Example. One permutation-group link has four isotypic states but one charged state [fdip-008C]
Example. One permutation-group link has four isotypic states but one charged state [fdip-008C]
Take one oriented edge with distinct source and target and \(G=S_3\). Then \(\mathcal H_{\mathrm {link}}=L^2(S_3)\) has complex dimension \(6\). Order the gauge factors as \(K=S_3^{\mathrm {target}}\times S_3^{\mathrm {source}}\); their function action is \[(R_{(h_t,h_s)}f)(U)=f(h_t^{-1}Uh_s).\] Let \(V=\{(z_1,z_2,z_3)\in \mathbb C^3:z_1+z_2+z_3=0\}\) carry the standard unitary permutation representation \(\rho \). Use the charge datum \[R=V_{\mathrm {target}}\otimes V_{\mathrm {source}}^*.\] It has dimension \(4\).
The standard representation is irreducible: a three-cycle has two distinct eigenlines on \(V\), with eigenvalues \(e^{2\pi i/3}\) and \(e^{-2\pi i/3}\), and a transposition exchanges them. Thus neither line is invariant under \(S_3\). The external tensor product \(R\) is also irreducible under \(K\). To see this, expand an endomorphism in a basis of \(\operatorname {End}(V^*)\). Commutation with the first factor and Schur's lemma make it \(I\otimes B\); commutation with the second factor makes \(B\) scalar. Any invariant subspace has a commuting orthogonal projector, so it is zero or all of \(R\).
The inverse matrix-coefficient map is \[\begin {aligned} \Psi :R&\longrightarrow L^2(S_3),\\ \Psi (v\otimes \lambda )(U) &=\sqrt 2\,\lambda (\rho (U^{-1})v). \end {aligned}\] Schur orthogonality makes this map isometric for normalized counting measure. Its order of duals is checked directly: \[R_{(h_t,h_s)}\Psi (v\otimes \lambda ) =\Psi (\rho (h_t)v\otimes \rho ^*(h_s)\lambda ).\] Thus its four-dimensional image is a copy of the specified charge representation, with the target and source conventions as stated.
The finite Peter–Weyl decomposition gives the bilateral decomposition \[L^2(S_3)\cong (\mathbf 1\otimes \mathbf 1) \oplus (\operatorname {sgn}\otimes \operatorname {sgn}) \oplus (V\otimes V^*).\] Indeed, the trivial, sign and standard representations are inequivalent, and their squared dimensions sum to \(1+1+4=6=|S_3|\), so their orthogonal coefficient spaces exhaust \(L^2(S_3)\). Hence the standard bilateral type has multiplicity one. By the charged-space theorem, \[\dim \mathcal H_R=4,\qquad \dim \mathcal H_{\mathrm {phys}}(R) =\dim \operatorname {Hom}_K(R,L^2(S_3))=1.\] The four-dimensional isotypic component is therefore not the charged physical space. Isotypic projectors separate representation types; invariant charge tensors supply the Gauss constraint.