that partitions in $\operatorname*{Par}\left[n\right]$ are the same as
nonnegative snakes. We leave the details to the reader (who can also find them
expanded in the detailed version [Grinbe20] of this paper).]

From the first Pieri rule ([GriRei20, (2.7.1)]999This also appears in
[MenRem15, Theorem 5.3], in [Stanle01, Theorem 7.15.7] and in [Egge19, Theorem
9.3]., applied to $k$ instead of $n$), we obtain

$s_{\lambda}h_{k}=\sum\limits_{\begin{subarray}{c}\lambda^{+}\in\operatorname*{Par};\\\
\lambda^{+}/\lambda\text{ is a horizontal
}k\text{-strip}\end{subarray}}s_{\lambda^{+}}=\sum\limits_{\begin{subarray}{c}\mu\in\operatorname*{Par};\\\
\mu/\lambda\text{ is a horizontal }k\text{-strip}\end{subarray}}s_{\mu}$

(here, we have renamed the summation index $\lambda^{+}$ as $\mu$).

Evaluating both sides of this equality at $x_{1},x_{2},\ldots,x_{n}$, we find

$\displaystyle\left(s_{\lambda}h_{k}\right)\left(x_{1},x_{2},\ldots,x_{n}\right)$
$\displaystyle=\sum\limits_{\begin{subarray}{c}\mu\in\operatorname*{Par};\\\
\mu/\lambda\text{ is a horizontal
}k\text{-strip}\end{subarray}}s_{\mu}\left(x_{1},x_{2},\ldots,x_{n}\right)$
$\displaystyle=\sum\limits_{\begin{subarray}{c}\mu\in\operatorname*{Par};\\\
\mu/\lambda\text{ is a horizontal }k\text{-strip;}\\\ \mu\text{ has length
}\leq
n\end{subarray}}s_{\mu}\left(x_{1},x_{2},\ldots,x_{n}\right)+\sum\limits_{\begin{subarray}{c}\mu\in\operatorname*{Par};\\\
\mu/\lambda\text{ is a horizontal }k\text{-strip;}\\\ \mu\text{ has length
}>n\end{subarray}}\ \
\underbrace{s_{\mu}\left(x_{1},x_{2},\ldots,x_{n}\right)}_{\begin{subarray}{c}=0\\\
\text{(by (\ref{pf.lem.alt.lrr.=0}),}\\\ \text{applied to }\mu\text{ instead
of }\lambda\text{)}\end{subarray}}$
$\displaystyle=\sum\limits_{\begin{subarray}{c}\mu\in\operatorname*{Par};\\\
\mu/\lambda\text{ is a horizontal }k\text{-strip;}\\\ \mu\text{ has length
}\leq
n\end{subarray}}s_{\mu}\left(x_{1},x_{2},\ldots,x_{n}\right)+\underbrace{\sum\limits_{\begin{subarray}{c}\mu\in\operatorname*{Par};\\\
\mu/\lambda\text{ is a horizontal }k\text{-strip;}\\\ \mu\text{ has length
}>n\end{subarray}}0}_{=0}$
$\displaystyle=\sum\limits_{\begin{subarray}{c}\mu\in\operatorname*{Par};\\\
\mu/\lambda\text{ is a horizontal }k\text{-strip;}\\\ \mu\text{ has length
}\leq n\end{subarray}}s_{\mu}\left(x_{1},x_{2},\ldots,x_{n}\right).$

In view of

$\displaystyle\left(s_{\lambda}h_{k}\right)\left(x_{1},x_{2},\ldots,x_{n}\right)$
$\displaystyle=\underbrace{s_{\lambda}\left(x_{1},x_{2},\ldots,x_{n}\right)}_{\begin{subarray}{c}=\overline{s}_{\lambda}\\\
\text{(by Lemma
\ref{lem.alt.sbar})}\end{subarray}}\cdot\underbrace{h_{k}\left(x_{1},x_{2},\ldots,x_{n}\right)}_{\begin{subarray}{c}=h_{k}^{+}\\\
\text{(since
}h_{k}^{+}=h_{k}\left(x_{1},x_{2},\ldots,x_{n}\right)\text{)}\end{subarray}}$
$\displaystyle=\overline{s}_{\lambda}\cdot
h_{k}^{+}=h_{k}^{+}\cdot\overline{s}_{\lambda},$

we can rewrite this as

$h_{k}^{+}\cdot\overline{s}_{\lambda}=\sum\limits_{\begin{subarray}{c}\mu\in\operatorname*{Par};\\\
\mu/\lambda\text{ is a horizontal }k\text{-strip;}\\\ \mu\text{ has length
}\leq n\end{subarray}}s_{\mu}\left(x_{1},x_{2},\ldots,x_{n}\right).$ (44)

But we have the following equality of summation signs:

$\displaystyle\sum\limits_{\begin{subarray}{c}\mu\in\operatorname*{Par};\\\
\mu/\lambda\text{ is a horizontal }k\text{-strip;}\\\ \mu\text{ has length
}\leq n\end{subarray}}$
$\displaystyle=\sum\limits_{\begin{subarray}{c}\mu\in\operatorname*{Par};\\\
\mu\text{ has length }\leq n\text{;}\\\ \mu/\lambda\text{ is a horizontal
}k\text{-strip}\end{subarray}}=\sum\limits_{\begin{subarray}{c}\mu\in\operatorname*{Par}\left[n\right];\\\
\mu/\lambda\text{ is a horizontal }k\text{-strip}\end{subarray}}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{since the
partitions }\mu\in\operatorname*{Par}\text{ that have length }\leq n\\\
\text{are precisely the partitions
}\mu\in\operatorname*{Par}\left[n\right]\end{array}\right).$

Thus, we can rewrite (44) as

$\displaystyle h_{k}^{+}\cdot\overline{s}_{\lambda}$
$\displaystyle=\sum\limits_{\begin{subarray}{c}\mu\in\operatorname*{Par}\left[n\right];\\\
\mu/\lambda\text{ is a horizontal
}k\text{-strip}\end{subarray}}\underbrace{s_{\mu}\left(x_{1},x_{2},\ldots,x_{n}\right)}_{\begin{subarray}{c}=\overline{s}_{\mu}\\\
\text{(since Lemma \ref{lem.alt.sbar}}\\\ \text{(applied to }\mu\text{ instead
of }\lambda\text{)}\\\ \text{yields
}\overline{s}_{\mu}=s_{\mu}\left(x_{1},x_{2},\ldots,x_{n}\right)\text{)}\end{subarray}}$
$\displaystyle=\underbrace{\sum\limits_{\begin{subarray}{c}\mu\in\operatorname*{Par}\left[n\right];\\\
\mu/\lambda\text{ is a horizontal
}k\text{-strip}\end{subarray}}}_{\begin{subarray}{c}=\sum\limits_{\begin{subarray}{c}\mu\text{
is a snake;}\\\ \mu\rightharpoonup\lambda;\
\left|\mu\right|-\left|\lambda\right|=k\end{subarray}}\\\ \text{(by Claim
1)}\end{subarray}}\overline{s}_{\mu}=\sum\limits_{\begin{subarray}{c}\mu\text{
is a snake;}\\\ \mu\rightharpoonup\lambda;\
\left|\mu\right|-\left|\lambda\right|=k\end{subarray}}\overline{s}_{\mu}.$

This proves (43). Thus, Proposition 4.34 is proved under the assumption that
$\lambda$ is nonnegative. This completes Step 1.

Step 2: Let us now prove Proposition 4.34 in the general case.

The snake $\lambda$ may or may not be nonnegative. However, there exists some
integer $d$ such that the snake $\lambda+d$ is nonnegative101010Indeed, this
can be proved in the same way as it was proved during Step 2 of the proof of
Lemma 4.13 above.. Consider this $d$.

The map
$\left\\{\text{snakes}\right\\}\rightarrow\left\\{\text{snakes}\right\\},\
\mu\mapsto\mu+d$ is a bijection. (Indeed, its inverse is the map
$\left\\{\text{snakes}\right\\}\rightarrow\left\\{\text{snakes}\right\\},\
\mu\mapsto\mu-d$.) Moreover, every snake $\mu$ satisfies

$\displaystyle\underbrace{\left|\mu+d\right|}_{\begin{subarray}{c}=\left|\mu\right|+nd\\\
\text{(by Proposition \ref{prop.size-snake.easies}
{(b)})}\end{subarray}}-\underbrace{\left|\lambda+d\right|}_{\begin{subarray}{c}=\left|\lambda\right|+nd\\\
\text{(by Proposition \ref{prop.size-snake.easies} {(b)})}\end{subarray}}$
$\displaystyle=\left(\left|\mu\right|+nd\right)-\left(\left|\lambda\right|+nd\right)$
$\displaystyle=\left|\mu\right|-\left|\lambda\right|.$ (45)

For any snake $\mu$, we have the logical equivalence

$\left(\mu\rightharpoonup\lambda\right)\ \Longleftrightarrow\
\left(\mu+d\rightharpoonup\lambda+d\right)$

(by Proposition 4.32 (c)). In other words, for any snake $\mu$, we have the
logical equivalence

$\left(\mu+d\rightharpoonup\lambda+d\right)\ \Longleftrightarrow\
\left(\mu\rightharpoonup\lambda\right).$

Hence, we have the following equality of summation signs:

$\sum\limits_{\begin{subarray}{c}\mu\text{ is a snake;}\\\
\mu+d\rightharpoonup\lambda+d;\
\left|\mu+d\right|-\left|\lambda+d\right|=k\end{subarray}}=\sum\limits_{\begin{subarray}{c}\mu\text{
is a snake;}\\\ \mu\rightharpoonup\lambda;\
\left|\mu+d\right|-\left|\lambda+d\right|=k\end{subarray}}=\sum\limits_{\begin{subarray}{c}\mu\text{
is a snake;}\\\ \mu\rightharpoonup\lambda;\
\left|\mu\right|-\left|\lambda\right|=k\end{subarray}}$ (46)

(by (45)).

The snake $\lambda+d$ is nonnegative; thus, we can apply Proposition 4.34 to
$\lambda+d$ instead of $\lambda$ (because in Step 1, we have proved that
Proposition 4.34 holds in the particular case when $\lambda$ is nonnegative).
Thus we conclude that

$\displaystyle h_{k}^{+}\cdot\overline{s}_{\lambda+d}$
$\displaystyle=\sum\limits_{\begin{subarray}{c}\mu\text{ is a snake;}\\\
\mu\rightharpoonup\lambda+d;\
\left|\mu\right|-\left|\lambda+d\right|=k\end{subarray}}\overline{s}_{\mu}=\underbrace{\sum\limits_{\begin{subarray}{c}\mu\text{
is a snake;}\\\ \mu+d\rightharpoonup\lambda+d;\
\left|\mu+d\right|-\left|\lambda+d\right|=k\end{subarray}}}_{\begin{subarray}{c}=\sum\limits_{\begin{subarray}{c}\mu\text{
is a snake;}\\\ \mu\rightharpoonup\lambda;\
\left|\mu\right|-\left|\lambda\right|=k\end{subarray}}\\\ \text{(by
(\ref{pf.prop.alt.pieri1.s2.sumeq}))}\end{subarray}}\underbrace{\overline{s}_{\mu+d}}_{\begin{subarray}{c}=x_{\Pi}^{d}\overline{s}_{\mu}\\\
\text{(by Lemma \ref{lem.alt.smove},}\\\ \text{applied to }\mu\text{ instead
of }\lambda\text{)}\end{subarray}}$ $\displaystyle\ \ \ \ \ \ \ \ \ \
\left(\begin{array}[c]{c}\text{here, we have substituted }\mu+d\text{ for
}\mu\text{ in the sum,}\\\ \text{since the map
}\left\\{\text{snakes}\right\\}\rightarrow\left\\{\text{snakes}\right\\},\
\mu\mapsto\mu+d\\\ \text{is a bijection}\end{array}\right)$
$\displaystyle=\sum\limits_{\begin{subarray}{c}\mu\text{ is a snake;}\\\
\mu\rightharpoonup\lambda;\
\left|\mu\right|-\left|\lambda\right|=k\end{subarray}}x_{\Pi}^{d}\overline{s}_{\mu}=x_{\Pi}^{d}\sum\limits_{\begin{subarray}{c}\mu\text{
is a snake;}\\\ \mu\rightharpoonup\lambda;\
\left|\mu\right|-\left|\lambda\right|=k\end{subarray}}\overline{s}_{\mu}.$

Comparing this with

$h_{k}^{+}\cdot\underbrace{\overline{s}_{\lambda+d}}_{\begin{subarray}{c}=x_{\Pi}^{d}\overline{s}_{\lambda}\\\
\text{(by Lemma \ref{lem.alt.smove})}\end{subarray}}=h_{k}^{+}\cdot
x_{\Pi}^{d}\overline{s}_{\lambda},$

we obtain

$h_{k}^{+}\cdot
x_{\Pi}^{d}\overline{s}_{\lambda}=x_{\Pi}^{d}\sum\limits_{\begin{subarray}{c}\mu\text{
is a snake;}\\\ \mu\rightharpoonup\lambda;\
\left|\mu\right|-\left|\lambda\right|=k\end{subarray}}\overline{s}_{\mu}.$

We can divide both sides of this equality by $x_{\Pi}^{d}$ (since
$x_{\Pi}^{d}\in\mathcal{L}$ is invertible (because $x_{\Pi}\in\mathcal{L}$ is
invertible)), and thus obtain

$h_{k}^{+}\cdot\overline{s}_{\lambda}=\sum\limits_{\begin{subarray}{c}\mu\text{
is a snake;}\\\ \mu\rightharpoonup\lambda;\
\left|\mu\right|-\left|\lambda\right|=k\end{subarray}}\overline{s}_{\mu}.$

This proves Proposition 4.34. Thus, Step 2 is complete, and Proposition 4.34
is proven.

Using Lemma 4.24, we can “turn Proposition 4.34 upside down”, obtaining the
following analogous result for $h_{k}^{-}$ instead of $h_{k}^{+}$:

######  Proposition 4.36.

Let $\lambda$ be a snake. Let $k\in\mathbb{Z}$. Then,

$h_{k}^{-}\cdot\overline{s}_{\lambda}=\sum\limits_{\begin{subarray}{c}\mu\text{
is a snake;}\\\ \lambda\rightharpoonup\mu;\
\left|\lambda\right|-\left|\mu\right|=k\end{subarray}}\overline{s}_{\mu}.$
(47)

######  Proof 4.37 (Proof of Proposition 4.36.).

It is easy to see (from Definition 4.2 (d)) that
$\left(\lambda^{\vee}\right)^{\vee}=\lambda$. Likewise,
$\left(\mu^{\vee}\right)^{\vee}=\mu$ for any snake $\mu$. Hence, the map
$\left\\{\text{snakes}\right\\}\rightarrow\left\\{\text{snakes}\right\\},\
\mu\mapsto\mu^{\vee}$ (which is well-defined because of Proposition 4.4 (b))
is inverse to itself. Thus, this map is a bijection.

If $\mu$ is a snake, then

$\text{we have }\mu\rightharpoonup\lambda\text{ if and only if
}\lambda^{\vee}\rightharpoonup\mu^{\vee}$ (48)

(by Proposition 4.32 (b)). Moreover, if $\mu$ is a snake, then

$\displaystyle\underbrace{\left|\lambda^{\vee}\right|}_{\begin{subarray}{c}=-\left|\lambda\right|\\\
\text{(by Proposition \ref{prop.size-snake.easies}
{(c)})}\end{subarray}}-\underbrace{\left|\mu^{\vee}\right|}_{\begin{subarray}{c}=-\left|\mu\right|\\\
\text{(by Proposition \ref{prop.size-snake.easies} {(c)})}\end{subarray}}$
$\displaystyle=\left(-\left|\lambda\right|\right)-\left(-\left|\mu\right|\right)=\left|\mu\right|-\left|\lambda\right|.$
(49)

But comparing the definitions of $h_{k}^{-}$ and $h_{k}^{+}$ easily yields

$h_{k}^{-}=h_{k}^{+}\left(x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right).$

Also, Lemma 4.24 yields
$\overline{s}_{\lambda^{\vee}}=\overline{s}_{\lambda}\left(x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)$.
Multiplying these two equalities, we obtain

$\displaystyle h_{k}^{-}\cdot\overline{s}_{\lambda^{\vee}}$
$\displaystyle=h_{k}^{+}\left(x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)\cdot\overline{s}_{\lambda}\left(x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)$
$\displaystyle=\underbrace{\left(h_{k}^{+}\cdot\overline{s}_{\lambda}\right)}_{\begin{subarray}{c}=\sum\limits_{\begin{subarray}{c}\mu\text{
is a snake;}\\\ \mu\rightharpoonup\lambda;\
\left|\mu\right|-\left|\lambda\right|=k\end{subarray}}\overline{s}_{\mu}\\\
\text{(by Proposition
\ref{prop.alt.pieri1})}\end{subarray}}\left(x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)$
$\displaystyle=\underbrace{\sum\limits_{\begin{subarray}{c}\mu\text{ is a
snake;}\\\ \mu\rightharpoonup\lambda;\
\left|\mu\right|-\left|\lambda\right|=k\end{subarray}}}_{\begin{subarray}{c}=\sum\limits_{\begin{subarray}{c}\mu\text{
is a snake;}\\\ \lambda^{\vee}\rightharpoonup\mu^{\vee};\
\left|\mu\right|-\left|\lambda\right|=k\end{subarray}}\\\ \text{(by
(\ref{pf.prop.alt.pieri2.1}))}\end{subarray}}\overline{s}_{\mu}\left(x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)$
$\displaystyle=\sum\limits_{\begin{subarray}{c}\mu\text{ is a snake;}\\\
\lambda^{\vee}\rightharpoonup\mu^{\vee};\
\left|\mu\right|-\left|\lambda\right|=k\end{subarray}}\overline{s}_{\mu}\left(x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right).$

Comparing this with

$\displaystyle\sum\limits_{\begin{subarray}{c}\mu\text{ is a snake;}\\\
\lambda^{\vee}\rightharpoonup\mu;\
\left|\lambda^{\vee}\right|-\left|\mu\right|=k\end{subarray}}\overline{s}_{\mu}$
$\displaystyle=\underbrace{\sum\limits_{\begin{subarray}{c}\mu\text{ is a
snake;}\\\ \lambda^{\vee}\rightharpoonup\mu^{\vee};\
\left|\lambda^{\vee}\right|-\left|\mu^{\vee}\right|=k\end{subarray}}}_{\begin{subarray}{c}=\sum\limits_{\begin{subarray}{c}\mu\text{
is a snake;}\\\ \lambda^{\vee}\rightharpoonup\mu^{\vee};\
\left|\mu\right|-\left|\lambda\right|=k\end{subarray}}\\\ \text{(by
(\ref{pf.prop.alt.pieri2.2}))}\end{subarray}}\ \
\underbrace{\overline{s}_{\mu^{\vee}}}_{\begin{subarray}{c}=\overline{s}_{\mu}\left(x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)\\\
\text{(by Lemma \ref{lem.alt.inverses},}\\\ \text{applied to }\mu\text{
instead of }\lambda\text{)}\end{subarray}}$ $\displaystyle\ \ \ \ \ \ \ \ \ \
\left(\begin{array}[c]{c}\text{here, we have substituted }\mu^{\vee}\text{ for
}\mu\text{ in the sum, since}\\\ \text{the map
}\left\\{\text{snakes}\right\\}\rightarrow\left\\{\text{snakes}\right\\},\
\mu\mapsto\mu^{\vee}\text{ is a bijection}\end{array}\right)$
$\displaystyle=\sum\limits_{\begin{subarray}{c}\mu\text{ is a snake;}\\\
\lambda^{\vee}\rightharpoonup\mu^{\vee};\
\left|\mu\right|-\left|\lambda\right|=k\end{subarray}}\overline{s}_{\mu}\left(x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right),$

we obtain

$h_{k}^{-}\cdot\overline{s}_{\lambda^{\vee}}=\sum\limits_{\begin{subarray}{c}\mu\text{
is a snake;}\\\ \lambda^{\vee}\rightharpoonup\mu;\
\left|\lambda^{\vee}\right|-\left|\mu\right|=k\end{subarray}}\overline{s}_{\mu}.$

We have proved this equality for any snake $\lambda$. Thus, we can apply it to
$\lambda^{\vee}$ instead of $\lambda$ (since Proposition 4.4 (b) shows that
$\lambda^{\vee}$ is a snake). We thus obtain

$h_{k}^{-}\cdot\overline{s}_{\left(\lambda^{\vee}\right)^{\vee}}=\sum\limits_{\begin{subarray}{c}\mu\text{
is a snake;}\\\ \left(\lambda^{\vee}\right)^{\vee}\rightharpoonup\mu;\
\left|\left(\lambda^{\vee}\right)^{\vee}\right|-\left|\mu\right|=k\end{subarray}}\overline{s}_{\mu}.$

In view of $\left(\lambda^{\vee}\right)^{\vee}=\lambda$, this can be rewritten
as follows:

$h_{k}^{-}\cdot\overline{s}_{\lambda}=\sum\limits_{\begin{subarray}{c}\mu\text{
is a snake;}\\\ \lambda\rightharpoonup\mu;\
\left|\lambda\right|-\left|\mu\right|=k\end{subarray}}\overline{s}_{\mu}.$

This proves Proposition 4.36.

###  4.3 Computing $\overline{s}_{\alpha}$

######  Convention 4.2.

From now on, for the rest of Section 4, we assume that $n\geq 2$.

Our next goal is to obtain a simple formula for the Schur polynomial
$\overline{s}_{\alpha}\left(x_{1},x_{2},\ldots,x_{n}\right)$, where $\alpha$
is as in Theorem 2.2. The first step is the following definition:

######  Definition 4.38.

Let $a,b\in\mathbb{N}$. Then, $b\ominus a$ will denote the snake
$\left(b,0^{n-2},-a\right)$. (This is indeed a well-defined snake, since
$n\geq 2$ and since $b\geq 0\geq-a$.)

######  Proposition 4.39.

Let $a,b\in\mathbb{Z}$. Then,

$h_{a}^{-}h_{b}^{+}=\sum\limits_{k=0}^{\min\left\\{a,b\right\\}}\overline{s}_{\left(b-k\right)\ominus\left(a-k\right)}.$
(50)

######  Proof 4.40 (Proof of Proposition 4.39.).

We must prove the equality (50). If (at least) one of the integers $a$ and $b$
is negative, then this equality boils down to $0=0$ 111111Indeed, its left
hand side is $0$ in this case because every negative integer $k$ satisfies
$h_{k}^{-}=0$ and $h_{k}^{+}=0$; but its right hand side is also $0$ in this
case, because the negativity of $\min\left\\{a,b\right\\}$ causes the sum to
become empty.. Hence, for the rest of this proof, we WLOG assume that none of
the integers $a$ and $b$ is negative. Hence, $a,b\in\mathbb{N}$.

Note that each $k\in\left\\{0,1,\ldots,\min\left\\{a,b\right\\}\right\\}$
satisfies $b-k\in\mathbb{N}$ (since $k\leq\min\left\\{a,b\right\\}\leq b$) and
$a-k\in\mathbb{N}$ (likewise). Hence, the snakes
$\left(b-k\right)\ominus\left(a-k\right)$ on the right hand side of the
equality (50) are well-defined.

Lemma 4.27 (applied to $k=b$) yields that the partition $\left(b\right)$ is a
nonnegative snake (when regarded as the $n$-tuple
$\left(b,0,0,\ldots,0\right)$), and satisfies
$\overline{s}_{\left(b\right)}=h_{b}^{+}$.

Now, Proposition 4.36 (applied to $\lambda=\left(b\right)$ and $k=a$) yields

$h_{a}^{-}\cdot\overline{s}_{\left(b\right)}=\sum\limits_{\begin{subarray}{c}\mu\text{
is a snake;}\\\ \left(b\right)\rightharpoonup\mu;\
\left|\left(b\right)\right|-\left|\mu\right|=a\end{subarray}}\overline{s}_{\mu}.$

In view of $\overline{s}_{\left(b\right)}=h_{b}^{+}$ and
$\left|\left(b\right)\right|=b$, we can rewrite this as

$h_{a}^{-}\cdot h_{b}^{+}=\sum\limits_{\begin{subarray}{c}\mu\text{ is a
snake;}\\\ \left(b\right)\rightharpoonup\mu;\
b-\left|\mu\right|=a\end{subarray}}\overline{s}_{\mu}.$ (51)

Now, we claim the following:

> Claim 1: The snakes $\mu$ satisfying $\left(b\right)\rightharpoonup\mu$ and
> $b-\left|\mu\right|=a$ are precisely the snakes of the form
> $\left(b-k\right)\ominus\left(a-k\right)$ for
> $k\in\left\\{0,1,\ldots,\min\left\\{a,b\right\\}\right\\}$.

[Proof of Claim 1: The proof of this claim is straightforward (and can be
found elaborated in the detailed version [Grinbe20] of this paper).]

Now, Claim 1 shows that

$\sum\limits_{\begin{subarray}{c}\mu\text{ is a snake;}\\\
\left(b\right)\rightharpoonup\mu;\
b-\left|\mu\right|=a\end{subarray}}\overline{s}_{\mu}=\sum\limits_{k\in\left\\{0,1,\ldots,\min\left\\{a,b\right\\}\right\\}}\overline{s}_{\left(b-k\right)\ominus\left(a-k\right)}$

(indeed, it is clear that the snakes $\left(b-k\right)\ominus\left(a-k\right)$
for $k\in\left\\{0,1,\ldots,\min\left\\{a,b\right\\}\right\\}$ are all
distinct). Hence, (51) becomes

$h_{a}^{-}\cdot h_{b}^{+}=\sum\limits_{\begin{subarray}{c}\mu\text{ is a
snake;}\\\ \left(b\right)\rightharpoonup\mu;\
b-\left|\mu\right|=a\end{subarray}}\overline{s}_{\mu}=\sum\limits_{k\in\left\\{0,1,\ldots,\min\left\\{a,b\right\\}\right\\}}\overline{s}_{\left(b-k\right)\ominus\left(a-k\right)}=\sum\limits_{k=0}^{\min\left\\{a,b\right\\}}\overline{s}_{\left(b-k\right)\ominus\left(a-k\right)}.$

This proves Proposition 4.39.

######  Proposition 4.41.

Let $a,b\in\mathbb{N}$. Then,

$\overline{s}_{b\ominus a}=h_{a}^{-}h_{b}^{+}-h_{a-1}^{-}h_{b-1}^{+}.$

(Recall that every negative integer $k$ satisfies $h_{k}^{-}=0$ and
$h_{k}^{+}=0$.)

######  Proof 4.42 (Proof of Proposition 4.41.).

We have $\min\left\\{a,b\right\\}\in\mathbb{N}$ (since $a,b\in\mathbb{N}$), so
that $\min\left\\{a,b\right\\}\geq 0$.

Proposition 4.39 yields

$\displaystyle h_{a}^{-}h_{b}^{+}$
$\displaystyle=\sum\limits_{k=0}^{\min\left\\{a,b\right\\}}\overline{s}_{\left(b-k\right)\ominus\left(a-k\right)}=\underbrace{\overline{s}_{\left(b-0\right)\ominus\left(a-0\right)}}_{=\overline{s}_{b\ominus
a}}+\sum\limits_{k=1}^{\min\left\\{a,b\right\\}}\overline{s}_{\left(b-k\right)\ominus\left(a-k\right)}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{here, we have
split off the addend for }k=0\text{ from the sum,}\\\ \text{since
}\min\left\\{a,b\right\\}\geq 0\end{array}\right)$ (54)
$\displaystyle=\overline{s}_{b\ominus
a}+\sum\limits_{k=1}^{\min\left\\{a,b\right\\}}\overline{s}_{\left(b-k\right)\ominus\left(a-k\right)}.$
(55)

Proposition 4.39 (applied to $a-1$ and $b-1$ instead of $a$ and $b$) yields

$\displaystyle h_{a-1}^{-}h_{b-1}^{+}$
$\displaystyle=\sum\limits_{k=0}^{\min\left\\{a-1,b-1\right\\}}\overline{s}_{\left(\left(b-1\right)-k\right)\ominus\left(\left(a-1\right)-k\right)}=\sum\limits_{k=0}^{\min\left\\{a,b\right\\}-1}\overline{s}_{\left(\left(b-1\right)-k\right)\ominus\left(\left(a-1\right)-k\right)}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\text{since
}\min\left\\{a-1,b-1\right\\}=\min\left\\{a,b\right\\}-1\right)$
$\displaystyle=\sum\limits_{k=1}^{\min\left\\{a,b\right\\}}\overline{s}_{\left(\left(b-1\right)-\left(k-1\right)\right)\ominus\left(\left(a-1\right)-\left(k-1\right)\right)}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\text{here, we have substituted
}k-1\text{ for }k\text{ in the sum}\right)$
$\displaystyle=\sum\limits_{k=1}^{\min\left\\{a,b\right\\}}\overline{s}_{\left(b-k\right)\ominus\left(a-k\right)}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\text{since
}\left(b-1\right)-\left(k-1\right)=b-k\text{ and
}\left(a-1\right)-\left(k-1\right)=a-k\right).$

Subtracting this equality from (55), we obtain

$h_{a}^{-}h_{b}^{+}-h_{a-1}^{-}h_{b-1}^{+}=\left(\overline{s}_{b\ominus
a}+\sum\limits_{k=1}^{\min\left\\{a,b\right\\}}\overline{s}_{\left(b-k\right)\ominus\left(a-k\right)}\right)-\sum\limits_{k=1}^{\min\left\\{a,b\right\\}}\overline{s}_{\left(b-k\right)\ominus\left(a-k\right)}=\overline{s}_{b\ominus
a}.$

This proves Proposition 4.41.

######  Corollary 4.43.

Let $a,b\in\mathbb{N}$. Define the partition
$\alpha=\left(a+b,a^{n-2}\right)$. Then, $\alpha$ is a nonnegative snake and
satisfies

$\overline{s}_{\alpha}=x_{\Pi}^{a}\cdot\left(h_{a}^{-}h_{b}^{+}-h_{a-1}^{-}h_{b-1}^{+}\right).$

######  Proof 4.44 (Proof of Corollary 4.43.).

From $a,b\in\mathbb{N}$, we obtain $a+\underbrace{b}_{\geq 0}\geq a\geq 0$.
Hence, $\alpha$ is a partition of length $\leq n$ (since $n-1\leq n$). In
other words, $\alpha$ is a nonnegative snake.

It is easy to see that $\alpha=\left(b\ominus a\right)+a$ (since
$\alpha=\left(a+b,a^{n-2}\right)=\left(a+b,\underbrace{a,a,\ldots,a}_{n-2\text{
times}},0\right)$ and $b\ominus
a=\left(b,0^{n-2},-a\right)=\left(b,\underbrace{0,0,\ldots,0}_{n-2\text{
times}},-a\right)$). Hence,

$\displaystyle\overline{s}_{\alpha}$
$\displaystyle=\overline{s}_{\left(b\ominus
a\right)+a}=x_{\Pi}^{a}\overline{s}_{b\ominus a}\ \ \ \ \ \ \ \ \ \
\left(\text{by Lemma \ref{lem.alt.smove}, applied to }\lambda=b\ominus a\text{
and }d=a\right)$
$\displaystyle=x_{\Pi}^{a}\cdot\left(h_{a}^{-}h_{b}^{+}-h_{a-1}^{-}h_{b-1}^{+}\right)\
\ \ \ \ \ \ \ \ \ \left(\text{by Proposition \ref{prop.ominus.s}}\right).$

This proves Corollary 4.43.

We notice that the expression $h_{a}^{-}h_{b}^{+}-h_{a-1}^{-}h_{b-1}^{+}$ in
Proposition 4.41 can be rewritten as
$\det\left(\begin{array}[c]{cc}h_{a}^{-}&h_{a-1}^{-}\\\
h_{b-1}^{+}&h_{b}^{+}\end{array}\right)$. This suggests a way to generalize
Proposition 4.41 (as well as the first Jacobi–Trudi formula [GriRei20,
(2.4.16)]). Such a generalization indeed exists, and has been proved by Koike
as well as by Hamel and King; see Proposition 5.1 below.

###  4.4 The sets $R_{\mu,a,b}\left(\gamma\right)$ and a formula for
$h_{a}^{-}h_{b}^{+}\overline{s}_{\mu}$

We shall next aim for a formula for $h_{a}^{-}h_{b}^{+}\overline{s}_{\mu}$
(for a snake $\mu$ and integers $a,b\in\mathbb{Z}$), which will be obtained in
a straightforward way by applying Propositions 4.34 and 4.36. We will need the
following definition:

######  Definition 4.45.

Let $\mu,\gamma\in\mathbb{Z}^{n}$ and $a,b\in\mathbb{Z}$. Then,
$R_{\mu,a,b}\left(\gamma\right)$ shall denote the set of all snakes $\nu$
satisfying the four conditions

$\mu\rightharpoonup\nu\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \
\left|\mu\right|-\left|\nu\right|=a\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \
\ \ \ \gamma\rightharpoonup\nu\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \
\ \left|\gamma\right|-\left|\nu\right|=b.$

######  Lemma 4.46.

Let $\mu,\gamma\in\mathbb{Z}^{n}$ and $a,b\in\mathbb{Z}$. Assume that $\gamma$
is not a snake. Then, $\left|R_{\mu,a,b}\left(\gamma\right)\right|=0$.

######  Proof 4.47 (Proof of Lemma 4.46.).

Let $\nu\in R_{\mu,a,b}\left(\gamma\right)$. We shall obtain a contradiction.

Indeed, we have $\nu\in R_{\mu,a,b}\left(\gamma\right)$. In other words, $\nu$
is a snake satisfying the four conditions

$\mu\rightharpoonup\nu\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \
\left|\mu\right|-\left|\nu\right|=a\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \
\ \ \ \gamma\rightharpoonup\nu\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \
\ \left|\gamma\right|-\left|\nu\right|=b$

(by the definition of $R_{\mu,a,b}\left(\gamma\right)$). Thus, in particular,
we have $\gamma\rightharpoonup\nu$. Hence, Proposition 4.32 (a) (applied to
$\gamma$ and $\nu$ instead of $\mu$ and $\lambda$) yields that both $\nu$ and
$\gamma$ are snakes. Hence, $\gamma$ is a snake. This contradicts the fact
that $\gamma$ is not a snake.

Now, forget that we fixed $\nu$. We thus have obtained a contradiction for
each $\nu\in R_{\mu,a,b}\left(\gamma\right)$. Hence, there exists no $\nu\in
R_{\mu,a,b}\left(\gamma\right)$. In other words, the set
$R_{\mu,a,b}\left(\gamma\right)$ is empty. Thus,
$\left|R_{\mu,a,b}\left(\gamma\right)\right|=0$. This proves Lemma 4.46.

######  Lemma 4.48.

Let $\mu$ be a snake. Let $a,b\in\mathbb{Z}$. Then,

$h_{a}^{-}h_{b}^{+}\overline{s}_{\mu}=\sum\limits_{\gamma\text{ is a
snake}}\left|R_{\mu,a,b}\left(\gamma\right)\right|\overline{s}_{\gamma}.$

######  Proof 4.49 (Proof of Lemma 4.48.).

Proposition 4.36 (with the letters $\lambda$, $k$ and $\mu$ renamed as $\mu$,
$a$ and $\nu$) says that

$h_{a}^{-}\cdot\overline{s}_{\mu}=\sum\limits_{\begin{subarray}{c}\nu\text{ is
a snake;}\\\ \mu\rightharpoonup\nu;\
\left|\mu\right|-\left|\nu\right|=a\end{subarray}}\overline{s}_{\nu}.$ (56)

Proposition 4.34 (with the letters $\lambda$, $k$ and $\mu$ renamed as $\nu$,
$b$ and $\gamma$) says that

$h_{b}^{+}\cdot\overline{s}_{\nu}=\sum\limits_{\begin{subarray}{c}\gamma\text{
is a snake;}\\\ \gamma\rightharpoonup\nu;\
\left|\gamma\right|-\left|\nu\right|=b\end{subarray}}\overline{s}_{\gamma}$
(57)

for each snake $\nu$.

Now,

$\displaystyle h_{a}^{-}h_{b}^{+}\overline{s}_{\mu}$
$\displaystyle=h_{b}^{+}\cdot\underbrace{h_{a}^{-}\cdot\overline{s}_{\mu}}_{\begin{subarray}{c}=\sum\limits_{\begin{subarray}{c}\nu\text{
is a snake;}\\\ \mu\rightharpoonup\nu;\
\left|\mu\right|-\left|\nu\right|=a\end{subarray}}\overline{s}_{\nu}\\\
\text{(by
(\ref{pf.lem.Rmab.formula.1}))}\end{subarray}}=h_{b}^{+}\cdot\sum\limits_{\begin{subarray}{c}\nu\text{
is a snake;}\\\ \mu\rightharpoonup\nu;\
\left|\mu\right|-\left|\nu\right|=a\end{subarray}}\overline{s}_{\nu}$
$\displaystyle=\sum\limits_{\begin{subarray}{c}\nu\text{ is a snake;}\\\
\mu\rightharpoonup\nu;\ \left|\mu\right|-\left|\nu\right|=a\end{subarray}}\ \
\
\underbrace{h_{b}^{+}\cdot\overline{s}_{\nu}}_{\begin{subarray}{c}=\sum\limits_{\begin{subarray}{c}\gamma\text{
is a snake;}\\\ \gamma\rightharpoonup\nu;\
\left|\gamma\right|-\left|\nu\right|=b\end{subarray}}\overline{s}_{\gamma}\\\
\text{(by
(\ref{pf.lem.Rmab.formula.2}))}\end{subarray}}=\underbrace{\sum\limits_{\begin{subarray}{c}\nu\text{
is a snake;}\\\ \mu\rightharpoonup\nu;\
\left|\mu\right|-\left|\nu\right|=a\end{subarray}}\ \ \
\sum\limits_{\begin{subarray}{c}\gamma\text{ is a snake;}\\\
\gamma\rightharpoonup\nu;\
\left|\gamma\right|-\left|\nu\right|=b\end{subarray}}}_{=\sum\limits_{\gamma\text{
is a snake}}\ \ \ \sum\limits_{\begin{subarray}{c}\nu\text{ is a snake;}\\\
\mu\rightharpoonup\nu;\ \left|\mu\right|-\left|\nu\right|=a;\\\
\gamma\rightharpoonup\nu;\
\left|\gamma\right|-\left|\nu\right|=b\end{subarray}}}\overline{s}_{\gamma}$
$\displaystyle=\sum\limits_{\gamma\text{ is a snake}}\ \ \
\underbrace{\sum\limits_{\begin{subarray}{c}\nu\text{ is a snake;}\\\
\mu\rightharpoonup\nu;\ \left|\mu\right|-\left|\nu\right|=a;\\\
\gamma\rightharpoonup\nu;\
\left|\gamma\right|-\left|\nu\right|=b\end{subarray}}}_{\begin{subarray}{c}=\sum\limits_{\nu\in
R_{\mu,a,b}\left(\gamma\right)}\\\ \text{(by the definition of
}R_{\mu,a,b}\left(\gamma\right)\text{)}\end{subarray}}\overline{s}_{\gamma}=\sum\limits_{\gamma\text{
is a snake}}\ \ \ \underbrace{\sum\limits_{\nu\in
R_{\mu,a,b}\left(\gamma\right)}\overline{s}_{\gamma}}_{=\left|R_{\mu,a,b}\left(\gamma\right)\right|\overline{s}_{\gamma}}$
$\displaystyle=\sum\limits_{\gamma\text{ is a
snake}}\left|R_{\mu,a,b}\left(\gamma\right)\right|\overline{s}_{\gamma}.$

This proves Lemma 4.48.

######  Corollary 4.50.

Let $\mu\in\operatorname*{Par}\left[n\right]$. Let $a,b\in\mathbb{N}$. Define
the partition $\alpha=\left(a+b,a^{n-2}\right)$. Then, every
$\lambda\in\mathbb{Z}^{n}$ satisfies

$c_{\alpha,\mu}^{\lambda}=\left|R_{\mu,a,b}\left(\lambda-a\right)\right|-\left|R_{\mu,a-1,b-1}\left(\lambda-a\right)\right|.$
(58)

Here, we understand $c_{\alpha,\mu}^{\lambda}$ to mean $0$ if $\lambda$ is not
a partition (i.e., if $\lambda$ is not a nonnegative snake).

######  Proof 4.51 (Proof of Corollary 4.50.).

Corollary 4.43 shows that $\alpha$ is a nonnegative snake and satisfies

$\overline{s}_{\alpha}=x_{\Pi}^{a}\cdot\left(h_{a}^{-}h_{b}^{+}-h_{a-1}^{-}h_{b-1}^{+}\right).$
(59)

Every snake $\gamma$ satisfies

$\overline{s}_{\gamma+a}=x_{\Pi}^{a}\overline{s}_{\gamma}$ (60)

(by Lemma 4.18, applied to $\gamma$ and $a$ instead of $\lambda$ and $d$).

We have $\alpha\in\operatorname*{Par}\left[n\right]$ (since $\alpha$ is a
nonnegative snake). Hence, Lemma 4.20 (applied to $\alpha$ and $\mu$ instead
of $\mu$ and $\nu$) yields

$\displaystyle\overline{s}_{\alpha}\overline{s}_{\mu}$
$\displaystyle=\underbrace{\sum\limits_{\lambda\in\operatorname*{Par}\left[n\right]}}_{\begin{subarray}{c}=\sum\limits_{\begin{subarray}{c}\lambda\text{
is a snake;}\\\ \lambda\text{ is nonnegative}\end{subarray}}\\\ \text{(since
the partitions }\lambda\in\operatorname*{Par}\left[n\right]\text{ are}\\\
\text{precisely the nonnegative
snakes)}\end{subarray}}c_{\alpha,\mu}^{\lambda}\overline{s}_{\lambda}=\sum\limits_{\begin{subarray}{c}\lambda\text{
is a snake;}\\\ \lambda\text{ is
nonnegative}\end{subarray}}c_{\alpha,\mu}^{\lambda}\overline{s}_{\lambda}$
$\displaystyle=\sum\limits_{\lambda\text{ is a
snake}}c_{\alpha,\mu}^{\lambda}\overline{s}_{\lambda}-\sum\limits_{\begin{subarray}{c}\lambda\text{
is a snake;}\\\ \lambda\text{ is not
nonnegative}\end{subarray}}\underbrace{c_{\alpha,\mu}^{\lambda}}_{\begin{subarray}{c}=0\\\
\text{(since we understand }c_{\alpha,\mu}^{\lambda}\\\ \text{to mean }0\text{
if }\lambda\text{ is not}\\\ \text{a nonnegative
snake)}\end{subarray}}\overline{s}_{\lambda}$
$\displaystyle=\sum\limits_{\lambda\text{ is a
snake}}c_{\alpha,\mu}^{\lambda}\overline{s}_{\lambda}-\underbrace{\sum\limits_{\begin{subarray}{c}\lambda\text{
is a snake;}\\\ \lambda\text{ is not
nonnegative}\end{subarray}}0\overline{s}_{\lambda}}_{=0}=\sum\limits_{\lambda\text{
is a snake}}c_{\alpha,\mu}^{\lambda}\overline{s}_{\lambda}.$

Hence,

$\displaystyle\sum\limits_{\lambda\text{ is a
snake}}c_{\alpha,\mu}^{\lambda}\overline{s}_{\lambda}$
$\displaystyle=\overline{s}_{\alpha}\overline{s}_{\mu}=x_{\Pi}^{a}\cdot\left(h_{a}^{-}h_{b}^{+}-h_{a-1}^{-}h_{b-1}^{+}\right)\overline{s}_{\mu}\
\ \ \ \ \ \ \ \ \ \left(\text{by (\ref{pf.cor.Rmab.lr.1})}\right)$
$\displaystyle=x_{\Pi}^{a}\cdot\underbrace{h_{a}^{-}h_{b}^{+}\overline{s}_{\mu}}_{\begin{subarray}{c}=\sum\limits_{\gamma\text{
is a
snake}}\left|R_{\mu,a,b}\left(\gamma\right)\right|\overline{s}_{\gamma}\\\
\text{(by Lemma
\ref{lem.Rmab.formula})}\end{subarray}}-x_{\Pi}^{a}\cdot\underbrace{h_{a-1}^{-}h_{b-1}^{+}\overline{s}_{\mu}}_{\begin{subarray}{c}=\sum\limits_{\gamma\text{
is a
snake}}\left|R_{\mu,a-1,b-1}\left(\gamma\right)\right|\overline{s}_{\gamma}\\\
\text{(by Lemma \ref{lem.Rmab.formula},}\\\ \text{applied to }a-1\text{ and
}b-1\\\ \text{instead of }a\text{ and }b\text{)}\end{subarray}}$
$\displaystyle=x_{\Pi}^{a}\cdot\sum\limits_{\gamma\text{ is a
snake}}\left|R_{\mu,a,b}\left(\gamma\right)\right|\overline{s}_{\gamma}-x_{\Pi}^{a}\cdot\sum\limits_{\gamma\text{
is a
snake}}\left|R_{\mu,a-1,b-1}\left(\gamma\right)\right|\overline{s}_{\gamma}$
$\displaystyle=\sum\limits_{\gamma\text{ is a
snake}}\left(\left|R_{\mu,a,b}\left(\gamma\right)\right|-\left|R_{\mu,a-1,b-1}\left(\gamma\right)\right|\right)\underbrace{x_{\Pi}^{a}\overline{s}_{\gamma}}_{\begin{subarray}{c}=\overline{s}_{\gamma+a}\\\
\text{(by (\ref{pf.cor.Rmab.lr.2}))}\end{subarray}}$
$\displaystyle=\sum\limits_{\gamma\text{ is a
snake}}\left(\left|R_{\mu,a,b}\left(\gamma\right)\right|-\left|R_{\mu,a-1,b-1}\left(\gamma\right)\right|\right)\overline{s}_{\gamma+a}$
$\displaystyle=\sum\limits_{\lambda\text{ is a
snake}}\left(\left|R_{\mu,a,b}\left(\lambda-a\right)\right|-\left|R_{\mu,a-1,b-1}\left(\lambda-a\right)\right|\right)\underbrace{\overline{s}_{\left(\lambda-a\right)+a}}_{=\overline{s}_{\lambda}}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{here, we have
substituted }\lambda-a\text{ for }\gamma\text{ in the sum,}\\\ \text{since the
map }\left\\{\text{snakes}\right\\}\rightarrow\left\\{\text{snakes}\right\\},\
\lambda\mapsto\lambda-a\\\ \text{is a bijection}\end{array}\right)$
$\displaystyle=\sum\limits_{\lambda\text{ is a
snake}}\left(\left|R_{\mu,a,b}\left(\lambda-a\right)\right|-\left|R_{\mu,a-1,b-1}\left(\lambda-a\right)\right|\right)\overline{s}_{\lambda}.$

We can compare coefficients on both sides of this equality (since Lemma 4.22
shows that the family
$\left(\overline{s}_{\lambda}\right)_{\lambda\in\left\\{\text{snakes}\right\\}}$
of elements of $\mathcal{L}$ is $\mathbf{k}$-linearly independent), and thus
conclude that

$c_{\alpha,\mu}^{\lambda}=\left|R_{\mu,a,b}\left(\lambda-a\right)\right|-\left|R_{\mu,a-1,b-1}\left(\lambda-a\right)\right|\
\ \ \ \ \ \ \ \ \ \text{for every snake }\lambda\text{.}$

This proves (58) in the case when $\lambda$ is a snake.

However, it is easy to see that (58) also holds in the case when $\lambda$ is
not a snake121212Proof. Let $\lambda\in\mathbb{Z}^{n}$ be such that $\lambda$
is not a snake. We must show that (58) holds for this $\lambda$. We have
assumed that $\lambda$ is not a snake. Hence, $\lambda-a$ is not a snake
(because it is easy to see from Definition 4.2 that $\lambda$ is a snake if
and only if $\lambda-a$ is a snake). Thus, Lemma 4.46 (applied to
$\gamma=\lambda-a$) yields $\left|R_{\mu,a,b}\left(\lambda-a\right)\right|=0$.
Also, Lemma 4.46 (applied to $\lambda-a$, $a-1$ and $b-1$ instead of $\gamma$,
$a$ and $b$) yields $\left|R_{\mu,a-1,b-1}\left(\lambda-a\right)\right|=0$. On
the other hand, $\lambda$ is not a snake, and thus not a nonnegative snake.
Hence, $c_{\alpha,\mu}^{\lambda}=0$ (since we have defined
$c_{\alpha,\mu}^{\lambda}$ to be $0$ if $\lambda$ is not a nonnegative snake).
Comparing this with
$\underbrace{\left|R_{\mu,a,b}\left(\lambda-a\right)\right|}_{=0}-\underbrace{\left|R_{\mu,a-1,b-1}\left(\lambda-a\right)\right|}_{=0}=0$,
we obtain
$c_{\alpha,\mu}^{\lambda}=\left|R_{\mu,a,b}\left(\lambda-a\right)\right|-\left|R_{\mu,a-1,b-1}\left(\lambda-a\right)\right|$.
In other words, (58) holds. Thus, we have shown that (58) holds in the case
when $\lambda$ is not a snake.. Thus, (58) always holds. This proves Corollary
4.50.

###  4.5 The map $\mathbf{f}_{\mu}$

######  Convention 4.3.

For the whole Subsection 4.5, we shall use Convention 3.3 (not only for
$n$-tuples $a\in\mathbb{K}^{n}$, but for any $n$-tuples $a$). This convention
does not conflict with Convention 4.1, because both conventions define
$\gamma_{i}$ in the same way when $\gamma$ is an $n$-tuple and
$i\in\left\\{1,2,\ldots,n\right\\}$ (whereas the latter convention does not
define $\gamma_{i}$ for any other values of $i$).

Convention 3.3 does conflict with our old convention (from Section 1) to
identify partitions with finite tuples: Indeed, if we let $\gamma$ be the
$n$-tuple $\left(\underbrace{1,1,\ldots,1}_{n\text{ times}}\right)$, then
Convention 3.3 yields $\gamma_{n+1}=\gamma_{1}=1$ when we regard $\gamma$ as
an $n$-tuple, but we get $\gamma_{n+1}=0$ if we regard $\gamma$ as a
partition. We shall resolve this conflict by agreeing not to identify
partitions with finite tuples in Subsection 4.5. (Thus, in particular, we will
not identify a nonnegative snake
$\left(\mu_{1},\mu_{2},\ldots,\mu_{n}\right)\in\mathbb{Z}^{n}$ with its
corresponding partition
$\left(\mu_{1},\mu_{2},\ldots,\mu_{n},0,0,0,\ldots\right)\in\operatorname*{Par}\left[n\right]$.)

Let us now apply the results of Section 3. The abelian group
$\left(\mathbb{Z},+,0\right)$ of integers is totally ordered (in the usual
way). Thus, Example 3.3 (applied to
$\left(\mathbb{A},\ast,e\right)=\left(\mathbb{Z},+,0\right)$) shows that there
is a semifield $\left(\mathbb{Z},\min,+,0\right)$ (that is, a semifield with
ground set $\mathbb{Z}$, addition $\min$, multiplication $+$ and unity $0$),
called the _min tropical semifield_ of $\left(\mathbb{Z},+,0\right)$. We have
the following little dictionary between various operations on this semifield
$\left(\mathbb{Z},\min,+,0\right)$ and familiar operations on integers:

  * •

The addition operation of the semifield $\left(\mathbb{Z},\min,+,0\right)$ is
the binary operation $\min$ on $\mathbb{Z}$. That is, for any
$a,b\in\mathbb{Z}$, the sum $a+b$ understood with respect to the semifield
$\left(\mathbb{Z},\min,+,0\right)$ is precisely the integer
$\min\left\\{a,b\right\\}$.

  * •

Thus, (nonempty) finite sums in the semifield
$\left(\mathbb{Z},\min,+,0\right)$ are minima of finite sets of integers. That
is, if $r\in\mathbb{N}$, and if $a_{0},a_{1},\ldots,a_{r}$ are any $r+1$
integers, then the sum $\sum\limits_{k=0}^{r}a_{k}$ understood with respect to
the semifield $\left(\mathbb{Z},\min,+,0\right)$ is
$\min\left\\{a_{0},a_{1},\ldots,a_{r}\right\\}=\min\left\\{a_{k}\ \mid\
k\in\left\\{0,1,\ldots,r\right\\}\right\\}$.

  * •

Furthermore, the multiplication operation of the semifield
$\left(\mathbb{Z},\min,+,0\right)$ is the addition $+$ of integers. That is,
for any $a,b\in\mathbb{Z}$, the product $ab$ understood with respect to the
semifield $\left(\mathbb{Z},\min,+,0\right)$ is precisely the sum $a+b$
understood with respect to the integer ring $\mathbb{Z}$. Meanwhile, the unity
of the semifield $\left(\mathbb{Z},\min,+,0\right)$ is the integer $0$.

  * •

Thus, the division operation of the semifield
$\left(\mathbb{Z},\min,+,0\right)$ is the subtraction $-$ of integers. That
is, for any $a,b\in\mathbb{Z}$, the quotient $\dfrac{a}{b}$ understood with
respect to the semifield $\left(\mathbb{Z},\min,+,0\right)$ is precisely the
difference $a-b$ understood with respect to the integer ring $\mathbb{Z}$.

  * •

For the same reason, squaring an element of the semifield
$\left(\mathbb{Z},\min,+,0\right)$ is tantamount to doubling it as an integer.
That is, for any $a\in\mathbb{Z}$, the square $a^{2}$ understood with respect
to the semifield $\left(\mathbb{Z},\min,+,0\right)$ is the product $2a$
understood with respect to the integer ring $\mathbb{Z}$.

  * •

For the same reason, taking reciprocals in the semifield
$\left(\mathbb{Z},\min,+,0\right)$ is tantamount to negation of integers. That
is, for any $a\in\mathbb{Z}$, the reciprocal $\dfrac{1}{a}$ understood with
respect to the semifield $\left(\mathbb{Z},\min,+,0\right)$ is the integer
$-a$ understood with respect to the integer ring $\mathbb{Z}$.

  * •

For the same reason, finite products in the semifield
$\left(\mathbb{Z},\min,+,0\right)$ are sums of integers. That is, if
$r\in\mathbb{N}$, and if $a_{1},a_{2},\ldots,a_{r}$ are any $r$ integers, then
the product $\prod\limits_{k=1}^{r}a_{k}$ understood with respect to the
semifield $\left(\mathbb{Z},\min,+,0\right)$ is the sum
$\sum\limits_{k=1}^{r}a_{k}$ understood with respect to the integer ring
$\mathbb{Z}$.

Thus, applying Definition 3.5 to $\mathbb{K}=\left(\mathbb{Z},\min,+,0\right)$
(and renaming everything131313Namely, we are  • renaming the (fixed) $n$-tuple
$u$ as $\mu$; • renaming the (variable) $n$-tuple $x$ as $\gamma$ (in order to
avoid a clash with the variables $x_{1},x_{2},\ldots,x_{n}$); • renaming the
elements $t_{r,j}$ as $\tau_{r,j}$; • renaming the $n$-tuple $y$ as $\eta$. ),
we obtain the following:

######  Definition 4.52.

Fix any $n$-tuple $\mu\in\mathbb{Z}^{n}$.

We define a map $\mathbf{f}_{\mu}:\mathbb{Z}^{n}\rightarrow\mathbb{Z}^{n}$ as
follows:

Let $\gamma\in\mathbb{Z}^{n}$ be an $n$-tuple. For each $j\in\mathbb{Z}$ and
$r\in\mathbb{N}$, define an element $\tau_{r,j}\in\mathbb{Z}$ by

$\displaystyle\tau_{r,j}$
$\displaystyle=\min\left\\{\underbrace{\gamma_{j+1}+\gamma_{j+2}+\cdots+\gamma_{j+k}}_{=\sum\limits_{i=1}^{k}\gamma_{j+i}}+\underbrace{\mu_{j+k+1}+\mu_{j+k+2}+\cdots+\mu_{j+r}}_{=\sum\limits_{i=k+1}^{r}\mu_{j+i}}\right.$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.\ \mid\
k\in\left\\{0,1,\ldots,r\right\\}\right\\}.$

Define $\eta\in\mathbb{Z}^{n}$ by setting

$\eta_{i}=\mu_{i}+\left(\mu_{i-1}+\tau_{n-1,i-1}\right)-\left(\gamma_{i+1}+\tau_{n-1,i+1}\right)\
\ \ \ \ \ \ \ \ \ \text{for each }i\in\left\\{1,2,\ldots,n\right\\}.$

Set $\mathbf{f}_{\mu}\left(\gamma\right)=\eta$.

Applying Theorem 3.8 to $\mathbb{K}=\left(\mathbb{Z},\min,+,0\right)$ (and
renaming everything141414Namely, we are  • renaming the (fixed) $n$-tuple $u$
as $\mu$; • renaming the (variable) $n$-tuple $x$ as $\gamma$; • renaming the
$n$-tuple $y$ as $\eta$. , and using our above dictionary), we thus obtain the
following:

######  Theorem 4.53.

Fix any $n$-tuple $\mu\in\mathbb{Z}^{n}$.

  1. (a)

The map $\mathbf{f}_{\mu}$ is an involution (i.e., we have
$\mathbf{f}_{\mu}\circ\mathbf{f}_{\mu}=\operatorname*{id}$).

  2. (b)

Let $\gamma\in\mathbb{Z}^{n}$ and $\eta\in\mathbb{Z}^{n}$ be such that
$\eta=\mathbf{f}_{\mu}\left(\gamma\right)$. Then,

$\left(\eta_{1}+\eta_{2}+\cdots+\eta_{n}\right)+\left(\gamma_{1}+\gamma_{2}+\cdots+\gamma_{n}\right)=2\left(\mu_{1}+\mu_{2}+\cdots+\mu_{n}\right).$

  3. (c)

Let $\gamma\in\mathbb{Z}^{n}$ and $\eta\in\mathbb{Z}^{n}$ be such that
$\eta=\mathbf{f}_{\mu}\left(\gamma\right)$. Then,

$\min\left\\{\mu_{i},\gamma_{i}\right\\}+\min\left\\{-\mu_{i+1},-\gamma_{i+1}\right\\}=\min\left\\{\mu_{i},\eta_{i}\right\\}+\min\left\\{-\mu_{i+1},-\eta_{i+1}\right\\}$

for each $i\in\mathbb{Z}$.

  4. (d)

Let $\gamma\in\mathbb{Z}^{n}$ and $\eta\in\mathbb{Z}^{n}$ be such that
$\eta=\mathbf{f}_{\mu}\left(\gamma\right)$. Then,

$\sum\limits_{i=1}^{n}\left(\min\left\\{\mu_{i},\gamma_{i}\right\\}-\gamma_{i}\right)=\sum\limits_{i=1}^{n}\left(\min\left\\{\mu_{i},\eta_{i}\right\\}-\mu_{i}\right).$

We obtain the following corollaries from Theorem 4.53:

######  Corollary 4.54.

Fix any $n$-tuple $\mu\in\mathbb{Z}^{n}$. Let $\gamma\in\mathbb{Z}^{n}$ and
$\eta\in\mathbb{Z}^{n}$ be such that
$\eta=\mathbf{f}_{\mu}\left(\gamma\right)$. Then:

  1. (a)

We have
$\left|\eta\right|-\left|\mu\right|=\left|\mu\right|-\left|\gamma\right|$.

  2. (b)

We have

$\min\left\\{\mu_{i},\eta_{i}\right\\}-\min\left\\{\mu_{i},\gamma_{i}\right\\}=\max\left\\{\mu_{i+1},\eta_{i+1}\right\\}-\max\left\\{\mu_{i+1},\gamma_{i+1}\right\\}$

for each $i\in\left\\{1,2,\ldots,n-1\right\\}$.

  3. (c)

We have

$\sum\limits_{i=1}^{n}\left(\mu_{i}-\min\left\\{\mu_{i},\eta_{i}\right\\}+\min\left\\{\mu_{i},\gamma_{i}\right\\}\right)=\sum\limits_{i=1}^{n}\gamma_{i}.$

  4. (d)

We have $\gamma=\mathbf{f}_{\mu}\left(\eta\right)$.

######  Proof 4.55 (Proof of Corollary 4.54.).

(a) Theorem 4.53 (b) yields

$\left(\eta_{1}+\eta_{2}+\cdots+\eta_{n}\right)+\left(\gamma_{1}+\gamma_{2}+\cdots+\gamma_{n}\right)=2\left(\mu_{1}+\mu_{2}+\cdots+\mu_{n}\right).$

In view of the equalities

$\displaystyle\left|\eta\right|$
$\displaystyle=\eta_{1}+\eta_{2}+\cdots+\eta_{n},\ \ \ \ \ \ \ \ \ \
\left|\gamma\right|=\gamma_{1}+\gamma_{2}+\cdots+\gamma_{n}$
$\displaystyle\text{and}\ \ \ \ \ \ \ \ \ \ \left|\mu\right|$
$\displaystyle=\mu_{1}+\mu_{2}+\cdots+\mu_{n},$

we can rewrite this as
$\left|\eta\right|+\left|\gamma\right|=2\left|\mu\right|$. Equivalently,
$\left|\eta\right|-\left|\mu\right|=\left|\mu\right|-\left|\gamma\right|$.
This proves Corollary 4.54 (a).

(b) Let $i\in\left\\{1,2,\ldots,n-1\right\\}$. Then, Theorem 4.53 (c) yields

$\min\left\\{\mu_{i},\gamma_{i}\right\\}+\min\left\\{-\mu_{i+1},-\gamma_{i+1}\right\\}=\min\left\\{\mu_{i},\eta_{i}\right\\}+\min\left\\{-\mu_{i+1},-\eta_{i+1}\right\\}.$

In view of
$\min\left\\{-\mu_{i+1},-\gamma_{i+1}\right\\}=-\max\left\\{\mu_{i+1},\gamma_{i+1}\right\\}$
and
$\min\left\\{-\mu_{i+1},-\eta_{i+1}\right\\}=-\max\left\\{\mu_{i+1},\eta_{i+1}\right\\}$,
we can rewrite this as

$\min\left\\{\mu_{i},\gamma_{i}\right\\}+\left(-\max\left\\{\mu_{i+1},\gamma_{i+1}\right\\}\right)=\min\left\\{\mu_{i},\eta_{i}\right\\}+\left(-\max\left\\{\mu_{i+1},\eta_{i+1}\right\\}\right).$

In other words,

$\min\left\\{\mu_{i},\gamma_{i}\right\\}-\max\left\\{\mu_{i+1},\gamma_{i+1}\right\\}=\min\left\\{\mu_{i},\eta_{i}\right\\}-\max\left\\{\mu_{i+1},\eta_{i+1}\right\\}.$

Equivalently,

$\min\left\\{\mu_{i},\eta_{i}\right\\}-\min\left\\{\mu_{i},\gamma_{i}\right\\}=\max\left\\{\mu_{i+1},\eta_{i+1}\right\\}-\max\left\\{\mu_{i+1},\gamma_{i+1}\right\\}.$

This proves Corollary 4.54 (b).

(c) Theorem 4.53 (d) yields

$\sum\limits_{i=1}^{n}\left(\min\left\\{\mu_{i},\gamma_{i}\right\\}-\gamma_{i}\right)=\sum\limits_{i=1}^{n}\left(\min\left\\{\mu_{i},\eta_{i}\right\\}-\mu_{i}\right).$
(61)

Now, we have

$\displaystyle\sum\limits_{i=1}^{n}\underbrace{\left(\mu_{i}-\min\left\\{\mu_{i},\eta_{i}\right\\}+\min\left\\{\mu_{i},\gamma_{i}\right\\}\right)}_{=\min\left\\{\mu_{i},\gamma_{i}\right\\}-\left(\min\left\\{\mu_{i},\eta_{i}\right\\}-\mu_{i}\right)}$
$\displaystyle=\sum\limits_{i=1}^{n}\left(\min\left\\{\mu_{i},\gamma_{i}\right\\}-\left(\min\left\\{\mu_{i},\eta_{i}\right\\}-\mu_{i}\right)\right)$
$\displaystyle=\sum\limits_{i=1}^{n}\min\left\\{\mu_{i},\gamma_{i}\right\\}-\underbrace{\sum\limits_{i=1}^{n}\left(\min\left\\{\mu_{i},\eta_{i}\right\\}-\mu_{i}\right)}_{\begin{subarray}{c}=\sum\limits_{i=1}^{n}\left(\min\left\\{\mu_{i},\gamma_{i}\right\\}-\gamma_{i}\right)\\\
\text{(by (\ref{pf.cor.thm.fZ.full2.c.1}))}\end{subarray}}$
$\displaystyle=\sum\limits_{i=1}^{n}\min\left\\{\mu_{i},\gamma_{i}\right\\}-\sum\limits_{i=1}^{n}\left(\min\left\\{\mu_{i},\gamma_{i}\right\\}-\gamma_{i}\right)$
$\displaystyle=\sum\limits_{i=1}^{n}\underbrace{\left(\min\left\\{\mu_{i},\gamma_{i}\right\\}-\left(\min\left\\{\mu_{i},\gamma_{i}\right\\}-\gamma_{i}\right)\right)}_{=\gamma_{i}}=\sum\limits_{i=1}^{n}\gamma_{i}.$

This proves Corollary 4.54 (c).

(d) Theorem 4.53 (a) shows that
$\mathbf{f}_{\mu}\circ\mathbf{f}_{\mu}=\operatorname*{id}$. But recall that
$\eta=\mathbf{f}_{\mu}\left(\gamma\right)$. Applying the map
$\mathbf{f}_{\mu}$ to both sides of this equality, we obtain

$\mathbf{f}_{\mu}\left(\eta\right)=\mathbf{f}_{\mu}\left(\mathbf{f}_{\mu}\left(\gamma\right)\right)=\underbrace{\left(\mathbf{f}_{\mu}\circ\mathbf{f}_{\mu}\right)}_{=\operatorname*{id}}\left(\gamma\right)=\gamma.$

This proves Corollary 4.54 (d).

We are now ready to prove the key lemma:

######  Lemma 4.56.

Fix any $n$-tuple $\mu\in\mathbb{Z}^{n}$. Let $\gamma\in\mathbb{Z}^{n}$. Let
$a,b\in\mathbb{Z}$. Then,

$\left|R_{\mu,b,a}\left(\mathbf{f}_{\mu}\left(\gamma\right)\right)\right|=\left|R_{\mu,a,b}\left(\gamma\right)\right|.$

######  Proof 4.57 (Proof of Lemma 4.56.).

Define $\eta\in\mathbb{Z}^{n}$ by $\eta=\mathbf{f}_{\mu}\left(\gamma\right)$.
We must then prove that
$\left|R_{\mu,b,a}\left(\eta\right)\right|=\left|R_{\mu,a,b}\left(\gamma\right)\right|$.

We know that $R_{\mu,a,b}\left(\gamma\right)$ is the set of all snakes $\nu$
satisfying the four conditions

$\mu\rightharpoonup\nu\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \
\left|\mu\right|-\left|\nu\right|=a\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \
\ \ \ \gamma\rightharpoonup\nu\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \
\ \left|\gamma\right|-\left|\nu\right|=b.$

Likewise, $R_{\mu,b,a}\left(\eta\right)$ is the set of all snakes $\nu$
satisfying the four conditions

$\mu\rightharpoonup\nu\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \
\left|\mu\right|-\left|\nu\right|=b\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \
\ \ \ \eta\rightharpoonup\nu\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \
\left|\eta\right|-\left|\nu\right|=a.$

Now, fix $\nu\in R_{\mu,a,b}\left(\gamma\right)$. Thus, $\nu$ is a snake
satisfying the four conditions

$\mu\rightharpoonup\nu\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \
\left|\mu\right|-\left|\nu\right|=a\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \
\ \ \ \gamma\rightharpoonup\nu\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \
\ \left|\gamma\right|-\left|\nu\right|=b$

(by the definition of $R_{\mu,a,b}\left(\gamma\right)$). In particular, we
have $\mu\rightharpoonup\nu$. In other words, we have

$\mu_{1}\geq\nu_{1}\geq\mu_{2}\geq\nu_{2}\geq\cdots\geq\mu_{n}\geq\nu_{n}$
(62)

(by the definition of “$\mu\rightharpoonup\nu$”). Likewise, from
$\gamma\rightharpoonup\nu$, we obtain

$\gamma_{1}\geq\nu_{1}\geq\gamma_{2}\geq\nu_{2}\geq\cdots\geq\gamma_{n}\geq\nu_{n}.$
(63)

We define an $n$-tuple $\zeta\in\mathbb{Z}^{n}$ by setting

$\zeta_{i}=\min\left\\{\mu_{i},\eta_{i}\right\\}-\min\left\\{\mu_{i},\gamma_{i}\right\\}+\nu_{i}\
\ \ \ \ \ \ \ \ \ \text{for each }i\in\left\\{1,2,\ldots,n\right\\}.$

We shall prove that $\zeta\in R_{\mu,b,a}\left(\eta\right)$.

We begin by proving several auxiliary claims:

> Claim 1: We have $\min\left\\{\mu_{i},\eta_{i}\right\\}\geq\zeta_{i}$ for
> each $i\in\left\\{1,2,\ldots,n\right\\}$.

[Proof of Claim 1: Let $i\in\left\\{1,2,\ldots,n\right\\}$. From (62), we
obtain $\mu_{i}\geq\nu_{i}$. From (63), we obtain $\gamma_{i}\geq\nu_{i}$.
Combining $\mu_{i}\geq\nu_{i}$ and $\gamma_{i}\geq\nu_{i}$, we obtain
$\min\left\\{\mu_{i},\gamma_{i}\right\\}\geq\nu_{i}$. Now, the definition of
$\zeta$ yields

$\zeta_{i}=\min\left\\{\mu_{i},\eta_{i}\right\\}-\underbrace{\min\left\\{\mu_{i},\gamma_{i}\right\\}}_{\geq\nu_{i}}+\nu_{i}\leq\min\left\\{\mu_{i},\eta_{i}\right\\}-\nu_{i}+\nu_{i}=\min\left\\{\mu_{i},\eta_{i}\right\\}.$

In other words, $\min\left\\{\mu_{i},\eta_{i}\right\\}\geq\zeta_{i}$. This
proves Claim 1.]

> Claim 2: We have $\zeta_{i}\geq\max\left\\{\mu_{i+1},\eta_{i+1}\right\\}$
> for each $i\in\left\\{1,2,\ldots,n-1\right\\}$.

[Proof of Claim 2: Let $i\in\left\\{1,2,\ldots,n-1\right\\}$. From (62), we
obtain $\nu_{i}\geq\mu_{i+1}$. From (63), we obtain $\nu_{i}\geq\gamma_{i+1}$.
Combining $\nu_{i}\geq\mu_{i+1}$ and $\nu_{i}\geq\gamma_{i+1}$, we obtain
$\nu_{i}\geq\max\left\\{\mu_{i+1},\gamma_{i+1}\right\\}$. Now, the definition
of $\zeta$ yields

$\displaystyle\zeta_{i}$
$\displaystyle=\underbrace{\min\left\\{\mu_{i},\eta_{i}\right\\}-\min\left\\{\mu_{i},\gamma_{i}\right\\}}_{\begin{subarray}{c}=\max\left\\{\mu_{i+1},\eta_{i+1}\right\\}-\max\left\\{\mu_{i+1},\gamma_{i+1}\right\\}\\\
\text{(by Corollary \ref{cor.thm.fZ.full2}
{(b)})}\end{subarray}}+\underbrace{\nu_{i}}_{\geq\max\left\\{\mu_{i+1},\gamma_{i+1}\right\\}}$
$\displaystyle\geq\max\left\\{\mu_{i+1},\eta_{i+1}\right\\}-\max\left\\{\mu_{i+1},\gamma_{i+1}\right\\}+\max\left\\{\mu_{i+1},\gamma_{i+1}\right\\}=\max\left\\{\mu_{i+1},\eta_{i+1}\right\\}.$

This proves Claim 2.]

> Claim 3: The $n$-tuple $\zeta$ is a snake and satisfies
> $\mu\rightharpoonup\zeta$ and $\eta\rightharpoonup\zeta$.

[Proof of Claim 3: For each $i\in\left\\{1,2,\ldots,n\right\\}$, we have
$\mu_{i}\geq\zeta_{i}$ (since
$\mu_{i}\geq\min\left\\{\mu_{i},\eta_{i}\right\\}\geq\zeta_{i}$ (by Claim 1)).
For each $i\in\left\\{1,2,\ldots,n-1\right\\}$, we have
$\zeta_{i}\geq\mu_{i+1}$ (since Claim 2 yields
$\zeta_{i}\geq\max\left\\{\mu_{i+1},\eta_{i+1}\right\\}\geq\mu_{i+1}$).
Combining the preceding two sentences, we obtain

$\mu_{1}\geq\zeta_{1}\geq\mu_{2}\geq\zeta_{2}\geq\cdots\geq\mu_{n}\geq\zeta_{n}.$

In other words, $\mu\rightharpoonup\zeta$ (by the definition of
“$\mu\rightharpoonup\zeta$”).

For each $i\in\left\\{1,2,\ldots,n\right\\}$, we have $\eta_{i}\geq\zeta_{i}$
(since $\eta_{i}\geq\min\left\\{\mu_{i},\eta_{i}\right\\}\geq\zeta_{i}$ (by
Claim 1)). For each $i\in\left\\{1,2,\ldots,n-1\right\\}$, we have
$\zeta_{i}\geq\eta_{i+1}$ (since Claim 2 yields
$\zeta_{i}\geq\max\left\\{\mu_{i+1},\eta_{i+1}\right\\}\geq\eta_{i+1}$).
Combining the preceding two sentences, we obtain

$\eta_{1}\geq\zeta_{1}\geq\eta_{2}\geq\zeta_{2}\geq\cdots\geq\eta_{n}\geq\zeta_{n}.$

In other words, $\eta\rightharpoonup\zeta$ (by the definition of
“$\eta\rightharpoonup\zeta$”). Hence, Proposition 4.32 (a) (applied to $\eta$
and $\zeta$ instead of $\mu$ and $\lambda$) yields that both $\zeta$ and
$\eta$ are snakes. Hence, $\zeta$ is a snake. This completes the proof of
Claim 3.]

> Claim 4: We have $\left|\mu\right|-\left|\zeta\right|=b$ and
> $\left|\eta\right|-\left|\zeta\right|=a$.

[Proof of Claim 4: We have

$\displaystyle\left|\mu\right|$
$\displaystyle=\mu_{1}+\mu_{2}+\cdots+\mu_{n}=\sum\limits_{i=1}^{n}\mu_{i}\ \
\ \ \ \ \ \ \ \ \text{and}$ $\displaystyle\left|\zeta\right|$
$\displaystyle=\zeta_{1}+\zeta_{2}+\cdots+\zeta_{n}=\sum\limits_{i=1}^{n}\zeta_{i}.$

Subtracting these two equalities from one another, we find

$\displaystyle\left|\mu\right|-\left|\zeta\right|$
$\displaystyle=\sum\limits_{i=1}^{n}\mu_{i}-\sum\limits_{i=1}^{n}\underbrace{\zeta_{i}}_{\begin{subarray}{c}=\min\left\\{\mu_{i},\eta_{i}\right\\}-\min\left\\{\mu_{i},\gamma_{i}\right\\}+\nu_{i}\\\
\text{(by the definition of }\zeta\text{)}\end{subarray}}$
$\displaystyle=\sum\limits_{i=1}^{n}\mu_{i}-\sum\limits_{i=1}^{n}\left(\min\left\\{\mu_{i},\eta_{i}\right\\}-\min\left\\{\mu_{i},\gamma_{i}\right\\}+\nu_{i}\right)$
$\displaystyle=\sum\limits_{i=1}^{n}\underbrace{\left(\mu_{i}-\left(\min\left\\{\mu_{i},\eta_{i}\right\\}-\min\left\\{\mu_{i},\gamma_{i}\right\\}+\nu_{i}\right)\right)}_{=\left(\mu_{i}-\min\left\\{\mu_{i},\eta_{i}\right\\}+\min\left\\{\mu_{i},\gamma_{i}\right\\}\right)-\nu_{i}}$
$\displaystyle=\sum\limits_{i=1}^{n}\left(\left(\mu_{i}-\min\left\\{\mu_{i},\eta_{i}\right\\}+\min\left\\{\mu_{i},\gamma_{i}\right\\}\right)-\nu_{i}\right)$
$\displaystyle=\underbrace{\sum\limits_{i=1}^{n}\left(\mu_{i}-\min\left\\{\mu_{i},\eta_{i}\right\\}+\min\left\\{\mu_{i},\gamma_{i}\right\\}\right)}_{\begin{subarray}{c}=\sum\limits_{i=1}^{n}\gamma_{i}\\\
\text{(by Corollary \ref{cor.thm.fZ.full2}
{(c)})}\end{subarray}}-\sum\limits_{i=1}^{n}\nu_{i}$
$\displaystyle=\underbrace{\sum\limits_{i=1}^{n}\gamma_{i}}_{=\gamma_{1}+\gamma_{2}+\cdots+\gamma_{n}=\left|\gamma\right|}-\underbrace{\sum\limits_{i=1}^{n}\nu_{i}}_{=\nu_{1}+\nu_{2}+\cdots+\nu_{n}=\left|\nu\right|}=\left|\gamma\right|-\left|\nu\right|=b.$

Furthermore,

$\left|\mu\right|-\left|\gamma\right|=\underbrace{\left(\left|\mu\right|-\left|\nu\right|\right)}_{=a}-\underbrace{\left(\left|\gamma\right|-\left|\nu\right|\right)}_{=b}=a-b$

and

$\left|\eta\right|-\left|\zeta\right|=\underbrace{\left(\left|\eta\right|-\left|\mu\right|\right)}_{\begin{subarray}{c}=\left|\mu\right|-\left|\gamma\right|\\\
\text{(by Corollary \ref{cor.thm.fZ.full2}
{(a)})}\end{subarray}}+\underbrace{\left(\left|\mu\right|-\left|\zeta\right|\right)}_{=b}=\underbrace{\left|\mu\right|-\left|\gamma\right|}_{=a-b}+b=\left(a-b\right)+b=a.$

Thus, Claim 4 is proven.]

We have now shown (in Claim 3 and Claim 4) that $\zeta$ is a snake satisfying
the four conditions

$\mu\rightharpoonup\zeta\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \ \
\left|\mu\right|-\left|\zeta\right|=b\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \
\ \ \ \ \eta\rightharpoonup\zeta\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \
\ \ \left|\eta\right|-\left|\zeta\right|=a.$

In other words, $\zeta\in R_{\mu,b,a}\left(\eta\right)$ (by the definition of
$R_{\mu,b,a}\left(\eta\right)$).

Forget that we fixed $\nu$. Thus, for each $\nu\in
R_{\mu,a,b}\left(\gamma\right)$, we have constructed a $\zeta\in
R_{\mu,b,a}\left(\eta\right)$. Let us denote this $\zeta$ by
$\widetilde{\nu}$. We thus have defined a map

$\displaystyle R_{\mu,a,b}\left(\gamma\right)$ $\displaystyle\rightarrow
R_{\mu,b,a}\left(\eta\right),$ $\displaystyle\nu$
$\displaystyle\mapsto\widetilde{\nu}.$

Let us denote this map by $\mathbf{g}_{\gamma,a,b}$. Its definition shows that

$\displaystyle\left(\underbrace{\mathbf{g}_{\gamma,a,b}\left(\nu\right)}_{=\widetilde{\nu}}\right)_{i}$
$\displaystyle=\widetilde{\nu}_{i}=\min\left\\{\mu_{i},\eta_{i}\right\\}-\min\left\\{\mu_{i},\gamma_{i}\right\\}+\nu_{i}$
(64) $\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\begin{array}[c]{c}\text{since
}\widetilde{\nu}\text{ was defined as the }n\text{-tuple }\zeta\text{, whose
entries}\\\ \text{are given by
}\zeta_{i}=\min\left\\{\mu_{i},\eta_{i}\right\\}-\min\left\\{\mu_{i},\gamma_{i}\right\\}+\nu_{i}\end{array}\right)$
(67)

for each $\nu\in R_{\mu,a,b}\left(\gamma\right)$ and each
$i\in\left\\{1,2,\ldots,n\right\\}$.

However, from $\eta=\mathbf{f}_{\mu}\left(\gamma\right)$, we obtain
$\gamma=\mathbf{f}_{\mu}\left(\eta\right)$ (by Corollary 4.54 (d)). The
relation between $\gamma$ and $\eta$ is thus symmetric. Hence, in the same way
as we defined a map
$\mathbf{g}_{\gamma,a,b}:R_{\mu,a,b}\left(\gamma\right)\rightarrow
R_{\mu,b,a}\left(\eta\right)$, we can define a map
$\mathbf{g}_{\eta,b,a}:R_{\mu,b,a}\left(\eta\right)\rightarrow
R_{\mu,a,b}\left(\gamma\right)$ (by repeating the above construction of
$\mathbf{g}_{\gamma,a,b}$ with $b$, $a$, $\eta$ and $\gamma$ taking the roles
of $a$, $b$, $\gamma$ and $\eta$, respectively). The resulting map
$\mathbf{g}_{\eta,b,a}$ satisfies

$\left(\mathbf{g}_{\eta,b,a}\left(\nu\right)\right)_{i}=\min\left\\{\mu_{i},\gamma_{i}\right\\}-\min\left\\{\mu_{i},\eta_{i}\right\\}+\nu_{i}$
(68)

for each $\nu\in R_{\mu,b,a}\left(\eta\right)$ and each
$i\in\left\\{1,2,\ldots,n\right\\}$. (Indeed, this can be proved just as we
proved (64), but with $b$, $a$, $\eta$ and $\gamma$ taking the roles of $a$,
$b$, $\gamma$ and $\eta$.)

Now it is easy to see (using (64) and (68)) that the two maps
$\mathbf{g}_{\gamma,a,b}$ and $\mathbf{g}_{\eta,b,a}$ are mutually inverse.
Hence, these two maps are invertible, i.e., are bijections.

Thus, there exists a bijection from $R_{\mu,a,b}\left(\gamma\right)$ to
$R_{\mu,b,a}\left(\eta\right)$ (namely, $\mathbf{g}_{\gamma,a,b}$). This
yields
$\left|R_{\mu,a,b}\left(\gamma\right)\right|=\left|R_{\mu,b,a}\left(\eta\right)\right|=\left|R_{\mu,b,a}\left(\mathbf{f}_{\mu}\left(\gamma\right)\right)\right|$
(since $\eta=\mathbf{f}_{\mu}\left(\gamma\right)$). This proves Lemma 4.56.

Having learned a lot about the map $\mathbf{f}_{\mu}$, let us now connect it
to the map $\varphi$ defined in Theorem 2.2. For this, we shall use the
following lemma:

######  Lemma 4.58.

Fix any $n$-tuple $\mu\in\mathbb{Z}^{n}$.

Let $\nu\in\mathbb{Z}^{n}$ be an $n$-tuple. For each $j\in\mathbb{Z}$, let

$\displaystyle\tau_{j}$
$\displaystyle=\min\left\\{\underbrace{\nu_{j+1}+\nu_{j+2}+\cdots+\nu_{j+k}}_{=\sum\limits_{i=1}^{k}\nu_{j+i}}+\underbrace{\mu_{j+k+1}+\mu_{j+k+2}+\cdots+\mu_{j+n-1}}_{=\sum\limits_{i=k+1}^{n-1}\mu_{j+i}}\right.$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.\ \mid\
k\in\left\\{0,1,\ldots,n-1\right\\}\right\\}.$

Let $\eta\in\mathbb{Z}^{n}$ be such that

$\eta_{i}=\mu_{i}+\left(\mu_{i-1}+\tau_{i-1}\right)-\left(\nu_{i+1}+\tau_{i+1}\right)\
\ \ \ \ \ \ \ \ \ \text{for each }i\in\left\\{1,2,\ldots,n\right\\}.$

Then, $\mathbf{f}_{\mu}\left(\nu\right)=\eta$.

######  Proof 4.59 (Proof of Lemma 4.58.).

Lemma 4.58 is obtained when we apply Lemma 3.11 to
$\mathbb{K}=\left(\mathbb{Z},\min,+,0\right)$ (and rename
everything151515Namely, we are  • renaming the (fixed) $n$-tuple $u$ as $\mu$;
• renaming the $n$-tuple $x$ as $\nu$; • renaming the elements $q_{j}$ as
$\tau_{j}$; • renaming the $n$-tuple $z$ as $\eta$. , and use our above
dictionary again).

We can now connect the map $\mathbf{f}_{\mu}$ with the map $\varphi$ from
Theorem 2.2:

######  Lemma 4.60.

Let $a,b\in\mathbb{N}$. Fix any $n$-tuple $\mu\in\mathbb{Z}^{n}$. Define a map
$\varphi:\mathbb{Z}^{n}\rightarrow\mathbb{Z}^{n}$ as in Theorem 2.2. Then,

$\varphi\left(\omega\right)=\mathbf{f}_{\mu}\left(\omega-a\right)+b\ \ \ \ \ \
\ \ \ \ \text{for each }\omega\in\mathbb{Z}^{n}.$

######  Proof 4.61 (Proof of Lemma 4.60.).

Let $\omega\in\mathbb{Z}^{n}$.

Define an $n$-tuple
$\nu=\left(\nu_{1},\nu_{2},\ldots,\nu_{n}\right)\in\mathbb{Z}^{n}$ by
$\nu_{i}=\omega_{i}-a$ for each $i\in\left\\{1,2,\ldots,n\right\\}$. Thus,
$\nu=\omega-a$.

For each $i\in\mathbb{Z}$, we let $i\\#$ denote the unique element of
$\left\\{1,2,\ldots,n\right\\}$ congruent to $i$ modulo $n$.

For each $j\in\mathbb{Z}$, set

$\displaystyle\tau_{j}$
$\displaystyle=\min\left\\{\left(\nu_{\left(j+1\right)\\#}+\nu_{\left(j+2\right)\\#}+\cdots+\nu_{\left(j+k\right)\\#}\right)\right.$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\left.+\left(\mu_{\left(j+k+1\right)\\#}+\mu_{\left(j+k+2\right)\\#}+\cdots+\mu_{\left(j+n-1\right)\\#}\right)\right.$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.\ \mid\
k\in\left\\{0,1,\ldots,n-1\right\\}\vphantom{\left(\nu_{\left(j+1\right)\\#}\right)}\right\\}.$
(69)

Define an $n$-tuple
$\eta=\left(\eta_{1},\eta_{2},\ldots,\eta_{n}\right)\in\mathbb{Z}^{n}$ by
setting

$\eta_{i}=\mu_{i\\#}+\left(\mu_{\left(i-1\right)\\#}+\tau_{\left(i-1\right)\\#}\right)-\left(\nu_{\left(i+1\right)\\#}+\tau_{\left(i+1\right)\\#}\right)\
\ \ \ \ \ \ \ \ \ \text{for each }i\in\left\\{1,2,\ldots,n\right\\}.$

The definition of $\varphi$ then yields

$\varphi\left(\omega\right)=\left(\eta_{1}+b,\eta_{2}+b,\ldots,\eta_{n}+b\right)=\eta+b.$
(70)

Our plan is now to show that $\mathbf{f}_{\mu}\left(\nu\right)=\eta$. We shall
achieve this by applying Lemma 4.58; but in order to do so, we need to show
that the assumptions of Lemma 4.58 are satisfied.

We shall do this piece by piece. First, we make the following two claims,
which both follow from Convention 3.3:

> Claim 1: We have $\nu_{p\\#}=\nu_{p}$ for each $p\in\mathbb{Z}$.

> Claim 2: We have $\mu_{p\\#}=\mu_{p}$ for each $p\in\mathbb{Z}$.

The next claim is an easy consequence of Claims 1 and 2:

> Claim 3: For each $j\in\mathbb{Z}$, we have
>
> $\displaystyle\tau_{j}$
> $\displaystyle=\min\left\\{\underbrace{\nu_{j+1}+\nu_{j+2}+\cdots+\nu_{j+k}}_{=\sum\limits_{i=1}^{k}\nu_{j+i}}+\underbrace{\mu_{j+k+1}+\mu_{j+k+2}+\cdots+\mu_{j+n-1}}_{=\sum\limits_{i=k+1}^{n-1}\mu_{j+i}}\right.$
> $\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.\ \mid\
> k\in\left\\{0,1,\ldots,n-1\right\\}\right\\}.$

The next claim is an easy fact from elementary number theory:

> Claim 4: We have $\left(p\\#+q\right)\\#=\left(p+q\right)\\#$ for any
> $p\in\mathbb{Z}$ and $q\in\mathbb{Z}$.

Using Claim 4, we easily obtain the following:

> Claim 5: We have $\tau_{p\\#}=\tau_{p}$ for each $p\in\mathbb{Z}$.

Now, let $i\in\left\\{1,2,\ldots,n\right\\}$. Then, the definition of $\eta$
yields

$\displaystyle\eta_{i}$
$\displaystyle=\underbrace{\mu_{i\\#}}_{\begin{subarray}{c}=\mu_{i}\\\
\text{(by Claim
2)}\end{subarray}}+\left(\underbrace{\mu_{\left(i-1\right)\\#}}_{\begin{subarray}{c}=\mu_{i-1}\\\
\text{(by Claim
2)}\end{subarray}}+\underbrace{\tau_{\left(i-1\right)\\#}}_{\begin{subarray}{c}=\tau_{i-1}\\\
\text{(by Claim
5)}\end{subarray}}\right)-\left(\underbrace{\nu_{\left(i+1\right)\\#}}_{\begin{subarray}{c}=\nu_{i+1}\\\
\text{(by Claim
1)}\end{subarray}}+\underbrace{\tau_{\left(i+1\right)\\#}}_{\begin{subarray}{c}=\tau_{i+1}\\\
\text{(by Claim 5)}\end{subarray}}\right)$
$\displaystyle=\mu_{i}+\left(\mu_{i-1}+\tau_{i-1}\right)-\left(\nu_{i+1}+\tau_{i+1}\right).$

Now, forget that we fixed $i$. We thus have proved that

$\eta_{i}=\mu_{i}+\left(\mu_{i-1}+\tau_{i-1}\right)-\left(\nu_{i+1}+\tau_{i+1}\right)\
\ \ \ \ \ \ \ \ \ \text{for each }i\in\left\\{1,2,\ldots,n\right\\}.$

Combining this with Claim 3, we conclude that the assumptions of Lemma 4.58
are satisfied. Hence, Lemma 4.58 yields
$\mathbf{f}_{\mu}\left(\nu\right)=\eta$. In view of $\nu=\omega-a$, this
rewrites as $\mathbf{f}_{\mu}\left(\omega-a\right)=\eta$. Hence,
$\eta=\mathbf{f}_{\mu}\left(\omega-a\right)$, so that (70) becomes

$\varphi\left(\omega\right)=\underbrace{\eta}_{=\mathbf{f}_{\mu}\left(\omega-a\right)}+b=\mathbf{f}_{\mu}\left(\omega-a\right)+b.$

This proves Lemma 4.60.

###  4.6 The finale

Now, let us again use the convention (from Section 1) by which we identify
partitions with finite tuples (and therefore identify partitions in
$\operatorname*{Par}\left[n\right]$ with nonnegative snakes). This is no
longer problematic, since we are not using Convention 3.3 any more.

######  Lemma 4.62.

Let $a,b\in\mathbb{N}$. Define the two partitions
$\alpha=\left(a+b,a^{n-2}\right)$ and $\beta=\left(a+b,b^{n-2}\right)$.

Fix any partition $\mu\in\operatorname*{Par}\left[n\right]$. Consider the map
$\mathbf{f}_{\mu}:\mathbb{Z}^{n}\rightarrow\mathbb{Z}^{n}$ defined in
Definition 4.52.

Then, for any $\lambda\in\mathbb{Z}^{n}$, we have

$c_{\alpha,\mu}^{\lambda+a}=c_{\beta,\mu}^{\mathbf{f}_{\mu}\left(\lambda\right)+b}.$

Here, we understand $c_{\alpha,\mu}^{\lambda+a}$ to mean $0$ if $\lambda+a$ is
not a partition, and likewise we understand
$c_{\beta,\mu}^{\mathbf{f}_{\mu}\left(\lambda\right)+b}$ to mean $0$ if
$\mathbf{f}_{\mu}\left(\lambda\right)+b$ is not a partition.

######  Proof 4.63 (Proof of Lemma 4.62.).

Let $\lambda\in\mathbb{Z}^{n}$. Corollary 4.50 (applied to $\lambda+a$ instead
of $\lambda$) yields

$\displaystyle c_{\alpha,\mu}^{\lambda+a}$
$\displaystyle=\left|R_{\mu,a,b}\left(\left(\lambda+a\right)-a\right)\right|-\left|R_{\mu,a-1,b-1}\left(\left(\lambda+a\right)-a\right)\right|$
$\displaystyle=\left|R_{\mu,a,b}\left(\lambda\right)\right|-\left|R_{\mu,a-1,b-1}\left(\lambda\right)\right|$
(71)

(since $\left(\lambda+a\right)-a=\lambda$). On the other hand,
$\beta=\left(\underbrace{a+b}_{=b+a},b^{n-2}\right)=\left(b+a,b^{n-2}\right)$.
Hence, Corollary 4.50 (applied to $b$, $a$, $\beta$ and
$\mathbf{f}_{\mu}\left(\lambda\right)+b$ instead of $a$, $b$, $\alpha$ and
$\lambda$) yields

$\displaystyle c_{\beta,\mu}^{\mathbf{f}_{\mu}\left(\lambda\right)+b}$
$\displaystyle=\left|R_{\mu,b,a}\left(\left(\mathbf{f}_{\mu}\left(\lambda\right)+b\right)-b\right)\right|-\left|R_{\mu,b-1,a-1}\left(\left(\mathbf{f}_{\mu}\left(\lambda\right)+b\right)-b\right)\right|$
$\displaystyle=\underbrace{\left|R_{\mu,b,a}\left(\mathbf{f}_{\mu}\left(\lambda\right)\right)\right|}_{\begin{subarray}{c}=\left|R_{\mu,a,b}\left(\lambda\right)\right|\\\
\text{(by Lemma \ref{lem.fZ.lr},}\\\ \text{applied to
}\gamma=\lambda\text{)}\end{subarray}}-\underbrace{\left|R_{\mu,b-1,a-1}\left(\mathbf{f}_{\mu}\left(\lambda\right)\right)\right|}_{\begin{subarray}{c}=\left|R_{\mu,a-1,b-1}\left(\lambda\right)\right|\\\
\text{(by Lemma \ref{lem.fZ.lr},}\\\ \text{applied to }\lambda\text{,
}a-1\text{ and }b-1\\\ \text{instead of }\gamma\text{, }a\text{ and
}b\text{)}\end{subarray}}$ $\displaystyle\ \ \ \ \ \ \ \ \ \ \left(\text{since
}\left(\mathbf{f}_{\mu}\left(\lambda\right)+b\right)-b=\mathbf{f}_{\mu}\left(\lambda\right)\right)$
$\displaystyle=\left|R_{\mu,a,b}\left(\lambda\right)\right|-\left|R_{\mu,a-1,b-1}\left(\lambda\right)\right|.$

Comparing this with (71), we find
$c_{\alpha,\mu}^{\lambda+a}=c_{\beta,\mu}^{\mathbf{f}_{\mu}\left(\lambda\right)+b}$.
This proves Lemma 4.62.

We are now ready to prove Theorem 2.2:

######  Proof 4.64 (Proof of Theorem 2.2.).

The map $\mathbf{f}_{\mu}$ is an involution (by Theorem 4.53 (a)), thus a
bijection.

Let $\mathbf{a}^{-}:\mathbb{Z}^{n}\rightarrow\mathbb{Z}^{n}$ be the map that
sends each $\omega\in\mathbb{Z}^{n}$ to $\omega-a$. This map $\mathbf{a}^{-}$
is clearly a bijection.

Let $\mathbf{b}^{+}:\mathbb{Z}^{n}\rightarrow\mathbb{Z}^{n}$ be the map that
sends each $\omega\in\mathbb{Z}^{n}$ to $\omega+b$. This map $\mathbf{b}^{+}$
is clearly a bijection.

From Lemma 4.60, we can easily see that

$\varphi=\mathbf{b}^{+}\circ\mathbf{f}_{\mu}\circ\mathbf{a}^{-}.$

Recall that the maps $\mathbf{b}^{+}$, $\mathbf{f}_{\mu}$ and $\mathbf{a}^{-}$
are bijections. Hence, their composition
$\mathbf{b}^{+}\circ\mathbf{f}_{\mu}\circ\mathbf{a}^{-}$ is a bijection as
well. In other words, $\varphi$ is a bijection (since
$\varphi=\mathbf{b}^{+}\circ\mathbf{f}_{\mu}\circ\mathbf{a}^{-}$). This proves
Theorem 2.2 (a).

(b) Let $\omega\in\mathbb{Z}^{n}$. Then, Lemma 4.60 yields
$\varphi\left(\omega\right)=\mathbf{f}_{\mu}\left(\omega-a\right)+b$. Hence,
$\mathbf{f}_{\mu}\left(\omega-a\right)+b=\varphi\left(\omega\right)$. But
Lemma 4.62 (applied to $\lambda=\omega-a$) yields

$c_{\alpha,\mu}^{\left(\omega-a\right)+a}=c_{\beta,\mu}^{\mathbf{f}_{\mu}\left(\omega-a\right)+b}=c_{\beta,\mu}^{\varphi\left(\omega\right)}\
\ \ \ \ \ \ \ \ \ \left(\text{since
}\mathbf{f}_{\mu}\left(\omega-a\right)+b=\varphi\left(\omega\right)\right).$

In view of $\left(\omega-a\right)+a=\omega$, this rewrites as
$c_{\alpha,\mu}^{\omega}=c_{\beta,\mu}^{\varphi\left(\omega\right)}$. This
proves Theorem 2.2 (b).

##  5 Final remarks

###  5.1 Aside: A Jacobi–Trudi formula for Schur Laurent polynomials

As mentioned above, Proposition 4.41 has the following generalization, which
can be obtained from an identity of Koike [Koike89, Proposition 2.8] (via the
map $\widetilde{\pi}_{n}$ from [Koike89] and the correspondence between Schur
Laurent polynomials and rational representations of
$\operatorname{GL}\left(n\right)$), which has later been extended by Hamel and
King [HamKin11] (see [HamKin11, (6) and (10)] for the connection):161616We
recall Definition 4.1, Definition 4.2 (a), Definition 4.6, Definition 4.15 and
Definition 4.26, as well as the conventions made in Section 1, for the
notations used in this proposition.

######  Proposition 5.1.

Let $p,q\in\mathbb{N}$ with $p+q\leq n$. Let
$\mathbf{a}=\left(a_{1},a_{2},\ldots,a_{p}\right)$ and
$\mathbf{b}=\left(b_{1},b_{2},\ldots,b_{q}\right)$ be two partitions. Let
$\mathbf{b}\ominus\mathbf{a}$ denote the snake
$\left(b_{1},b_{2},\ldots,b_{q},0^{n-p-q},-a_{p},-a_{p-1},\ldots,-a_{1}\right)$.
Let $M$ be the $\left(p+q\right)\times\left(p+q\right)$-matrix

$\displaystyle\left(\begin{cases}h_{a_{p-i+1}+i-j}^{-},&\text{if }i\leq p;\\\
h_{b_{i-p}-i+j}^{+},&\text{if
}i>p\end{cases}\right)_{i,j\in\left\\{1,2,\ldots,p+q\right\\}}$
$\displaystyle=\left(\begin{array}[c]{cccccccc}h_{a_{p}}^{-}&h_{a_{p}-1}^{-}&\cdots&h_{a_{p}-p+1}^{-}&h_{a_{p}-p}^{-}&h_{a_{p}-p-1}^{-}&\cdots&h_{a_{p}-p-q+1}^{-}\\\
h_{a_{p-1}+1}^{-}&h_{a_{p-1}}^{-}&\cdots&h_{a_{p-1}-p+2}^{-}&h_{a_{p-1}-p+1}^{-}&h_{a_{p-1}-p}^{-}&\cdots&h_{a_{p-1}-p-q+2}^{-}\\\
\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\\
h_{a_{1}+p-1}^{-}&h_{a_{1}+p-2}^{-}&\cdots&h_{a_{1}}^{-}&h_{a_{1}-1}^{-}&h_{a_{1}-2}^{-}&\cdots&h_{a_{1}-q}^{-}\\\
h_{b_{1}-p}^{+}&h_{b_{1}-p+1}^{+}&\cdots&h_{b_{1}-1}^{+}&h_{b_{1}}^{+}&h_{b_{1}+1}^{+}&\cdots&h_{b_{1}+q-1}^{+}\\\
h_{b_{2}-p-1}^{+}&h_{b_{2}-p}^{+}&\cdots&h_{b_{2}-2}^{+}&h_{b_{2}-1}^{+}&h_{b_{2}}^{+}&\cdots&h_{b_{2}+q-2}^{+}\\\
\vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\\
h_{b_{q}-p-q+1}^{+}&h_{b_{q}-p-q+2}^{+}&\cdots&h_{b_{q}-q}^{+}&h_{b_{q}-q+1}^{+}&h_{b_{q}-q+2}^{+}&\cdots&h_{b_{q}}^{+}\end{array}\right).$

Then,

$\overline{s}_{\mathbf{b}\ominus\mathbf{a}}=\det M.$

######  Remark 5.2.

The analogous generalization of the second Jacobi–Trudi formula ([GriRei20,
(2.4.17)]) can easily be proved (although we leave both stating and proving it
to the reader). What makes it easy is the (fairly obvious) fact that the
elementary symmetric functions $e_{k}$ satisfy

$e_{k}\left(x_{1}^{-1},x_{2}^{-1},\ldots,x_{n}^{-1}\right)=x_{\Pi}^{-1}e_{n-k}\left(x_{1},x_{2},\ldots,x_{n}\right)$

for all $k\in\mathbb{Z}$. (See [GriRei20, Definition 2.2.1] for the definition
of $e_{k}$.)

Proposition 5.1 generalizes Proposition 4.41 (which corresponds to the
particular case when $p=1$ and $q=1$) as well as the first Jacobi–Trudi
formula [GriRei20, (2.4.16)] (which corresponds to the particular case $p=0$).

We notice that what we called $\mathbf{b}\ominus\mathbf{a}$ in Proposition 5.1
has been called $\left[\mathbf{b},\mathbf{a}\right]$ in [Stembr87].

We thank Grigori Olshanski for informing us of the provenance of Proposition
5.1.

###  5.2 Questions on $\mathbf{f}_{u}$

We shall now pose several questions about the birational involution
$\mathbf{f}_{u}$ studied in Section 3. Convention 3.1, Convention 3.2 and
Convention 3.3 will be used throughout Subsection 5.2.

Most of our questions are meant to attempt seeing the involution
$\mathbf{f}_{u}$ from different directions. The first one is inspired by what
is now known as the “toggle approach” to dynamical combinatorics (see, e.g.,
[Roby15]), but is really an application of the age-old “divide and conquer”
paradigm to complicated maps:

######  Question 5.3.

Is there an equivalent definition of $\mathbf{f}_{u}$ as a composition of
toggles? (A toggle here means a birational map
$\mathbb{K}^{n}\rightarrow\mathbb{K}^{n}$ that changes only one entry of the
$n$-tuple. An example for a birational map that can be defined as a
composition of toggles is birational rowmotion – see, e.g., [EinPro13].
Cluster mutations, as in the theory of cluster algebras, are another example
of toggles.)

Another set of questions concern the uniqueness of $\mathbf{f}_{u}$. While we
defined the map $\mathbf{f}_{u}$ explicitly, all we have then used are the
properties listed in Theorem 3.8. Thus, it is a natural question to ask
whether these properties characterize $\mathbf{f}_{u}$ uniquely. A pointwise
version of this question can be asked as well: Given $x\in\mathbb{K}^{n}$ and
$y\in\mathbb{K}^{n}$ satisfying some of the equalities in parts (b), (c) and
(d) of Theorem 3.8, does it follow that $y=\mathbf{f}_{u}\left(x\right)$ ?
(Keep in mind that $u$ is fixed.)

Depending on which equalities we require, we may of course get different
answers. Let us first ask what happens if we require the equalities from
Theorem 3.8 (c) only:

######  Question 5.4.

Given $x\in\mathbb{K}^{n}$ and $y\in\mathbb{K}^{n}$ satisfying

$\left(u_{i}+x_{i}\right)\left(\dfrac{1}{u_{i+1}}+\dfrac{1}{x_{i+1}}\right)=\left(u_{i}+y_{i}\right)\left(\dfrac{1}{u_{i+1}}+\dfrac{1}{y_{i+1}}\right)$
(72)

for all $i\in\mathbb{Z}$. Does it follow that $y=\mathbf{f}_{u}\left(x\right)$
or $y=x$ ?

Note that the “or $y=x$” part is needed here, since $y=x$ is obviously a
solution to the equations (72).

The following example shows that the answer to Question 5.4 is “no” if
$\mathbb{K}$ is the min tropical semifield $\left(\mathbb{Z},\min,+,0\right)$
of the totally ordered abelian group $\mathbb{Z}$.

######  Example 5.5.

Let $k,g\in\mathbb{N}$ with $g\geq k$. Let
$\mathbb{K}=\left(\mathbb{Z},\min,+,0\right)$ and $n=3$ and
$u=\left(0,0,g\right)$ and $x=\left(1,2,0\right)$. Set
$y=\left(k+1,2,k\right)$ (where the “$+$” sign in “$k+1$” stands for addition
of integers, not addition in $\mathbb{K}$). Then, the equations (72) hold in
$\mathbb{K}$ for all $i\in\mathbb{Z}$. (Restated in terms of standard
operations on integers, this is saying that

$\min\left\\{u_{i},x_{i}\right\\}+\min\left\\{-u_{i+1},-x_{i+1}\right\\}=\min\left\\{u_{i},y_{i}\right\\}+\min\left\\{-u_{i+1},-y_{i+1}\right\\}$

for all $i\in\mathbb{Z}$.) This is straightforward to verify, and shows that
for a given $x$ there can be an arbitrarily high (finite) number of
$y\in\mathbb{K}^{n}$ satisfying the equations (72) for all $i\in\mathbb{Z}$.
(We do not know if this number can be infinite.)

However, the answer to Question 5.4 is “yes” if $\mathbb{K}=\mathbb{Q}_{+}$
and, more generally, if the semifield $\mathbb{K}$ embeds into an integral
domain:

######  Proposition 5.6.

Assume that there is an integral domain $\mathbb{L}$ such that the semifield
$\mathbb{K}$ is a subsemifield of $\mathbb{L}$ (in the sense that
$\mathbb{K}\subseteq\mathbb{L}$ and that the operations $+$ and $\cdot$ of
$\mathbb{K}$ are restrictions of those of $\mathbb{L}$, whereas the unity of
$\mathbb{K}$ is the unity of $\mathbb{L}$). Let $x\in\mathbb{K}^{n}$. Then,
the only $n$-tuples $y\in\mathbb{K}^{n}$ satisfying the equations (72) for all
$i\in\mathbb{Z}$ are $y=\mathbf{f}_{u}\left(x\right)$ and $y=x$.

See the detailed version [Grinbe20] of this paper for a rough outline of the
proof of Proposition 5.6.

Another avatar of the uniqueness question is the following:

######  Question 5.7.

Given $x\in\mathbb{K}^{n}$ and $y\in\mathbb{K}^{n}$ satisfying both (72) for
all $i\in\mathbb{Z}$ and

$y_{1}y_{2}\cdots y_{n}\cdot x_{1}x_{2}\cdots x_{n}=\left(u_{1}u_{2}\cdots
u_{n}\right)^{2}.$ (73)

Does it follow that $y=\mathbf{f}_{u}\left(x\right)$ ?

The answer to this question is definitely “yes” when
$\mathbb{K}=\mathbb{Q}_{+}$, by essentially the same argument that was used in
Remark 3.18. Again, however, the answer is “no” when
$\mathbb{K}=\left(\mathbb{Z},\min,+,0\right)$. For example, if
$\mathbb{K}=\left(\mathbb{Z},\min,+,0\right)$ and $n=4$ and
$u=\left(2,1,1,0\right)$ and $x=\left(1,1,1,1\right)$, then the two $n$-tuples
$\left(1,1,1,1\right)$ and $\left(2,2,0,0\right)$ both can be taken as $y$ in
Question 5.7, but clearly cannot both equal $\mathbf{f}_{u}\left(x\right)$.
(It appears, based on some experiments, that the answer might be “yes” for
$n=3$.)

An even stronger version of Question 5.7 holds when
$\mathbb{K}=\mathbb{Q}_{+}$:

######  Proposition 5.8.

Assume that $\mathbb{K}=\mathbb{Q}_{+}$. Let $x\in\mathbb{K}^{n}$ and
$y\in\mathbb{K}^{n}$. Assume that (72) holds for all
$i\in\left\\{1,2,\ldots,n-1\right\\}$, and assume that (73) holds. Then,
$y=\mathbf{f}_{u}\left(x\right)$.

The proof of Proposition 5.8 is sketched in the detailed version [Grinbe20] of
this paper.

Another question concerns Lemma 3.9:

######  Question 5.9.

What is the “real meaning” of some of the more complicated parts of Lemma 3.9?
In particular, Lemma 3.9 (g) reminds of the Plücker relation for minors of a
$2\times m$-matrix; can it be viewed that way? (Such a proof would not be
superior to the one given above, as it wouldn’t be subtraction-free and thus
wouldn’t work natively over arbitrary semifields. But it would shine more
light on the lemma.)

###  5.3 On the genesis of $\varphi$ (and $\mathbf{f}_{u}$)

As we mentioned in the introduction to this paper, Pelletier and Ressayre did
not conjecture Theorem 2.2 in this exact form; instead, they conjectured the
existence of a mysterious bijection $\varphi$ that satisfies Theorem 2.2 (b).
Our definition of $\varphi$ appears ex caelis oblatus; while we have seen that
our $\varphi$ duly plays its part, it is far from clear how we have found it
in the first place. The following few paragraphs are meant to demystify this
process.

We were looking for a bijection
$\varphi:\mathbb{Z}^{n}\rightarrow\mathbb{Z}^{n}$ satisfying Theorem 2.2 (b).
In other words, we were looking for a way to match171717“Matching” means
“perfect matching” here – i.e., every coefficient on either side should get a
unique partner. the nonzero coefficients in the product
$s_{\alpha}\left(x_{1},x_{2},\ldots,x_{n}\right)\cdot
s_{\mu}\left(x_{1},x_{2},\ldots,x_{n}\right)$ (when expanded in the basis
$\left(s_{\lambda}\left(x_{1},x_{2},\ldots,x_{n}\right)\right)_{\lambda\in\operatorname*{Par}\left[n\right]}$
of the $\mathbf{k}$-module of symmetric polynomials in
$x_{1},x_{2},\ldots,x_{n}$) with the nonzero coefficients in the product
$s_{\beta}\left(x_{1},x_{2},\ldots,x_{n}\right)\cdot
s_{\mu}\left(x_{1},x_{2},\ldots,x_{n}\right)$ in such a way that matching
coefficients are equal.

The first step towards this goal was the discovery of the formula
$s_{\alpha}\left(x_{1},x_{2},\ldots,x_{n}\right)=x_{\Pi}^{a}\cdot\left(h_{a}^{-}h_{b}^{+}-h_{a-1}^{-}h_{b-1}^{+}\right)$:
our Corollary 4.43. We originally proved this formula combinatorially, by
analyzing the structure of semistandard tableaux of shape $\alpha$. 181818Each
of the first $a$ columns of such a tableau would have the form
$\left(1,2,\ldots,i-1,i+1,\ldots,n\right)$ for some
$i\in\left\\{1,2,\ldots,n\right\\}$, and these numbers $i$ would weakly
increase as one moves right. The proof of Corollary 4.43 given above (using
the Pieri rule) was an afterthought.

Corollary 4.43 was a visible step in the right direction, as it moved the
problem from the world of Littlewood–Richardson coefficients into the simpler
world of Pieri rules. Indeed, instead of expanding
$s_{\alpha}\left(x_{1},x_{2},\ldots,x_{n}\right)\cdot
s_{\mu}\left(x_{1},x_{2},\ldots,x_{n}\right)$, we now only had to expand
$x_{\Pi}^{a}\cdot\left(h_{a}^{-}h_{b}^{+}-h_{a-1}^{-}h_{b-1}^{+}\right)\cdot
s_{\mu}\left(x_{1},x_{2},\ldots,x_{n}\right)$, which looked like an expansion
that the Pieri rule could help with (to be fully honest, we only knew the
Pieri rule for multiplying by $h_{k}^{+}$; but we soon would find one for
multiplying by $h_{k}^{-}$). The $x_{\Pi}^{a}$ factor was clearly a mere
distraction, but in order to get rid of it, we had to extend our polynomial
ring to the ring $\mathcal{L}$ of Laurent polynomials (since
$\left(h_{a}^{-}h_{b}^{+}-h_{a-1}^{-}h_{b-1}^{+}\right)\cdot
s_{\mu}\left(x_{1},x_{2},\ldots,x_{n}\right)$ is, in general, not a
polynomial). This extension had already been done by Stembridge in [Stembr87],
and all we had to do was rename “staircases” as “snakes”, define Schur Laurent
polynomials (by generalizing the alternant formula for Schur polynomials in
the most obvious way), extend some basic properties of Schur polynomials to
Schur Laurent polynomials, and find the “upside-down” Pieri rule (Proposition
4.36). None of this was difficult; in particular, the “upside-down” Pieri rule
followed easily from the usual Pieri rule using Lemma 4.24 (which is our main
device for turning things “upside down”). The Schur polynomial
$s_{\nu}\left(x_{1},x_{2},\ldots,x_{n}\right)$ was generalized to the Schur
Laurent polynomial $\overline{s}_{\nu}$.

Thus our problem was reduced to matching the nonzero coefficients in the
product
$\left(h_{a}^{-}h_{b}^{+}-h_{a-1}^{-}h_{b-1}^{+}\right)\cdot\overline{s}_{\mu}$
with the nonzero coefficients in the product
$\left(h_{b}^{-}h_{a}^{+}-h_{b-1}^{-}h_{a-1}^{+}\right)\cdot\overline{s}_{\mu}$.
The products could both be expressed using the Pieri rules, but the
differences were still a distraction. At this point, we made a fortunate
guess: We hoped it would suffice to match the nonzero coefficients in the
product $h_{a}^{-}h_{b}^{+}\cdot\overline{s}_{\mu}$ with the nonzero
coefficients in the product $h_{b}^{-}h_{a}^{+}\cdot\overline{s}_{\mu}$. More
precisely, we hoped to find such a matching that would not depend on $a$ and
$b$; then it would also provide a matching between the nonzero coefficients in
the product $h_{a-1}^{-}h_{b-1}^{+}\cdot\overline{s}_{\mu}$ and the nonzero
coefficients in the product $h_{b-1}^{-}h_{a-1}^{+}\cdot\overline{s}_{\mu}$,
and therefore by taking differences we would obtain a matching between the
nonzero coefficients in the product
$\left(h_{a}^{-}h_{b}^{+}-h_{a-1}^{-}h_{b-1}^{+}\right)\cdot\overline{s}_{\mu}$
and the nonzero coefficients in the product
$\left(h_{b}^{-}h_{a}^{+}-h_{b-1}^{-}h_{a-1}^{+}\right)\cdot\overline{s}_{\mu}$.

Thus we needed to expand $h_{a}^{-}h_{b}^{+}\cdot\overline{s}_{\mu}$. Using
the Pieri rules, this was straightforward – the answer is in Lemma 4.48. Our
problem was to connect this result with what we would similarly obtain from
expanding $h_{b}^{-}h_{a}^{+}\cdot\overline{s}_{\mu}$. In other words, we
wanted to construct a bijection
$\mathbf{f}_{\mu}:\mathbb{Z}^{n}\rightarrow\mathbb{Z}^{n}$ that would satisfy

$\left|R_{\mu,b,a}\left(\mathbf{f}_{\mu}\left(\gamma\right)\right)\right|=\left|R_{\mu,a,b}\left(\gamma\right)\right|\
\ \ \ \ \ \ \ \ \ \text{for any }a,b\in\mathbb{Z}\text{ and
}\gamma\in\mathbb{Z}^{n}.$

Fixing $\gamma\in\mathbb{Z}^{n}$, we thus were looking for an $n$-tuple $\eta$
(our $\mathbf{f}_{\mu}\left(\gamma\right)$-to-be) that would satisfy
$\left|R_{\mu,b,a}\left(\eta\right)\right|=\left|R_{\mu,a,b}\left(\gamma\right)\right|$.

In the case when $b=0$ (in which case this equality would be equivalent to
saying “$\eta\rightharpoonup\mu$ if and only if $\mu\rightharpoonup\gamma$”),
we found such an $\eta$ directly, by setting

$\eta_{i}=\mu_{i}+\mu_{i+1}-\gamma_{i+1}\ \ \ \ \ \ \ \ \ \ \text{for each
}i\in\left\\{1,2,\ldots,n\right\\},$

where indices are cyclic modulo $n$ (so that $\mu_{0}=\mu_{n}$ and
$\nu_{0}=\nu_{n}$). This formula surprised us with its cyclic symmetry (which
was not expected from the original problem, and which foreshadowed the
usefulness of Convention 3.3, although we thought nothing of it at that
point). Nevertheless, the formula failed in various examples for $b>0$, and we
could not easily fix it.

We tried to be more systematic. It was easy to rewrite the definition of
$R_{\mu,a,b}\left(\gamma\right)$ as

$\displaystyle R_{\mu,a,b}\left(\gamma\right)$
$\displaystyle=\left\\{\nu\in\mathbb{Z}^{n}\ \mid\
\left(\min\left\\{\mu_{i},\gamma_{i}\right\\}\geq\nu_{i}\geq\max\left\\{\mu_{i+1},\gamma_{i+1}\right\\}\text{
for each }i\in\left\\{1,2,\ldots,n-1\right\\}\right)\right.$ $\displaystyle\ \
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left.\text{and
}\left|\mu\right|-\left|\nu\right|=a\text{ and
}\left|\gamma\right|-\left|\nu\right|=b\right\\}.$

Thus, the size of this set would depend only

  * •

on the differences
$\min\left\\{\mu_{i},\gamma_{i}\right\\}-\max\left\\{\mu_{i+1},\gamma_{i+1}\right\\}$
for $i\in\left\\{1,2,\ldots,n-1\right\\}$ (each of which differences would
determine the “breathing space” for the corresponding $\nu_{i}$),

  * •

on the difference $\left|\mu\right|-\left|\gamma\right|$ (which would have to
equal $a-b$ in order for the two conditions
$\left|\mu\right|-\left|\nu\right|=a$ and
$\left|\gamma\right|-\left|\nu\right|=b$ to be satisfiable simultaneously),

  * •

as well as on something else we could not quite pinpoint (in order for
$\left|\mu\right|-\left|\nu\right|=a$ and
$\left|\gamma\right|-\left|\nu\right|=b$ to actually hold, as opposed to
merely $\left|\mu\right|-\left|\gamma\right|=a-b$).

Analogous observations held for $R_{\mu,b,a}\left(\eta\right)$. With Occam’s
razor in hand, we suspected that
$\left|R_{\mu,b,a}\left(\eta\right)\right|=\left|R_{\mu,a,b}\left(\gamma\right)\right|$
could best be achieved by requiring these differences to be the same for
$\left(\mu,a,b,\gamma\right)$ as for $\left(\mu,b,a,\eta\right)$. Thus, in
particular, we hoped to have

$\displaystyle\min\left\\{\mu_{i},\gamma_{i}\right\\}-\max\left\\{\mu_{i+1},\gamma_{i+1}\right\\}$
$\displaystyle=\min\left\\{\mu_{i},\eta_{i}\right\\}-\max\left\\{\mu_{i+1},\eta_{i+1}\right\\}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \text{for each
}i\in\left\\{1,2,\ldots,n-1\right\\}$

and

$\left|\mu\right|-\left|\gamma\right|=\left|\eta\right|-\left|\mu\right|.$

(Due to the “mystery ingredient”, this would likely neither be necessary nor
sufficient for
$\left|R_{\mu,b,a}\left(\eta\right)\right|=\left|R_{\mu,a,b}\left(\gamma\right)\right|$,
but it looked like the right tree to bark up.) This is a system of equations
that whose solution is neither unique nor straightforward. However, the system
was a beacon rather than a destination to us, so we merely needed something
like a good solution.

Systems of equations involving sums, differences, minima and maxima belong to
tropical geometry – a discipline we were not expert in and could not hope to
master quickly. However, we were aware of a surprisingly successful strategy
for taming such systems: detropicalization. The mainstay of this strategy is
the observation (made above in Example 3.3) that the binary operations $\min$,
$\max$, $+$ and $-$ are the addition, the “harmonic addition”191919Harmonic
addition is a binary operation defined on any semifield. It sends any pair
$\left(a,b\right)$ of elements of the semifield to
$\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}=\dfrac{ab}{a+b}$., the multiplication
and the division of a certain semifield (the min tropical semifield of
$\left(\mathbb{Z},+,0\right)$, or of whatever totally ordered abelian group
our numbers belong to). Thus, even if we could not solve our system, we could
generalize it to arbitrary semifields by replacing $\min$, $\max$, $+$ and $-$
by addition, “harmonic addition”, multiplication and division, respectively.
Thus our system would become

$\displaystyle\left(\mu_{i}+\gamma_{i}\right)/\dfrac{1}{\dfrac{1}{\mu_{i+1}}+\dfrac{1}{\gamma_{i+1}}}$
$\displaystyle=\left(\mu_{i}+\eta_{i}\right)/\dfrac{1}{\dfrac{1}{\mu_{i+1}}+\dfrac{1}{\eta_{i+1}}}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \text{for each
}i\in\left\\{1,2,\ldots,n-1\right\\}$

and

$\dfrac{\mu_{1}\mu_{2}\cdots\mu_{n}}{\gamma_{1}\gamma_{2}\cdots\gamma_{n}}=\dfrac{\eta_{1}\eta_{2}\cdots\eta_{n}}{\mu_{1}\mu_{2}\cdots\mu_{n}}.$

Renaming $\mu$, $\gamma$ and $\eta$ as $u$, $x$ and $y$, and simplifying the
fractions somewhat, we rewrote this as

$\displaystyle\left(u_{i}+x_{i}\right)\left(\dfrac{1}{u_{i+1}}+\dfrac{1}{x_{i+1}}\right)$
$\displaystyle=\left(u_{i}+y_{i}\right)\left(\dfrac{1}{u_{i+1}}+\dfrac{1}{y_{i+1}}\right)$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \text{for each
}i\in\left\\{1,2,\ldots,n-1\right\\}$

and

$y_{1}y_{2}\cdots y_{n}\cdot x_{1}x_{2}\cdots x_{n}=\left(u_{1}u_{2}\cdots
u_{n}\right)^{2}.$

This new system was a system of polynomial equations (at least after clearing
denominators), so we did the obvious thing: We left it to the computer for
small values of $n$ (specifically, $n=2$, $n=3$ and $n=4$) and looked at the
results. For $n=3$, the computer (SageMath’s solve function, to be precise)
laid out the following two solutions:

  * •

Solution 1:

$\displaystyle y_{1}$
$\displaystyle=\dfrac{u_{1}\left(u_{1}u_{2}u_{3}+x_{1}u_{2}u_{3}+x_{1}x_{2}u_{3}+x_{1}x_{2}x_{3}\right)}{u_{1}x_{2}u_{3}-x_{1}x_{2}x_{3}},$
$\displaystyle y_{2}$ $\displaystyle=\dfrac{-u_{1}u_{2}u_{3}}{x_{1}x_{3}},$
$\displaystyle y_{3}$
$\displaystyle=\dfrac{u_{2}u_{3}\left(x_{1}x_{3}-u_{1}u_{3}\right)}{u_{1}u_{2}u_{3}+x_{1}u_{2}u_{3}+x_{1}x_{2}u_{3}+x_{1}x_{2}x_{3}}.$

  * •

Solution 2:

$\displaystyle y_{1}$
$\displaystyle=\dfrac{u_{1}u_{3}\left(u_{1}u_{2}+x_{1}u_{2}+x_{1}x_{2}\right)}{x_{2}\left(u_{1}u_{3}+u_{1}x_{3}+x_{1}x_{3}\right)},$
$\displaystyle y_{2}$
$\displaystyle=\dfrac{u_{1}u_{2}\left(u_{2}u_{3}+x_{2}u_{3}+x_{2}x_{3}\right)}{x_{3}\left(u_{1}u_{2}+x_{1}u_{2}+x_{1}x_{2}\right)},$
$\displaystyle y_{3}$
$\displaystyle=\dfrac{u_{2}u_{3}\left(u_{1}u_{3}+u_{1}x_{3}+x_{1}x_{3}\right)}{x_{1}\left(u_{2}u_{3}+x_{2}u_{3}+x_{2}x_{3}\right)}.$

The computer did not know that we were trying to work over a semifield (which
had no subtraction), but we did, so we immediately discarded Solution 1 as
useless due to the minus signs. The question was whether Solution 2 would be
of any use. The omens were favorable: There were no minus signs; the
(unexpected, but not unwelcome) cyclic symmetry reared its head again;
finally, the nontrivial factors (such as $u_{1}u_{2}+x_{1}u_{2}+x_{1}x_{2}$)
had a structure that appeared in the definition of the geometric crystal
R-matrix (see, e.g., [Etingo03, (5)] or [NouYam02, (4.19)]) – a known
successful case of detropicalization.

Solution 2 turned out to be generalizable indeed. Proving that the general
formula indeed produced a solution to our system (parts (b) and (c) of Theorem
3.8) was not completely trivial, but not hard either. (The first few parts of
Lemma 3.9 were discovered along the way.) Thus we had a candidate for the map
$\mathbf{f}_{\mu}$ (and thus for the map $\varphi$, which was obtained from
$\mathbf{f}_{\mu}$ by shifting by $a$ and $b$, corresponding to the
$x_{\Pi}^{a}$ factor that we had dropped).

Why was this map $\mathbf{f}_{\mu}$ a bijection? Again, we believed that the
easiest way lay through the birational realm (i.e., we had to detropicalize).
Computer experiments suggested that $\mathbf{f}_{\mu}$ was not only a
bijection but actually an involution (part (a) of Theorem 3.8). The first
proof of this we found was the one sketched in Remark 3.18; the alternative,
computational proof that we gave first was found afterwards.

Having found our bijection $\mathbf{f}_{\mu}$, we had to retrace our steps.
Most of this was straightforward. The equality
$\left|R_{\mu,b,a}\left(\mathbf{f}_{\mu}\left(\gamma\right)\right)\right|=\left|R_{\mu,a,b}\left(\gamma\right)\right|$
still had to be proved, but this turned out to be rather easy (part (d) of
Theorem 3.8 was discovered along the way, as the missing ingredient from our
previous analysis of the size of $R_{\mu,a,b}\left(\gamma\right)$). The way
the proof was written up in the end was mostly decided by concerns of
readability rather than authenticity; we believe that, had we followed the
logic of its discovery in our writeup, we would have lost more in clarity than
would be gained in motivation. We placed the study of the birational map
$\mathbf{f}_{u}$ (Section 3) in front due to its self-contained nature and
possible applicability to different problems; likewise, Section 4 begins with
general properties of Schur Laurent polynomials and slowly progresses towards
more technical lemmas tailored for the proof of Theorem 2.2. We would not be
too surprised if some unnecessary detours were made along our way (Lemma 3.9
appears a particularly likely place for such), for which we apologize in
advance (any simplifications are appreciated).

###  5.4 The birational $R$-matrix connection

In this section, we shall connect the map $\mathbf{f}_{u}$ from our Definition
3.5 with the birational R-matrix $\eta$ defined in [LamPyl12, §6] and studied
further (e.g.) in [CheLin20].

We fix a positive integer $n$ and a semifield $\mathbb{K}$. We shall use
Convention 3.1 and Convention 3.3. Let us recall the definition of the
birational R-matrix $\eta$ (no relation to the $\eta$ in Theorem 2.2):

######  Definition 5.10.

We define a map
$\eta:\mathbb{K}^{n}\times\mathbb{K}^{n}\rightarrow\mathbb{K}^{n}\times\mathbb{K}^{n}$
as follows:

Let $a\in\mathbb{K}^{n}$ and $b\in\mathbb{K}^{n}$ be two $n$-tuples. For any
$i\in\mathbb{Z}$, define an element $\kappa_{i}\left(a,b\right)\in\mathbb{K}$
by

$\kappa_{i}\left(a,b\right)=\sum\limits_{j=i}^{i+n-1}\underbrace{b_{i+1}b_{i+2}\cdots
b_{j}}_{=\prod\limits_{p=i+1}^{j}b_{p}}\cdot\underbrace{a_{j+1}a_{j+2}\cdots
a_{i+n-1}}_{=\prod\limits_{p=j+1}^{i+n-1}a_{p}}.$

Define $a^{\prime}\in\mathbb{K}^{n}$ and $b^{\prime}\in\mathbb{K}^{n}$ by
setting

$a_{i}^{\prime}=\dfrac{a_{i-1}\kappa_{i-1}\left(a,b\right)}{\kappa_{i}\left(a,b\right)}\
\ \ \ \ \ \ \ \ \ \text{for each }i\in\left\\{1,2,\ldots,n\right\\}$

and

$b_{i}^{\prime}=\dfrac{b_{i+1}\kappa_{i+1}\left(a,b\right)}{\kappa_{i}\left(a,b\right)}\
\ \ \ \ \ \ \ \ \ \text{for each }i\in\left\\{1,2,\ldots,n\right\\}.$

Set $\eta\left(a,b\right)=\left(a^{\prime},b^{\prime}\right)$.

The map $\eta$ we just defined is known as a birational R-matrix; related maps
have previously appeared in the literature ([BraKaz00, Lemma 8.6], [Yamada01,
Definition 2.1], [Etingo03, Proposition 3.1]). In particular, the map $R$ from
[Etingo03, Proposition 3.1] is equivalent to $\eta$ (at least up to technical
issues of where it is defined202020Namely: We have defined our map $\eta$ as a
literal map
$\mathbb{K}^{n}\times\mathbb{K}^{n}\rightarrow\mathbb{K}^{n}\times\mathbb{K}^{n}$
for any semifield $\mathbb{K}$, whereas [Etingo03, Proposition 3.1] defines
$R$ as a birational map
$\left(\mathbb{C}^{\times}\right)^{n}\times\left(\mathbb{C}^{\times}\right)^{n}\dashrightarrow\left(\mathbb{C}^{\times}\right)^{n}\times\left(\mathbb{C}^{\times}\right)^{n}$.
Neither of these two settings generalizes the other, but it is not hard to
transfer identities from one to the other (as long as they are subtraction-
free, i.e., no minus signs appear in them).). Indeed, it is not hard to see
that the map $\eta$ from Definition 5.10 becomes the map $R$ from [Etingo03,
Proposition 3.1] if we set $x_{i}=b_{i+1}$ and $y_{i}=a_{i}$ and
$x_{i}^{\prime}=b_{i}^{\prime}$ and $y_{i}^{\prime}=a_{i+1}^{\prime}$ (that
is, if we define $x_{i},y_{i},x_{i}^{\prime},y_{i}^{\prime}$ this way, then
the equalities [Etingo03, (8), (9) and (10)] are satisfied, so that we have
$R\left(x,y\right)=\left(x^{\prime},y^{\prime}\right)$ where $R$ is as defined
in [Etingo03, Proposition 3.1]). This birational R-matrix $R$ has its origins
in the theory of geometric crystals and total positivity. A related map is the
transformation $\left(x,a\right)\mapsto\left(y,b\right)$ in [NouYam02, §2.2]
(see also [Zygour18]).

Now, we shall see that the map $\eta$ is intimately related to our map
$\mathbf{f}_{u}$ (even though $\mathbf{f}_{u}$ transforms a single $n$-tuple
$x$ into a single $n$-tuple $y$ using the fixed $n$-tuple $u$, while $\eta$
takes a pair of two $n$-tuples to another such pair). In order to state this
relation, we define some more notation:

######  Definition 5.11.

If $a\in\mathbb{K}^{n}$ and $b\in\mathbb{K}^{n}$ are two $n$-tuples, then we
define two new $n$-tuples $ab\in\mathbb{K}^{n}$ and
$\dfrac{a}{b}\in\mathbb{K}^{n}$ by setting

$\left(ab\right)_{i}=a_{i}b_{i}\ \ \ \ \ \ \ \ \ \ \text{and}\ \ \ \ \ \ \ \ \
\ \left(\dfrac{a}{b}\right)_{i}=\dfrac{a_{i}}{b_{i}}\ \ \ \ \ \ \ \ \ \
\text{for each }i\in\left\\{1,2,\ldots,n\right\\}.$

We can now express the map $\mathbf{f}_{u}$ from Definition 3.5 through the
map $\eta$ from Definition 5.10 as follows:

######  Theorem 5.12.

Let $u\in\mathbb{K}^{n}$ and $x\in\mathbb{K}^{n}$ be two $n$-tuples. Let
$\left(a^{\prime},b^{\prime}\right)=\eta\left(u,x\right)$. Then,

$\mathbf{f}_{u}\left(x\right)=u\dfrac{a^{\prime}}{b^{\prime}}.$

######  Proof 5.13 (Proof of Theorem 5.12.).

Set $a=u$ and $b=x$. We shall use the notations $\kappa_{i}\left(a,b\right)$
from Definition 5.10 and the notations $t_{r,j}$ and $y$ from Definition 3.5.
Then, $\mathbf{f}_{u}\left(x\right)=y$ (by Definition 3.5).

For each $i\in\mathbb{Z}$, we have

$\displaystyle\kappa_{i}\left(a,b\right)$
$\displaystyle=\sum\limits_{j=i}^{i+n-1}b_{i+1}b_{i+2}\cdots b_{j}\cdot
a_{j+1}a_{j+2}\cdots a_{i+n-1}\ \ \ \ \ \ \ \ \ \ \left(\text{by the
definition of }\kappa_{i}\left(a,b\right)\right)$
$\displaystyle=\sum\limits_{j=i}^{i+n-1}x_{i+1}x_{i+2}\cdots x_{j}\cdot
u_{j+1}u_{j+2}\cdots u_{i+n-1}\ \ \ \ \ \ \ \ \ \ \left(\text{since }a=u\text{
and }b=x\right)$ $\displaystyle=\sum\limits_{k=0}^{n-1}x_{i+1}x_{i+2}\cdots
x_{i+k}\cdot u_{i+k+1}u_{i+k+2}\cdots u_{i+n-1}$ $\displaystyle\ \ \ \ \ \ \ \
\ \ \ \ \ \ \ \ \ \ \ \ \left(\text{here, we have substituted }i+k\text{ for
}j\text{ in the sum}\right)$ $\displaystyle=t_{n-1,i}$ (74)

(since the definition of $t_{n-1,i}$ yields
$t_{n-1,i}=\sum\limits_{k=0}^{n-1}x_{i+1}x_{i+2}\cdots x_{i+k}\cdot
u_{i+k+1}u_{i+k+2}\cdots u_{i+n-1}$).

However,
$\left(a^{\prime},b^{\prime}\right)=\eta\left(u,x\right)=\eta\left(a,b\right)$
(sicne $u=a$ and $x=b$). Hence, Definition 5.10 yields that

$a_{i}^{\prime}=\dfrac{a_{i-1}\kappa_{i-1}\left(a,b\right)}{\kappa_{i}\left(a,b\right)}\
\ \ \ \ \ \ \ \ \ \text{for each }i\in\left\\{1,2,\ldots,n\right\\}$

and

$b_{i}^{\prime}=\dfrac{b_{i+1}\kappa_{i+1}\left(a,b\right)}{\kappa_{i}\left(a,b\right)}\
\ \ \ \ \ \ \ \ \ \text{for each }i\in\left\\{1,2,\ldots,n\right\\}.$

Hence, for each $i\in\left\\{1,2,\ldots,n\right\\}$, we have

$\displaystyle a_{i}^{\prime}/b_{i}^{\prime}$
$\displaystyle=\dfrac{a_{i-1}\kappa_{i-1}\left(a,b\right)}{\kappa_{i}\left(a,b\right)}/\dfrac{b_{i+1}\kappa_{i+1}\left(a,b\right)}{\kappa_{i}\left(a,b\right)}=\dfrac{a_{i-1}\kappa_{i-1}\left(a,b\right)}{b_{i+1}\kappa_{i+1}\left(a,b\right)}=\dfrac{a_{i-1}t_{n-1,i-1}}{b_{i+1}t_{n-1,i+1}}$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
\left(\begin{array}[c]{c}\text{since
}\kappa_{i-1}\left(a,b\right)=t_{n-1,i-1}\text{ (by (\ref{pf.thm.eta-f1.k=t}),
applied to }i-1\text{ instead of }i\text{)}\\\ \text{and
}\kappa_{i+1}\left(a,b\right)=t_{n-1,i+1}\text{ (by (\ref{pf.thm.eta-f1.k=t}),
applied to }i+1\text{ instead of }i\text{)}\end{array}\right)$ (77)
$\displaystyle=\dfrac{u_{i-1}t_{n-1,i-1}}{x_{i+1}t_{n-1,i+1}}\ \ \ \ \ \ \ \ \
\ \left(\text{since }a=u\text{ and }b=x\right)$ $\displaystyle=y_{i}/u_{i}$
(78)

(since the definition of $y$ yields
$y_{i}=u_{i}\cdot\dfrac{u_{i-1}t_{n-1,i-1}}{x_{i+1}t_{n-1,i+1}}$). Now, for
each $i\in\left\\{1,2,\ldots,n\right\\}$, we have

$\displaystyle\left(u\dfrac{a^{\prime}}{b^{\prime}}\right)_{i}$
$\displaystyle=u_{i}\dfrac{a_{i}^{\prime}}{b_{i}^{\prime}}\ \ \ \ \ \ \ \ \ \
\left(\text{by Definition \ref{def.n-tup-frac}}\right)$
$\displaystyle=u_{i}\cdot\underbrace{a_{i}^{\prime}/b_{i}^{\prime}}_{\begin{subarray}{c}=y_{i}/u_{i}\\\
\text{(by (\ref{pf.thm.eta-f1.fracs}))}\end{subarray}}=u_{i}\cdot
y_{i}/u_{i}=y_{i}.$

In other words, $u\dfrac{a^{\prime}}{b^{\prime}}=y$. Comparing this with
$\mathbf{f}_{u}\left(x\right)=y$, we obtain
$\mathbf{f}_{u}\left(x\right)=u\dfrac{a^{\prime}}{b^{\prime}}$. This proves
Theorem 5.12.

We finish by stating some “gauge-invariance” properties for $\mathbf{f}_{u}$
and $\eta$:

######  Proposition 5.14.

Let $g,u,x\in\mathbb{K}^{n}$. Then,
$\mathbf{f}_{gu}\left(gx\right)=g\mathbf{f}_{u}\left(x\right)$.

######  Proposition 5.15.

Let $g,a,b\in\mathbb{K}^{n}$. Let
$\left(a^{\prime},b^{\prime}\right)=\eta\left(a,b\right)$. Then,
$\left(ga^{\prime},gb^{\prime}\right)=\eta\left(ga,gb\right)$.

Both of these propositions can be proved by fairly simple computations, which
are left to the reader.

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