Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2007 / B / 6

For an integer partition $\lambda =({\lambda }_1,\dots ,{\lambda }_r)$, its Young diagram has ${\lambda }_i$ left-justified cells in row $i$. A Young tableau of shape $\lambda$ is a filling of its cells. Under the usual standard numbering convention, a standard Young tableau uses each integer $1,\dots ,n$ once and increases strictly along every row and down every column. For a different fixed ordered set of distinct labels, the same inequalities define the corresponding ordered-label version, which becomes standard after replacing its labels by their ranks.

First convert the printed cycle to one-line notation:Apply row insertion to these values, recording the insertion time in each new cell. The successive insertion-tableau row lists areTherefore the Robinson-Schensted-Knuth correspondence givesBoth have shape $(4,2,1,1)$.

First prove preservation of the row and column inequalities for distinct ordered labels. If $b>c_l$, row insertion appends it to the first row, and the inequalities remain valid. Otherwise let $j$ be the first position with $c_j>b$. Replace $c_j$ by $b$ and insert the bumped $c_j$ in the second row. Its first larger entry is some $d_k$ with $k\le j$, because $d_j>c_j$. Replace $d_k$ by $c_j$ and put the bumped $d_k$ in the new third-row cell.

The first-larger-entry choices preserve strict row order. At column $k$, if $k<j$ the first-row entry is $c_k<c_j$, while if $k=j$ it is the new $b<c_j$. At column $j$ when $k<j$, the unchanged lower entry satisfies $d_j>c_j>b$. Every other old column keeps its previous inequalities. The new third-row entry is larger than the second-row first entry: for $k=1$ this is $d_1>c_j$, while for $k>1$ it follows from $d_k>d_1$. Thus the new shape $(l,l,1)$ is a Young diagram with strictly increasing rows and columns. This proves the insertion assertion for the ordered-label convention.

There is a numbering qualification in the literal standard convention of part (a). If $T$ uses exactly $1,\dots ,2l$, an arbitrary missing natural number $b$ need not be $2l+1$. For example, inserting six into rows $(1,2);(3,4)$ produces $(1,2,6);(3,4)$, which has the order inequalities but not the labels $1,\dots ,5$. Thus standardness with consecutive labels requires $b=2l+1$, or increasing relabelling afterward. The printed wording is retained in the private transcription; row insertion itself preserves the needed order structure.

The Robinson-Schensted-Knuth correspondence is a bijection between permutations and ordered pairs of standard Young tableaux of the same shape. Hence the class with shape $\lambda$ has size $(f^{\lambda }{)}^2$. For $\lambda =(4,2,1,1)$, the hook lengths arewith product $448$. The hook-length formula gives $f^{\lambda }=8!/448=90$. Therefore