9
$\begingroup$

It is well known via the RSK-correspondence that the length of the longest decreasing subsequence in a permutation $\pi \in S_n$ is the length of the longest column of the insertion tableau of $\pi$. (The insertion tableau and the recording tableau produced by this algorithm have the same shape.)

Robinson–Schensted correspondence - Wikipedia

My question is what else can be gleaned from the RSK correspondence in terms of, say,

a) the length of the next longest decreasing subsequence in $\pi$?

b) the number of longest decreasing subsequences in $\pi$, given the fact that there is exactly one column of maximum length?

c) can we say more about the above two questions if we knew that $\pi$ was an involution? (If $\pi$ happens to be an involution, then insertion tableau and recording tableau produced are equal.)

$\endgroup$
4
  • $\begingroup$ You probably know this, but I figure it's worth reminding people that π is an involution if and only if P and Q (the tableaux under RSK) are equal. $\endgroup$ Commented Nov 10, 2010 at 14:13
  • $\begingroup$ Agreed. I edited accordingly. $\endgroup$ Commented Nov 11, 2010 at 2:32
  • $\begingroup$ @Vasu-vineet, while you do have a link to a page explaining the Robinson-Schensten-(Knuth) RSK correspondence, you never specifically state the full name of the algorithm. Others reading this page would be a bit mystified and may not bother clicking through to read the details of concept which establishes a bijective correspondence between elements of the symmetric group $S_n$ and pairs of standard Young tableaux of the same shape. Also, your wikipedia link has malformed characters in it and doesn't work. Try en.wikipedia.org/wiki/Robinson-Schensted_algorithm $\endgroup$ Commented Nov 11, 2010 at 2:56
  • $\begingroup$ The condition that there is one column of maximum length is equivalent to $\pi$ not having two disjoint decreasing subsequences of maximum length. $\endgroup$ Commented May 14 at 14:14

2 Answers 2

7
$\begingroup$

For a version of a), see C. Greene, An extension of Schensted's theorem, Advances in Mathematics, 1974. Note: the result is beautiful, but the statement is a bit delicate.

$\endgroup$
3
1
$\begingroup$

To complete Igor's answer (whose C. Greene's quotation is perfect), for a permutation $\pi \in S_n$ with $T$ its insertion tableau. Let $$ \lambda_1\geq \cdots \geq \lambda_q $$ be the shape of $T$ (sequence of row lengths). Then

  • $\lambda_1$ the length of the longest increasing subsequence
  • $\lambda_1+\lambda_2$ is the length of the longest subsequence which is the shuffle of two increasing subsequences
  • ...
  • $\lambda_1+\cdots \lambda_k$ is the length of the longest subsequence which is the shuffle of $k$ increasing subsequences.

    Marcel-Paul Schützenberger used to call these numbers (during my Ph. D. supervision) "Greene Invariants".

    For example for the permutation $13582467$ with insertion tableau
    $$ T = \begin{bmatrix} 1 & 3 & 5 & 8\cr 2 & 4 \cr 6 \cr 7\cr \end{bmatrix} $$ of shape $(\lambda_1,\lambda_2,\lambda_3,\lambda_4)=(4,2,1,1)$, the Greene invariants are $(4,6,7,8)$. Reversing the orders (reading the permutation from right to left or reversing numbers $k\mapsto n+1-k$) you can get dual statements.

  • $\endgroup$

    You must log in to answer this question.

    Not the answer you're looking for? Browse other questions tagged .