Unitary exchange blocks for collective spins

On elementary tensors, applying $\rho (\pi )$ after ${\mathrm{\Delta }}_N(U)$ gives $U{v}_{{\pi }^{-1}(1)}\otimes \cdots \otimes U{v}_{{\pi }^{-1}(N)}$, which is also the result of applying ${\mathrm{\Delta }}_N(U)$ after $\rho (\pi )$. The product formula and [F6] make $\rho (\pi )$ unitary because it permutes the factors, and make $U^{\otimes N}$ unitary because $⟨Uv,Uw{⟩}_V=⟨v,w{⟩}_V$ for every factor. For $N=0$ both actions are the identity on $\mathbb{C}$.

The normalized trace formula is an inner product on $M_{\lambda }$. Choose an orthonormal basis $(s_k{)}_{k=1}^{d_{\lambda }}$ of $S^{\lambda }$ using [F9]. The adjoint identity [F7], conjugate symmetry, and the diagonal trace formula [F12] give $d_{\lambda }^{-1}\mathrm{Tr}(g^{\ast }f)=d_{\lambda }^{-1}{\sum }_k⟨f(s_k),g(s_k){⟩}_{H_N}\mathrm{.}$ This expression is linear in $f$, conjugate-linear in $g$, conjugate symmetric, and for $g=f$ is a sum of nonnegative squared norms. If it is zero, every $f(s_k)=0$, hence $f=0$; so it is positive definite. The product inner product on $S^{\lambda }\otimes M_{\lambda }$ is well-defined by its finite product basis from [F6] and the tensor basis theorem, and is positive definite by the same coordinate formula.

By [F2] and [F3], the irreducible $S_N$-types occurring in $H_N$ are exactly the modules $S^{\lambda }$ with $\lambda ⊢N$ and $\mathrm{\ell }(\lambda )\le d$. Their multiplicity spaces are $M_{\lambda }={\mathrm{Hom}}_{S_N}(S^{\lambda },H_N)$, and the algebraic Schur–Weyl decomposition places $S_N$ on the Specht factor and $GL(V)$ on the multiplicity factor. This gives the algebraic partition and factor- placement inputs for the local evaluation argument.

By [F6], $H_N$ is finite-dimensional. In any nonzero invariant subspace of the current remainder, choose an invariant subspace $W$ of least positive dimension; it is irreducible. Its orthogonal complement in that remainder is invariant: for $x\perp W$ and $w\in W$, $⟨\rho (\pi )x,w⟩=⟨x,\rho (\pi {)}^{-1}w⟩=0$ because $W$ is invariant under $\rho (\pi {)}^{-1}$ and the action is unitary. [F8] gives the orthogonal splitting at each stage. Repeating in the complement, whose dimension strictly decreases, yields a finite orthogonal decomposition of all $H_N$ into irreducible copies. By [F2], each copy has a unique Specht type. The orthogonal projection onto one copy commutes with $S_N$, since that copy and the sum of the other invariant copies are invariant. Its restriction to an inequivalent irreducible copy is an intertwiner, so [F4] forces it to vanish. Hence copies of distinct Specht types are orthogonal.

Since $S_N$ is finite by [F11], choose a basis $(v_j)$ of the finite- dimensional $S^{\lambda }$ and define its coordinate product $⟨\sum a_j{v}_j,\sum b_j{v}_j{⟩}_0:=\sum a_j\overline{b_j}$. Average this product over $S_N$ by $⟨v,w⟩:=\mathrm{\mid }{S}_N{\mathrm{\mid }}^{-1}{\sum }_{\pi \in S_N}⟨\rho (\pi )v,\rho (\pi )w{⟩}_0$. Each summand on $(v,v)$ is nonnegative and the identity term is positive for $v\mathrm{\ne }0$, while right multiplication reindexes the finite sum; hence this is a positive definite invariant Hermitian product. This proves existence; now use the invariant product fixed in the Statement. For each irreducible copy $K_a$ of type $S^{\lambda }$, choose an equivariant isomorphism $j_a:S^{\lambda }\to K_a$. Its adjoint exists by [F10], and $j_a^{\ast }{j}_a$ commutes with $S_N$: the adjoint identity [F7], equivariance of $j_a$, and unitarity of both representations give $\rho (\pi {)}^{-1}{j}_a^{\ast }{j}_a\rho (\pi )=j_a^{\ast }{j}_a$. By [F5] this is $c_{a}I$; injectivity gives $c_a>0$, so replacing $j_a$ by $c_a^{-1\mathrm{/}2}{j}_a$ makes it unitary. For any $f\in M_{\lambda }$, decompose $f={\sum }_b{P}_{b}f$ using the orthogonal projections onto all irreducible copies in $H_N$; each $P_{b}f$ is an intertwiner because $P_b$ commutes with $S_N$ by step 2.1. If $K_b$ has type different from $S^{\lambda }$, then $P_{b}f=0$ by [F4] and [F2]. If $K_b$ has type $S^{\lambda }$, then $j_b^{-1}{P}_{b}f:S^{\lambda }\to S^{\lambda }$ is an intertwiner, hence is a scalar by [F5]. Thus the maps $j_a$ span $M_{\lambda }$. Their ranges are orthogonal, so $d_{\lambda }^{-1}\mathrm{Tr}(j_b^{\ast }{j}_a)={\delta }_{ab}$; they are an orthonormal basis in the stated normalized Hilbert-Schmidt inner product. By [F9], take an orthonormal basis of each $S^{\lambda }$; evaluation sends the tensor product basis to the union of the corresponding orthonormal bases of all copies $K_a$. This union is an orthonormal basis of $H_N$, so $\mathrm{\Phi }$ is unitary.

For $\pi \in S_N$, evaluation gives $\mathrm{\Phi }({\rho }_{\lambda }(\pi )s\otimes f)=f({\rho }_{\lambda }(\pi )s)=\rho (\pi )f(s)=\rho (\pi )\mathrm{\Phi }(s\otimes f)$ because $f$ is an intertwiner. For $U\in U(d)$, ${\mathrm{\Delta }}_N(U)\circ f$ is again an intertwiner by [F1], and evaluation gives $\mathrm{\Phi }(s\otimes ({\mathrm{\Delta }}_N(U)\circ f))={\mathrm{\Delta }}_N(U)f(s)={\mathrm{\Delta }}_N(U)\mathrm{\Phi }(s\otimes f)$. Thus the actions have the asserted factor placement. Since ${\mathrm{\Delta }}_N(U)$ is unitary and $\mathrm{\Phi }$ is unitary, the postcomposition action is unitary on each multiplicity factor: on the orthogonal summand $S^{\lambda }\otimes M_{\lambda }$ the operator is $I\otimes T_U$, and evaluating on any unit vector of the nonzero $S^{\lambda }$ factor shows that $T_U$ preserves inner products; $T_{U^{-1}}$ is its inverse.