## Tableaux and Words

In our context, a word is a sequence of positive integers; concatenation of words is denoted by $w * w'.$ Given a skew SSYT $T,$ the corresponding word $w(T)$ is obtained by taking the tableau entries from left to right, then bottom to top. For example the following skew SSYT gives $w(T) = (1, 4, 2, 4, 5, 2, 3, 4, 5, 1, 1, 4, 6).$

The following are quite easy to verify.

• If $T$ is an SSYT, then it can be recovered from $w(T)$.
• We pick the longest weakly increasing sequence starting from the first number; this forms the last row. Repeat with the next number to obtain the second-to-last row, etc. E.g. given (4, 5, 3, 3, 4, 6, 1, 2, 3, 3, 5, 7), we get a three-row SSYT comprising of (1, 2, 3, 3, 5, 7), (3, 3, 4, 6) and (4, 5).
• Not every word occurs as $w(T)$ from some SSYT $T$.
• E.g. (1, 2, 1, 2) does not come from an SSYT.
• However, every word occurs as $w(T)$ for some skew SSYT. E.g. just place the $n$ elements of the sequence diagonally: 1st row and $n$-th square, 2nd row and $(n-1)$-th square etc.
• In general, a skew SSYT $T$ cannot be uniquely recovered from $w(T).$
• E.g. for (1, 2), we can either take the SSYT (1, 2), or the skew diagram (2,1)/(1) and fill in 1, 2.

## Words and Row Insertion

Next we formulate rules such that row-inserting (bumping) an entry $m$ into an SSYT $T$ is equivalent to concatenation:

$w(T \leftarrow m) \equiv w(T) * (m).$

Suppose we have a row $(v_1, \ldots, v_n)$ in the SSYT with $m$ added to the end, and assume

$v_1 \le v_2 \le \ldots \le v_{i-1} \le m < v_{i} \le \ldots \le v_n,$

so that $m$ bumps off $v_{i}$ and we get $(v_1, \ldots, v_n, m) \mapsto (v_i, v_1,\ldots, v_{i-1}, m, v_{i+1}, \ldots, v_n)$. Break this down into a series of steps:

\begin{aligned} (v_1, \ldots, \boxed{v_{n-1}, v_n, m}) &\mapsto (v_1, \ldots, v_{n-1}, m, v_n) \\&\ldots \\&\mapsto (v_1, \ldots, v_{i-1}, v_i, m, \ldots, v_n).\end{aligned}

From this we obtain the first rule:

Rule R1. For any consecutive $(a,b,c)$ in the word satisfying $c , we switch $(a,b,c) \mapsto (a,c,b).$

Next, to shift $v_i$ to the beginning of the row, we break it down as follows:

\begin{aligned}(v_1, \ldots, \boxed{v_{i-1}, v_i, m}, \ldots, v_n) &\mapsto(v_1, \ldots, v_{i}, v_{i-1}, m, \ldots, v_n) \\ &\ldots\\ &\mapsto (v_i, v_1, \ldots, v_{i-1}, m, \ldots, v_n).\end{aligned}

From here, we obtain the second rule:

Rule R2. For any consecutive $(a,b,c)$ in the word satisfying $a \le c , we switch $(a,b,c) \mapsto (b,a,c).$

### Example

Suppose we insert $(1, 2, 2, 3, 4, 5, 5) \leftarrow 3$. Applying R1 and R2 gives:

\begin{aligned} &(1, 2, 3, 4, \overset{R1}{\boxed{5, 5, 3}}) \mapsto (1, 2, 3, \overset{R1}{\boxed{4, 5, 3}}, 5)\mapsto (1, 2, \overset{R2}{\boxed{3, 4, 3}}, 5, 5)\\ \mapsto &(1, \overset{R2}{\boxed{2, 4, 3}}, 3, 5, 5) \mapsto (\overset{R2}{\boxed{1, 4, 2}}, 3, 3, 5, 5)\mapsto (4, 1, 2, 3, 3, 5, 5).\\ \end{aligned}

Now we wish to revert the inserting process, so we will consider reverse of R1 and R2 as well, denoted by R1′ and R2′ respectively.

## Summary of Rules

Given a consecutive triplet $(a,b,c)$ in the word, we consider the following distinct cases:

• $a \le b \le c$ or $a > b > c$: no change;
• $c < a \le b$: switch to $(a, c, b)$ by R1;
• $a \le c < b$: switch to $(b, a, c)$ by R2;
• $b < a \le c$: switch to $(a, c, b)$ by R1′;
• $b \le c < a$: switch to $(b, a, c)$ by R2′.

One checks that these cases are disjoint and cover all possibilities. Two words are said to be Knuth-equivalent, written as $w\equiv w',$ if we can obtain one from the other via the above 4 operations.

### Mnemonic

The following diagram helps us in recalling R1 and R2.

\begin{aligned} c < a\le b\ : \quad &\left(\ \boxed{\begin{matrix}a & b\end{matrix}}\ \leftarrow\ c\ \right)\ = \boxed{\begin{matrix} c & b \\ a\end{matrix}} \\ a\le c < b\ :\quad &\left(\ \boxed{\begin{matrix}a & b\end{matrix}}\ \leftarrow\ c\ \right)\ = \boxed{\begin{matrix} a & c \\ b\end{matrix}} \end{aligned}.

## Consequences of Equivalence

From our construction of R1 and R2, we immediately have:

Proposition 1. $w(T \leftarrow m) \equiv w(T) * (m)$.

The following result is less trivial.

Proposition 2. If $T'$ is obtained by sliding an inner corner of skew SSYT $T,$ then $w(T') \equiv w(T).$

Proof

Clearly, horizontal sliding does not change the word, so we will only consider vertical sliding. By trimming off excess elements in the two rows:

we may assume the two rows have the same length. We will prove that the corresponding words are Knuth-equivalent via induction on p. Write $\vec n$ for $(n_1, \ldots, n_q),$ $\vec m$ for $(m_1, \ldots, m_p)$ etc. First consider the case $p=0,$ i.e. we need to show $x \vec n \vec v \equiv \vec n x \vec v.$

• Imagine performing row insertion for $v_1$ into $x \vec n.$ Since $x \le v_1 this bumps out $n_1$ and we transform $x\vec n v_1$ to $n_1 x v_1 n_2 n_3 \ldots.$
• Inserting $v_2$ into this row then bumps out $n_2,$ and we obtain $n_1 n_2 x v_1 v_2 n_3 \ldots.$ Moving on, we eventually obtain $\vec n x \vec v$ as desired.

Now assume $p>0.$ Write $\vec u' = (u_2, \ldots, u_p)$ and  $\vec m' = (m_2, \ldots, m_p)$ so that $\vec u = (u_1) * \vec u'$ and $\vec m = (m_1) * \vec m'.$ We need to show

$m_1 \vec m' \boxed{x \vec n u_1 \vec u'} \vec v\ \overset{?}\equiv\ m_1 \vec m' \boxed{\vec n u_1 \vec u' x} \vec v.$

Inserting $u_1$ into the row $m_1 \vec m' x \vec n$ bumps off $m_1$ which gives $m_1 u_1 \vec m' x \vec n.$ Thus the LHS is equivalent to $m_1 u_1 \vec m' x \vec n \vec u' \vec v.$ By induction hypothesis for $p-1,$ this is equivalent to $m_1 u_1 \vec m' \vec n \vec u' x \vec v.$

On the other hand, on the RHS above we insert $u_1$ into the row $m_1 \vec m' \vec n$ to give $m_1 u_1 \vec m' \vec n.$ Hence the RHS is equivalent to $m_1 u_1 \vec m' \vec n \vec u' x \vec v.$ This completes the proof. ♦

Corollary. Every word $w$ is Knuth-equivalent to $w(T)$ for some SSYT $T.$

Proof

Indeed write $w = w(T')$ for some skew SSYT $T'.$ Apply the sliding algorithm to it to obtain an SSYT $T.$ By proposition 2, $w(T') \equiv w(T).$ ♦

This $T$ will later turn out to be unique. 🙂

Lemma. If the skew tableau $T$ is cut into $T'$ and $T''$ by a horizontal or vertical line (where $T'$ is respectively left or above), then $w(T) \equiv w(T') * w(T'').$
In the case of horizontal cut, we in fact have $w(T) = w(T') * w(T'')$ so the result is trivial. Now suppose we have a vertical cut as follows.
Start removing the inner corners from $T',$ starting from the rightmost column. E.g. in the above diagram, we remove the corner labelled $a,$ then the one above etc. Note that upon removing $a,$ the column beneath it gets shifted up one unit. Eventually, we obtain the LHS diagram. By proposition 2, the above two tableaux have equivalent words and the tableau $T_1$ on the right gives $w(T_1) = w(T') * w(T'').$ ♦