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For the sum-zero standard representation of the symmetric group, the vectors $v_j=e_1+\cdots +e_j-j{e}_{j+1}$ form a Gelfand–Tsetlin basis. The Young–Jucys–Murphy element $X_i$ has eigenvalue $i-1$ for $i\le j$, $-1$ for $i=j+1$, and $i-2$ for $i>j+1$. Normalizing by $\sqrt{j(j+1)}$ gives an orthonormal basis corresponding to tableaux of shape $(n-1,1)$ whose second-row entry is $j+1$.

Take $m=2$ and $\lambda =(2,1)$, so $\lambda +\delta =(3,1)$. A transposition in $S_3$ has one singleton cycle and one two-cycle. The Frobenius alternant character formula therefore givesIn the coordinate permutation representation on ${\mathbb{C}}^3$, a transposition fixes one basis vector, so its character is $1$. This representation is the direct sum of the invariant line of constant vectors and the standard representation of the symmetric group, whose vectors have coordinate sum zero. Subtracting the trivial character gives ${\chi }_V((12))=1-1=0$, as required.

Let $e_1,\dots ,e_n$ be the coordinate vectors and chooseThese vectors lie in the sum-zero standard representation of the symmetric group. They are mutually orthogonal for the usual Hermitian inner product, with $\Vert v_j{\Vert }^2=j(j+1)$, so they form a basis of its $(n-1)$-dimensional space.

If $i\le j$, every transposition in $X_i$ exchanges two coordinates equal to $1$, so each fixes $v_j$. If $i=j+1$, summing the $j$ transpositions gives coordinate $-1$ in each of positions $1,\dots ,j$ and coordinate $j$ in position $j+1$, hence $X_{j+1}{v}_j=-v_j$. Finally, if $i>j+1$, the $i$th coordinate of $v_j$ is zero. In the sum of the $i-1$ transpositions, any coordinate in its support retains its original value in $i-2$ terms and becomes zero in the remaining term; the $i$th output coordinate is the sum of the original first $i-1$ coordinates, which is zero.

Thus all actions are diagonal in this basis:This supplies an explicit Gelfand–Tsetlin basis of the standard 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.

The implication is false. ChooseThe inequalities $0\le 4<9$ and the relation $13=1\cdot 9+4$ hold. At cell $(1,4)$, the hook length is $12-4+1=9=a{p}^k$.

For the partition of an integer $(12,1)$, the Hook-length formula giveswhich is divisible by $3$. This also agrees with the degree of the standard representation of the symmetric group $S_{13}$.

The obstruction is visible in the core of a partition: removing the horizontal rim hook of length nine leaves $\gamma =(3,1)$, which has no hook length nine and hence is the $9$-core. Its degree is ${\chi }^{(3,1)}(1)=3$. The large hook ensures the required weight of a partition here, but the core degree is not coprime to $p$.

The normis a positive integer, because it is the sum of the squares of the multiplicities of the irreducible constituents. The character restriction norm bound and $[G:K]=3$ make it at most three. It is therefore $1$, $2$, or $3$.

All three values occur:

Permuting the gaps, or equivalently relabelling the particles, leaves both quadratic forms invariant. The three-dimensional permutation representation splits as the rigid-rotation line spanned by $(1,1,1)$ and the two-dimensional sum-zero standard representation of the symmetric group.

The two displayed vibrational vectors merely choose a basis of that sum-zero plane. A permutation generally mixes them, but their common frequency leaves the whole plane invariant. This is precisely the degenerate normal modes from permutation symmetry required by the physical symmetry.

The standard representation of the symmetric group $S_4$ restricts irreducibly to $A_4$. Its character has value $3$ at the identity and $-1$ on each nonidentity element of the normal Klein four-group. Restricted to that subgroup, it is the sum of the three nontrivial linear characters.