {c}$
under (6.4).
Proof. It follows from almost the same argument as in [Reference Kwon22, Theorem 4.9].
Example 6.4. Let
$\mathfrak {g}=\mathfrak {c}_{2|5}$
. The matrix
$\mathbf {m}\in \mathbf {M}^{\mathrm {sym}}$
and
$\tilde {f}_0(\mathbf {m})$
are given by

Then
$P(\mathbf {m})$
and
$P(\tilde {f}_0\mathbf {m})$
are given by

where we can also see that
$\tilde {f}_0(P(\mathbf {m}))=P(\tilde {f}_0\mathbf {m})$
.
6.4.2 The case when
$\mathfrak {g}=\mathfrak {d}_{m|n}$
In this case, we give a crystal theoretic super-analogue of the bijection of type D [Reference Burge2], which proves (6.2).
We first remark that the set
$\mathbf {M}^{\mathfrak {d}_{m|n}}$
is in bijection with the following set of biwords with letters in
$\mathbb {I}$
;
where
$(i,j) <' (k,l)$
if either one of the following conditions is satisfied: (i)
$i < k$
, (ii)
$i = k \in \mathbb {I}_{\overline {0}}$
and
$j> l$
, (iii)
$i = k \in \mathbb {I}_{\overline {1}}$
and
$j < l$
. Here we identify
$\left ( \mathbf {i}, \mathbf {j} \right )\in {\Omega ^{\mathfrak {d}_{m|n}}}$
with
$\mathbf {m}(\mathbf { i}, \mathbf {j}) = (m_{ij}) \in \mathbf {M}^{\mathfrak {d}_{m|n}}$
, the unique element of
$\mathbf {M}^{\mathfrak {d}_{m|n}}$
such that
$m_{ij} = \left | \left \{ k \, | \, (i_k, j_k) = (i, j) \right \} \right |$
.
For a biletter
$\binom {\,i\,}{\,j\,}$
and
$T \in SST_{m|n}(\lambda ^\pi )$
for
$i,j \in \mathbb {I}$
and
$\lambda \in \mathscr {P}_{\mathfrak {d}_{m|n}}$
, let
be the tableau obtained by adding
at the corner of
$T \leftarrow j$
located above the box in the skew Young diagram
$\mathrm { sh}\left (T \leftarrow j \right ) / \mathrm {sh}\left ( T \right )$
. It is an analogue of the insertion in [Reference Burge2] with respect to
$\mathbb {I}$
-semistandard tableaux.
For
$(\mathbf {i}, \mathbf {j}) \in {\Omega ^{\mathfrak {d}_{m|n}}}$
with
$\mathbf {i} = i_1 \dots i_r$
and
$\mathbf {j} = j_1 \dots j_r$
, we define

Example 6.5. Suppose that
$m=4$
and
$n=4$
and let
$(\mathbf {i}, \mathbf {j}) \in \Omega ^{\mathfrak {d}_{4|4}}$
be given by
Let
$P_k=P(i_k\dots i_6,j_k\dots j_6)$
for
$k=1,\dots ,6$
. Then we have

since
, and by iterating the preceding procedure, we have that
$P_1=P(\mathbf {i}, \mathbf {j})$
is equal to

and
$P(\mathbf {i}, \mathbf {j}) \in SST_{4|4}(\delta ^\pi )$
where
$\delta = (5,5,1,1) \in \mathscr {P}_{\mathfrak {d}_{4|4}}$
.
Now we have the following, which proves (6.2).
Theorem 6.6. The map

where
$\mathbf {m} = \mathbf {m}(\mathbf {i}, \mathbf {j})$
for
$(\mathbf {i}, \mathbf {j}) \in {\Omega ^{\mathfrak {d}_{m|n}}}$
, is an isomorphism of
$\mathfrak {d}_{m|n}$
-crystals.
When
$n=0$
, it is proved in [Reference Jang and Kwon9, Theorem 4.6]. But unlike the case of
$\mathfrak {b}_{m|n}$
and
$\mathfrak {c}_{m|n}$
, the result when
$n=0$
does not admit an immediate generalization to the case of
$n\ge 1$
, whose proof requires additional nontrivial (combinatorial) arguments. The proof is given in Section 7.
Example 6.7. Suppose that
$m=4$
and
$n=4$
and
$(\mathbf {i}, \mathbf {j})$
is as in Example 6.5. We have
$\tilde {f}_0 (\mathbf {m}(\mathbf {i}, \mathbf {j}))= \mathbf {m}(\mathbf {i}', \mathbf {j}')$
, where

and
$P(\mathbf {i}',\mathbf {j}')$
is given by

which is equal to
$\tilde {f}_0(P(\mathbf {i},\mathbf {j}))$
.
6.5 Separation and the embedding of
$\mathscr {B}(V(\lambda ,\ell ))$
into
$\mathscr {B}(P(\lambda ))$
For
$\lambda \in \mathscr {P}_{m|n}$
, set
So it remains to describe the embedding of
$\mathbf {T}^{\mathfrak {g}}(\lambda ,\ell )$
into
$\mathbf {V}^{\mathfrak {g}}_\lambda $
in order to have (5.7).
Let
$(\lambda ,\ell )\in \mathcal {P}(\mathfrak {g})$
be given. For
$\mathbf {T} \in \mathbf {T}^{\mathfrak {g}}(\lambda , \ell )$
, we apply the combinatorial algorithms (called separation) in [Reference Kwon22, Section 5.1] for
$\mathfrak {g}=\mathfrak {b}_{m|n}$
,
$\mathfrak {c}_{m|n}$
, and in [Reference Jang and Kwon11, Sections 4.2 and 4.3] for
$\mathfrak {g}=\mathfrak {d}_{m|n}$
, which indeed do not depend on
$n\ge 0$
(cf. Remark 6.1).
Then we obtain an
$\mathbb {I}$
-semistandard tableau
$\overline {\mathbf {T}}$
of skew shape satisfying the following properties;
-
(S1) $\overline {\mathbf {T}}$
is equivalent to
$\mathbf {T}$
as elements of
$\mathfrak {l}$
-crystals, -
(S2) $\overline {\mathbf {T}}^{\texttt {tail}} \in SST_{m|n}(\lambda )$
and
$\overline {\mathbf {T}}^{\texttt {body}} \in SST_{m|n}(\delta ^{\pi })$
for some
$\delta \in \mathscr {P}_{\mathfrak {g}}$
, -
(S3) $\mathbf {T}$
is equivalent to
$\overline {\mathbf {T}}^{\texttt {body}} \otimes \overline {\mathbf {T}}^{\texttt {tail}}$
as elements of
$\mathfrak {l}$
-crystals,
where
$\overline {\mathbf {T}}^{\texttt {body}}$
and
$\overline {\mathbf {T}}^{\texttt {tail}}$
are subtableaux obtained by dividing
$\overline {\mathbf {T}}$
with respect to a certain horizontal line L (see Example 6.2 and 6.9).
Remark 6.8. Let
$\mathbf {T} = (T_l, \dots , T_1, T_0)\in \mathbf {T}^{\mathfrak {g}}(\lambda , \ell )$
be given. Recall that each
$T_i$
is an
$\mathbb {I}$
-semistandard tableau of skew shape with at most two columns. Roughly speaking,
$\overline {\mathbf {T}}$
is obtained from
$\mathbf {T}$
by applying a sequence of jeu-de-taquin (for
$\mathbb {I}$
-semistandard tableaux) as far as possible so that no subtableau below the horizontal line L is movable to the left.
Example 6.9. (1) Let
$\mathfrak {g}=\mathfrak {c}_{2|5}$
and
$\mathbf {T}=(T_3,T_2,T_1)$
as in Example 6.2 (1). We first regard
$\mathbf {T}$
as the following sequence of single columned tableaux

We then apply jeu-de-taquin as far as possible so that no box below L is movable to the left to get

and hence

Thus we obtain two semistandard tableaux
$\overline {\mathbf {T}}^{\mathrm {body}}$
and
$\overline {\mathbf {T}}^{\mathrm {tail}}$
of shape
$\delta ^{\pi }$
and
$\mu $
, respectively, where
$\delta = (6,4,4,2)$
and
$\lambda = (3,2,1,1)$
.
(2) Let
$\mathfrak {g}=\mathfrak {d}_{4|4}$
and
$\mathbf {T}=(T_4,T_3,T_2,T_1)$
as in Example 6.2 (2). By applying the separation in [Reference Jang and Kwon11, Section 4.2] to
$\mathbf {T}$
, we have

Hence we obtain two semistandard tableaux
$\overline {\mathbf {T}}^{\mathrm {body}}$
and
$\overline {\mathbf {T}}^{\mathrm {tail}}$
of shape
$\delta ^{\pi }$
and
$\mu $
, respectively, where
$\delta = (5,5,1,1)$
and
$\lambda = (4,4,2)$
.
For
$\Lambda \in P$
, let
$T_{\Lambda }=\{\,t_{\Lambda }\,\}$
be a
$\mathfrak {g}$
-crystal with
$\mathrm {wt}(t_\Lambda )=\Lambda $
, and
$\varepsilon _i(t_\Lambda )=\varphi _i(t_\Lambda )=-\infty $
. Then we have the following.
Theorem 6.10. For
$(\lambda , \ell ) \in \mathcal {P}(\mathfrak {g})$
, the map

is an injective morphism of
$\mathfrak {g}$
-crystals.
Proof. It follows from the same arguments as in [Reference Kwon22, Theorem 5.7] for
$\mathfrak {g} = \mathfrak {b}_{m|n}, \mathfrak {c}_{m|n}$
, and [Reference Jang and Kwon11, Theorem 4.12] for
$\mathfrak {g} = \mathfrak {d}_{m|n}$
simply by replacing
$\mathbb {I}_{\overline 0}$
-semistandard tableaux in [Reference Jang and Kwon11, Reference Kwon22] with
$\mathbb {I}$
-semistandard ones.
Summarizing, we have the following realization of (5.7):

where
$\kappa ^{\mathfrak {g}}(\mathbf {m})=\overline {\mathbf {T}}^{\texttt {body}}$
.
Example 6.11. (1) Let
$\mathfrak {g}={\mathfrak c}_{2|5}$
and
$\mathbf {T}$
be as in Example 6.2(1). Then the corresponding
${\mathfrak m} \in \mathbf {M}^{\mathfrak {g}}$
is given in Example 6.4, and
$\overline {\mathbf {T}}^{\texttt {tail}}$
is given in Example 6.9(1).
(2) Let
$\mathfrak {g}={\mathfrak d}_{4|4}$
and
$\mathbf {T}$
be as in Example 6.2(2). Then the corresponding
${\mathfrak m} \in \mathbf {M}^{\mathfrak {g}}$
is associated with
$(\mathbf {i}, \mathbf {j})$
in Example 6.5, and
$\overline {\mathbf {T}}^{\texttt {tail}}$
is given in Example 6.9(2).
Remark 6.12. Let us give a proof of Lemma 3.6. Let
$\xi =\sum _{i\in I}c_i\alpha _i \in Q^-$
be given. Since the action of
$\tilde {f}_0$
is given by adding a vertical domino, we see that the number of boxes in
$\mathbf {T}$
above the horizontal line L is bounded by
$-2c_0$
for any
$(\lambda ,\ell )\in \mathscr {P}(\mathfrak {g})$
and
$\mathbf {T}\in \mathbf {T}^{\mathfrak {g}}(\lambda , \ell )_{\Lambda (\lambda ,\ell )+\xi }$
.
Let
$(\lambda ,\ell )\in \mathscr {P}(\mathfrak {g})$
be given. From the above observation, we may choose a sufficiently large t such that for any
$\mathbf {T}\in \mathbf {T}^{\mathfrak {g}}(\lambda , \ell )_{\Lambda (\lambda ,\ell +t)+\xi }$
, all boxes in
$\mathbf {T}$
above L belong to the columns which are strictly to the right of those the boxes below L belong to. This implies that
$\overline {\mathbf {T}}^{\texttt {body}}$
(resp.
$\overline {\mathbf {T}}^{\texttt {tail}}$
) is simply obtained by sliding the boxes above L (resp. below L) to the right (resp. left) separately. This yields a bijection
and hence a bijection from
$\mathscr {B}(V(\lambda ,\ell ))_{\Lambda (\lambda ,\ell +t)+\xi }\otimes T_{-(\ell +t)\Lambda _0}$
to
$\mathscr {B}(P(\lambda ))_{\Lambda (\lambda ,0)+\xi }$
. This proves the lemma.
7 Proof of Theorem 6.6
Throughout this section, we assume that
$\mathfrak {g} = \mathfrak {d}_{m|n}$
. For simplicity, we omit
$\mathfrak {g}$
as super- or subscripts in certain notations if there is no confusion. For example, we simply write
$\mathbf {M} = \mathbf {M}^{\mathfrak {d}_{m|n}}$
and
$\mathbf {V} = \mathbf {V}^{\mathfrak {d}_{m|n}}$
. But we denote
$\kappa ^{\mathfrak {d}_{m|n}}$
by
$\kappa ^\searrow $
following [Reference Jang and Kwon9] to avoid notational confusion. This section is organized as follows. In Section 7.1, we check that
$\kappa ^\searrow $
is a well-defined weight-preserving bijection between
$\mathbf {M}$
and
$\mathbf {V}$
. In Section 7.2, we investigate a relation between Burge and RSK correspondences, which plays a crucial role in the proof of Theorem 6.6, while we present the proofs of some delicate combinatorial arguments in Section 7.4. Then in Section 7.3, we give the proof of Theorem 6.6.
Convention. Throughout Section 7, we adopt the following conventions.
-
(1) For $T \in SST_{m|n}(\lambda ^\pi )$
, we denote by
$T_i$
(resp.
$T^i$
) the i-th column (resp. row) of T from the right (resp. bottom) so that the length of
$T_i$
(resp.
$T^i$
) is greater than or equal to that of
$T_{i+1}$
(resp.
$T^{i+1}$
). -
(2) For two tableaux $S \in SST_{m|n}(\lambda ^\pi )$
and
$T \in SST_{m|n}(\mu ^\pi )$
, we say that T is a subtableau of S, denoted by
$T \subset S$
, if
$\mu \subset \lambda $
and each letter of T is exactly the same as that of S on
$\mu ^\pi $
. When T is a subtableau of S, we define
$S / T$
to be the skew tableau of shape
$(\lambda / \mu )^\pi $
obtained by removing T inside S. We often write
$S = (S / T) \sqcup T$
.
7.1 Well-definedness of Burge correspondence
For
$T \in SST_{m|n}(\lambda ^\pi )$
and
$a \in \mathbb {I}$
, let
$i_k$
be the letter bumped out from
$T_k$
along the insertion of a into T to have
$\left ( T \leftarrow a \right )$
, where we set
$i_0 = a$
. The sequence of boxes including
$i_k$
’s in T is called the bumping route of a in T, and is denoted by
$R_T(a)$
. We denote by
$B_T(a)$
the box corresponding to
$\mathrm {sh}\left ( T \leftarrow a \right )^\pi / \mathrm {sh}\left ( T \right )^\pi $
. We say that a route R is strictly below (resp. weakly below) a route
$R'$
if in each column which contains a box of
$R'$
, R has a box which is below (resp. below or equal to) the box in
$R'$
. The notions of strictly above (resp. weakly above) are defined analogously. The following lemma is a super analogue of [Reference Fulton7, Appendix A.2], and its proof is straightforward.
Lemma 7.1 (Column Bumping Lemma).
Let
$T \in SST_{m|n}(\lambda ^\pi )$
be given. For
$a, b \in \mathbb {I}$
, put
$U = \left ( T \leftarrow a \right )$
and
$V = \left ( U \leftarrow b \right )$
.
-
(1) If $a, b \in \mathbb {I}_{\overline {0}}$
with
$a \le b$
, then
$R_U(b)$
is weakly below
$R_T(a)$
, and
$B_U(b)$
is southwest of
$B_T(a)$
. -
(2) If $a, b \in \mathbb {I}_{\overline {0}}$
with
$a> b$
, then
$R_U(b)$
is strictly above
$R_T(a)$
, and
$B_U(b)$
is northeast of
$B_T(a)$
. -
(3) If $a \in \mathbb {I}_{\overline {0}}$
and
$b \in \mathbb {I}_{\overline {1}}$
, then
$R_U(b)$
is weakly below
$R_T(a)$
and
$B_U(b)$
is southwest of
$B_T(a)$
. -
(4) If $a \in \mathbb {I}_{\overline {1}}$
and
$b \in \mathbb {I}_{\overline {0}}$
, then
$R_U(b)$
is strictly above
$R_T(a)$
and
$B_U(b)$
is northeast of
$B_T(a)$
. -
(5) If $a, b \in \mathbb {I}_{\overline {1}}$
with
$a < b$
, then
$R_U(b)$
is weakly below
$R_T(a)$
and
$B_U(b)$
is southwest of
$B_T(a)$
. -
(6) If $a, b \in \mathbb {I}_{\overline {1}}$
with
$a \ge b$
, then
$R_U(b)$
is strictly above
$R_T(a)$
and
$B_U(b)$
is northeast of
$B_T(a)$
.
Proposition 7.2 (cf. [Reference Burge2, Reference Jang and Kwon9, Reference Sundaram25]).
Let
$\mathbf {c} \in \mathbf {M}$
be given.
-
(1) $\kappa ^\searrow \left ( \mathbf {c} \right ) \in \mathbf {V}$
. -
(2) $\kappa ^\searrow $
is a weight-preserving bijection between
$\mathbf {M}$
and
$\mathbf {V}$
.
Proof. (1) can be proved directly by induction on the sum of
$c_{(i,j)}$
in
$\mathbf {c}$
, and applying Lemma 7.1 at each step. To prove (2), we describe the reverse map of
$\kappa ^\searrow $
. First, associate
$\mathbf {c}$
with
$(\mathbf {i}, \mathbf {j}) \in {\Omega ^{\mathfrak {d}_{m|n}}}$
for some
$\mathbf {i}=i_1 \dots i_r$
,
$\mathbf {j}=j_1 \dots j_r$
and
$r \in \mathbb {Z}_{\ge 0}$
, and consider the positions of
$i_k$
’s in
$\kappa ^\searrow (\mathbf {c})=P(\mathbf {i},\mathbf {j})$
. Suppose
$i_k = i_{k+1} = i \in \mathbb {I}_{\overline 0}$
. Then, by (6.5), we have
$j_k\geq j_{k+1}$
(
$j_k> j_{k+1}$
if
$j_k \in \mathbb {I}_{\overline 1}$
). Applying Lemma 7.1 (1), (3) and (5), it follows that
$i_k$
is located southwest of
$i_{k+1}$
in
$\kappa ^\searrow ({\mathbf {c}})$
. Hence, the leftmost i of
$\kappa ^\searrow ({\mathbf {c}})$
is the most recent one. Similarly, if we assume
$i\in \mathbb {I}_{\overline 1}$
, then the topmost i is the most recent one by Lemma 7.1 (5).
Now let
$T \in \mathbf {V}$
be given and let i denote the smallest entry in T. If
$i \in \mathbb {I}_{\overline 0}$
(resp.
$i \in \mathbb {I}_{\overline 1}$
), remove the leftmost (resp. topmost) i and then apply the reverse column insertion with respect to the entry immediately below it. In this way, the removed entry is identified with
$i_1$
and the element bumped out is
$j_1$
. Repeating this procedure until the tableau becomes empty yields
$(\mathbf {i}, \mathbf {j}) \in {\Omega ^{\mathfrak {d}_{m|n}}}$
with
$\mathbf {i}=i_1 \dots i_r$
and
$\mathbf {j}=j_1 \dots j_r$
. Therefore,
$\kappa ^\searrow $
is a bijective map.
7.2 Relation between Burge and RSK correspondences
We investigate the relation between Burge and RSK correspondences generalizing the work in the case when
$n = 0$
[Reference Jang and Kwon9].
Let us follow the convention in [Reference Jang, Kwon and Uruno13, Section 3.2]. For
$\mathbf {c} \in \mathbf {M}$
and
$i \in \mathbb {I}$
, we use the following notations:
-
•
: the triangular-shaped arrangement of
$(k,l) \in \Phi (\mathfrak {u})$
for
$i \le k \le l \le m+n$
, -
•
: the complement of
in
, -
• ${\triangle _i}$
: the subset of
given by $$ \begin{align*} \begin{cases} \{ (i, i+1) \} & \text{if }i < m, \\ \{ (i, i+1), (i+1, i+1) \} & \text{if }i = m, \\ \{ (i, i), (i, i+1), (i+1, i+1) \} & \text{if }i> m, \end{cases} \end{align*} $$
-
•
: the complement of
in
, -
• $\mathbf {c}_X$
: the element of
$\mathbf {M}$
whose components on X coincide with those of
$\mathbf {c}$
, and
$0$
elsewhere, -
• $\mathbf {M}_X$
: the subset of
$\mathbf {M}$
consisting of
$\mathbf {c} \in \mathbf {M}$
such that every component of
$\mathbf {c}_{X^c}$
is
$0$
,
where X is a subset of
and
$X^c$
is the complement of X in
.
Example 7.3. Let
$\mathfrak {g}=\mathfrak {d}_{2|4}$
. Then
is decomposed as
as illustrated in the following diagram:

When we consider
$\tilde {e}_i$
or
$\tilde {f}_i$
on
$\mathbf {c}$
for
$i \in I \setminus \{ 0 \}$
, we often decompose
$\mathbf {c} $
into
. Thus we first focus on
in this subsection.
Let
$T \in SST_{m|n}(\mu ^\pi )$
and
$\mathbf {c} \in \mathbf {M}$
with
$\mathbf {c} = \mathbf {c}(\mathbf {a} ,\mathbf {b})$
, where
$\mu \in \mathscr {P}_{m|n}$
and
$(\mathbf {a},\mathbf {b}) = (a_1 \dots a_s\,,\, b_1 \dots b_s) \in {\Omega ^{\mathfrak {d}_{m|n}}}$
. Then we define
-
• ${\texttt P}(T, \mathbf {c}) = \left ( \cdots \left ( T \leftarrow b_s \right ) \leftarrow \cdots \leftarrow b_1 \right )$
, -
• ${\texttt Q}(T, \mathbf {c}) \in SST\left ( \left ( \lambda / \mu \right )^\pi \right )$
, where the subtableau of
${\texttt Q}(T, \mathbf { c})$
corresponding to
$\mathrm {sh}({\texttt P}_k) / \mathrm {sh}({\texttt P}_{k+1})$
is filled with
$a_k$
for
$1 \le k \le s$
,
where
$\lambda = \mathrm {sh}\left ( {\texttt P}(T, \mathbf {c}) \right )^\pi $
and
${\texttt P}_k = \left ( \left ( T \leftarrow b_s \right ) \leftarrow \cdots \leftarrow b_k \right )$
with
${\texttt P}_{s+1} = T$
.
Fix
$i \in I \setminus \{ 0, m \}$
. Let
be given. We define
where
. For a tableau U, let
$U_{\geq k}$
be the subtableau of S consisting of entries greater than or equal to k.
Lemma 7.4. We have
${\texttt P}_i(\mathbf {c})=\kappa ^\searrow (\mathbf {c})_{\geq i+2}$
.
Proof. We proceed by induction on s, the length of the biword corresponding to
$\mathbf {c}_{\lozenge _i}$
. When
$s = 0$
, our assertion clearly holds. Let
$\mathbf {1}_{(a_1, b_1)} \in \mathbf {M}$
be defined by
$c_{(a_1, b_1)} = 1$
and
$c_\gamma = 0$
for all
. By the induction hypothesis, we have
${\texttt P}_i(\mathbf {c}-\mathbf {1}_{(a_1, b_1)}) = \kappa ^\searrow \left ( \mathbf {c} - \mathbf {1}_{(a_1, b_1)} \right )_{\geq i+2}$
, where
$\mathbf {c} - \mathbf {1}_{(a_1, b_1)}$
is understood as the component-wise subtraction. Then we can show that the bumping route of
$({\texttt P}_i\left ( \mathbf {c} - \mathbf {1}_{(a_1, b_1)} \right ) \leftarrow b_1)$
is contained in the bumping route of
$(\kappa ^\searrow \left ( \mathbf {c} - \mathbf {1}_{(a_1, b_1)} \right ) \leftarrow b_1)$
, which yields the desired statement.
The following lemma is one of the key ingredients in the proof of Theorem 6.6.
Lemma 7.5. Let us identify
$\mathbf {c}$
with
. If
and
$\tilde {f}_i \mathbf {c}_{\lozenge _i} \neq \mathbf {0}$
, then we have
-
(1) ${\texttt P}_i( \tilde {f}_i \mathbf {c} ) = {\texttt P}_i( \mathbf {c} )$
, -
(2) ${\texttt Q}_i( \tilde {f}_i \mathbf {c} ) = \tilde {f}_i {\texttt Q}_i( \mathbf {c} )$
.
Proof. Recall that
$i \in I \setminus \{ 0, m \}$
. Note that for
$1 \leq i<m$
, (7.1) always holds, as both
$\mathbf {c}_{\triangle _i}$
and
have a trivial crystal structure with respect to
$\tilde {e}_i$
and
$\tilde {f}_i$
[Reference Jang, Kwon and Uruno13, Section 4.2].
Let
and
$\nu = \mathrm {sh}\left ( T \right )$
. Each element of
$\mathbf {M}_{\lozenge _i}$
can be regarded as an element of
$\mathbf {M}$
in [Reference Kwon16, Section 3.2], and the action of
$\tilde {f}_i$
on
$\mathbf {M}_{\lozenge _i}$
coincides with that on
$\mathbf {M}$
by [Reference Jang, Kwon and Uruno13, Section 4.2]. Under this identification, statements (1) and (2) follow by applying the same argument as in [Reference Kwon17, Proposition 4.6], where [Reference Kwon17, Proposition 3.3] is replaced by [Reference Kwon16, Theorem 3.11] (see also the paragraph of [Reference Kwon16, Theorem 3.11] above). This is because the map
$(\mathbf {c}_{\lozenge _i},\emptyset ,T) \mapsto \left ( {\texttt P}_i(\mathbf {c}), {\texttt Q}_i(\mathbf {c}) \right )$
for
$\mathbf {c}_{\lozenge _i} \in \mathbf {M}_{\lozenge _i}$
is a crystal morphism, which may be regarded as a super-analogue of the inverse of
$\Psi _{\emptyset , \nu; s}$
in [Reference Kwon17, Proposition 4.6] for
${s \ge \ell (\nu )}$
.
We now present another construction of
$\kappa ^\searrow (\mathbf {c})$
in terms of
${\texttt P}_i(\mathbf {c})$
,
${\texttt Q}_i(\mathbf {c})$
and
$\mathbf {c}_{\triangle _i}$
in the following algorithms. Recall that
$\kappa ^\searrow (\mathbf {c})$
and
${\texttt P}_i(\mathbf {c})$
only differ by their subtableaux with entries i and
$i+1$
. We show that
$\kappa ^\searrow (\mathbf {c})$
can be given by gluing
${\texttt P}_i(\mathbf {c})$
and a tableau which is determined by
${\texttt Q}_i(\mathbf {c})$
and
$\mathbf {c}_{\triangle _i}$
.
Algorithm 7.6 for the case of
$1\leq i<m$
is essentially identical to that in [Reference Jang and Kwon9, Section 5.4], whereas Algorithm 7.8 provides its super-analogue. Let
${\texttt P}(\mathbf {c})={\texttt P}_i(\mathbf {c})$
and
${\texttt Q}(\mathbf {c})={\texttt Q}_i(\mathbf {c})$
for simplicity.
Algorithm 7.6. Suppose that
$1 \leq i<m$
. We construct a tableau
${\texttt T}(\mathbf {c})$
as follows:
-
(1) Let $\lambda =\mathrm {sh}({\texttt P}(\mathbf {c}))^\pi $
and
$L=\lfloor \ell (\lambda )/2 \rfloor $
. For
$1 \le l \le L$
, let
${\texttt Q}(\mathbf {c})^l$
be the subtableau consisting of the
$(2l-1)$
-th and
$2l$
-th rows of
${\texttt Q}(\mathbf {c})$
from the bottom row. Apply jeu de taquin to
${\texttt Q}(\mathbf {c})^l$
, to obtain a normal-shaped tableau, which we denote by
$\overline {{\texttt Q}}(\mathbf {c})^l$
. -
(2) Let ${\texttt P}(\mathbf {c})^l$
be the subtableau of
${\texttt P}(\mathbf {c})$
corresponding to the same positions as
${\texttt Q}(\mathbf {c})^l$
. We also denote by
${\texttt P}(\mathbf {c})^{\ge l}$
the subtableau of
${\texttt P}(\mathbf {c})$
lying strictly above the
$(2l-2)$
-th row from the bottom, with the convention
${\texttt P}(\mathbf {c})^{\ge 1} = {\texttt P}(\mathbf {c})$
. Define
${\texttt T}_L(\mathbf {c})$
to be the two-rowed tableau obtained by gluing
$\overline {{\texttt Q}}(\mathbf {c})^{L}$
to
${\texttt P}(\mathbf {c})^{L}$
and
$c_{(i,i+1)}$
copies of
to the left of the resulting tableau. -
(3) Next, we construct a tableau from ${\texttt P}(\mathbf {c})^{\ge L-1}$
by replacing the subtableau
${\texttt P}(\mathbf {c})^{\ge L}$
with
${\texttt T}_L(\mathbf {c})$
, and then gluing
$\overline {{\texttt Q}}(\mathbf {c})^{L-1}$
to the corresponding subtableau
${\texttt P}(\mathbf {c})^{L-1}$
. If the number of columns in
${\texttt T}_L(\mathbf {c})$
exceeds
$\mu _{2L-3}-\mu _{2L-1}$
, where
$\mu =\mathrm {sh}(T)^\pi $
, then move dominos
down to the next two rows as many as it exceeds, and denote the resulting tableau by
${\texttt T}_{L-1}(\mathbf {c})$
. -
(4) Repeat step (3) to obtain ${\texttt T}_{L-2}(\mathbf {c}), \dots , {\texttt T}_1(\mathbf {c})$
, and finally set
${\texttt T}(\mathbf {c})={\texttt T}_1(\mathbf {c})$
.
Example 7.7. Suppose
$\mathfrak {g} = \mathfrak {d}_{2|4}$
and set
$i=1$
. Let us consider
where
Let us compute
${\texttt T}(\mathbf {c})$
following Algorithm 7.6. By (6.6),
is given by

where
$\mu =(3,3,3,3,1,1)$
, and

Then
${\texttt Q}(\mathbf {c})^l$
’s and
$\overline {{\texttt Q}}(\mathbf {c})^l$
’s are given by


Thus
${\texttt T}_4(\mathbf {c})$
,
${\texttt T}_3(\mathbf {c})$
,
${\texttt T}_2(\mathbf {c})$
and
${\texttt T}_1(\mathbf {c})$
are as follows:


where the red
’s in
${\texttt T}_{4}(\mathbf {c})$
are the
$c_{(1,2)}=2$
copies of dominos described in Algorithm 7.6 (2), while the red
’s in
${\texttt T}_{l}(\mathbf {c})$
for
$1\leq l<4$
are those moved down from
${\texttt T}_{l+1}(\mathbf { c})$
as described in Algorithm 7.6 (3). One can check that
$\kappa ^\searrow (\mathbf {c})$
coincides with
${\texttt T}(\mathbf {c})$
by (6.6).
For
$\mathsf {z} \in \mathbb {Z}_+$
, let
$\underline {\mathsf {z}} \in \mathbb {Z}_{\ge 0}$
and
$\overline {\mathsf {z}} \in \{ 0, 1 \} $
be such that
$\mathsf {z} = 2\underline {\mathsf {z}} + \overline {\mathsf {z}}$
.
Algorithm 7.8. Suppose that
$i>m$
. We construct a tableau
${\texttt T}(\mathbf {c})$
as follows:
-
(1) Let ${\boldsymbol {\mathsf {T}}}(\mathbf {c})=(\mathsf {x}_k; \mathsf {y}_k)_{k \ge -1}$
be a two-rowed array as follows: (7.3) $$ \begin{align} {\boldsymbol{\mathsf{T}}}(\mathbf{c}) = \begin{array}{c|ccccccc} & \cdots & k & \cdots & 2 & 1 & 0 & -1 \\ \hline \mathsf{x}_k & \cdots & \mathsf{a}_k & \cdots & \mathsf{a}_2 & \mathsf{a}_1 & c_{(i , i+1)} + 2 c_{(i,i)} & 0 \\ \mathsf{y}_k & \cdots & \mathsf{b}_k & \cdots & \mathsf{b}_2 & \mathsf{b}_1 & 2 c_{(i+1, i+1)} + \underline{c_{(i,i+1)}} & 2 \overline{c_{(i,i+1)}} \end{array}, \end{align} $$where $\mathsf {a}_k$
(resp.
$\mathsf {b}_k$
) is the number of i’s (resp.
$i+1$
’s) in the k-th column of
${\texttt Q}(\mathbf {c})$
from the right.
-
(2) Define a two-rowed array $\mathcal {F}({\boldsymbol {\mathsf {T}}}(\mathbf {c}))= (\widetilde {\mathsf {X}}_k; \widetilde {\mathsf {Y}}_k)_{k \ge -1}$
by the following steps: for each
$k\ge -1$
-
(I) Replace $\mathsf {y}_k$
in
${\boldsymbol {\mathsf {T}}}(\mathbf {c})$
by
$\overline {\mathsf {y}_k} + 2\underline {\mathsf { y}_{k-1}}$
. -
(II) Set $\mathsf {z}_k = \min \left \{\, \mathsf {x}_k,\, \overline {\mathsf {y}_k} + 2\underline {\mathsf {y}_{k-1}} \,\right \}$
. Then replace
$\mathsf {x}_k$
(resp.
$\overline {\mathsf {y}_k} + 2\underline {\mathsf {y}_{k-1}}$
) by
$\mathsf {x}_k - \mathsf {z}_k + \mathsf {z}_{k-1}$
(resp.
$\overline {\mathsf {y}_k} + 2\underline {\mathsf {y}_{k-1}} - \mathsf {z}_k + \mathsf {z}_{k-1}$
), say
$\mathsf {X}_k$
(resp.
$\widetilde {\mathsf {Y}}_k$
). -
(III) Replace $\mathsf {X}_k$
by
$\overline {\mathsf {X}_k} + 2\underline {\mathsf {X}_{k-1}}$
, say
$\widetilde {\mathsf {X}}_k$
.
Here we assume that $\mathsf {x}_{-2}=\mathsf {y}_{-2}=\mathsf {X}_{-2}=0$
as a convention. It is easy to see that
$\widetilde {\mathsf {X}}_k=\widetilde {\mathsf {Y}}_k =0 $
unless
$ 1 \leq k \leq n$
, since
$\mathsf {x}_k=\mathsf {y}_k=0$
for
$k \geq n-2$
. -
-
(3) Consider $\mathcal {F}({\boldsymbol {\mathsf {T}}}(\mathbf {c}))= (\widetilde {\mathsf {X}}_k; \widetilde {\mathsf {Y}}_k)_{k \ge -1}$
. For each
$k\ge 1$
, attach to the top of the k-th column of
${\texttt P}(\mathbf {c})$
a tableau
$C_k \in {SST}_{\{i,i+1\}}((1^{c_k}))$
, where
$c_k=\widetilde {\mathsf {X}}_k+\widetilde {\mathsf {Y}}_k$
and the number of i’s (resp.
$i+1$
’s) in
$C_k$
is
$\widetilde {\mathsf {X}}_k$
(resp.
$\widetilde {\mathsf {Y}}_k$
). We denote the resulting tableau by
${\texttt T}(\mathbf {c})$
.
Example 7.9. Suppose
$\mathfrak {g} = \mathfrak {d}_{2|4}$
and set
$i=3$
. Let us consider
where
By (6.6),
$\kappa ^\searrow (\mathbf {c})$
coincides with (7.2).
Let us compute
${\texttt T}(\mathbf {c})$
following Algorithm 7.8. We have
and

Since
$\mathbf {c}_{{\triangle _i}}=(c_{(3, 3)},c_{(3, 4)}, c_{(4, 4)})=(0,1,1)$
, we have
${\boldsymbol {\mathsf {T}}}(\mathbf {c})$
and
$\mathcal {F}({\boldsymbol {\mathsf {T}}}(\mathbf {c}))$
as shown below.
Then
${\texttt T}(\mathbf {c})$
is obtained by gluing

to
, which coincides with (7.2).
Now we have the following.
Proposition 7.10. For
$i \in I \setminus \{ 0 , m \}$
and
, we have
$\kappa ^\searrow (\mathbf {c}) = {\texttt T}(\mathbf {c})$
.
For the case where
$1 \leq i < m$
, which corresponds to the nonsuper case, the argument is essentially identical to that of [Reference Jang and Kwon9, Lemma 5.14(1)]. The key distinction between the super and nonsuper cases lies in the presence of insertion letters
$j> m$
of odd parity. However, by the definition of
$\mathbf {M}$
(recall (6.5)), each letter
$j> m$
can appear at most once. It then follows from Lemma 7.1 that the inserted boxes for a fixed i still form a horizontal strip as in the nonsuper case. On the other hand, the proof for
$i> m$
is more involved, since the inserted boxes no longer form a horizontal strip. Thus, we will present it in Section 7.4.1.
Remark 7.11. When
$i> m$
, one may check that
$\kappa ^\searrow ( \mathbf {c}_{\triangle _i} )={\texttt T}(\mathbf {c}_{\triangle _i})$
is one of the following forms:

where
$\mathsf {a} = c_{(i,i)}$
,
$\mathsf {b} = c_{(i,i+1)}$
and
$\mathsf {c} = c_{(i+1,i+1)}$
. Recall from [Reference Jang, Kwon and Uruno13, Remark 4.14] that
Then it is straightforward to check that
$\kappa ^\searrow ( \tilde {f}_i \mathbf {c}_{\triangle _i} ) = \tilde {f}_i \kappa ^\searrow ( \mathbf {c}_{\triangle _i} )$
by comparing (7.4) and (7.5).
Lemma 7.12. For
$i \in I \setminus \{ 0, m \}$
and
, we have
${\texttt T}( \tilde {f}_i \mathbf {c} ) = \tilde {f}_i {\texttt T}( \mathbf {c} )$
.
For the case of
$1 \leq i < m$
, by the same reasoning stated below Proposition 7.10, the assertion follows from an argument similar to that in [Reference Jang and Kwon9, Lemma 5.14(2)]Footnote
1
, using Lemma 7.5. For the case where
$i>m$
, the proof is presented in Section 7.4.2.
Corollary 7.13. For
$i \in I \setminus \{ 0 \}$
and
, we have
$\kappa ^\searrow ( \tilde {f}_i \mathbf {c} ) = \tilde {f}_i \kappa ^\searrow ( \mathbf {c} )$
.
Proof. For
$i \ne m$
, it follows immediately from Proposition 7.10 and Lemma 7.12. Now assume
$i = m$
. For
, define
and set
$k_a=\infty $
for
$a=0,1$
if no such k exists. If
$k_0=\infty $
, then
$\tilde {f}_m \mathbf {c}=\mathbf {0}$
and
$\tilde {f}_m \kappa ^\searrow (\mathbf {c})=\mathbf {0}$
since there is no m in
$\kappa ^\searrow (\mathbf {c})$
. So we may assume
$k_0<\infty $
and consider the following two cases.
Case 1. Suppose
$k_1 < k_0$
. By the tensor product rule for
$\tilde {f}_m$
and [Reference Jang, Kwon and Uruno13, Remark 4.14], we have
$\tilde {f}_m \mathbf {c} = \mathbf {0}$
. On the other hand, by Lemma 7.1(5) and (6), all entries in
$\kappa ^\searrow (\mathbf {c})$
equal to m lie southwest of the entry
$m+1$
obtained by the insertion of
$\binom {m+1}{k_1}$
. Hence
$\tilde {f}_m \kappa ^\searrow ( \mathbf {c} ) = \mathbf {0}$
.
Case 2. Suppose
$k_1 \ge k_0$
. By [Reference Jang, Kwon and Uruno13, Remark 4.14],
$\tilde {f}_m(\mathbf {c})$
is obtained from
$\mathbf {c}$
by decreasing
$\mathbf {c}_{(m,k)}$
by
$1$
and increasing
$\mathbf {c}_{(m+1,k)}$
by
$1$
. On the other hand, since
$k_1 \ge k_0$
, Lemma 7.1 (6) implies that the northeasternmost m in
$\kappa ^\searrow (\mathbf {c})$
lies northeast of all
$m+1$
’s. Thus,
$\tilde {f}_m \kappa ^\searrow (\mathbf {c})$
is obtained from
$\kappa ^\searrow (\mathbf {c})$
by changing this m to
$m+1$
. If we apply the reverse algorithm for the inverse of
$\kappa ^\searrow $
to
$\tilde {f}_m \kappa ^\searrow (\mathbf {c})$
(see the proof of Proposition 7.2) the entry
$m+1$
obtained by applying
$\tilde {f}_m$
still bumps out
$k_0$
. This implies
$\tilde {f}_m \kappa ^\searrow (\mathbf {c}) = \kappa ^\searrow (\tilde {f}_m \mathbf {c})$
.
Example 7.14. Let us illustrate Case 1 and Case 2 in the proof of Corollary 7.13. Suppose that
$\mathfrak {g} = \mathfrak {d}_{4|3}$
and
$i=4$
.
Case 1. Let us consider
whose biword is given by

Then
$k_0 = 6$
,
$k_1 = 5$
, and we have

where
(resp.
) is the recording letter from the insertion of
(resp.
). Thus
$\tilde {f}_4 \kappa ^\searrow ( \mathbf {c} ) = \mathbf {0}$
.
Case 2. Let us consider
and
$\tilde {f}_4 \mathbf {c}$
whose biwords are given by

respectively, where
$k_0 = 6$
,
$k_1 = 6$
. We have

from which we see that
$\tilde {f}_4 \kappa ^\searrow (\mathbf {c}) = \kappa ^\searrow (\tilde {f}_4 \mathbf {c})$
.
7.3 Proof of Theorem 6.6
Let
$\mathbf {c} \in \mathbf {M}$
and
$i \in I$
be given. Let
${\mathcal U}(\mathfrak {g}_i)$
denote the subalgebra of
${\mathcal U}(\mathfrak {g})$
generated by
$e_i,f_i$
and
$k_\mu $
for
$\mu \in P$
. By Proposition 7.2, it is enough to show that
$\kappa ^\searrow ( \tilde {f}_i \mathbf {c} ) = \tilde {f}_i \kappa ^\searrow ( \mathbf {c} )$
.
We first suppose that
$i \neq 0$
. The map
is an isomorphism of
${\mathcal U}(\mathfrak {g}_i)$
-crystals with respect to
$\tilde {e}_i$
and
$\tilde {f}_i$
[Reference Jang, Kwon and Uruno13, Section 4.2]. By definition, one may write

where
$(\mathbf {a}, \mathbf {b}) \in {\Omega ^{\mathfrak {d}_{m|n}}}$
such that
. Put
$\mathbf {V}_{\ge i}$
to be the subcrystal of
$\mathbf {V}$
(with respect to
$\tilde {e}_i$
and
$\tilde {f}_i$
) consisting of
$T \in \mathbf {V}$
all of whose entries are greater than or equal to i. The map
$\kappa ^\searrow $
can be factorized as follows:

where
$\kappa ^\searrow _2$
is defined by
. Note that
$\kappa ^\searrow _2$
is a
${\mathcal U}(\mathfrak {g}_i)$
-crystal isomorphism by Corollary 7.13. Since there are no letters i and
$i+1$
in
$\mathbf {a}$
, if we ignore the letters smaller than i, then
is essentially equal to Schensted’s column insertion. Then it follows from [Reference Benkart, Kang and Kashiwara1, Section 4.4] (especially, [Reference Benkart, Kang and Kashiwara1, Theorem 4.1]) that
$\kappa ^\searrow _3$
is an isomorphism of
${\mathcal U}(\mathfrak {g}_i)$
-crystals. Hence we have
$\kappa ^\searrow (\tilde {f}_i \mathbf {c}) = \tilde {f}_i \kappa ^\searrow ( \mathbf {c} )$
for all
$i \neq 0$
.
Next suppose that
$i = 0$
. By the argument in the last paragraph of [Reference Jang and Kwon9, Section 5.4], we have the following commuting diagram:

where we regard
by letting
$c_{(0,k)} = 0$
for all
$k \ge 1$
, and define
$({\texttt P}(\mathbf {c}),{\texttt Q}(\mathbf {c}))=({\texttt P}_0(\mathbf {c}),{\texttt Q}_0(\mathbf {c}))$
in the same way as in
$i\ne 0$
. First, we can show that the map
$(\dagger )$
commutes with
$\tilde {f}_0$
by the same argument as in the proof of [Reference Kwon17, Theorem 3.6], where the set of
$({\texttt P}(\mathbf {c}),{\texttt Q}(\mathbf {c}))$
has a
${\mathcal U}(\mathfrak {g})$
-crystal structure (in particular, with respect to
$\tilde {e}_0$
and
$\tilde {f}_0$
) in the sense of [Reference Kwon17, Section 3.2]. Second, we can easily check that the map
$(\ddagger )$
commutes with
$\tilde {f}_0$
by comparing the actions of
$\tilde {f}_0$
on both sides. Finally, we know
$\kappa ^\searrow (\mathbf {c}) = {\texttt T}(\mathbf {c})$
by Proposition 7.10. This completes the proof of Theorem 6.6.
Remark 7.15. It is known that
$\mathcal {N}$
is decomposed into irreducible polynomial
${\mathcal U}(\mathfrak {l})$
-modules:
Since the crystals of the above modules (as
$\mathfrak {l}$
-crystals) are realized as
$\mathbf {M}$
and
$\mathbf {V}$
, respectively, the map
$\kappa ^\searrow $
can be viewed as a crystal-theoretic interpretation of the above decomposition.
7.4 Proof of Proposition 7.10 and Lemma 7.12
The purpose of this section is to complete the proofs of Proposition 7.10 and Lemma 7.12 for the case of
$i> m$
. To prove these, it suffices to consider only the numbers of i and
$i+1$
in each column of the tableau
${\texttt Q}(\mathbf {c})$
or
$\kappa ^\searrow (\mathbf {c})$
, which we represent as a two-rowed array with nonnegative integers as
${\boldsymbol {\mathsf {T}}}(\mathbf {c})$
in (7.3). We assume
$i>m$
throughout Section 7.4.
Here we introduce some conventions. We assume that all two-rowed arrays considered here have only finitely many nonzero entries. For a given two-rowed array
we often write
${\boldsymbol {\mathsf {X}}} = (\mathsf {x}_k\,; \mathsf {y}_k)_{k \geq -1}= (\mathsf {x}_k\,; \mathsf {y}_k)$
. For
${\boldsymbol {\mathsf {X}}} = (\mathsf {q}_k\,; \mathsf {r}_k)$
and
${\boldsymbol {\mathsf {Y}}} = (\mathsf {s}_k\,; \mathsf {t}_k)$
, we define
${\boldsymbol {\mathsf {X}}} \pm {\boldsymbol {\mathsf {Y}}} = (\mathsf {q}_k \pm \mathsf {s}_k\,; \mathsf {r}_k \pm \mathsf {t}_k)$
. For
$a \in \mathbb {Z}_{\geq 0}$
and
$ k \geq -1 $
, we denote
${\boldsymbol {\mathsf {x}}}_k(a)$
and
${\boldsymbol {\mathsf {y}}}_k(a)$
by
Recall that for
$\mathsf {z} \in \mathbb {Z}_+$
, we define
$\underline {\mathsf {z}} \in \mathbb {Z}_+$
and
$\overline {\mathsf {z}} \in \{ 0, 1 \} $
by
$\mathsf {z} = 2\underline {\mathsf {z}} + \overline {\mathsf {z}}$
.
7.4.1 Proof of Proposition 7.10
For
, we prove in this subsection that
$\kappa ^\searrow (\mathbf {c})={\texttt T}(\mathbf {c})$
. Recall from Lemma 7.4 that
${\texttt P}(\mathbf {c})$
is precisely the subtableau of
$\kappa ^\searrow (\mathbf {c})$
consisting of entries greater than or equal to
$i+2$
. Hence, by Algorithm 7.8, it remains to show that the number of i’s (resp.
$i+1$
’s) in each column of the tableau
$\kappa ^\searrow (\mathbf {c})$
coincides with
$\widetilde {\mathsf {X}}_k$
(resp.
$\widetilde {\mathsf {Y}}_k$
) in
$\mathcal {F}({\boldsymbol {\mathsf {T}}}(\mathbf {c}))= (\widetilde {\mathsf {X}}_k; \widetilde {\mathsf {Y}}_k)_{k \in \mathbb {Z}}$
described in Algorithm 7.8 (2).
For convenience, we set
for all
$ j> m+n$
. The case of
$i=m+n-1$
is trivial, and the case of
$ m+n-2$
is already treated in (7.4), since
. Let
$({\mathbf {a}}, {\mathbf {b}}) = (a_1 \dots a_s, b_1 \dots b_s) \in {\Omega ^{\mathfrak {d}_{m|n}}}$
be such that
. By (6.6), we have

For
$1 \leq t \leq s$
, we define
$(\mathbf {a}_t , \mathbf {b}_t)= (a_t \dots a_s , b_t \dots b_s)$
and set
. By convention, we let
. We proceed by induction on t in decreasing order to prove that
$\kappa ^\searrow (\mathbf {c}_t)={\texttt T}(\mathbf {c}_t)$
.
Note that
$\kappa ^\searrow (\mathbf {c}_t)= \left ( \kappa ^\searrow (\mathbf {c}_{t+1}) \,{\overset {\,\,\mathrm {B}}{\longleftarrow }}\, \binom {\,a_t\,}{\,b_t\,} \right )$
. Suppose that
$\kappa ^\searrow (\mathbf {c}_{t+1}) = {\texttt T}(\mathbf {c}_{t+1})$
. For simplicity, we write
${\texttt P}_t = {\texttt P}(\mathbf {c}_t)$
and
${\texttt Q}_t = {\texttt Q}(\mathbf {c}_t)$
, and let
To show
$\kappa ^\searrow (\mathbf {c}_{t}) = {\texttt T}(\mathbf {c}_{t})$
, we analyze
${\texttt T}(\mathbf {c}_{t})$
by considering the difference
$\mathcal {F}({\boldsymbol {\mathsf {T}}}_t) - \mathcal {F}({\boldsymbol {\mathsf {T}}}_{t+1})$
, since this difference captures the change in the positions of i and
$i+1$
between
${\texttt T}(\mathbf {c}_t)$
and
${\texttt T}(\mathbf {c}_{t+1})$
.
If
$(a_t,b_t)$
lies on
${\lozenge _i}$
, then let k be the column index of the unique box in
${\texttt Q}_t \setminus {\texttt Q}_{t+1}$
, which is filled with
$a_t$
. Then it follows from the definition of
${\boldsymbol {\mathsf {T}}}_t$
in (7.3) that
Now we consider five cases according to the position of
$(a_t, b_t)$
in
as follows.
Case 1. Suppose that
$a_t=i+1$
and
$b_t>i+1$
. Then all the boxes of
${\texttt Q}_{t+1}$
are filled with
$i+1$
, and, by Lemma 7.1 (6), their column indices are greater than or equal to k. Thus, we have
and
${\boldsymbol {\mathsf {T}}}_t = {\boldsymbol {\mathsf {T}}}_{t+1} + {\boldsymbol {\mathsf {y}}}_k(1)$
. We have
$\widetilde {\mathsf {Y}}_{t+1,k} = \overline {\mathsf {y}_{t+1,k}}$
by Algorithm 7.8 (2), which denotes the number of
$i+1$
’s in the k-th column of
$\kappa ^\searrow (\mathbf {c}_{t+1})={\texttt T}(\mathbf {c}_{t+1}) $
. Consider the following two subcases, illustrating how the bumping route behaves differently.
Case 1-1. If
$\widetilde {\mathsf {Y}}_{t+1,k}=0$
, then the bumping route of
$b_t$
in
$(\kappa ^\searrow (\mathbf {c}_{t+1}) \leftarrow b_t)$
coincides with that in
$({\texttt P}_{t+1} \leftarrow b_t)$
. Indeed, the only difference between
$\kappa ^\searrow (\mathbf {c}_{t+1})$
and
${\texttt P}_{t+1}$
lies in their subtableaux of entries
$i+1$
, all of which are located in the columns strictly to the left of the k-th column of
$\kappa ^\searrow (\mathbf {c}_{t+1})$
. Hence
$\kappa ^\searrow (\mathbf {c}_t)= \left ( \kappa ^\searrow (\mathbf {c}_{t+1}) \,{\overset {\,\,\mathrm {B}}{\longleftarrow }}\, \binom {\,a_t\,}{\,b_t\,} \right )$
is obtained from
$\kappa ^\searrow (\mathbf {c}_{t+1})$
by adding
$b_t$
and then
$i+1$
directly above it at the top of the k-th column. On the other hand, we have
$\mathcal {F}({\boldsymbol {\mathsf {T}}}_t) - \mathcal {F}({\boldsymbol {\mathsf {T}}}_{t+1}) = {\boldsymbol {\mathsf {y}}}_k(1) $
since
$\widetilde {\mathsf {Y}}_{t+1,k}=0$
. This shows that a new letter
$i+1$
appears in the k-th column of
${\texttt T}(\mathbf {c}_t)$
, and thus
$\kappa ^\searrow (\mathbf {c}_{t}) = {\texttt T}(\mathbf {c}_{t})$
.
Case 1-2. If
$\widetilde {\mathsf {Y}}_{t+1,k}=1$
, then there is exactly one
$i+1$
in the k-th column of
$\kappa ^\searrow (\mathbf {c}_{t+1})$
. Hence the bumping route of
$b_t$
in
$\left ( \kappa ^\searrow ( \mathbf {c}_{t+1} ) \leftarrow b_t \right )$
does not terminate in the k-th column. Instead, it bumps out the
$i+1$
from the k-th column into the
$(k+1)$
-th column, thereby placing a new
$i+1$
at the top of the
$(k+1)$
-th column. Consequently, one
$i+1$
is removed from the k-th column while two
$i+1$
’s are added to the top of the
$(k+1)$
-th column of
$\kappa ^\searrow (\mathbf {c}_{t+1})$
. On the other hand, since
$\mathsf {y}_{t+1,k}=1 $
, it immediately implies
$\mathcal {F}({\boldsymbol {\mathsf {T}}}_t)-\mathcal {F}({\boldsymbol {\mathsf {T}}}_{t+1})={\boldsymbol {\mathsf {y}}}_{k+1}(2)+{\boldsymbol {\mathsf {y}}}_k(-1)$
. This shows that
$\kappa ^\searrow (\mathbf {c}_{t}) = {\texttt T}(\mathbf {c}_{t})$
.
Case 2. Suppose that
$a_t=i+1$
and
$b_t=i+1$
. In this case, we have
and hence
${\boldsymbol {\mathsf {T}}}_t = {\boldsymbol {\mathsf {T}}}_{t+1} + {\boldsymbol {\mathsf {y}}}_0(2)$
. Since
$\mathsf {y}_{t+1,0}$
is even, it follows that
$\widetilde {\mathsf {Y}}_{t+1,0}=0$
. Consequently,
$\mathcal {F}({\boldsymbol {\mathsf {T}}}_{t+1})-\mathcal {F}({\boldsymbol {\mathsf {T}}}_{t})={\boldsymbol {\mathsf {y}}}_1(2)$
. On the other hand, we have
$\kappa ^\searrow (\mathbf {c}_t)=\left ( \kappa ^\searrow (\mathbf {c}_{t+1}) \,{\overset {\,\,\mathrm {B}}{\longleftarrow }}\, \binom {\,i+1\,}{\,i+1\,} \right )$
. Since all letters in
$\kappa ^\searrow (\mathbf {c}_{t+1})$
are greater than or equal to
$i+1$
, the bumping route of
$i+1$
in
$\left ( \kappa ^\searrow (\mathbf { c}_{t+1}) \leftarrow i+1 \right )$
terminates in the first column. Therefore
$\kappa ^\searrow (\mathbf {c}_t)$
is obtained from
$\kappa ^\searrow (\mathbf {c}_{t+1})$
by adding two entries
$i+1$
at the top of the first column of
$\kappa ^\searrow (\mathbf {c}_{t+1})$
. This shows that
$\kappa ^\searrow (\mathbf {c}_{t}) = {\texttt T}(\mathbf {c}_{t})$
.
Case 3. Suppose that
$a_t=i$
and
$b_t>i+1$
. In this case, we have
and hence
${\boldsymbol {\mathsf {T}}}_t = {\boldsymbol {\mathsf {T}}}_{t+1} + {\boldsymbol {\mathsf {x}}}_k(1)$
. Consider the following four subcases, illustrating how the bumping route behaves differently.
Case 3-1. If
$\widetilde {\mathsf {Y}}_{t+1,k}=\widetilde {\mathsf {X}}_{t+1,k}=0$
, then
$\mathcal {F}({\boldsymbol {\mathsf {T}}}_{t})-\mathcal {F}({\boldsymbol {\mathsf {T}}}_{t+1})={\boldsymbol {\mathsf {x}}}_k(1)$
. Since the bumping route of
$b_t$
in
$\left ( \kappa ^\searrow (\mathbf {c}_{t+1}) \leftarrow b_t \right )$
terminates in the k-th column,
$\kappa ^\searrow (\mathbf {c}_t)$
is obtained from
$\kappa ^\searrow (\mathbf {c}_{t+1})$
by adding one i to the k-th column.
Case 3-2. If
$\widetilde {\mathsf {Y}}_{t+1,k}=0$
and
$\widetilde {\mathsf {X}}_{t+1,k}=1$
, then
$\mathcal {F}({\boldsymbol {\mathsf {T}}}_{t})-\mathcal {F}({\boldsymbol {\mathsf {T}}}_{t+1})={\boldsymbol {\mathsf {x}}}_{k+1}(2)+{\boldsymbol {\mathsf {x}}}_k(-1)$
. In this case, the bumping route of
$b_t$
in
$\left ( \kappa ^\searrow (\mathbf {c}_{t+1}) \leftarrow b_t \right )$
terminates in the
$(k+1)$
-th column, since there is one i on top of the k-th column of
$\kappa ^\searrow (\mathbf {c}_{t+1})$
. As a result, this i is moved from the k-th column to the
$(k+1)$
-th column, and another i is added to the
$(k+1)$
-th column.
Case 3-3. If
$\widetilde {\mathsf {Y}}_{t+1,k}\geq 1$
and
$\widetilde {\mathsf {X}}_{t+1,k+1}=0$
, then
$\mathcal {F}({\boldsymbol {\mathsf {T}}}_{t})-\mathcal {F}({\boldsymbol {\mathsf {T}}}_{t+1})={\boldsymbol {\mathsf {x}}}_{k+1}(1)+{\boldsymbol {\mathsf {y}}}_{k+1}(1)+{\boldsymbol {\mathsf {y}}}_k(-1)$
. Here the bumping route of
$b_t$
in
$\left (\kappa ^\searrow (\mathbf {c}_{t+1}) \leftarrow b_t \right )$
terminates in the
$(k+1)$
-th column, since there is at least one
$i+1$
on top of the k-th column of
$\kappa ^\searrow (\mathbf {c}_{t+1})$
. Consequently one
$i+1$
is moved from the k-th column to the
$(k+1)$
-th column, and one i is added to the
$(k+1)$
-th column.
Case 3-4. If
$\widetilde {\mathsf {Y}}_{t+1,k}\geq 1$
and
$\widetilde {\mathsf {X}}_{t+1,k+1} \geq 1$
, then
$\mathcal {F}({\boldsymbol {\mathsf {T}}}_{t})-\mathcal {F}({\boldsymbol {\mathsf {T}}}_{t+1})={\boldsymbol {\mathsf {x}}}_{k+2}(2)+{\boldsymbol {\mathsf {x}}}_{k+1}(-1)+{\boldsymbol {\mathsf {y}}}_{k+1}(1)+{\boldsymbol {\mathsf {y}}}_k(-1)$
. In this case, the bumping route of
$b_t$
in
$\left (\kappa ^\searrow (\mathbf {c}_{t+1}) \leftarrow b_t \right )$
terminates in the
$(k+2)$
-th column, since one
$i+1$
in the k-th column of
$\kappa ^\searrow (\mathbf {c}_{t+1})$
moves to the
$(k+1)$
-th column, and this
$i+1$
bumps out one i from the top of the
$(k+1)$
-th column to the
$(k+2)$
-th column. As a result, one
$i+1$
is moved from the k-th column to the
$(k+1)$
-th column, one i is moved from the
$(k+1)$
-th column to the
$(k+2)$
-th column, and another i is added to the
$(k+2)$
-th column. From Case 3-1–Case 3-4, we conclude that
$\kappa ^\searrow (\mathbf {c}_t) = {\texttt T}(\mathbf {c}_t)$
.
Case 4. Suppose that
$a_t=i$
and
$b_t=i+1$
. In this case, we have
where
$\mathsf {y}_{t+1,-1} = 2 \underline {\mathsf {x}_{t+1,0}}$
. We consider two subcases.
Case 4-1. If
$\widetilde {\mathsf {X}}_{t+1,1} =0 $
, then
$\mathcal {F}({\boldsymbol {\mathsf {T}}}_{t})-\mathcal {F}({\boldsymbol {\mathsf {T}}}_{t+1})={\boldsymbol {\mathsf {x}}}_{1}(1)+{\boldsymbol {\mathsf {y}}}_{1}(1)$
. In this case, the bumping route of
$i+1$
in
$\left (\kappa ^\searrow (\mathbf {c}_{t+1}) \leftarrow i+1 \right )$
terminates in the first column. Thus,
$\kappa ^\searrow (\mathbf {c}_t)$
is obtained from
$\kappa ^\searrow (\mathbf {c}_{t+1})$
by adding two
$i+1$
’s on top of the first column.
Case 4-2. If
$\widetilde {\mathsf {X}}_{t+1,1} =1 $
, then
$\mathcal {F}({\boldsymbol {\mathsf {T}}}_{t})-\mathcal {F}({\boldsymbol {\mathsf {T}}}_{t+1})={\boldsymbol {\mathsf {x}}}_{2}(2)+{\boldsymbol {\mathsf {x}}}_{1}(-1)+{\boldsymbol {\mathsf {y}}}_{2}(1)$
, which implies
$\kappa ^\searrow (\mathbf {c}_{t}) = {\texttt T}(\mathbf {c}_{t})$
. In this case, the bumping route of
$i+1$
in
$\left ( \kappa ^\searrow (\mathbf {c}_{t+1}) \leftarrow {i+1} \right ) $
terminates in the second column, since there is one i on top of the first column. Thus,
$\kappa ^\searrow (\mathbf {c}_t)$
is obtained from
$\kappa ^\searrow (\mathbf {c}_{t+1})$
by adding one
$i+1$
to the first column, bumping out i from the first column to the second column, and then adding another i to the second column. From Case 4-1 and Case 4-2, we conclude that
$\kappa ^\searrow (\mathbf {c}_t) = {\texttt T}(\mathbf {c}_t)$
.
Case 5. Suppose that
$a_t=i$
and
$b_t=i$
. Then
$\mathcal {F}({\boldsymbol {\mathsf {T}}}_{t})-\mathcal {F}({\boldsymbol {\mathsf {T}}}_{t+1})={\boldsymbol {\mathsf {x}}}_{1}(2)$
. As in Case 2,
$\kappa ^\searrow (\mathbf {c}_t)$
is obtained from
$\kappa ^\searrow (\mathbf {c}_{t+1})$
by adding two i’s on top of the first column. Hence
$\kappa ^\searrow (\mathbf {c}_{t}) = {\texttt T}(\mathbf {c}_{t})$
.
By Case 1–Case 5, we have
$\kappa ^\searrow (\mathbf {c}) = {\texttt T}(\mathbf {c})$
by induction on s. This completes the proof.
Example 7.16. We continue Example 7.9. Suppose
$\mathfrak {g} = \mathfrak {d}_{2|4}$
and set
$i=3$
. Let
be given by
and
. Consider
$\kappa ^\searrow (\mathbf {c}_2)= \left ( \kappa ^\searrow (\mathbf {c}_{3}) \,{\overset {\,\,\mathrm {B}}{\longleftarrow }}\, \binom {\,3\,}{\,5\,} \right )$
where
$\mathbf {c}_3= \mathbf {c}(\, 33445 \,, \, 66465 \,) $
and
$\mathbf {c}_2= \mathbf {c}( \, 333445 \,, \,566465 \,) $
. We have

On the other hand,
${\boldsymbol {\mathsf {T}}}_t$
and
$\mathcal {F}({\boldsymbol {\mathsf {T}}}_t)$
for
$t=2,3$
are given by
This corresponds to Case 3-4 in the proof above. We have
$\mathcal {F}({\boldsymbol {\mathsf {T}}}_{2})-\mathcal {F}({\boldsymbol {\mathsf {T}}}_{3})={\boldsymbol {\mathsf {x}}}_{3}(2)+{\boldsymbol {\mathsf {x}}}_{2}(-1)+{\boldsymbol {\mathsf {y}}}_{2}(1)+{\boldsymbol {\mathsf {y}}}_1(-1)$
, which shows the difference of the column indices of
$3$
’s and
$4$
’s between
$\kappa ^\searrow (\mathbf {c}_3)$
and
$\kappa ^\searrow (\mathbf {c}_2)$
.
7.4.2 Proof of Lemma 7.12
In this subsection, every two-rowed array
${\boldsymbol {\mathsf {X}}}=(\mathsf {x}_k \,; \mathsf {y}_k)_{k \in \mathbb {Z}}$
is assumed to have nonnegative integer entries and finite support, that is,
$\mathsf {x}_k=\mathsf {y}_k=0$
for all but finitely many
$k \in \mathbb {Z}$
. The two-rowed array
${\boldsymbol {\mathsf {T}}}(\mathbf {c}) $
defined in (7.3) also satisfies this condition once we extend it by setting
$\mathsf {x}_k=\mathsf {y}_k=0$
for
$k \leq -2$
.
Let us define the actions of
$\tilde {e}_i$
and
$\tilde {f}_i$
on
${\boldsymbol {\mathsf {X}}}$
, motivated by their actions on a tableau. We define a sequence of
$+$
and
$-$
as follows:
where the superscript means the multiplicity of each
$\pm $
, which has finitely many
$\pm $
’s since
${{\boldsymbol {\mathsf {X}}}}$
has finite support. We write
$\pm ^0 = \cdot $
by convention. We call
$\sigma _i({\boldsymbol {\mathsf {X}}})$
the i-signature (or simply signature) of
${\boldsymbol {\mathsf {X}}}$
. Write
$\sigma _i({\boldsymbol {\mathsf {X}}}) = (\xi _1 \, \xi _2 \, \dots )$
, where
$\xi _s \in \left \{ -,\,+ \right \}$
. We replace a pair
$(\xi _r, \xi _t) = (+,-)$
by
$(\,\cdot \,, \,\cdot \,)$
, where
$r < t$
and
$\xi _s = \cdot $
for
$ r < s < t$
, and then repeat this process until we get a sequence
$\overline {\sigma }_i({\boldsymbol {\mathsf {X}}})$
with no
$-$
placed to the right of
$+$
. We call
$\overline {\sigma }_i({\boldsymbol {\mathsf {X}}})$
the reduced i-signature of
${\boldsymbol {\mathsf {X}}}$
. Assume that the rightmost
$-$
(resp. leftmost
$+$
) in
$\overline {\sigma }_i({\boldsymbol {\mathsf {X}}})$
comes from
$-^{\mathsf {y}_k}$
(resp.
$+^{\mathsf { x}_k}$
), then
$\tilde {e}_i {\boldsymbol {\mathsf {X}}} = {\boldsymbol {\mathsf {X}}} + {\boldsymbol {\mathsf {x}}}_k(1) - {\boldsymbol {\mathsf {y}}}_k(1)$
(resp.
$\tilde {f}_i {\boldsymbol {\mathsf {X}}} = {\boldsymbol {\mathsf {X}}} - {\boldsymbol {\mathsf {x}}}_k(1) + {\boldsymbol {\mathsf {y}}}_k(1)$
). In this way, we regard any two-rowed array of the form (7.8) with nonnegative integers as an element of a
${\mathcal U}(\mathfrak {g}_i)$
-crystal.
To prove Lemma 7.12, it suffices to show that
for all
.
Lemma 7.17. For
, we have
Proof. If
$\mathbf {c} \in \mathbf {M}_{\triangle _i}$
, then (7.10) follows by direct calculation (see Remark 7.11). If
, then (7.10) holds by Lemma 7.5(2). When
$\tilde {f}_i$
acts on
, we may regard
$\mathbf {c}$
as
, viewed as an element of a
${\mathcal U}(\mathfrak {g}_i)$
-crystal. Recalling (7.3), we obtain
as an element of a
${\mathcal U}(\mathfrak {g}_i)$
-crystal, where
${\boldsymbol {\mathsf {X}}}_{\geq 1}$
(resp.
${\boldsymbol {\mathsf {X}}}_{\leq 0}$
) denotes the subarray of
${\boldsymbol {\mathsf {X}}}$
consisting of the columns with positive (resp. nonpositive) indices. This completes the proof.
The next lemma follows easily from Algorithm 7.8 (2).
Lemma 7.18. For
$k \in \mathbb {Z} $
and
$a \in \mathbb {Z}_{\geq 0}$
, let
We have
-
(1) $\mathcal {F}({\boldsymbol {\mathsf {X}}}_k(2a)) = {\boldsymbol {\mathsf {X}}}_{k+2}(2a)$
, -
(2) $\mathcal {F}({\boldsymbol {\mathsf {X}}} + {\boldsymbol {\mathsf {X}}}_k(2a)) = \mathcal {F}({\boldsymbol {\mathsf {X}}}) + {\boldsymbol {\mathsf {X}}}_{k+2}(2a)$
, -
(3) $\tilde {f}_i \left ( {\boldsymbol {\mathsf {X}}} + {\boldsymbol {\mathsf {X}}}_k(2a) \right ) = \tilde {f}_i \left ( {\boldsymbol {\mathsf {X}}} \right ) + {\boldsymbol {\mathsf {X}}}_k(2a)$
,
for any two-rowed array
${\boldsymbol {\mathsf {X}}}$
with nonnegative integers.
For
${\boldsymbol {\mathsf {X}}} = (\mathsf {x}_k\,; \mathsf {y}_k)_{k \in \mathbb {Z}}$
with nonnegative integer entries, we say that
${\boldsymbol {\mathsf {X}}}$
is reduced if
$ \min \left \{\, \mathsf {x}_{k+1},\, \overline {\mathsf {y}_k} \,\right \} \leq 1$
for all
$k\in \mathbb {Z}$
. Then every
${\boldsymbol {\mathsf {X}}}$
can be written as
for some
$a_k \in \mathbb {Z}_{\geq 0} $
and a unique reduced two-rowed array
${\boldsymbol {\mathsf {X}}}'$
, where the summation is finite since
${\boldsymbol {\mathsf {X}}}$
has finite support. Hence, by Lemma 7.18, to prove (7.9), it suffices to consider reduced two-rowed arrays
${\boldsymbol {\mathsf {X}}}$
only.
Remark 7.19. Let
${\boldsymbol {\mathsf {X}}} = (\mathsf {x}_k\,; \mathsf {y}_k)_{k \in \mathbb {Z}}$
be a reduced two-rowed array. Then one can check that either
$0 \leq \mathsf { x}_{k+1} \leq 1 $
or
$0 \leq \mathsf {y}_k \leq 1 $
for all
$k \in \mathbb {Z}$
. Also we have
$\mathsf {z}_k=\min \left \{\, \mathsf {x}_k,\, \overline {\mathsf {y}_k} + 2\underline {\mathsf {y}_{k-1}} \,\right \} = 0$
or
$1$
(recall Algorithm 7.8 (2)-(II)); If
$0 \leq \mathsf {x}_k \leq 1 $
, then clearly
$\mathsf {z}_k \leq \mathsf {x}_k \leq 1 $
. If
$ 0\leq \mathsf {y}_{k-1} \leq 1 $
, then
$2\underline {\mathsf {y}_{k-1}}=0$
and
$\mathsf {z}_k \leq \overline {\mathsf {y}_k} + 2\underline {\mathsf {y}_{k-1}} \leq 1$
. These facts will be frequently used in the subsequent proofs.
The following lemma enables us to describe
$\tilde {f}_i \mathcal {F}({\boldsymbol {\mathsf {X}}})$
from
$\tilde {f}_i {\boldsymbol {\mathsf {X}}}$
for a reduced
${\boldsymbol {\mathsf {X}}}$
.
Lemma 7.20. Let
${\boldsymbol {\mathsf {X}}} = (\mathsf {x}_k\,;\mathsf {y}_k)_{k \in \mathbb {Z}}$
be a reduced two-rowed array.
-
(1) If $\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
has an adjacent pair
$+-$
coming from
$+^{{\boldsymbol {\mathsf {x}}}_k} -^{{\boldsymbol {\mathsf {y}}}_{k-1}}$
, then
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
has either the same adjacent pair or
$++--$
, where the inner pair
$+-$
comes from
$+^{{\boldsymbol {\mathsf {x}}}_{k+1}} -^{{\boldsymbol {\mathsf {y}}}_{k}}$
, and the outer pair
$+-$
comes from
$ +^{{\boldsymbol {\mathsf {x}}}_k} -^{{\boldsymbol {\mathsf {y}}}_{k-1}}$
, both as seen in
$\sigma _i({\boldsymbol {\mathsf {X}}})$
. -
(2) The $+$
’s in
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
coming from
$+^{{\boldsymbol {\mathsf {x}}}_k}$
in
$\sigma _i({\boldsymbol {\mathsf {X}}})$
cannot appear to the left of any
$-$
’s in
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
coming from
$-^{{\boldsymbol {\mathsf {y}}}_l}$
in
$\sigma _i({\boldsymbol {\mathsf {X}}})$
for all
$l \ge k+1$
. If such
$+$
’s appear to the left of the
$-$
’s coming from
$-^{{\boldsymbol {\mathsf {y}}}_k}$
, then this
$-$
must be unique and canceled by the adjacent
$+$
arising from
$+^{\mathsf {x}_{k+1}}$
. -
(3) The $-$
’s in
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
coming from
$-^{{\boldsymbol {\mathsf {y}}}_{k-1}}$
in
$\sigma _i({\boldsymbol {\mathsf {X}}})$
cannot appear to the left of any
$+$
’s in
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
coming from
$+^{{\boldsymbol {\mathsf {x}}}_l}$
in
$\sigma _i({\boldsymbol {\mathsf {X}}})$
for all
$l \ge k+1$
. If such
$-$
’s appear to the left of the
$+$
’s coming from
$+^{{\boldsymbol {\mathsf {x}}}_k}$
, then this
$-$
must be unique and canceled by the adjacent
$+$
coming from
$+^{\mathsf {x}_{k}}$
. -
(4) We have
$$ \begin{align*}\overline{\sigma}_i( \mathcal{F}( {\boldsymbol{\mathsf{X}}} ) ) = \overline{\sigma}_i( {\boldsymbol{\mathsf{X}}} ).\end{align*} $$Furthermore, the leftmost $+$
in
$\overline {\sigma }_i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
comes from the leftmost
$+$
in
$\overline {\sigma }_i({\boldsymbol {\mathsf {X}}})$
.
Example 7.21. (1) Let
${\boldsymbol {\mathsf {X}}}=(\mathsf {x}_k\,;\mathsf {y}_k)_{k \in \mathbb {Z}}$
be given by
Then the i-signature of
${\boldsymbol {\mathsf {X}}}$
is
$\sigma _i({\boldsymbol {\mathsf {X}}})=(-^2 +^3 -^1 +^2)$
, and the reduced signature is
$\overline {\sigma }_i({\boldsymbol {\mathsf {X}}})=(-^2 +^2 (+ -) +^2)=(-^2 +^4)$
, where the canceled
$(+-)$
comes from
$+^{\mathsf {x}_1=3} $
and
$-^{\mathsf {y}_0=1}$
. Next, we consider
$\mathcal {F}({\boldsymbol {\mathsf {X}}})$
. Applying (I), we obtain
Applying (II) then gives
Finally, applying (III), we find that
$\mathcal {F}({\boldsymbol {\mathsf {X}}})=(\widetilde {\mathsf {X}}_k\,;\widetilde {\mathsf {Y}}_k)_{k \in \mathbb {Z}}$
is given by
We then compute
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))=(-^2 +^4 -^1 +^1)$
and
$\overline {\sigma }_i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))=(-^2 +^3 (+ -) +^1)=(-^2 +^4) = \overline {\sigma }_i({\boldsymbol {\mathsf {X}}})$
, showing that Lemma 7.20 (4) holds for this choice of
${\boldsymbol {\mathsf {X}}}$
.
We observe that
$+^{\widetilde {\mathsf {X}}_2=4}$
in
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
is made of
$+^3$
from
$+^{\mathsf {x}_1=3}$
and
$+^1$
in
$\sigma _i({\boldsymbol {\mathsf {X}}})$
from
$+^{\mathsf {x}_0=2}$
. Here, the canceled
$+$
in
$\overline {\sigma }_i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
, which is one of the
$+$
’s in
$+^{\widetilde {\mathsf {X}}_2=4}$
, is interpreted as the one coming from
$+^{\mathsf {x}_1}$
in
$\sigma _i({\boldsymbol {\mathsf {X}}})$
. This indicates that the canceled pair
$(+-)$
in
$ \overline {\sigma }_i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
originates from the same canceled pair in
$\overline {\sigma }_i({\boldsymbol {\mathsf {X}}})$
, thereby verifying that Lemma 7.20 (1) holds for this example.
On the other hand, there exists a
$+$
in
${\sigma }_i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
coming from
$+^{\mathsf {x}_1}$
in
${\sigma }_i({\boldsymbol {\mathsf {X}}})$
which appears to the left of a
$-$
in
${\sigma }_i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
coming from
$-^{\mathsf {y}_1}$
in
$\sigma _i({\boldsymbol {\mathsf {X}}})$
. However, this
$+$
can be regarded as canceled with
$-$
in
${\sigma }_i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
coming from
$-^{\mathsf {y}_0}$
in
$\sigma _i({\boldsymbol {\mathsf {X}}})$
, thereby satisfying Lemma 7.20 (2). Accordingly, this
$+$
is canceled in
$\overline {\sigma }_i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
.
Finally, one can similarly verify that Lemma 7.20 (3) also holds for this example.
(2) Let
${\boldsymbol {\mathsf {X}}}=(\mathsf {x}_k\,;\mathsf {y}_k)_{k \in \mathbb {Z}}$
be given by
Then
$\mathcal {F}({\boldsymbol {\mathsf {X}}})=(\widetilde {\mathsf {X}}_k\,;\widetilde {\mathsf {Y}}_k)_{k \in \mathbb {Z}}$
is given by
We then compute
$\sigma _i({\boldsymbol {\mathsf {X}}})=(+-+-)$
,
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))=(++--)$
, and
$\overline {\sigma _i}({\boldsymbol {\mathsf {X}}})=\overline {\sigma _i}(\mathcal {F}({\boldsymbol {\mathsf {X}}}))=(\cdot )$
. Here, we interpret the second
$+$
in
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
as coming from
$+^{\mathsf {x}_1}$
in
$\sigma _i({\boldsymbol {\mathsf {X}}})$
. Then the inner pair
$+-$
in
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))=(++--)$
comes from
$+^{\mathsf {x}_1}$
and
$-^{\mathsf {y}_0}$
, while the outer pair
$+-$
comes from
$+^{\mathsf {x}_0}$
and
$-^{\mathsf {y}_{-1}}$
. This provides an example of the second case described in (1) of Lemma 7.20.
Proof. We will prove (1), (2) and (4) since the proof of (3) is essentially identical to that of (2).
(1) Assume that there exists an adjacent pair
$+-$
coming from
$+^{{\boldsymbol {\mathsf {x}}}_k}$
and
$-^{{\boldsymbol {\mathsf {y}}}_{k-1}}$
in
$\sigma _i({\boldsymbol {\mathsf {X}}})$
. Since
${\boldsymbol {\mathsf {X}}}$
is reduced, we have
$\mathrm {min}(\mathsf {x}_k, \mathsf {y}_{k-1})=1$
. Hence, there are two possible cases: either
$\mathsf {x}_k \geq \mathsf {y}_{k-1} = 1$
or
$\mathsf {x}_k = 1 < \mathsf {y}_{k-1}$
.
Case 1. Suppose
$\mathsf {x}_k \geq \mathsf {y}_{k-1} = 1$
. Note that under
$\mathcal {F}$
, each
$+$
or
$-$
in
$\sigma _i({\boldsymbol {\mathsf {X}}})$
may be shifted by at most two positions. In particular, the
$+$
(resp.
$-$
) coming from
$+^{\mathsf {x}_k}$
(resp.
$-^{\mathsf {y}_{k-1}}$
) may be moved to
$+^{\widetilde {\mathsf {X}}_k}$
,
$+^{\widetilde {\mathsf {X}}_{k+1}}$
, or
$+^{\widetilde {\mathsf {X}}_{k+2}}$
(resp.
$-^{\widetilde {\mathsf {Y}}_{k-1}}$
,
$-^{\widetilde {\mathsf {Y}}_{k}}$
, or
$-^{\widetilde {\mathsf {Y}}_{k+1}}$
) in
$ \sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
. We first focus on the
$-=-^{\mathsf {y}_{k-1}}$
in
$ \sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
. By Algorithm 7.8 (2), this
$-$
can be moved by at most one position, since
$\mathsf {y}_{k-1} = 1$
.
Let us consider two subcases:
Case 1-1. Suppose that
$-=-^{\mathsf {y}_{k-1}}$
is moved to
$-^{\widetilde {\mathsf {Y}}_{k}}$
in
$ \sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
by
$\mathcal {F}$
. This implies
$\mathsf {z}_{k-1}=1=\mathrm {min}\{\mathsf {x}_{k-1},1+2\underline {\mathsf {y}_{k-2}}\}$
, and hence
$\mathsf {x}_{k-1}\geq 1 $
. We now examine the positions of the
$+$
’s in
$ \sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
that come from
$+^{\mathsf {x}_{k}}$
. If
$\mathsf {X}_k=\mathsf {x}_k-\mathsf {z}_k+1 \geq 2$
, then at least one
$+$
from
$+^{\mathsf {x}_k}$
is moved to
$+^{\widetilde {\mathsf {X}_{k+1}}}$
, so (1) holds. Otherwise, suppose
$\mathsf {x}_k-\mathsf {z}_k=0$
. Then
$\mathsf {x}_k=\mathsf {z}_k=1$
and
$\overline {\mathsf {y}_k}=\mathsf {z}_k=1$
, since
$\underline {\mathsf {y}_{k-1}}=0$
. If
$\overline {\mathsf {X}_{k+1}}=\overline {\mathsf {x}_{k+1}-\mathsf {z}_{k+1}+1}=1$
, then one
$+$
from
$+^{\mathsf {x}_k}$
remains at
$+^{\widetilde {\mathsf { X}_{k+1}}}$
, and again (1) holds. Otherwise, suppose
$\overline {\mathsf {x}_{k+1}-\mathsf {z}_{k+1}+1}=0$
. Then
$\mathsf {x}_{k+1}-\mathsf {z}_{k+1}\geq 1$
and
$\mathsf {y}_{k+1}-\mathsf {z}_{k+1}=0$
. Thus, in
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
, we obtain:
-
• $+^{\widetilde {\mathsf {X}_{k+2}}}$
contains at least two
$+$
’s, each coming from
$+^{\mathsf {x}_{k}}$
and
$+^{\mathsf {x}_{k+1}}$
, -
• $\widetilde {\mathsf {Y}_{k+1}}=1$
, originating from
$\mathsf {y}_k=1$
, -
• $\widetilde {\mathsf {X}_{k+1}}=0$
, -
• $\widetilde {\mathsf {Y}_{k}}=1$
, originating from
$\mathsf {y}_{k-1}=1$
.
This is precisely the second case of (1), namely (
$++--$
), verifying (1) in this subcase.
Case 1-2. Suppose instead that
$-=-^{\mathsf {y}_{k-1}}$
is not moved under
$\mathcal {F}$
. This implies
$\mathsf {z}_{k-1}=0=\mathrm {min}(\mathsf {x}_{k-1},1+2\underline {\mathsf {y}_{k-2}})$
, hence
$\mathsf {x}_{k-1}=0 $
. If
$\overline {\mathsf {x}_k-\mathsf {z}_k}=1$
, then one
$+$
from
$+^{\mathsf {x}_k}$
remains at
$+^{\widetilde {\mathsf {X}_{k}}}$
, verifying (1) in this subcase. Otherwise, suppose
$\overline {\mathsf {x}_k-\mathsf {z}_k}=0$
. Then
$\mathsf {x}_k=\mathsf {z}_k=1$
and
$\overline {\mathsf {y}_k}-\mathsf {z}_k=0$
. In
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
we then have:
-
• $+^{\widetilde {\mathsf {X}_{k+1}}}$
contains at least one
$+$
coming from
$+^{\mathsf {x}_{k}}$
, -
• $\widetilde {\mathsf {Y}_{k}}=0$
, since
$\overline {\mathsf {y}_k}-\mathsf {z}_k=0$
, -
• $\widetilde {\mathsf {X}_{k}}=0$
, since
$\overline {\mathsf {x}_k-\mathsf {z}_k}=0$
and
$\mathsf {X}_{k-1}=\mathsf {z}_{k-2} \leq 1$
, -
• $-^{\widetilde {\mathsf {Y}_{k-1}}}$
contains at least one
$-$
coming from
$-^{\mathsf {y}_{k-1}}$
.
Thus,
$+=+^{\mathsf {x}_{k}}$
and
$-=-^{\mathsf {y}_{k-1}}$
remain adjacent in
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
, verifying (1).
Case 2. Suppose
$\mathsf {x}_k = 1 < \mathsf {y}_{k-1}$
. We focus on the
$+$
coming from
$+^{\mathsf {x}_{k}}$
in
$ \sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
. Since
$\underline {\mathsf {y}_{k-1}}\geq 1$
, we have
$\mathsf {z}_k=\mathrm {\mathsf {x}_k,\overline {\mathsf {y}_k}+2\underline {\mathsf {y}_{k-1}}}=1$
. Therefore, this
$+$
is moved to either
$+^{\widetilde {\mathsf {X}}_{k+2}}$
or
$+^{\widetilde {\mathsf {X}}_{k+1}}$
under
$\mathcal {F}$
.
Suppose first that such a
$+$
in
$+^{\mathsf {x}_{k}}$
is moved to
$+^{\widetilde {\mathsf {X}}_{k+2}}$
. This means
$\mathsf {z}_k=1$
, so the
$+$
is moved to
$+^{\widetilde {\mathsf {X}}_{k+1}}$
by (II), and since
$\overline {\mathsf {x}_{k+1}-\mathsf {z}_{k+1}+1}=0$
, it is further moved to
$+^{\widetilde {\mathsf {X}}_{k+2}}$
by (III). From
$\overline {\mathsf {x}_{k+1}-\mathsf {z}_{k+1}+1}=0$
we deduce
$\mathsf {x}_{k+1}-\mathsf {z}_{k+1} \geq 1$
, which implies
${\overline {\mathsf { y}_{k+1}}}+2\underline {\mathsf {y}_{k}}-\mathsf {z}_{k+1}=0$
, hence
$\underline {\mathsf {y}_{k}}=0$
. Thus, in
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
we have the following:
-
• $+^{\widetilde {\mathsf {X}_{k+2}}}$
contains at least two
$+$
’s, each coming from
$+^{\mathsf {x}_{k}}$
and
$+^{\mathsf {x}_{k+1}}$
, -
• $\widetilde {\mathsf {Y}_{k+1}}=1=\mathsf {z}_k$
, originating from either
$\mathsf {y}_k$
or
$\mathsf {y}_{k-1}$
, -
• $\widetilde {\mathsf {X}_{k+1}}=0$
, since
$\overline {\mathsf {x}_{k+1}-\mathsf {z}_{k+1}+1}=0$
and
$\mathsf {X}_k=\mathsf {z}_{k-1}\leq 1$
, -
• $\widetilde {\mathsf {Y}_{k}}=\overline {\mathsf {y}_k}+2\underline {\mathsf {y}_{k-1}}-1+\mathsf {z}_{k-1}$
.
If
$\widetilde {\mathsf {Y}_{k+1}}=1$
originates from
$\mathsf {y}_{k-1}$
, then (1) holds. Otherwise, suppose it originates from
$\mathsf {y}_{k}$
, so that
$\overline {\mathsf {y}_k}=1$
. Then
$-^{\widetilde {\mathsf {Y}_{k+1}}}$
coming from
$-^{\mathsf {y}_k}$
is canceled with one
$+$
in
$+^{\widetilde {\mathsf {X}_{k+2}}}$
coming from
$+^{\mathsf { x}_{k+1}}$
. Since
$ 2\underline {\mathsf {y}_{k-1}} \geq 2$
, we have
$\widetilde {\mathsf {Y}_{k}}\geq 1$
, so
$-^{\widetilde {\mathsf {Y}_{k}}}$
contains at least one
$-$
from
$-^{\mathsf {y}_{k-1}}$
. Hence (1) holds.
Next suppose that such a
$+$
in
$+^{\mathsf {x}_{k}}$
is moved to
$+^{\widetilde {\mathsf {X}}_{k+1}}$
under
$\mathcal {F}$
. Then for this
$+$
to be moved by (II), we must have
$\mathsf {z}_k=1$
and
$\overline {\mathsf {x}_{k+1}-\mathsf {z}_{k+1}}=0$
. Since
$\widetilde {\mathsf {Y}_k}=\overline {\mathsf {y}_k} + 2 \underline {\mathsf {y}_{k-1}} - 1 + \mathsf {z}_{k-1} $
and
$ 2\underline {\mathsf {y}_{k-1}} \geq 2$
, it follows that
$-^{\widetilde {\mathsf {Y}_k}} $
contains at least one
$-$
from
$-^{\mathsf {y}_{k-1}}$
. Hence (1) holds.
(2) Let us refer to Algorithm 7.8 (2)-(I),(II),(III) simply as (I),(II),(III). Let us focus on the location of
$+^{\mathsf {x}_k}$
under (I),(II) and (III). According to Algorithm 7.8 (2),
$+^{\mathsf {x}_k}$
can be moved at most two positions to the left, so it cannot be positioned to the left of
$-^{\mathsf {y}_l}$
for any
$l \geq k+2$
. Thus it suffices to compare its position with
$-^{\mathsf {y}_{k+1}}$
and
$-^{\mathsf {y}_k}$
. If all
$+$
’s of
$+^{\mathsf {x}_k}$
remain to the left of
$+^{{\widetilde {\mathsf {X}}_{k}}}$
in
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
, then there is nothing to prove. Otherwise, we consider the following two cases.
Case 1. There is a
$+$
of
$+^{\mathsf {x}_k}$
in
$\sigma _i({\boldsymbol {\mathsf {X}}})$
that is moved to
$+^{{\widetilde {\mathsf {X}}_{k+2}}}$
in
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
.
First, at least one
$+$
of
$+^{\mathsf {x}_k}$
is moved by (II), so that
$\mathsf {z}_k=1$
by Remark 7.19. This
$+$
is then moved again by (III), which implies
$\mathsf {X}_{k+1} = \mathsf {x}_{k+1} - \mathsf {z}_{k+1} + \mathsf {z}_k \ge 2$
and
$\overline {\mathsf {X}_k} =0$
. Since
$\mathsf {x}_{k+1} - \mathsf {z}_{k+1} \ge 1 $
, we have
$\overline {\mathsf {y}_{k+1}}+2\underline {\mathsf {y}_k} -\mathsf {z}_{k+1}=0 $
by the definition of
$\mathsf {z}_{k+1}$
.
From this, every
$-^{\mathsf {y}_{k+1}}$
in
$\sigma _i({\boldsymbol {\mathsf {X}}})$
is moved to either
$-^{{\widetilde {\mathsf {Y}}_{k+2}}}$
or
$-^{{\widetilde {\mathsf {Y}}_{k+3}}}$
in
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
. Hence, no
$+$
coming from
$+^{\mathsf {x}_k}$
appears to the left of any
$-$
coming from
$-^{\mathsf {y}_{k+1}}$
in
$\sigma _i({\boldsymbol {\mathsf {X}}})$
.
Next, consider the position of
$-$
’s coming from
$-^{{\boldsymbol {\mathsf {y}}}_{k}}$
. We have
$2 \underline {\mathsf {y}_k}=0$
since
$\overline {\mathsf {y}_{k+1}}+2\underline {\mathsf {y}_k} -\mathsf {z}_{k+1}=0 $
and
$\mathsf {z}_{k+1} \leq 1$
. Assume
$\mathsf {y}_k=1$
, since if
$\mathsf {y}_k=0$
there is nothing to prove. This
$-=-^{\mathsf {y}_k}$
is moved to
$-^{{\widetilde {\mathsf {Y}}_{k+1}}}$
by (II) since
$\mathsf {z}_k=1$
, so there exists a
$+$
coming from
$+^{\mathsf {x}_{k}}$
that appears to the left of this
$-=-^{\mathsf {y}_{k}}$
in
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
.
We then show that this
$-$
is canceled by
$+$
coming from
$+^{\mathsf {x}_{k+1}}$
. Since
$\mathsf {x}_{k+1} - \mathsf {z}_{k+1} + \mathsf {z}_k \ge 2$
we have
$\mathsf {x}_{k+1}\geq 1$
, so there exists an adjacent pair
$+-$
coming from
$+^{\mathsf {x}_k} -^{\mathsf {y}_{k-1}}$
in
$\sigma _i({\boldsymbol {\mathsf {X}}})$
. The
$+$
from
$+^{\mathsf {x}_{k+1}}$
ends at
$+^{\widetilde {\mathsf {X}}_{k+2}}$
in
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
, since
$\mathsf {X}_{k+1} = \mathsf {x}_{k+1} - \mathsf {z}_{k+1} + \mathsf {z}_k \ge 2$
. Therefore the
$-$
coming from
$ -^{\mathsf {y}_k}$
is canceled by the adjacent
$+$
coming from
$ +^{\mathsf {x}_{k+1}}$
.
Case 2. There is a
$+$
from
$+^{\mathsf {x}_k}$
in
$\sigma _i({\boldsymbol {\mathsf {X}}})$
that is moved to
$+^{{\widetilde {\mathsf {X}}_{k+1}}}$
in
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
, and no
$+$
is moved to
$+^{{\widetilde {\mathsf {X}}_{k+1}}}$
in
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
. To derive a contradiction, assume that a
$-$
coming from
$-^{\mathsf {y}_k}$
remains at
$-^{\widetilde {\mathsf {Y}_{k}}}$
in
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
. Then
$\overline {\mathsf {y}_k}=1$
and
$\mathsf {z}_k=0$
. By the definition of
$\mathsf {z}_k$
, we have
$\mathsf {x}_k=0$
, so there is nothing to prove.
(4) Let
$\sigma _i'$
be the sequence obtained from
$\sigma _i({\boldsymbol {\mathsf {X}}})$
by canceling only the adjacent pairs
$+-$
coming from
$+^{\mathsf {x}_k} -^{\mathsf {y}_{k-1}}$
, for all
$k \in \mathbb {Z}$
. Next, let
$\sigma _i"$
be the sequence obtained from
$\sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
by canceling only the pairs
$+-$
induced by the adjacent pairs
$+-$
in
$\sigma _i({\boldsymbol {\mathsf {X}}})$
, which is possible by (1). By (2) and (3), we have
$\sigma _i'=\sigma _i"$
, since any uncanceled
$+$
’s in
$\sigma _i"$
come from
$+^{\mathsf {x}_k}$
located to the left of
$+^{\mathsf {x}_{k-1}}$
and
$-^{\mathsf {y}_{k-1}}$
, and to the right of
$+^{\mathsf {x}_{k+1}}$
and
$-^{\mathsf {y}_{k}}$
. Therefore, the reduced sequences
$\overline {\sigma }_i({\boldsymbol {\mathsf {X}}})$
and
$\overline {\sigma }_i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
coincide. This proves (4).
Now we are ready to prove (7.9).
For
, we have
$\mathcal {F}\left ( \tilde {f}_i \left ( {\boldsymbol {\mathsf {T}}}(\mathbf {c}) \right ) \right ) = \tilde {f}_i\, \mathcal {F}\left ( {\boldsymbol {\mathsf {T}}}(\mathbf {c}) \right )$
. By Lemma 7.17 and 7.18, it suffices to verify that
$\mathcal {F}(\tilde {f}_i({\boldsymbol {\mathsf {X}}}))=\tilde {f}_i \mathcal {F}({\boldsymbol {\mathsf {X}}})$
for any reduced two-rowed array
${\boldsymbol {\mathsf {X}}}$
. Assume that
$\tilde {f}_i {\boldsymbol {\mathsf {X}}} = {\boldsymbol {\mathsf {X}}} - {\boldsymbol {\mathsf {x}}}_k(1) + {\boldsymbol {\mathsf {y}}}_k(1)$
. In other words, the left-most uncanceled
$+$
in
$\overline {\sigma _i}({\boldsymbol {\mathsf {X}}})$
comes from
$+^{\mathsf {x}_k}$
in
$\sigma _i({\boldsymbol {\mathsf {X}}})$
. Then
$\mathsf {x}_{k}> \mathsf {y}_{k-1}$
, since at least one
$+$
coming from
$+^{\mathsf {x}_k}$
is not canceled, so
$\mathsf {y}_{k-1} \leq 1$
. Similarly,
$\mathsf {x}_{k+1} \leq \mathsf {y}_{k}$
, since the left-most uncanceled
$+$
comes from
$+^{\mathsf {x}_k}$
, hence
$\mathsf {x}_{k+1} \leq 1$
.
Now, depending on the number of positions an uncanceled
$+$
coming from
$+^{\mathsf {x}_k}$
is moved under
$\mathcal {F}$
, we consider the following three cases.
Case 1. Suppose that there is a
$+$
of
$+^{\mathsf {x}_{k}}$
moved to
$+^{\widetilde {\mathsf {X}}_{k+2}}$
in
$ \sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
by
$\mathcal {F}$
. Then
$\mathsf {z}_k = 1$
and
$\overline {\mathsf {x}_{k+1}-\mathsf {z}_{k+1}+1}=0$
. Since
$\underline {\mathsf {y}_{k-1}}=0$
and
$\mathsf {z}_k=1$
, we have
$\overline {\mathsf {y}_k}=1$
. Moreover, as
$\mathsf {x}_{k+1}-\mathsf {z}_{k+1}\geq 1$
and
$\mathsf {x}_{k+1} \leq 1$
, we deduce
$\mathsf {x}_{k+1}=1$
,
$\mathsf {z}_{k+1}=0$
, and
$\overline {\mathsf { y}_{k+1}}=\underline {\mathsf {y}_k}=0$
. Hence,
-
• $\widetilde {\mathsf {X}_{k+2}}=\overline {(\mathsf {x}_{k+2} - \mathsf {z}_{k+2})}+2$
, where
$+2$
is the sum of
$\mathsf {x}_{k+1}=1$
and
$1$
from
$\mathsf { x}_{k}$
, -
• $\widetilde {\mathsf {Y}_{k+1}}=1$
, originating from
$\mathsf {y}_k=1$
, -
• $\widetilde {\mathsf {X}_{k+1}}=2\underline {(\mathsf {x}_k-1+\mathsf {z}_{k-1})}$
, -
• $\widetilde {\mathsf {Y}_{k}}=\mathsf {z}_{k-1}$
.
We claim that the
$+$
in
$+^{\widetilde {\mathsf {X}_{k+2}}}$
coming from
$+^{\mathsf {x}_k}$
is not canceled. Indeed,
$\overline {(\mathsf {x}_{k+2} - \mathsf {z}_{k+2})}=0$
by Lemma 7.20(4). The
$+$
in
$+^{\widetilde {\mathsf {X}_{k+2}}}$
coming from
$+^{\mathsf {x}_{k+1}}$
is canceled with
$-$
in
$-^{\widetilde {\mathsf {Y}_{k+1}}}$
. Consider two subcases: If
$\mathsf {y}_{k-1}=0$
, the claim follows by Lemma 7.20(4). If
$\mathsf {y}_{k-1}=1$
, then
$\mathsf {z}_{k-1}=1$
and
$\mathsf {x}_k-1+\mathsf {z}_{k-1} \geq 2$
, so
$-^{\widetilde {\mathsf {Y}_{k}}}$
is canceled by
$+^{\widetilde {\mathsf {X}_{k+1}}}$
, and the claim holds by Lemma 7.20(4).
Thus the claim is verified.
Summarizing, in Case 1, we have
and
with
$\widetilde {\mathsf {X}_k}=\overline {\mathsf {x}_k-\mathsf {z}_k+\mathsf {z}_{k-1}} + 2\underline {(\mathsf {x}_{k-1}-\mathsf {z}_{k-1}+\mathsf {z}_{k-2})}$
and
$\widetilde {\mathsf {Y}_{k}}=\overline {\mathsf {y}_{k}}+2\underline {\mathsf {y}_{k-1}}-\mathsf {z}_{k}+\mathsf {z}_{k-1}$
. Then, by the claim,
On the other hand, by assumption
Hence we conclude
$\tilde {f}_i \mathcal {F}({\boldsymbol {\mathsf {X}}}) = \mathcal {F}(\tilde {f}_i {\boldsymbol {\mathsf {X}}})$
.
Case 2. Suppose that the
$+$
of
$+^{\mathsf {x}_{k}}$
is maximally moved to
$+^{\widetilde {\mathsf {X}}_{k+1}}$
in
$ \sigma _i(\mathcal {F}({\boldsymbol {\mathsf {X}}}))$
by
$\mathcal {F}$
. Then there are two possibilities: the
$+$
is moved by (II) or by (III) as follows.
Case 2-1. Suppose that the
$+$
is moved by (II). Then
$\mathsf {z}_k=1$
and
$\overline {(\mathsf {x}_{k+1}-\mathsf {z}_{k+1}+1)}=1$
. Since
$\mathsf {z}_k=1$
and
$\underline {\mathsf {y}_{k-1}}=0$
, we have
$\overline {\mathsf {y}_k}=1$
.
We claim that there exists a
$+$
in
$+^{\widetilde {\mathsf {X}}_{k+1}}$
coming from
$+^{\mathsf {x}_{k}}$
which is not canceled. By Lemma 7.20, these
$+$
’s can only be canceled with
$-$
’s coming from
$-^{\mathsf {y}_{k-1}}$
. If
$\mathsf {y}_{k-1}=1$
and
$\mathsf {z}_{k-1}=1$
, then one
$+$
is canceled with one
$-$
, but since
$\mathsf {x}_k \geq 2$
,
$\mathsf {X}_k=\mathsf {x}_k-\mathsf {z}_k+\mathsf {z}_{k-1}\geq 2$
, so
$+^{\widetilde {\mathsf {X}}_{k+1}}$
contains at least two
$+$
’s coming from
$+^{\mathsf {x}_k}$
. Hence, the claim holds, which implies
$\tilde {f}_i \mathcal {F}({\boldsymbol {\mathsf {X}}})= \mathcal {F}(\tilde {f}_i {\boldsymbol {\mathsf {X}}})$
.
Case 2-2. Suppose that the
$+$
is moved by (III). Then
$\mathsf {z}_k=0$
and
$\mathsf {x}_{k}+\mathsf {z}_{k-1}\geq 2$
. We have
$\overline {\mathsf {y}_{k}}=0$
since
$\mathsf {z}_k=0$
.
We claim that
$\mathsf {x}_k \geq 2$
. Indeed, if
$\mathsf {x}_k=1 $
, then
$\mathsf {y}_{k-1}=0$
. Since
$+^{\mathsf {x}_k}$
is not canceled in
$\overline {\sigma _i}({\boldsymbol {\mathsf {X}}})$
, we have
$\mathsf {x}_{k-1} \geq \mathsf {y}_{k-2}$
, so
$\mathsf {y}_{k-2} \leq 1$
, and then
$\mathsf {z}_{k-1}=\mathrm {\mathsf {x}_{k-1},\overline {\mathsf {y}_{k-1}}+2\underline {\mathsf {y}_{k-2}}}=0$
, contradicting
$\mathsf {x}_{k}+\mathsf {z}_{k-1}\geq 2$
. Hence
$\mathsf {x}_k \geq 2$
, and
$+^{\widetilde {\mathsf {X}}_{k+1}}$
contains an uncanceled
$+$
from
$+^{\mathsf {x}_k}$
, giving
$\tilde {f}_i \mathcal {F}({\boldsymbol {\mathsf {X}}})= \mathcal {F}(\tilde {f}_i {\boldsymbol {\mathsf {X}}})$
.
Case 3. Suppose that none of the
$+$
’s in
$+^{\mathsf {x}_k}$
is moved under
$\mathcal {F}$
. Then
$\mathsf {z}_k=0$
and
$\mathsf {x}_k+\mathsf {z}_{k-1} \leq 1$
. Consequently, we have
$\overline {\mathsf {y}_{k}}=0$
. Since
$\mathsf {x}_k+\mathsf {z}_{k-1} \leq 1$
, we have
$\mathsf {x}_k=1$
,
$\mathsf {z}_{k-1}=0$
, and
$\mathsf {y}_{k-1}=0$
. Hence,
$\tilde {f}_i \mathcal {F}({\boldsymbol {\mathsf {X}}})= \mathcal {F}(\tilde {f}_i {\boldsymbol {\mathsf {X}}})$
.
Example 7.22. We continue Example 7.21(1). Then
so that
On the other hand, we have
where
$\mathcal {F}(\tilde {f}_i {\boldsymbol {\mathsf {X}}})$
coincides with
$\tilde {f}_i \mathcal {F}({\boldsymbol {\mathsf {X}}})$
. This is an example of Case 2-2 in the above proof.
Acknowledgments
We would like to sincerely thank the anonymous referees for helpful and detailed comments.
Funding statement
The first author is supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIT) (No. RS-2023-00277388). The second author is supported by the National Research Foundation of Korea(NRF) grant funded by the Korea government(MSIT) (No.2020R1A5A1016126) and the Global-LAMP Program of the National Research Foundation of Korea (NRF) grant funded by the Ministry of Education (No. RS-2023-00301976) The third author is supported by the National Center for Theoretical Sciences.
Competing interests
The authors have no competing interests to declare.
Data availability statement
Data sharing is not applicable to this article as no datasets were generated or analyzed during the current study.





