[proofplan] We compare each Jucys-Murphy element with the difference of two central transposition class sums. The branching rule identifies a seminormal basis vector $v_T$ with the unique path of shapes obtained by restricting along $S_1\le S_2\le\cdots\le S_n$, and [Schur's lemma](/theorems/2414) makes the relevant central sums act by scalars on each Specht summand. The scalar difference is the difference between the sums of contents of two consecutive shapes, hence the content of the single box added at that stage. [/proofplan] [step:Express $X_k$ as a difference of central transposition class sums] Define $\Omega_0\in\mathbb C[S_n]$ by $\Omega_0=0$. For each integer $m$ with $1\le m\le n$, define the transposition class sum $\Omega_m\in\mathbb C[S_m]\subseteq\mathbb C[S_n]$ by \begin{align*} \Omega_m=\sum_{1\le i<j\le m}(i\ j), \end{align*} where the sum is empty when $m=1$, so $\Omega_1=0$. If $2\le k\le n$, then the summands of $\Omega_k$ are precisely the transpositions $(i\ j)$ with $1\le i<j\le k$, while the summands of $\Omega_{k-1}$ are precisely those with $1\le i<j\le k-1$. Subtracting cancels exactly the transpositions not involving $k$ and gives \begin{align*} \Omega_k-\Omega_{k-1}=\sum_{i=1}^{k-1}(i\ k)=X_k. \end{align*} For $k=1$, the identity reads $X_1=\Omega_1-\Omega_0$, and both sides are $0$ by definition. [/step] [step:Attach to each seminormal vector its branching path of shapes] Let $S_0$ denote the trivial subgroup of $S_n$, and let $\mathbb C[S_0]$ denote its complex group algebra inside $\mathbb C[S_n]$. For each integer $m$ with $1\le m\le n$, let $S_m$ denote the symmetric group on $\{1,\dots,m\}$, embedded in $S_n$ as the subgroup fixing $m+1,\dots,n$, and let $\mathbb C[S_m]$ denote its complex group algebra inside $\mathbb C[S_n]$. Let $\varnothing$ denote the empty partition of $0$, and let $S^{\varnothing}$ denote the one-dimensional trivial $\mathbb C[S_0]$-module. For a standard Young tableau $T$ of shape $\lambda$, define $\lambda^{(0)}(T)=\varnothing$, and for $1\le k\le n$ define $\lambda^{(k)}(T)$ to be the Young diagram formed by the boxes of $T$ whose entries lie in $\{1,\dots,k\}$. Since $T$ is standard, $\lambda^{(k)}(T)$ is a partition of $k$, and $\lambda^{(k-1)}(T)$ is obtained from $\lambda^{(k)}(T)$ by deleting the unique box occupied by $k$. By the branching rule [citetheorem:8477], applied over the characteristic-zero field $\mathbb C$ and to the standard embedding $S_{k-1}\le S_k$, the restriction of a Specht module from $S_k$ to $S_{k-1}$ is multiplicity-free and decomposes as the direct sum of the Specht modules obtained by deleting one removable box. The Young seminormal basis is the basis adapted to the resulting multiplicity-free chain \begin{align*} S_1\le S_2\le\cdots\le S_n. \end{align*} Thus, for each $T$ and each $k$, the line $\mathbb C v_T$ lies in the copy of $S^{\lambda^{(k)}(T)}$ occurring along this branching path when $S^\lambda$ is restricted to $S_k$. [guided] We first translate the tableau label into representation-theoretic data, including the boundary case at level $0$. Let $S_0$ be the trivial subgroup of $S_n$, let $\mathbb C[S_0]$ be its complex group algebra inside $\mathbb C[S_n]$, let $\varnothing$ be the empty partition of $0$, and let $S^{\varnothing}$ be the one-dimensional trivial $\mathbb C[S_0]$-module. For a standard Young tableau $T$, define $\lambda^{(0)}(T)=\varnothing$, and for $1\le k\le n$ define $\lambda^{(k)}(T)$ to be the set of boxes occupied by the entries $1,\dots,k$. Because entries increase along rows and columns, this set is again a Young diagram; hence it is a partition of $k$. Passing from $\lambda^{(k-1)}(T)$ to $\lambda^{(k)}(T)$ adds exactly one box, namely the box occupied by $k$ in $T$; for $k=1$, this means passing from the empty diagram to the single box containing $1$. The point of introducing these shapes is that the Young seminormal basis is adapted to the chain of symmetric groups \begin{align*} S_1\le S_2\le\cdots\le S_n. \end{align*} At each restriction step, the branching rule [citetheorem:8477] applies because we are working over the characteristic-zero field $\mathbb C$, with Specht modules for symmetric groups, and with the standard embedding $S_{k-1}\le S_k$ that fixes $k$. It says that restricting $S^{\nu}$ from $S_k$ to $S_{k-1}$ gives a direct sum of Specht modules $S^\mu$, one for each partition $\mu$ obtained from $\nu$ by deleting a removable box, and each such summand appears with multiplicity one. This multiplicity-one condition is what makes the tableau path meaningful. The vector $v_T$ is chosen so that, at level $k$, it belongs to the summand indexed by the shape $\lambda^{(k)}(T)$. Therefore the action of any central element of $\mathbb C[S_k]$ on this component can be computed from the central character of the Specht module $S^{\lambda^{(k)}(T)}$. [/guided] [/step] [step:Compute the scalar action of $\Omega_k$ on the branching component] We use the central-character formula for the transposition class sum on Specht modules. The precise external result is: if $m\ge 0$, if $\nu\vdash m$, and if $S^\nu$ is the complex Specht module for $S_m$, then the central element \begin{align*} \Omega_m=\sum_{1\le i<j\le m}(i\ j), \end{align*} with the empty sum interpreted as $0$ for $m=0$ and $m=1$, acts on $S^\nu$ as multiplication by \begin{align*} a(\nu)=\sum_{b\in\nu} c(b), \end{align*} where $c(b)$ is the column index of the box $b$ minus its row index. Its hypotheses match the present setting because every branching component below is a complex Specht module $S^{\lambda^{(m)}(T)}$ for $S_m$, and $\lambda^{(m)}(T)\vdash m$ for $0\le m\le n$. Fix a standard tableau $T$ of shape $\lambda$ and an integer $k$ with $1\le k\le n$. The preceding formula applies to the $S^{\lambda^{(k)}(T)}$ branching component for $S_k$, so $\Omega_k$ acts there by the scalar $a(\lambda^{(k)}(T))$. Likewise, it applies to the $S^{\lambda^{(k-1)}(T)}$ branching component for $S_{k-1}$ when $k\ge 2$, and to the trivial module $S^{\varnothing}$ for $S_0$ when $k=1$; in both cases, $\Omega_{k-1}$ acts by the scalar $a(\lambda^{(k-1)}(T))$. Therefore \begin{align*} (\Omega_k-\Omega_{k-1})v_T=\left(a(\lambda^{(k)}(T))-a(\lambda^{(k-1)}(T))\right)v_T. \end{align*} The two diagrams $\lambda^{(k)}(T)$ and $\lambda^{(k-1)}(T)$ differ by exactly the box occupied by $k$ in $T$, so the difference of the two content sums is precisely $c_k(T)$. Hence \begin{align*} (\Omega_k-\Omega_{k-1})v_T=c_k(T)v_T. \end{align*} [/step] [step:Replace the class-sum difference by the Jucys-Murphy element] By the identity established above, \begin{align*} X_k=\Omega_k-\Omega_{k-1}. \end{align*} Substituting this into the previous scalar-action formula gives, for every standard Young tableau $T$ of shape $\lambda$ and every $1\le k\le n$, \begin{align*} X_kv_T=c_k(T)v_T. \end{align*} This is the desired content eigenvalue formula for the Jucys-Murphy elements. [/step]