Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 103 / 5

For a standard Young tableau $T$, let $c_T(r)=j-i$ when its entry $r$ lies in cell $(i,j)$. Its content vector of a standard Young tableau is $C(T)=(c_T(1),\dots ,c_T(n))$. We use the following sign convention for axial distance in a Young tableau:This is the axial distance from $j$ to $i$. In particular, write $d=c_T(r+1)-c_T(r)$ for the adjacent transposition $s_r=(rr+1)$.

Here is Young orthogonal form, with one orthonormal basis vector $w_T$ for each standard Young tableau of a fixed shape:If $s_{r}T$ is not standard, omit that term. In this case $r,r+1$ occupy adjacent cells in one row or column, so $d=1$ or $d=-1$, respectively, and the scalar action is $+1$ or $-1$. On a standard pair $T,R=s_{r}T$, $|d|\ge 2$ and the matrix isThis is a real orthogonal involution.

To prove the formula, explicitly use the permitted Young seminormal form theorem: the Specht module has a tableau basis $v_T$ whose vectors are joint eigenvectors of the Young–Jucys–Murphy elements, and whose actions, on an admissible pair oriented so that tableau length increases from $T$ to $R$, areNonadmissible swaps have the scalar actions just described. The theorem supplies these as one globally consistent representation, not as independently chosen two-dimensional blocks.

Average a positive definite Hermitian inner product over $S_n$. Distinct tableau lines are orthogonal because their commuting self-adjoint Young–Jucys–Murphy elements have distinct joint weights. Also $s_r$ is self-adjoint, so the two seminormal equations implyUpon setting $w_T=v_T/\Vert v_T\Vert$, the coefficient from $w_T$ to $w_R$ becomes $\Vert v_R\Vert /\Vert v_T\Vert =\sqrt{1-d^{-2}}$; the reverse coefficient is the same. All phases are consistent because the seminormal basis was global, and the group relations remain true under this change of basis. This proves Young orthogonal form.

For $\lambda =(n)$, there is exactly one standard Young tableau, the row $1,2,\dots ,n$. Its content vector of a standard Young tableau isEvery adjacent distance is $1$, so $s_r{w}_T=w_T$ for every $r$. Thus $\boxed{S^{(n)}\cong {\mathbb{C}}_{\mathrm{triv}}}$ is the trivial representation.

For $\lambda =(1^n)$, there is exactly one standard Young tableau, the column with entries $1,2,\dots ,n$ from top to bottom. Its content vector of a standard Young tableau isEvery adjacent distance is $-1$, so Young orthogonal form gives $s_r{w}_T=-w_T$ for every $r$. Thereforethe sign representation.

Assume $n\ge 2$. The standard Young tableaux of shape $(n-1,1)$ are $T_k$, for $2\le k\le n$: the sole second-row entry is $k$, and the first row isThis lists all of them, since $1$ must occupy the first box and the rest of the first row is forced to increase. Their contents areWrite $w_k=w_{T_k}$. The only nontrivial adjacent swaps exchange the lower entry with $k-1$ or $k+1$. Substituting the content differences into Young orthogonal form yields all generator actions:In the first line at $k=2$, the square-root coefficient is zero and the action is $-w_2$, so no vector $w_1$ is needed. The second line is used only when $k<n$, since the generator $s_n$ does not exist.

The corresponding Specht module is the sum-zero standard representation of the symmetric group:A concrete orthonormal realization isPermuting coordinates gives exactly the actions above, linking this orthogonal tableau basis with the Gelfand–Tsetlin basis found in Question 1. Here the printed $s_j$ are the adjacent Coxeter generators, rather than products of all generators.