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  "body": "\n## Statement\n\nLet $n \\ge 1$ and let $S_n$ be the [[symmetric group|0515A6D2Y1]] of [[permutations|05RWFKVTJ7]] of $\\set{1, \\dots, n}$.\nRow insertion defines a map $\\sigma \\mapsto (P(\\sigma), Q(\\sigma))$ from $S_n$ to pairs of standard [[Young tableaux|Q6TH9PH2R7]] of the same shape, a [[partition|0JV30XN73Z]] of $n$.\n\n1. This map is a bijection onto the set of all pairs $(P, Q)$ of standard Young tableaux of the same shape with $n$ cells.\n2. For every $\\sigma \\in S_n$, $P(\\sigma^{-1}) = Q(\\sigma)$ and $Q(\\sigma^{-1}) = P(\\sigma)$.\n3. The length of the first row of the common shape equals the length of a longest increasing subsequence of $\\sigma(1), \\dots, \\sigma(n)$.\n4. The number of rows of the common shape equals the length of a longest decreasing subsequence of $\\sigma(1), \\dots, \\sigma(n)$.\n\n## Remarks\n\nRow insertion of a number $x$ into a tableau works as follows.\nIf $x$ is larger than every entry of the first row, put $x$ at the end of that row.\nOtherwise $x$ replaces the smallest entry of the row that is larger than $x$, and the replaced entry is inserted into the next row in the same way.\nThe tableau $P(\\sigma)$ is obtained by inserting $\\sigma(1), \\dots, \\sigma(n)$ in order into the empty tableau.\nThe tableau $Q(\\sigma)$ records the step: it has entry $k$ in the cell that was created at step $k$.\nEach step can be undone from the position of the new cell, which gives the inverse map and proves part 1.\nParts 3 and 4 are due to Schensted.\n\nExample: for $\\sigma = 2\\,3\\,1$ in one-line notation, inserting $2$ and $3$ gives the row $2\\,3$.\nInserting $1$ bumps $2$ into the second row, so $P = \\begin{smallmatrix} 1 \u0026 3 \\\\ 2 \u0026 \\end{smallmatrix}$ and $Q = \\begin{smallmatrix} 1 \u0026 2 \\\\ 3 \u0026 \\end{smallmatrix}$.\nThe first row has length $2$, the length of the increasing subsequence $2\\,3$.\n\n## Explanations\n\n### L1 human\n\nThe correspondence turns a permutation into two tableaux of the same shape, and back again without loss.\nIt is like a card-sorting game in which each new card pushes a bigger card down to the next row.\nThe shape you end with records how long the increasing and decreasing runs hidden in the permutation can be.\n\n### L2 human\n\nCounting both sides of part 1 gives $\\sum_{\\lambda \\vdash n} (f^\\lambda)^2 = n!$, where $f^\\lambda$ is the number of standard tableaux of shape $\\lambda$.\nThis matches the identity that the squares of the dimensions of the irreducible representations of $S_n$ add up to $\\abs{S_n}$.\nBy part 2, involutions correspond to single tableaux, so the number of involutions in $S_n$ is $\\sum_{\\lambda \\vdash n} f^\\lambda$.\nThe [[Robinson–Schensted–Knuth correspondence|8JW8F4VVED]] extends the bijection to matrices with nonnegative integer entries and semistandard tableaux.\n",
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