Tableaux and Words
In our context, a word is a sequence of positive integers; concatenation of words is denoted by Given a skew SSYT the corresponding word is obtained by taking the tableau entries from left to right, then bottom to top. For example the following skew SSYT gives
The following are quite easy to verify.
- If is an SSYT, then it can be recovered from .
- 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 from some SSYT .
- E.g. (1, 2, 1, 2) does not come from an SSYT.
- However, every word occurs as for some skew SSYT. E.g. just place the elements of the sequence diagonally: 1st row and -th square, 2nd row and -th square etc.
- In general, a skew SSYT cannot be uniquely recovered from
- 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 into an SSYT is equivalent to concatenation:
Suppose we have a row in the SSYT with added to the end, and assume
so that bumps off and we get . Break this down into a series of steps:
From this we obtain the first rule:
Rule R1. For any consecutive in the word satisfying , we switch
Next, to shift to the beginning of the row, we break it down as follows:
From here, we obtain the second rule:
Rule R2. For any consecutive in the word satisfying , we switch
Suppose we insert . Applying R1 and R2 gives:
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 in the word, we consider the following distinct cases:
- or : no change;
- : switch to by R1;
- : switch to by R2;
- : switch to by R1′;
- : switch to by R2′.
One checks that these cases are disjoint and cover all possibilities. Two words are said to be Knuth-equivalent, written as if we can obtain one from the other via the above 4 operations.
The following diagram helps us in recalling R1 and R2.
Consequences of Equivalence
From our construction of R1 and R2, we immediately have:
Proposition 1. .
The following result is less trivial.
Proposition 2. If is obtained by sliding an inner corner of skew SSYT then
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 for for etc. First consider the case i.e. we need to show
- Imagine performing row insertion for into Since this bumps out and we transform to
- Inserting into this row then bumps out and we obtain Moving on, we eventually obtain as desired.
Now assume Write and so that and We need to show
Inserting into the row bumps off which gives Thus the LHS is equivalent to By induction hypothesis for this is equivalent to
On the other hand, on the RHS above we insert into the row to give Hence the RHS is equivalent to This completes the proof. ♦
Corollary. Every word is Knuth-equivalent to for some SSYT
Indeed write for some skew SSYT Apply the sliding algorithm to it to obtain an SSYT By proposition 2, ♦
This will later turn out to be unique. 🙂
Finally, the following result is interesting but will not be used.
Lemma. If the skew tableau is cut into and by a horizontal or vertical line (where is respectively left or above), then
In the case of horizontal cut, we in fact have so the result is trivial. Now suppose we have a vertical cut as follows.
Start removing the inner corners from starting from the rightmost column. E.g. in the above diagram, we remove the corner labelled then the one above etc. Note that upon removing 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 on the right gives ♦