Young's orthogonal form from the seminormal rescaling

Let $T$ have $T^{\mathrm{\prime }}=s_{i}T$ standard with $i,i+1$ in different rows and columns, in the phasing of [F3]. The vectors $v_T,v_{T^{\mathrm{\prime }}}$ are nonzero, so by [F2] their norms $\mathrm{\parallel }{v}_T\mathrm{\parallel }=\sqrt{⟨v_T,v_T⟩}$ and $\mathrm{\parallel }{v}_{T^{\mathrm{\prime }}}\mathrm{\parallel }$ are positive real numbers; define $u_T:=v_T\mathrm{/}\mathrm{\parallel }{v}_T\mathrm{\parallel }$ and $u_{T^{\mathrm{\prime }}}:=v_{T^{\mathrm{\prime }}}\mathrm{/}\mathrm{\parallel }{v}_{T^{\mathrm{\prime }}}\mathrm{\parallel }$.

Matrix in the unit basis. Each $X_k$ is self-adjoint by [F1], since it is a sum of transpositions, which are unitary involutions. Distinct Young lines have distinct joint content vectors by [F5], so some self-adjoint $X_k$ has different eigenvalues on them; the identity $⟨X_{k}v,w⟩=⟨v,X_{k}w⟩$ then makes them orthogonal. Thus all the unit vectors $u_T$ form an orthonormal basis. Write $A$ for the matrix of $s_i$ in $(v_T,v_{T^{\mathrm{\prime }}})$ and $B$ for its matrix in $(u_T,u_{T^{\mathrm{\prime }}})$. Since $u_T=v_T\mathrm{/}\mathrm{\parallel }{v}_T\mathrm{\parallel }$, set $D=\mathrm{diag}(\mathrm{\parallel }{v}_T{\mathrm{\parallel }}^{-1},\mathrm{\parallel }{v}_{T^{\mathrm{\prime }}}{\mathrm{\parallel }}^{-1})$; then $B=D^{-1}AD$. By [F3], $A=\left(\begin{array}{cc}{r}^{-1}& 1-r^{-2}\\ 1& -r^{-1}\end{array}\right),B=\left(\begin{array}{cc}{r}^{-1}& (1-r^{-2})\mathrm{\parallel }{v}_T\mathrm{\parallel }\mathrm{/}\mathrm{\parallel }{v}_{T^{\mathrm{\prime }}}\mathrm{\parallel }\\ \mathrm{\parallel }{v}_{T^{\mathrm{\prime }}}\mathrm{\parallel }\mathrm{/}\mathrm{\parallel }{v}_T\mathrm{\parallel }& -r^{-1}\end{array}\right)\mathrm{.}$

Same row and same column. If $i,i+1$ lie in the same row or column of a standard tableau, [F3] gives $s_i{v}_T=\pm v_T$ with the sign $+1$ in the row case and $-1$ in the column case; rescaling by a positive norm does not change these scalars.

Symmetry. By [F4] and [F1] the operator $s_i$ is a unitary involution, hence self-adjoint; in the orthonormal basis $(u_T,u_{T^{\mathrm{\prime }}})$ its matrix $B$ therefore satisfies $B=B^{\ast }$, the conjugate transpose of $B$. Since $B_{11}=r^{-1}$ and $B_{22}=-r^{-1}$ are real, this forces $B_{21}=\overline{B_{12}}$ and $\mathrm{\mid }{B}_{12}\mathrm{\mid }=\mathrm{\mid }{B}_{21}\mathrm{\mid }$.

The off-diagonal entries are positive and equal. By step 2.1, $B_{21}=\mathrm{\parallel }{v}_{T^{\mathrm{\prime }}}\mathrm{\parallel }\mathrm{/}\mathrm{\parallel }{v}_T\mathrm{\parallel }>0$, so self-adjointness in step 3.1 makes both off-diagonal entries the same positive real number $c$. The $(1,1)$ entry of $B^2=I$ gives $r^{-2}+c^2=1$, hence $c=\sqrt{1-r^{-2}}$, with $\mathrm{\mid }r\mathrm{\mid }>1$ by [F3]. In particular $\mathrm{\parallel }{v}_{T^{\mathrm{\prime }}}\mathrm{\parallel }\mathrm{/}\mathrm{\parallel }{v}_T\mathrm{\parallel }=\sqrt{1-r^{-2}}$, and $B$ is the displayed symmetric orthogonal matrix.

Orthogonality. The matrices $B$ of step 4.1 are real, symmetric and satisfy $B^2=I$, hence are orthogonal with determinant $-r^{-2}-(1-r^{-2})=-1$; the unit basis was chosen with positive square roots, and the common phase of the initial vector does not affect these matrices. In particular each $s_i$ acts by a real orthogonal involution on each $V^{\lambda }$ for the restricted inner product.

Representation. The matrices so obtained are the matrices of the actual elements $s_i\in S_n$ in a basis of $V^{\lambda }$, so they satisfy the Coxeter relations $s_i^2=1$ and $s_i{s}_{i+1}{s}_i=s_{i+1}{s}_i{s}_{i+1}$ and generate a representation equivalent to $V^{\lambda }$; this is Young's orthogonal representation.