Young's seminormal form from the Jucys-Murphy eigenlines

Swapping consecutive entries in different rows and columns preserves standardness: every neighbour other than the swapped entry is either smaller than both entries or larger than both. Such nodes must be incomparable in the northwest order, since comparable nodes in different rows and columns would have an intermediate node with entry strictly between $i$ and $i+1$. Their row and column differences therefore have opposite signs, so $\mathrm{\mid }{c}_T(i+1)-c_T(i)\mathrm{\mid }\ge 2$. In a common row or column the nodes are adjacent and the difference is $+1$ or $-1$ by [F4]. Thus the axial distance never vanishes.

For $0\mathrm{\ne }v\in L_T$, put $a=c_T(i)$, $b=c_T(i+1)$ and $r=b-a$. By [F3], $u:=(s_i-r^{-1})v$ is a joint eigenvector with the weight of $T$ having coordinates $i,i+1$ exchanged: $X_{i}u=bu$, $X_{i+1}u=au$, and the other eigenvalues are unchanged. If the nodes are in different rows and columns, [F2] identifies this weight with $T^{\mathrm{\prime }}=s_{i}T$. Moreover $u\mathrm{\ne }0$, since $u=0$ would imply $s_{i}v=r^{-1}v$ and then $v=s_i^{2}v=r^{-2}v$, contrary to $\mathrm{\mid }r\mathrm{\mid }>1$. Hence $s_i{L}_T\subseteq L_T\oplus L_{T^{\mathrm{\prime }}}$, and its component in $L_{T^{\mathrm{\prime }}}$ is nonzero.

A reduced admissible chain joins $T^{\lambda }$ to each $T$. To construct it in reverse, let the last node in row order carry $k$ in $T$. Swap $k$ with $k+1$, then $k+1$ with $k+2$, through $n$. Every value larger than the current value in that node is in a different row and column: entries in its own row or column are smaller by standardness. The swaps are therefore admissible by step 1.1. In the row-reading word each swap moves the larger of two consecutive values from an earlier position to the final position, decreasing its inversion count by exactly one; all other inversion comparisons are unchanged. Remove the final node and entry $n$ and repeat. The process reaches $T^{\lambda }$ after exactly $\mathrm{\ell }(T)$ swaps, since the row-reading word is the one-line notation of ${\sigma }_T$ and the final word has no inversions. Reversing this chain gives the asserted reduced chain.

In the same-row or same-column case, the exchanged weight is not a tableau weight. Indeed, uniqueness of reconstruction from contents fixes the prefix through $i-1$, while equality of the content multisets through $i+1$ fixes that prefix shape by [F5]. Thus the two new nodes must be the original nodes of $i,i+1$, with their entries exchanged. The node originally carrying $i+1$ cannot be added first, because its immediate left or upper neighbour is the still-absent node of $i$. This violates standardness. By [F2] the vector $u$ of step 2.1 is zero, giving $s_{i}v=r^{-1}v$, hence $+v$ in the row case and $-v$ in the column case.

Expand ${\sigma }_T{v}_0$ along such a reduced chain of length $L=\mathrm{\ell }(T)$ using [F1] and steps 2.1 and 3.1. At each factor a vector in a Young line either stays in that line or passes to its admissible neighbour; every neighbour changes the inversion count by one, and the component passing to it is nonzero by step 2.1. To reach a line of inversion count $L$ after $L$ factors, every factor must pass to the neighbour and increase the count. This unique sequence is the reduced chain to $T$, and its product of nonzero coefficients is nonzero. All other resulting lines have length less than $L$. Therefore $v_T=P_T{\sigma }_T{v}_0\mathrm{\ne }0$ and ${\sigma }_T{v}_0-v_T$ is a linear combination of lines $L_R$ with $\mathrm{\ell }(R)<\mathrm{\ell }(T)$. The permutation ${\sigma }_T$ is uniquely determined by the fillings, so this definition does not depend on a reduced expression.

Suppose $T^{\mathrm{\prime }}=s_{i}T$ is standard and $\mathrm{\ell }(T^{\mathrm{\prime }})=\mathrm{\ell }(T)+1$. The unique permutations obey ${\sigma }_{T^{\mathrm{\prime }}}=s_i{\sigma }_T$. By step 4.1, write ${\sigma }_T{v}_0=v_T+w$, with $w$ supported on lines of length strictly less than $\mathrm{\ell }(T)$. Steps 2.1 and 3.1 show that $s_{i}w$ is supported on lines of length at most $\mathrm{\ell }(T)$, so $P_{T^{\mathrm{\prime }}}{s}_{i}w=0$. Hence $v_{T^{\mathrm{\prime }}}=P_{T^{\mathrm{\prime }}}{s}_i{\sigma }_T{v}_0=P_{T^{\mathrm{\prime }}}{s}_i{v}_T$. Together with step 2.1 this gives $s_i{v}_T=v_{T^{\mathrm{\prime }}}+r^{-1}{v}_T$. Applying $s_i$ once more and using $s_i^2=1$ yields $s_i{v}_{T^{\mathrm{\prime }}}=(1-r^{-2})v_T-r^{-1}{v}_{T^{\mathrm{\prime }}}$.

If $T^{\mathrm{\prime }}$ is shorter, apply step 5.1 to the pair $(T^{\mathrm{\prime }},T)$ with axial distance $-r$. It gives $s_i{v}_{T^{\mathrm{\prime }}}=v_T-r^{-1}{v}_{T^{\mathrm{\prime }}}$ and $s_i{v}_T=(1-r^{-2})v_{T^{\mathrm{\prime }}}+r^{-1}{v}_T$, as stated. The vectors $v_T$ form a basis by [F1] and step 4.1. Steps 3.1, 5.1 and this reverse reading give rational matrices for every generator, with no zero denominator. They are matrices of the actual group action, so satisfy all Coxeter relations and define a rational representation whose complexification is the given Specht module. The chain construction and projections are finite; no arbitrary-index choice is used.