# Dualities in Refined Grothendieck Polynomials

Graham Hawkes  Department of Mathematics, Ben Gurion University of the Negev,
1 David Ben Gurion Blvd., Be’er Sheva, Israel<EMAIL_ADDRESS>
###### Abstract.

We give new proofs of the two types of duality for Grothendieck polynomials.
Our proofs extend to proofs of these dualities for the refined Grothendieck
polynomials. The second of these dualities was unknown for the refined case.

##  1\. Introduction

The Grassmannian is the set of $k$-dimensional hyperplanes in
$\mathbb{C}^{n}$. Lascoux and Schützenberger [7, 8] introduced Grothendieck
polynomials to represent the K-theory ring of the Grassmannian. In particular,
they correspond to the K-theory classes of structure sheaves of Schubert
varieties, and so they are indexed by permutations in $S_{n}$. By taking the
stable limit of $n\to\infty$, Fomin and Kirillov [3, 4] initiated the study of
stable Grothendieck functions. The subset of stable Grothendieck functions
corresponding to Grassmannian permutations can be indexed by partitions and
are called symmetric Grothendieck polynomials. They are Schur positive and
form a linearly independent set in the space of symmetric functions of any
degree. In particular, the lowest degree term of a symmetric Grothendieck
polynomial is the Schur polynomial indexed by the same partition.

For these reasons, it is natural to ask (1) how the involution, $\omega$, that
interchanges elementary and homogeneous symmetric functions, affects symmetric
Grothendieck polynomials, and (2) whether there is a set of polynomials
orthonormal (under the Hall inner product: $\langle
s_{\lambda},s_{\mu}\rangle=\delta_{\lambda,\mu}$) to the symmetric
Grothendieck polynomials. These considerations led to the establishment of the
four following versions of the symmetric Grothendieck polynomials and to the
invention of the following combinatorial models to study them.

  1. (1A)

Symmetric Grothendieck polynomials using set-valued tableaux,

  2. (1B)

Weak symmetric Grothendieck polynomials using multiset-valued tableaux,

  3. (2A)

Dual symmetric Grothendieck polynomials using reverse plane partitions, and

  4. (2B)

Dual weak symmetric Grothendieck polynomials using valued-set tableaux.

As the reader may guess from the names, (1A) goes to (1B) under $\omega$ and
(2A) goes to (2B) under $\omega$ while (1A) is dual under $\langle,\rangle$ to
(2A) and (1B) is dual under $\langle,\rangle$ to (2B). In other words we have
the following diagram:

$\begin{CD}1A@>{\omega}>{}>1B\\\
@V{}V{\langle,\rangle}V@V{}V{\langle,\rangle}V\\\ 2A@>{\omega}>{}>2B\end{CD}$

Crystal analysis of these objects appearing in [5], [11], and [16] all point
toward a meaningful refinement of these polynomials using an additional
parameter (in this paper $\mathbf{z}$). (For more on this, please see
definition 3.2, Theorem 3.3, and remark 3.9 of [17].) Each of these
refinements appears naturally in the underlying combinatorial object and
denoting the new refined versions as 1A, 1B, 2A, and 2B we still have:

$\begin{CD}\mathbf{1A}@>{\omega}>{}>\mathbf{1B}\\\
@V{}V{\langle,\rangle}V@V{}V{\langle,\rangle}V\\\
\mathbf{2A}@>{\omega}>{}>\mathbf{2B}\end{CD}$

As of yet, there has been little attempt to explain all these arrows in a
single combinatorial framework. In order to do this, we begin by realizing
each of the four combinatorial models as a particular instance of a certain
type of tableau. Then we use a standard RSK argument and a standard jeu de
taquin argument to find out what the fundamental combinatorial facts we need
to prove are. As it turns out, a single combinatorial fact (main fact 1.10)
explains both horizontal arrows, and another fact (main fact 1.13) explains
both vertical arrows.

###  1.1. Primed Tableaux

######  Definition 1.1.

Consider the alphabet $\\{1^{\prime}<1<2^{\prime}<2<3^{\prime}<3,\ldots\\}$.
An _overfull tableau_ of shape $\lambda$, or an element of $OT(\lambda)$ is a
filling of a Young diagram of shape $\lambda$ such that

  * •

Each box is nonempty and contains a set from
$\\{1^{\prime},2^{\prime},\ldots\\}$ and a multiset from $\\{1,2,\ldots\\}$.

  * •

Suppose there is an $a$ in box $A$ and a $b$ in box $B$. Suppose box $B$ lies
immediately to the right of box $B$. Then $a<b$ or else $a=b$ and both are
unprimed numbers.

  * •

Suppose there is an $a$ in box $A$ and a $c$ in box $C$. Suppose box $C$ lies
immediately below box $A$. Then $a<c$ or else $a=c$ and both are primed
numbers.

An _underfull tableau_ of shape $\lambda$, or an element of $UT(\lambda)$ is a
filling of a Young diagram of shape $\lambda$ such that

  * •

Each box is empty, contains one number from
$\\{1^{\prime},2^{\prime},\ldots\\}$, or, one number from $\\{1,2,\ldots\\}$.,
but no box in the leftmost column is empty.

  * •

Suppose there is an $a$ in box $A$ and a $b$ in box $B$. Suppose box $B$ lies
to the right of box $A$ in the same row and is the leftmost such box that is
nonempty. Then $a<b$ or else $a=b$ and both are unprimed numbers.

  * •

Suppose there is an $a$ in box $A$ and a $c$ in box $C$. Suppose box $C$ lies
in the row below box $A$ and weakly to its left, and, is the rightmost such
box that is nonempty. Then $a<c$ or else $a=c$ and both are primed numbers.

The _left weight_ of an $OT$ or a $UT$ is the vector whose $i^{th}$ coordinate
records the number of times $i$ appears in the tableau. The _right weight_ of
an $OT$ or a $UT$ is the vector whose $i^{th}$ coordinate records the number
of times $i^{\prime}$ appears in the tableau. The _overweight_ of an $OT$ of
shape $\lambda$ is the vector whose $i^{th}$ coordinate records the difference
between the number of entries in column $i$ and the number of boxes in column
$i$. The _underweight_ of a $UT$ of shape $\lambda$ is the vector whose
$i^{th}$ coordinate records the difference between the number of boxes in
column $(i+1)$ and the number of entries in column $(i+1)$. (By convention,
either of these differences is taken to be $0$ if the specified column is not
part of the tableau.) A tableau which is both an overfull tableau and an
underfull tableau (i.e., has exactly one entry per box) is called a primed
tableau and the set of all such tableaux of shape $\lambda$ is denoted by
$PT(\lambda)$.

######  Definition 1.2.

Let $\lambda$ be a partition and let $\mathbf{x}=(x_{1},x_{2},\ldots)$,
$\mathbf{y}=(y_{1},y_{2},\ldots)$, and $\mathbf{z}=(z_{1},z_{2},\ldots)$ be
infinite sets of indeterminants. We define polynomials:

$\displaystyle\mathfrak{G}_{\lambda}(\mathbf{x},\mathbf{y},\mathbf{z})$
$\displaystyle=$ $\displaystyle\sum_{T\in OT(\lambda)}\mathbf{x}^{\ell
w(T)}\mathbf{y}^{rw(T)}\mathbf{z}^{O(T)}(-1)^{|O(T)|}$
$\displaystyle\mathfrak{G}^{*}_{\lambda}(\mathbf{x},\mathbf{y},\mathbf{z})$
$\displaystyle=$ $\displaystyle\sum_{T\in UT(\lambda)}\mathbf{x}^{\ell
w(T)}\mathbf{y}^{rw(T)}\mathbf{z}^{U(T)}$

Here, $\ell w(T)$ is the left weight of $T$, $rw(T)$ is the right weight of
$T$, $O(T)$ is the overweight of $T$, and $U(T)$ is the underweight of $T$.

######  Example 1.3.

An overfull and an underfull tableau.

$P=\Yboxdim{24pt}\young({{1^{\prime}11}}{{12^{\prime}}}{{23^{\prime}}},{{2^{\prime}}}{{2}}{{3^{\prime}33}},{{2^{\prime}3^{\prime}}}{{3}})$

  * •

$P\in OT(3,3,2)$

  * •

$\ell w(P)=(2,1,3)$

  * •

$rw(P)=(1,3,3)$

  * •

$O(P)=(3,1,3)$

$Q=\Yboxdim{18pt}\young({{1^{\prime}}}{{}}{{1}}{{}}{{1}},{{1}}{{2^{\prime}}}{{}}{{2}}{{}},{{2^{\prime}}}{{}}{{3}},{{3}})$

  * •

$Q\in UT(4,4,3,1)$

  * •

$\ell w(Q)=(3,1,2)$

  * •

$rw(Q)=(1,2,0)$

  * •

$U(Q)=(2,1,1,1)$

######  Definition 1.4.

Let $\lambda$ be a partition with conjugate, $\lambda^{\prime}$. For the
partition $\lambda$ the

  1. (1A)

refined symmetric Grothendieck polynomial is
$\mathfrak{G}_{\lambda^{\prime}}(\mathbf{0},\mathbf{x},\mathbf{z})$.

  2. (1B)

refined weak symmetric Grothendieck polynomial is
$\mathfrak{G}_{\lambda}(\mathbf{x},\mathbf{0},\mathbf{z})$.

  3. (2A)

refined dual symmetric Grothendieck polynomial is
$\mathfrak{G}^{*}_{\lambda^{\prime}}(\mathbf{0},\mathbf{x},\mathbf{z})$.

  4. (2B)

refined dual weak symmetric Grothendieck polynomial is
$\mathfrak{G}^{*}_{\lambda}(\mathbf{x},\mathbf{0},\mathbf{z})$.

The nonrefined versions of these polynomials are obtained by setting
$\mathbf{z}=\mathbf{1}$.

It is not hard to see that these definitions coincide with the combinatorial
definitions of these polynomials given elsewhere. For the nonrefined versions
of each of these four polynomials see, respectively, Theorem 3.1 of [1],
definition 9.20 of [9], corollary 2 of [5], and definition 9.23 of [9]. To see
an example of the refined versions see definition 3.2 and Theorem 3.3 of [17].

######  Definition 1.5.

Let $\mu\subseteq\lambda$ be partitions with an equal number of rows. An _over
flagged tableau_ of shape $\lambda/\mu$, or an element of $OFT(\lambda/\mu)$,
is a filling of a Young diagram of shape $\lambda/\mu$ using the alphabet
$1<2<\cdots$ such that:

  * •

Each box in row $i$ of $\lambda/\mu$ contains one element from
$\\{1,2,\ldots,\mu_{i}\\}$.

  * •

Suppose box $A$ lies immediately to the left of box $B$. Suppose there is an
$a$ in $A$ and a $b$ in $B$. Then $a\geq b$.

  * •

Suppose box $A$ lies immediately to the above of box $C$. Suppose there is an
$a$ in $A$ and a $c$ in $C$. Then $a>c$.

The weight, $wt(P)$, of an $OFT$, $P$, is the vector whose $i^{th}$ coordinate
records the number of $i$s in the tableau.

Let $\mu\subseteq\lambda$ be partitions with an equal number of rows. An
_under flagged tableau_ of shape $\lambda/\mu$, or an element of
$UFT(\lambda/\mu)$ is a filling of a Young diagram of shape $\lambda/\mu$
using the alphabet $1^{\prime}<2^{\prime}<\cdots$ such that:

  * •

Each box in row $i$ of $\lambda/\mu$ contains one element from
$\\{1^{\prime},\ldots,(\lambda_{i}-1)^{\prime}\\}$.

  * •

Suppose box $A$ lies immediately to the left of box $B$. Suppose there is an
$a$ in $A$ and a $b$ in $B$. Then $a<b$.

  * •

Suppose box $A$ lies immediately to the above of box $C$. Suppose there is an
$a$ in $A$ and a $c$ in $C$. Then $a\leq c$.

The weight, $wt(P)$, of a $UFT$, $P$ is the vector whose $i^{th}$ coordinate
records the number of $i^{\prime}$s in the tableau.

######  Example 1.6.

An over flagged and an under flagged tableau.

$P=\Yboxdim{18pt}\young(****42,***321,**221,*111)$

  * •

$P\in OFT((6,6,5,4)/(4,3,2,1))$

  * •

$wt(P)=(5,3,1,1)$

$Q=\Yboxdim{18pt}\young(****{{1^{\prime}}}{{5^{\prime}}},***{{2^{\prime}}}{{3^{\prime}}}{{5^{\prime}}},**{{1^{\prime}}}{{2^{\prime}}}{{4^{\prime}}},*{{1^{\prime}}}{{2^{\prime}}}{{3^{\prime}}})$

  * •

$Q\in UFT((6,6,5,4)/(4,3,2,1))$

  * •

$wt(Q)=(3,3,2,1,2)$

If $\lambda\supseteq\mu$ have a different number of rows, set
$OFT(\lambda/\mu)=\emptyset=UFT(\lambda/\mu)$.

######  Lemma 1.7.

There are bijections between the following sets:

  1. (1)

$OT(\mu)$ and $\\{(P,Q):\exists\lambda\supseteq\mu:P\in PT(\lambda),Q\in
OFT(\lambda/\mu)\\}$.

  2. (2)

$UT(\lambda)$ and $\\{(P,Q):\exists\mu\subseteq\lambda:P\in PT(\mu),Q\in
UFT(\lambda/\mu)\\}$.

In case (1), if $T\rightarrow(P,Q)$ then $\ell w(T)=\ell w(P)$, $rw(T)=rw(P)$,
and $O(T)=wt(Q)$.
In case (2), if $T\rightarrow(P,Q)$ then $\ell w(T)=\ell w(P)$, $rw(T)=rw(P)$,
and $U(T)=wt(Q)$.

###### Proof.

We construct each bijection. We will use RSK column insertion in the first and
jeu de taquin in the second. The proper versions of these algorithms to use
are those that are obtained from the standard versions (for alphabets of
distinct letters) by beginning with the order
$1^{\prime}<1<2^{\prime}<2<\cdots$ and standardizing the unprimed entries of a
tableau left to right and the primed entries of a tableau top to bottom.

  1. (1)

Start with $T\in OT(\mu)$ and construct $(P,Q)$ as follows. Beginning with the
rightmost column, that is column $\mu_{1}$, and working to the leftmost
column, that is, column $1$, do as follows. From each box in the current
column, say column $i$, remove all but the smallest entry. Now, in weakly
decreasing order, insert the removed entries to the tableau formed by the
columns $i+1,i+2,\ldots$ using RSK insertion. Let $P$ be the resulting tableau
and suppose it has shape $\lambda$. Now construct a tableau $Q$ of shape
$\lambda/\mu$ by placing an $i$ in each box that corresponds to the position
of a box appended during the RSK insertions that occurred after removing
entries from column $i$.

  2. (2)

Start with $T\in UT(\lambda)$ and construct $(P,Q)$ as follows. Beginning with
the rightmost column, that is column $\lambda_{1}$, and working to the second
to leftmost column, that is, column $2$, do as follows. Starting with the
lowest and working to the highest, do the following for each empty box, $b$,
in column $i$. Consider the tableau, $R$, whose upper left corner box is $b$.
$R$ has exactly one empty box which is box $b$. Apply jdt into box $b$ and
continue until an outer corner box is removed from $R$. After this has been
done for each box $b$ for each column $i$ from $\lambda_{1}$ to $2$ define the
result to be $P$ and denote its shape by $\mu$. Now construct a tableau $Q$ of
shape $\lambda/\mu$ by placing a $(i-1)^{\prime}$ in each box that corresponds
to the position of a box removed after a jdt move into a box in column $i$.

The $P$-tableaux constructed are valid primed tableaux by the properties of
RSK and jdt. The $Q$-tableaux constructed are valid column strict and row
strict tableaux respectively, also by properties of RSK and jdt. On the other
hand, the flag conditions for an $OT$ or a $UT$ are ensured directly from the
construction. Moreover, the weight conditions $\ell w(T)=\ell w(P)$,
$rw(T)=rw(P)$ are immediate since RSK and jdt don’t affect the set of entries
appearing in a tableau, and the weight conditions $O(T)=wt(Q)$ and
$U(T)=wt(Q)$ follow by construction. Finally, it is not difficult to see how
to invert the maps using the fact that RSK and jdt are themselves invertible.

∎

######  Example 1.8.

An example of bijection $(1)$ of Lemma 1.7. In each step the newly colored
entry is to be RSK inserted into the column to its right. The final colored
tableau is $P$ and $Q$ is shown directly below it.

$\displaystyle\Yboxdim{23pt}\young({{1^{\prime}}}{{2^{\prime}3^{\prime}}},{{12^{\prime}3^{\prime}}}{{3{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}},{{3^{\prime}4^{\prime}4}})\longrightarrow\young({{1^{\prime}}}{{2^{\prime}{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}3^{\prime}}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}},{{12^{\prime}3^{\prime}}}{{3}},{{3^{\prime}4^{\prime}4}})\longrightarrow\young({{1^{\prime}}}{{2^{\prime}}}{{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}3^{\prime}}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}},{{12^{\prime}3^{\prime}}}{{3}},{{3^{\prime}4^{\prime}{\color[rgb]{0,1,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,0}4}}})\longrightarrow\young({{1^{\prime}}}{{2^{\prime}}}{{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}3^{\prime}}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}},{{12^{\prime}3^{\prime}}}{{3}},{{3^{\prime}{\color[rgb]{.75,.5,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,.5,.25}4^{\prime}}}}{{{\color[rgb]{0,1,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,0}4}}})$
$\displaystyle\Yboxdim{23pt}\longrightarrow\young({{1^{\prime}}}{{2^{\prime}}}{{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}3^{\prime}}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}},{{12^{\prime}{\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}3^{\prime}}}}{{3}}{{{\color[rgb]{0,1,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,0}4}}},{{3^{\prime}}}{{{\color[rgb]{.75,.5,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,.5,.25}4^{\prime}}}})\longrightarrow\young({{1^{\prime}}}{{2^{\prime}}}{{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}3^{\prime}}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}},{{1{\color[rgb]{.5,.5,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,.5,.5}\pgfsys@color@gray@stroke{.5}\pgfsys@color@gray@fill{.5}2^{\prime}}}}{{{\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}3^{\prime}}}}{{3}}{{{\color[rgb]{0,1,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,0}4}}},{{3^{\prime}}}{{{\color[rgb]{.75,.5,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,.5,.25}4^{\prime}}}})\longrightarrow\young({{1^{\prime}}}{{2^{\prime}}}{{{\color[rgb]{0,0,1}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,1}3^{\prime}}}}{{3}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}},{{1}}{{{\color[rgb]{.5,.5,.5}\definecolor[named]{pgfstrokecolor}{rgb}{.5,.5,.5}\pgfsys@color@gray@stroke{.5}\pgfsys@color@gray@fill{.5}2^{\prime}}}}{{{\color[rgb]{1,.5,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,.5,0}3^{\prime}}}}{{{\color[rgb]{0,1,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,1,0}4}}},{{3^{\prime}}}{{{\color[rgb]{.75,.5,.25}\definecolor[named]{pgfstrokecolor}{rgb}{.75,.5,.25}4^{\prime}}}})$
$\displaystyle\Yboxdim{23pt}\young(**221,**11,*1)$

######  Example 1.9.

An example of bijection $(2)$ of Lemma 1.7. In each step the jeu de taquin
algorithm is to be applied into the box marked by the
${\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\times}$.
The final tableau is $P$ and $Q$ is shown directly below it.

$\displaystyle\Yboxdim{12pt}\young({{1^{\prime}}}{{}}{{}}{{}}{{3^{\prime}}},{{2^{\prime}}}{{}}{{3^{\prime}}}{{}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\times}}},{{2}}{{3}}{{}}{{4}}{{4}},{{4}}{{}}{{4}})\longrightarrow\young({{1^{\prime}}}{{}}{{}}{{}}{{3^{\prime}}},{{2^{\prime}}}{{}}{{3^{\prime}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\times}}}{{4}},{{2}}{{3}}{{}}{{4}},{{4}}{{}}{{4}})\longrightarrow\young({{1^{\prime}}}{{}}{{}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\times}}}{{3^{\prime}}},{{2^{\prime}}}{{}}{{3^{\prime}}}{{4}}{{4}},{{2}}{{3}}{{}},{{4}}{{}}{{4}})\longrightarrow\young({{1^{\prime}}}{{}}{{}}{{3^{\prime}}}{{4}},{{2^{\prime}}}{{}}{{3^{\prime}}}{{4}},{{2}}{{3}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\times}}},{{4}}{{}}{{4}})\longrightarrow$
$\displaystyle\Yboxdim{12pt}\young({{1^{\prime}}}{{}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\times}}}{{3^{\prime}}}{{4}},{{2^{\prime}}}{{}}{{3^{\prime}}}{{4}},{{2}}{{3}}{{4}},{{4}}{{}})\rightarrow\young({{1^{\prime}}}{{}}{{3^{\prime}}}{{4}}{{4}},{{2^{\prime}}}{{}}{{3^{\prime}}},{{2}}{{3}}{{4}},{{4}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\times}}})\rightarrow\young({{1^{\prime}}}{{}}{{3^{\prime}}}{{4}}{{4}},{{2^{\prime}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\times}}}{{3^{\prime}}},{{2}}{{3}}{{4}},{{4}})\rightarrow\young({{1^{\prime}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\times}}}{{3^{\prime}}}{{4}}{{4}},{{2^{\prime}}}{{3^{\prime}}}{{4}},{{2}}{{3}},{{4}})\rightarrow\young({{1^{\prime}}}{{3^{\prime}}}{{4}}{{4}}{{4}},{{2^{\prime}}}{{3^{\prime}}},{{2}}{{3}},{{4}})$
$\displaystyle\Yboxdim{12pt}\young(*****,**{{1^{\prime}}}{{2^{\prime}}}{{3^{\prime}}},**{{1^{\prime}}}{{3^{\prime}}}{{4^{\prime}}},*{{1^{\prime}}}{{2^{\prime}}})$

The following two main facts are the main results of this paper. These facts
along with the results so far stated give a complete combinatorial
understanding of two types of duality appearing in Grothendieck polynomials.
These facts will be proven in the next section.

######  Main Fact 1.10.

Let $\omega_{\mathbf{x}}$ denote the involution on functions symmetric in
$\mathbf{x}$ over the ring $\mathbb{Z}[\mathbf{y},\mathbf{z}]$ defined by
$\omega_{\mathbf{x}}(s_{\lambda}(\mathbf{x}))=s_{\lambda^{\prime}}(\mathbf{x})$.
Similarly, let $\omega_{\mathbf{y}}$ denote the involution on functions
symmetric in $\mathbf{y}$ over the ring $\mathbb{Z}[\mathbf{x},\mathbf{z}]$
defined by
$\omega_{\mathbf{y}}(s_{\lambda}(\mathbf{y}))=s_{\lambda^{\prime}}(\mathbf{y})$.
Then we have that:

$\displaystyle\omega_{\mathbf{x}}\omega_{\mathbf{y}}\left(\sum_{P\in
PT(\lambda)}\mathbf{x}^{\ell w(P)}\mathbf{y}^{rw(P)}\right)=\sum_{P\in
PT(\lambda)}\mathbf{y}^{\ell w(P)}\mathbf{x}^{rw(P)}$

######  Corollary 1.11.

We have:

$\displaystyle\omega_{\mathbf{x}}\omega_{\mathbf{y}}\mathfrak{G}_{\mu}(\mathbf{x},\mathbf{y},\mathbf{z})=\mathfrak{G}_{\mu}(\mathbf{y},\mathbf{x},\mathbf{z})$
$\displaystyle\omega_{\mathbf{x}}\omega_{\mathbf{y}}\mathfrak{G}^{*}_{\lambda}(\mathbf{x},\mathbf{y},\mathbf{z})=\mathfrak{G}^{*}_{\lambda}(\mathbf{y},\mathbf{x},\mathbf{z})$

###### Proof.

Using Lemma 1.7 we may write:

$\displaystyle\mathfrak{G}_{\mu}(\mathbf{x},\mathbf{y},\mathbf{z})=\sum_{\lambda\supseteq\mu}$
$\displaystyle(-1)^{|\lambda|-|\mu|}$ $\displaystyle\sum_{Q\in
OFT(\lambda/\mu)}\mathbf{z}^{O(Q)}\sum_{P\in PT(\lambda)}\mathbf{x}^{\ell
w(P)}\mathbf{y}^{rw(P)}$
$\displaystyle\mathfrak{G}^{*}_{\lambda}(\mathbf{x},\mathbf{y},\mathbf{z})=\sum_{\mu\subseteq\lambda}$
$\displaystyle\sum_{Q\in UFT(\lambda/\mu)}\mathbf{z}^{U(Q)}\sum_{P\in
PT(\lambda)}\mathbf{x}^{\ell w(P)}\mathbf{y}^{rw(P)}$

The corollary now follows from main fact 1.10 and the linearity of
$\omega_{\mathbf{x}}$ and $\omega_{\mathbf{y}}$. ∎

######  Remark 1.12.

Setting $\mathbf{y}=\mathbf{0}$ in the first equation of corollary 1.11 we see
that the refined weak symmetric Grothendieck polynomial maps under
$\omega_{\mathbf{x}}$ to the refined symmetric Grothendieck polynomial of
conjugate shape. Setting $\mathbf{y}=\mathbf{0}$ in the second equation of
corollary 1.11 we see that the refined dual weak symmetric Grothendieck
polynomial maps under $\omega_{\mathbf{x}}$ to the refined dual symmetric
Grothendieck polynomial of conjugate shape. In other words this shows
$\textbf{1A}\xrightarrow{\omega}\textbf{1B}$ and
$\textbf{2A}\xrightarrow{\omega}\textbf{2B}$.

######  Main Fact 1.13.

Let $\mu\subset\lambda$ be partitions with the same number of rows. Then:

$\displaystyle\sum_{\mu\subseteq\rho\subseteq\lambda}\sum_{(P,Q)\in
T_{\rho}}(-1)^{|\rho|-|\mu|}\mathbf{z}^{wt(P)}\mathbf{z}^{wt(Q)}=0$

where $T_{\rho}$ is the set of all pairs $(P,Q)$ with $P\in OFT(\rho/\mu)$ and
$Q\in UFT(\lambda/\rho)$.

######  Corollary 1.14.

Let $\langle,\rangle$ denote the bilinear form on functions symmetric in
$\mathbf{x}$ over the ring $\mathbb{Z}[\mathbf{z}]$ defined by $\langle
s_{\mu}(\mathbf{x}),s_{\lambda}(\mathbf{x})\rangle=\delta_{\mu,\lambda}$ and
extended $\mathbb{Z}[\mathbf{z}]$-bilinearly. Then we have:

$\displaystyle\langle\mathfrak{G}_{\mu}(\mathbf{x},\mathbf{0},\mathbf{z}),\mathfrak{G}^{*}_{\lambda}(\mathbf{x},\mathbf{0},\mathbf{z})\rangle=\delta_{\mu,\lambda}$
$\displaystyle\langle\mathfrak{G}_{\mu}(\mathbf{0},\mathbf{x},\mathbf{z}),\mathfrak{G}^{*}_{\lambda}(\mathbf{0},\mathbf{x},\mathbf{z})\rangle=\delta_{\mu,\lambda}$

###### Proof.

Since a primed tableau with right weight (resp. leftweight) of $\mathbf{0}$ is
just a semistandard Young tableau (resp. conjugated semistandard Young
tableau) it follows from the formulas in the proof of corollary 1.11 that
$\langle\mathfrak{G}_{\mu}(\mathbf{x},\mathbf{0},\mathbf{z}),\mathfrak{G}^{*}_{\lambda}(\mathbf{x},\mathbf{0},\mathbf{z})\rangle$
is equal to:

$\displaystyle\langle\left(\sum_{\rho\supseteq\mu}(-1)^{|\rho|-|\mu|}\sum_{P\in
OFT(\rho/\mu)}\mathbf{z}^{O(P)}s_{\rho}(\mathbf{x})\right),\left(\sum_{\sigma\subseteq\lambda}\sum_{Q\in
UFT(\lambda/\sigma)}\mathbf{z}^{U(Q)}s_{\sigma}(\mathbf{x})\right)\rangle$

and that
$\langle\mathfrak{G}_{\mu}(\mathbf{0},\mathbf{x},\mathbf{z}),\mathfrak{G}^{*}_{\lambda}(\mathbf{0},\mathbf{x},\mathbf{z})\rangle$
is equal to:

$\displaystyle\langle\left(\sum_{\rho\supseteq\mu}(-1)^{|\rho|-|\mu|}\sum_{P\in
OFT(\rho/\mu)}\mathbf{z}^{O(P)}s_{\rho^{\prime}}(\mathbf{x})\right),\left(\sum_{\sigma\subseteq\lambda}\sum_{Q\in
UFT(\lambda/\sigma)}\mathbf{z}^{U(Q)}s_{\sigma^{\prime}}(\mathbf{x})\right)\rangle$

If $\mu=\lambda$ then it is clear there is exactly one nonzero term in the
expansion of these inner products which occurs when $\mu=\rho=\sigma$ and that
this term has coefficient $(-1)^{0}\mathbf{z}^{\mathbf{0}}=1$. If
$\mu\not\subseteq\lambda$ or $\mu\subset\lambda$ have different numbers of
rows then it is clear that all terms in the expansion of the inner products
must be $0$. Finally, if $\mu\subset\lambda$ then it follows from main fact
1.13 that the inner products evaluate to $0$. ∎

######  Remark 1.15.

Corollary 1.14 says that under $\langle,\rangle$ the refined weak symmetric
Grothendieck polynomial is dual to the refined dual weak symmetric
Grothendieck polynomial and that the refined symmetric Grothendieck polynomial
is dual to the refined dual symmetric Grothendieck polynomial. I.e.,
$\textbf{1A}\xrightarrow{\langle,\rangle}\textbf{2A}$ and
$\textbf{1B}\xrightarrow{\langle,\rangle}\textbf{2B}$.

##  2\. Proof of the Main Facts

######  Definition 2.1.

Let $\prec$ denote any total order on the set
$\\{1^{\prime},1,2^{\prime},2^{\prime},\ldots\\}$. We define a primed tableau
with respect to $\prec$ of shape $\lambda/\mu$ to be a filling of the shape
$\lambda/\mu$ such that:

  * •

Each box of $\lambda/\mu$ contains exactly one of
$\\{1^{\prime},1,2^{\prime},2^{\prime},\ldots\\}$.

  * •

The entries in the rows of $\lambda/\mu$ weakly increase under $\prec$ left to
right.

  * •

The entries in the columns of $\lambda/\mu$ weakly increase $\prec$ top to
bottom.

  * •

There is at most one $i$ in each column for each $i$.

  * •

There is at most one $i^{\prime}$ in each row for each $i$.

The set of such tableaux is denoted by $PT_{\prec}(\lambda/\mu)$. The left
weight of a $PT_{\prec}$ is the vector whose $i^{th}$ coordinate records the
number of times $i$ appears in the tableau. The right weight of a $PT_{\prec}$
is the vector whose $i^{th}$ coordinate records the number of times
$i^{\prime}$ appears in the tableau.

Note that if $\prec$ is the order $1^{\prime}\prec 1\prec 2^{\prime}\prec
2\prec\cdots$ then $PT_{\prec}(\lambda/\emptyset)=PT(\lambda)$.

######  Lemma 2.2.

The total order chosen in the definition above is irrelevant. In other words,
given any two total orderings $\prec$ and $\vartriangleleft$ there is a left
weight and right weight preserving bijection between $PT_{\prec}(\lambda/\mu)$
and $PT_{\vartriangleleft}(\lambda/\mu)$.

###### Proof.

First we prove the lemma in the special case that $\prec$ coincides with
$\vartriangleleft$ when restricted to $\\{1,2,\ldots\\}$ and when restricted
to $\\{1^{\prime},2^{\prime},\ldots\\}$. To do this it suffices to suppose
that there are some $i$ and $j$ (possibly equal) such that $i\prec j^{\prime}$
and $j^{\prime}\vartriangleleft i$ and all other pairs of letters have the
same relationship in both orders. From this assumption it follows that in both
orders there is no other letter between $i$ and $j^{\prime}$.

If $T\in PT_{\prec}(\lambda/\mu)$ or $T\in
PT_{\vartriangleleft}(\lambda/\mu)$, let $\mathcal{B}(T)$ denote all of the
boxes of $\lambda/\mu$ that contain an $i$ or a $j^{\prime}$. If
$\lambda=(\lambda_{1},\ldots,\lambda_{\ell})$ then label the the upper left to
lower right diagonals of $\lambda$ by
$\\{D_{-\ell+1},\ldots,D_{-1},D_{0},D_{1},\ldots,D_{\lambda_{1}-1}\\}$. It is
clear that either $\mathcal{B}(T)\cap D_{s}$ is a single box or is empty for
each $s$. If $p$ is minimal and $q$ is maximal such that the adjacent
diagonals $d_{p},d_{p+1},\ldots,d_{q-1},d_{q}$ each have nonempty intersection
with $\mathcal{B}(T)$ then we call
$B=\\{d_{p}\cap\mathcal{B}(T),\ldots,d_{q}\cap\mathcal{B}(T)\\}$ a connected
component of $\mathcal{B}(T)$.

Define a map $\nearrow$ from $PT_{\prec}(\lambda/\mu)$ to
$PT_{\vartriangleleft}(\lambda/\mu)$ as follows. Suppose that $T\in
PT_{\prec}(\lambda/\mu)$. Perform the following to each connected component
$B$ of $\mathcal{B}(T)$.

  * •

Remove the entry that appears in the upper rightmost box of $B$. Record what
you have removed.

  * •

Move every remaining $i$ one box to the right.

  * •

Move every remaining $j^{\prime}$ one box up.

  * •

Fill the lower leftmost box of $B$ with the entry recorded in step one.

Next define a map $\swarrow$ from $PT_{\vartriangleleft}(\lambda/\mu)$ to
$PT_{\prec}(\lambda/\mu)$ as follows. Suppose that $T\in
PT_{\vartriangleleft}(\lambda)$. Perform the following to each connected
component $B$ of $\mathcal{B}(T)$.

  * •

Remove the entry that appears in the lower leftmost box of $B$. Record what
you have removed.

  * •

Move every remaining $i$ one box to the left.

  * •

Move every remaining $j^{\prime}$ one box down.

  * •

Fill the upper rightmost box of $B$ with the entry recorded in step one.

It is not difficult to check that the maps $\nearrow$ and $\swarrow$ are well-
defined and preserve weights and are mutual inverses. This proves the special
case. Thus, to prove the general case we may assume, for instance, that for
all $i$ and all $j^{\prime}$ we have $i\prec j^{\prime}$ and
$i\vartriangleleft j^{\prime}$. But in this case a primed tableau may be
thought of as a pair of tableaux composed of the tableau formed by the
unprimed entries and the tableau formed by the primed entries. Thus we may
write:

$\displaystyle
PT_{\prec}(\lambda/\mu)=\bigcup_{\mu\subseteq\rho\subseteq\lambda}SSYT_{\prec}(\rho/\mu)\times
SSYT^{\prime}_{\prec}(\lambda/\rho)$

where $SSYT_{\prec}(\rho/\mu)$ is the set of skew semistandard Young tableaux
of shape $(\rho/\mu)$in the alphabet $\\{1,2,\cdots\\}$ under the order
$\prec$ as restricted to $\\{1,2,\cdots\\}$ and
$SSYT^{\prime}_{\prec}(\lambda/\rho)$ is the set of conjugate skew
semistandard Young tableaux of shape $(\lambda/\rho)$ in the alphabet
$\\{1^{\prime},2^{\prime},\cdots\\}$ under the order $\prec$ as restricted to
$\\{1^{\prime},2^{\prime},\cdots\\}$. But all skew Schur functions are
symmetric so this implies that $\prec$ may be replaced with $\vartriangleleft$
in the right hand side of the equation above, which makes it equal to
$PT_{\vartriangleleft}(\lambda/\mu)$ by the same logic with which we arrived
at this equation.

∎

######  Example 2.3.

Suppose that we are given the following orderings:

$\displaystyle 1^{\prime}\prec 1\prec 2\prec 3\prec 2^{\prime}\prec 4\prec
3^{\prime}\prec 4^{\prime}$ $\displaystyle 1^{\prime}\vartriangleleft
1\vartriangleleft 2\vartriangleleft 2^{\prime}\vartriangleleft
3\vartriangleleft 4\vartriangleleft 3^{\prime}\vartriangleleft 4^{\prime}$

Then under the map $\nearrow$ with $i=3$ and $j=2$ the tableau $S\in
PT_{\prec}(\lambda/\mu)$ on the left below is sent to the tableau $T\in
PT_{\vartriangleleft}(\lambda/\mu)$ on the right below. Note that
$\mathcal{B}(S)$ is composed of $3$ connected components of sizes $1$, $7$,
and $3$.

$\displaystyle
S=\young(::{{1^{\prime}}}122{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}2^{\prime}}}},{{1^{\prime}}}11{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}}{{3^{\prime}}}{{4^{\prime}}},{{1^{\prime}}}22{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}2^{\prime}}}}44{{3^{\prime}}},1{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}2^{\prime}}}}{{3^{\prime}}},24{{3^{\prime}}},{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}}{{3^{\prime}}})\nearrow\,\,\,\,\,\,\,\,\,\,\,\,\young(::{{1^{\prime}}}122{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}2^{\prime}}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}},{{1^{\prime}}}11{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}2^{\prime}}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}}{{3^{\prime}}}{{4^{\prime}}},{{1^{\prime}}}22{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}2^{\prime}}}}44{{3^{\prime}}},1{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}}{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}}{{3^{\prime}}},24{{3^{\prime}}},{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}3}}}{{3^{\prime}}})=T$

###### Proof of Main Fact 1.10.

Consider the orders:

$\displaystyle 1^{\prime}\prec 1\prec 2^{\prime}\prec 2\prec\cdots$
$\displaystyle 1\vartriangleleft 2\vartriangleleft\cdots
1^{\prime}\vartriangleleft 2^{\prime}\vartriangleleft\cdots$ $\displaystyle
1^{\prime}\blacktriangleleft 2^{\prime}\blacktriangleleft\cdots
1\blacktriangleleft 2\blacktriangleleft\cdots$

Since an element of $PT_{\vartriangleleft}(\lambda/\emptyset)$ can be thought
as a semistandard Young tableau of some shape $\mu\subseteq\lambda$ along with
a conjugate semistadard Young tableau (with its entries primed) of shape
$\lambda/\mu$, we see that:

$\displaystyle\sum_{P\in
PT_{\vartriangleleft}(\lambda/\emptyset)}\mathbf{x}^{\ell
w(P)}\mathbf{y}^{rw(P)}=\sum_{\mu\subseteq\lambda}s_{\mu}(\mathbf{x})s_{\lambda^{\prime}/\mu^{\prime}}(\mathbf{y})$

On the other hand, an element $PT_{\blacktriangleleft}(\lambda/\emptyset)$ can
be thought as a conjugate semistandard Young tableau (with its entries primed)
of some shape $\mu\subseteq\lambda$ along with a semistandard Young tableau of
shape $\lambda/\mu$, so that:

$\displaystyle\sum_{P\in
PT_{\blacktriangleleft}(\lambda/\emptyset)}\mathbf{y}^{\ell
w(P)}\mathbf{x}^{rw(P)}=\sum_{\mu\subseteq\lambda}s_{\mu^{\prime}}(\mathbf{x})s_{\lambda/\mu}(\mathbf{y})$

Since the right hand side of the equation above is obtained from the right
hand side of the equation before it by applying
$\omega_{\mathbf{x}}\omega_{\mathbf{y}}$ the same is true of the left hand
sides. Moreover by Lemma 2.2 both $PT_{\vartriangleleft}(\lambda/\emptyset)$
and $PT_{\blacktriangleleft}(\lambda/\emptyset)$ may be replaced by
$PT_{\prec}(\lambda/\emptyset)$ which is turn equivalent to $PT(\lambda)$.
This proves that:

$\displaystyle\omega_{\mathbf{x}}\omega_{\mathbf{y}}\left(\sum_{P\in
PT(\lambda)}\mathbf{x}^{\ell w(P)}\mathbf{y}^{rw(P)}\right)=\sum_{P\in
PT(\lambda)}\mathbf{y}^{\ell w(P)}\mathbf{x}^{rw(P)}$

∎

######  Definition 2.4.

Fix the order $\cdots\prec 2\prec 1\prec 1^{\prime}\prec
2^{\prime}\prec\cdots$. Define a primed flagged tableau of shape $\lambda/\mu$
to be a tableau, $P$, that is an element of $PT_{\prec}(\lambda/\mu)$ such
that:

  * •

If row $i$ of $P$ contains a $j$ then $j\leq\mu_{i}$

  * •

If row $i$ of $P$ contains a $j^{\prime}$ then $j<\lambda_{i}$.

Let $PFT(\lambda/\mu)$ denote the set of all primed flagged tableaux of shape
$\lambda/\mu$. We say a “row is $\prec$” if the entries are weakly increasing
left ro right under $\prec$ and there are no repeated primed entries. We say a
“column is $\prec$” if the entries are weakly increasing top to bottom under
$\prec$ and there are no repeated unprimed entries. The condition concerning
the nonprimed entries of row $i$ is called the nonprimed row $i$ flag
condition. The condition concerning the primed entries of row $i$ is called
the primed row $i$ flag condition. In other words, $P\in PFT(\lambda/\mu)$ if
and only all its rows and columns are $\prec$ and it satisfies the nonprimed
and primed row $i$ flag conditions for each $i$.

######  Lemma 2.5.

Let $\mu\subset\lambda$ be partitions with the same number of rows.

$\displaystyle\sum_{P\in
PFT_{\prec}(\lambda/\mu)}(-1)^{\\#(P)}\mathbf{z}^{\ell
w(P)}\mathbf{z}^{rw(P)}=0$

Where $\\#(P)$ is the number of nonprimed entries in $P$, $\ell w(P)$ is the
left weight of $P$, and $rw(P)$ is the right weight of $P$.

###### Proof.

We complete the proof by constructing a sign changing involution on
$PFT(\lambda/\mu)$ which preserves the value of $\ell w(P)+rw(P)$ and where
the sign is the value of $(-1)^{\\#(P)}$. The involution, $\iota$, is
described as follows:

Let $P\in PFT(\lambda/\mu)$. Let $m$ be the smallest integer such that $P$
contains an $m$ or an $m^{\prime}$. Now, let $i$ be minimal such that row $i$
of $P$ contains an $m$ or $m^{\prime}$. Finally, let $j$ be maximal such that
the box, $b$, of $P$, in row $i$ and column $j$ contains an $m$ or an
$m^{\prime}$. Define $\iota(P)$ to be the tableau obtained by:

  * •

If $b$ contains an $m$, replace it with an $m^{\prime}$.

  * •

If $b$ contains an $m^{\prime}$, replace it with an $m$.

First suppose that box $b$ of $P$ contains an $m$. If $\lambda/\mu$ has a box
to the right of $b$ then $P$ must have in this box an $s$ or an $s^{\prime}$
for some $s>m$ by the minimality of $m$ and the maximality of $j$. But it
cannot be an $s$ because row $i$ of $P$ is assumed to be $\prec$. Since
$m^{\prime}\prec s^{\prime}$ it follows that row $i$ of $\iota(P)$ is $\prec$.
If $\lambda/\mu$ has a box below $b$ then $P$ must have in this box a $t$ or a
$t^{\prime}$ for some $t\geq m$ by the minimality of $m$. But it cannot be a
$t$ because column $j$ of $P$ is assumed to be $\prec$. Since
$m^{\prime}\preceq t^{\prime}$ it follows that column $j$ of $\iota(P)$ is
$\prec$. Consequently, all rows and columns of $\iota(P)$ are $\prec$. Next,
we have $m\leq\mu_{i}$ because $P$ is assumed to satisfy the nonprimed row $i$
flag condition. But $\mu_{i}<\lambda_{i}$ since we know row $i$ of
$\lambda/\mu$ has at least one box. Thus $m<\lambda_{i}$ and it follows that
$\iota(P)$ satisfies the primed row $i$ flag condition. Consequently,
$\iota(P)$ satisfies all flag conditions. This shows that $\iota(P)\in
PFT(\lambda/\mu)$.

Now suppose that box $b$ of $P$ contains an $m^{\prime}$. If $\lambda/\mu$ has
a box above $b$ then $P$ must have in this box an $s$ or an $s^{\prime}$ for
some $s>m$ by the minimality of $m$ and the minimality of $i$. But it cannot
be an $s^{\prime}$ because column $j$ of $P$ is assumed to be $\prec$. Since
$s\prec m$ it follows that column $j$ of $\iota(P)$ is $\prec$. If
$\lambda/\mu$ has a box to the left of $b$ then $P$ must have in this box a
$t$ or a $t^{\prime}$ for some $t\geq m$ by the minimality of $m$. But it
cannot be a $t^{\prime}$ because row $i$ of $P$ is assumed to be $\prec$.
Since $t\preceq m$ it follows that row $i$ of $\iota(P)$ is $\prec$.
Consequently, all rows and columns of $\iota(P)$ are $\prec$. Next, suppose
that $m>\mu_{i}$. As stated, if $\lambda/\mu$ has a box to the left of $b$
then $P$ has in this box some $t$ with $t\geq m$. But this would contradict
the nonprimed row $i$ flag condition so $b$ must be the leftmost box in row
$i$ of $\lambda/\mu$. It follows from this and the fact that row $i$ of $P$ is
$\prec$ that the box in row $i$ and column $\lambda_{i}$ of $P$ contains
$n^{\prime}$ for some $n\geq\lambda_{i}-\mu_{i}-1+m$. But this means that
$n\geq\lambda_{i}$ since we assumed $m>\mu_{i}$. This would contradict the
primed row $i$ flag condition so we conclude that $m\leq\mu_{i}$. This means
that $\iota(P)$ satisfies the nonprimed row $i$ flag condition. Consequently,
$\iota(P)$ satisfies all flag conditions. This shows that $\iota(P)\in
PFT(\lambda/\mu)$.

It is clear that for all $P\in PFT(\lambda/\mu)$ we have $\iota^{2}(P)=P$ and
that $\ell w(P)+rw(P)=\ell w(\iota(P))+rw(\iota(P))$ and that
$(-1)^{\\#(P)}=-(-1)^{\\#(\iota(P))}$. This proves the lemma.

∎

######  Example 2.6.

The involution $\iota$ associates the following two tableaux.

$\displaystyle\Yboxdim{14pt}\young(*****544{{3^{\prime}}}{{5^{\prime}}},****433{{3^{\prime}}}{{6^{\prime}}}{{8^{\prime}}},****32{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}2}}}{{3^{\prime}}}{{6^{\prime}}},***32{{2^{\prime}}}{{3^{\prime}}}{{4^{\prime}}})\,\,\,\,\,\longleftrightarrow\,\,\,\,\,\young(*****544{{3^{\prime}}}{{5^{\prime}}},****433{{3^{\prime}}}{{6^{\prime}}}{{8^{\prime}}},****32{{{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}2^{\prime}}}}{{3^{\prime}}}{{6^{\prime}}},***32{{2^{\prime}}}{{3^{\prime}}}{{4^{\prime}}})$

In this case, $m=2$, $i=3$ and $j=7$.

###### Proof of Main Fact 1.13.

Let $\mu\subset\lambda$ be partitions with exactly $\ell$ rows. There is a
canonical bijection:

$\displaystyle\bigcup_{\mu\subseteq\rho\subseteq\lambda}OFT(\rho/\mu)\times
UFT(\lambda/\rho)\rightarrow PFT(\lambda/\mu)$

It is given by sending a pair $(P,Q)$ to the tableau obtained by superimposing
$P$ and $Q$ on a single Young diagram of shape $\lambda/\mu$. Moreover, if
$(P,Q)\rightarrow R$ then $\ell w(R)=wt(P)$ and $rw(R)=wt(Q)$ while
$\\#(R)=|\rho|-|\mu|$ where $\rho$ is the shape of $P$. Therefore writing
$T_{\rho}=OFT(\rho/\mu)\times UFT(\lambda/\rho)$, we have:

$\displaystyle\sum_{\mu\subseteq\rho\subseteq\lambda}\sum_{(P,Q)\in
T_{\rho}}(-1)^{|\rho|-|\mu|}\mathbf{z}^{wt(P)}\mathbf{z}^{wt(Q)}=\sum_{R\in
PFT(\lambda/\mu)}(-1)^{\\#(R)}\mathbf{z}^{\ell w(R)}\mathbf{z}^{rw(R)}$

But the latter is $0$ by lemma 2.5. This implies the statement of Main Fact
1.13. ∎

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