Finite-group Hopf algebra and lattice quantum-double sectors

For basis elements $d_{h,g},d_{k,l},d_{m,n}$, the product $(d_{h,g}{d}_{k,l})d_{m,n}$ is nonzero exactly when $h=gk{g}^{-1}$ and $h=(gl)m(gl{)}^{-1}$, while $d_{h,g}(d_{k,l}{d}_{m,n})$ is nonzero exactly when $k=lm{l}^{-1}$ and $h=gk{g}^{-1}$. These conditions are equivalent since $(gl)m(gl{)}^{-1}=g(lm{l}^{-1})g^{-1}$; when they hold both products equal $d_{h,gln}$. Bilinearity gives associativity on $D(G)$.

For each $d_{h,g}$, $1_D{d}_{h,g}=d_{h,g}$ because only the summand $d_{h,e}$ survives, and $d_{h,g}{1}_D=d_{h,g}$ because exactly the summand indexed by $g^{-1}hg$ survives. Thus $1_D$ is a two-sided unit.

Both iterated coproducts of $d_{h,g}$ are ${\sum }_{abc=h}{d}_{a,g}\otimes d_{b,g}\otimes d_{c,g}$, and applying either counit to $\mathrm{\Delta }(d_{h,g})$ leaves $d_{h,g}$, by selecting respectively the factor indexed by $e$. Hence $\mathrm{\Delta }$ is coassociative and $\epsilon$ is a counit. Also $\mathrm{\Delta }(1_D)={\sum }_{a,b}{d}_{a,e}\otimes d_{b,e}=1_D\otimes 1_D$ and $\epsilon (1_D)=1$.

For basis elements $d_{h,g},d_{k,l}$, the product $\mathrm{\Delta }(d_{h,g})\mathrm{\Delta }(d_{k,l})$ is zero unless $h=gk{g}^{-1}$. In that case its terms correspond bijectively to factorizations $rs=k$, through $(r,s)\mapsto (gr{g}^{-1},gs{g}^{-1})$, and their sum is $\mathrm{\Delta }(d_{h,gl})=\mathrm{\Delta }(d_{h,g}{d}_{k,l})$. Thus $\mathrm{\Delta }$ preserves multiplication. Moreover $\epsilon (d_{h,g}{d}_{k,l})=\epsilon (d_{h,g})\epsilon (d_{k,l})$: both sides equal $1$ exactly when $h=k=e$.

In $m(S\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(d_{h,g})$, a summand indexed by $ab=h$ is nonzero exactly when $b=a^{-1}$. This is impossible unless $h=e$; if $h=e$, the resulting sum is ${\sum }_a{d}_{g^{-1}{a}^{-1}g,e}=1_D$. Therefore $m(S\otimes \mathrm{i}\mathrm{d})\mathrm{\Delta }(d_{h,g})=\epsilon (d_{h,g})1_D$.

In $m(\mathrm{i}\mathrm{d}\otimes S)\mathrm{\Delta }(d_{h,g})$, the summand indexed by $ab=h$ is nonzero exactly when $a=b^{-1}$. Again it vanishes for $h\mathrm{\ne }e$, and for $h=e$ it is ${\sum }_a{d}_{a,e}=1_D$. Therefore $m(\mathrm{i}\mathrm{d}\otimes S)\mathrm{\Delta }(d_{h,g})=\epsilon (d_{h,g})1_D$, proving the Hopf algebra assertion by [F2].

The formula in [F1] gives $(d_{h,g}^{\ast }{)}^{\ast }=d_{h,g}$ and $1_D^{\ast }=1_D$. On basis products, $(d_{h,g}{d}_{k,l}{)}^{\ast }=d_{k,l}^{\ast }{d}_{h,g}^{\ast }$: both sides vanish unless $h=gk{g}^{-1}$, and in that case both equal $d_{l^{-1}kl,l^{-1}{g}^{-1}}$. Conjugate-linear extension makes $\ast$ an involution reversing multiplication.

The two expressions $\mathrm{\Delta }(d_{h,g}^{\ast })$ and $\mathrm{\Delta }(d_{h,g}{)}^{\ast }$ agree after the bijection $(a,b)\mapsto (g^{-1}ag,g^{-1}bg)$ between factorizations of $h$ and of $g^{-1}hg$. Also $\epsilon (d_{h,g}^{\ast })=\overline{\epsilon (d_{h,g})}$. Hence the stated involution is compatible with the coproduct and counit.

For a fixed vertex $v$, applying $U_v(y)$ and then $U_v(x)$ sends each edge label to $(xy)g_a$ at an outgoing incidence, to $g_a(xy{)}^{-1}$ at an incoming incidence, and to $(xy)g_a(xy{)}^{-1}$ on a loop based at $v$. All other labels remain fixed. Thus $U_v(x)U_v(y)=U_v(xy)$; each $U_v(x)$ is a permutation of an orthonormal basis and is unitary.

In the configuration basis, each $B_f$ is diagonal with entries in $\{0,1\}$. Therefore $B_f^{\ast }=B_f$ and $B_f^2=B_f$, so every $B_f$ is an orthogonal projection.

For distinct vertices $v,w$, their gauge permutations commute: on a shared edge they act at opposite ends, as left multiplication by one group element and right multiplication by the inverse of the other, and these operations commute; on all other edges at least one action is the identity. Thus $A_v{A}_w=A_w{A}_v$. Since all $B_f$ are diagonal in the same configuration basis, they commute pairwise.

Under a gauge change $k_u=x$ at $u=v$ and $k_u=e$ elsewhere, a face-boundary holonomy based at a vertex $u_0$ changes to $k_{u_0}{H}_f{k}_{u_0}^{-1}$: along the closed boundary walk the intermediate factors $k_u^{-1}{k}_u$ cancel. The condition $H_f=e$ is invariant under conjugation, so $U_v(x)$ preserves every eigenspace of $B_f$ and commutes with $B_f$. Averaging over $x$ gives $A_v{B}_f=B_f{A}_v$.

Taking adjoints in the finite average gives $A_v^{\ast }=\mathrm{\mid }G{\mathrm{\mid }}^{-1}{\sum }_x{U}_v(x^{-1})=A_v$. Also $A_v^2=\mathrm{\mid }G{\mathrm{\mid }}^{-2}{\sum }_{x,y}{U}_v(xy)=\mathrm{\mid }G{\mathrm{\mid }}^{-1}{\sum }_z{U}_v(z)=A_v$, since each $z\in G$ has exactly $\mathrm{\mid }G\mathrm{\mid }$ factorizations $xy=z$. Hence $A_v$ is an orthogonal projection.

Each complement $I-A_v$ and $I-B_f$ is an orthogonal projection, hence positive. Their finite sum $H_{\mathrm{\Sigma }}$ is positive.

For a vertex assignment $\mathbf{y}=(y_v{)}_{v\in V}\in G^V$, define the pure-gauge edge configuration $g_a=y_{s(a)}{y}_{t(a)}^{-1}$. The holonomy around every face telescopes to $e$, so each such basis vector lies in the $+1$ eigenspace of every $B_f$. The set $\mathcal{P}$ of distinct pure-gauge configurations is finite because $G^V$ is finite, and nonempty by taking every $y_v=e$. Each $U_v(x)$ permutes $\mathcal{P}$, because it replaces $y_v$ by $x{y}_v$ and leaves other vertex labels fixed. Therefore $\psi :={\sum }_{\mathbf{g}\in \mathcal{P}}\mathrm{\mid }\mathbf{g}⟩$ is nonzero, $U_v(x)\psi =\psi$ for every $v,x$, and consequently $A_v\psi =B_f\psi =\psi$ for every $v,f$. This gives a common zero vector of all positive terms in $H_{\mathrm{\Sigma }}$, proving frustration-freeness and ground energy zero.

On the group basis of $\mathbb{C}[H]$, define $\lambda (h)\mathrm{\mid }x⟩=\mathrm{\mid }hx⟩$. Then $\lambda (h)\lambda (k)\mathrm{\mid }x⟩=\mathrm{\mid }hkx⟩=\lambda (hk)\mathrm{\mid }x⟩$ and $\lambda (e)=I$, so [F2] makes this a finite-dimensional representation. The regular representation is nonzero because $e\in H$. For any nonzero finite-dimensional representation, if it has no nonzero proper invariant subspace, [F3] says it is irreducible; otherwise choose such a subspace $W$ and [F1] supplies an invariant complement $U$. Both $W$ and $U$ are nonzero and have smaller dimension, so induction decomposes them into irreducibles. Applying this induction to $\mathbb{C}[H]$ proves that it is a finite direct sum of irreducible representations.

Let $\pi :H\to \mathrm{GL}(V)$ be any irreducible representation and choose $0\mathrm{\ne }v\in V$. Define the linear map $\mathrm{\Phi }:\mathbb{C}[H]\to V$ by $\mathrm{\Phi }({\sum }_h{a}_{h}h)={\sum }_h{a}_h\pi (h)v$. For $k\in H$, $\mathrm{\Phi }(k{\sum }_h{a}_{h}h)={\sum }_h{a}_h\pi (kh)v=\pi (k)\mathrm{\Phi }({\sum }_h{a}_{h}h)$, so $\mathrm{\Phi }$ is equivariant. Its image contains $v=\mathrm{\Phi }(e)$, is invariant, and therefore equals $V$ by irreducibility. By [F1], $\ker \mathrm{\Phi }$ has an invariant complement $W$ in $\mathbb{C}[H]$; then $\mathrm{\Phi }{\mathrm{\mid }}_W$ is onto and injective, so $W\cong V$ is an irreducible summand of $\mathbb{C}[H]$. At least one irreducible summand $U$ in the fixed decomposition of step 1.1 has $\mathrm{\Phi }{\mathrm{\mid }}_U\mathrm{\ne }0$, since $\mathrm{\Phi }$ is onto a nonzero space. Its kernel and image are invariant, so irreducibility of $U$ and $V$ makes $\mathrm{\Phi }{\mathrm{\mid }}_U$ an isomorphism. Consequently every irreducible representation occurs among those finitely many summands.

For each irreducible representation $\pi$, choose any positive-definite inner product and average it as $⟨v,w{⟩}_H:={\sum }_{h\in H}⟨\pi (h)v,\pi (h)w⟩$. This is positive definite and is $H$-invariant by reindexing the finite sum, so $\pi (h)$ is unitary and ${\chi }_{\pi }(h^{-1})=\overline{{\chi }_{\pi }(h)}$. For two irreducibles $\pi ,\sigma$, define $P(T):=\mathrm{\mid }H{\mathrm{\mid }}^{-1}{\sum }_{h\in H}\sigma (h)T\pi (h^{-1})$ on ${\mathrm{Hom}}_{\mathbb{C}}(V_{\pi },V_{\sigma })$. Reindexing gives $\sigma (k)P(T)\pi (k^{-1})=P(T)$ for every $k$, so $P(T)$ intertwines the group actions; if $T$ already intertwines, every summand equals $T$. Hence $P$ is a projection with image exactly ${\mathrm{Hom}}_H(V_{\pi },V_{\sigma })$.

For matrices $A=\sigma (h)$ and $B=\pi (h^{-1})$ in chosen bases, the coefficient of a rectangular matrix unit $E_{ij}$ in $A{E}_{ij}B$ is $A_{ii}{B}_{jj}$. Summing these diagonal coefficients gives $\mathrm{tr}(T\mapsto ATB)={\sum }_{i,j}{A}_{ii}{B}_{jj}=\mathrm{tr}A\mathrm{tr}B={\chi }_{\sigma }(h)\overline{{\chi }_{\pi }(h)}$. Linearity of the diagonal sum yields $\mathrm{tr}P=\mathrm{\mid }H{\mathrm{\mid }}^{-1}{\sum }_h{\chi }_{\sigma }(h)\overline{{\chi }_{\pi }(h)}$. Since $P^2=P$, every $T$ splits as $P(T)+(T-P(T))\in \mathrm{im}P\oplus \ker P$; the intersection is zero because $P$ is the identity on its image. In a basis adapted to this direct sum, $P$ has matrix $\mathrm{diag}(I,0)$, so $\mathrm{tr}P=\dim \mathrm{im}P$. An intertwiner between irreducibles is either zero or an isomorphism because its kernel and image are invariant. By [F4]–[F5], $\mathbb{C}$ is algebraically closed, so [F6] gives every endomorphism $T$ of the nonzero finite-dimensional space $V_{\pi }$ an eigenvalue $\alpha$. For $T\in {\mathrm{End}}_H(V_{\pi })$, the eigenspace $\ker (T-\alpha I)$ is nonzero and $H$-invariant because $T$ intertwines the $H$-action; irreducibility forces it to be all of $V_{\pi }$, so $T=\alpha I$. Thus $\dim \mathrm{im}P$ is $1$ if $\pi \cong \sigma$ and $0$ otherwise, proving character orthogonality by the averaging calculation.

The character of $\lambda$ is ${\chi }_{\lambda }(e)=\mathrm{\mid }H\mathrm{\mid }$ and ${\chi }_{\lambda }(h)=0$ for $h\mathrm{\ne }e$, since left multiplication by $h\mathrm{\ne }e$ fixes no group-basis element. Define $⟨\alpha ,\beta ⟩:=\mathrm{\mid }H{\mathrm{\mid }}^{-1}{\sum }_h\overline{\alpha (h)}\beta (h)$. If $\lambda \cong {⨁}_{\pi }{m}_{\pi }\pi$ as in step 1.1, then character additivity and the orthogonality from step 3.1 give $m_{\pi }=⟨{\chi }_{\pi },{\chi }_{\lambda }⟩=\mathrm{\mid }H{\mathrm{\mid }}^{-1}{\sum }_h\overline{{\chi }_{\pi }(h)}{\chi }_{\lambda }(h)=\dim \pi$.

Taking dimensions of the regular decomposition yields $\mathrm{\mid }H\mathrm{\mid }=\dim \mathbb{C}[H]={\sum }_{\pi }{m}_{\pi }\dim \pi ={\sum }_{\pi }(\dim \pi {)}^2$. This proves the stated formula and finiteness.

In any irreducible $D(G)$-module $V$, put $p_h=\rho (d_{h,e})$ and $u_g=\rho ({\sum }_{r\in G}{d}_{r,g})$. The multiplication in [F1] gives $p_h{p}_k={\mathbf{1}}_{h=k}{p}_h$, ${\sum }_h{p}_h=I$, $u_g{u}_l=u_{gl}$, $u_g{p}_h{u}_g^{-1}=p_{gh{g}^{-1}}$, and $\rho (d_{h,g})=p_h{u}_g$. Hence $V={⨁}_{h\in G}{V}_h$ with $V_h=p_{h}V$, and $u_g$ maps $V_h$ isomorphically onto $V_{gh{g}^{-1}}$.

Conversely, fix a conjugacy class $C$, representative $c$, and irreducible representation $\pi :Z_G(c)\to \mathrm{GL}(W)$ as in [F3]–[F4]. For each $h\in C$ choose $t_h$ with $t_{h}c{t}_h^{-1}=h$ and take $t_c=e$. On $M_{C,\pi }:={⨁}_{h\in C}{W}_h$ define $p_a$ as projection onto $W_a$ (zero if $a\mathrm{\notin }C$), and define $u_g(h,w)=(gh{g}^{-1},\pi (t_{gh{g}^{-1}}^{-1}g{t}_h)w)$. With $h^{\mathrm{\prime }}=gh{g}^{-1}$, the factor $z=t_{h^{\mathrm{\prime }}}^{-1}g{t}_h$ centralizes $c$, since $zc{z}^{-1}=t_{h^{\mathrm{\prime }}}^{-1}gh{g}^{-1}{t}_{h^{\mathrm{\prime }}}=c$. For $k\in G$, the product of the two centralizer factors in $u_g{u}_k$ is $(t_{gkh{k}^{-1}{g}^{-1}}^{-1}g{t}_{kh{k}^{-1}})(t_{kh{k}^{-1}}^{-1}k{t}_h)=t_{gkh{k}^{-1}{g}^{-1}}^{-1}gk{t}_h$, so $u_g{u}_k=u_{gk}$. The fiber definition also gives $u_g{p}_a{u}_g^{-1}=p_{ga{g}^{-1}}$.

The set of $h$ for which $V_h\mathrm{\ne }0$ is a union of conjugacy classes. For any such class $C$, the subspace ${⨁}_{h\in C}{V}_h$ is invariant under every $p_a$ and $u_g$, hence under every $d_{a,g}$. Irreducibility therefore forces the support to be one conjugacy class $C$. Fix $c\in C$.

Define $\rho (d_{a,g})=p_a{u}_g$. The identities of step 1.2 imply $\rho (d_{a,g})\rho (d_{b,l})={\mathbf{1}}_{a=gb{g}^{-1}}\rho (d_{a,gl})$, exactly the multiplication in [F1], and $\rho (1_D)={\sum }_a{p}_a=I$. Thus $M_{C,\pi }$ is a unital $D(G)$-module. If $N$ is a nonzero invariant subspace, the projections $p_h$ decompose it into fibers because they are pairwise orthogonal and sum to $I$; $N_c$ is $\pi$-invariant, so by [F4] it is either zero or all of $W_c$. In the first case, for each $h$ the isomorphism $u_{t_h}$ carries $N_c$ onto $N_h$, so every fiber vanishes. In the second case, those images fill every fiber. Thus $N=0$ or $N=M_{C,\pi }$, proving irreducibility.

By [F3], the restrictions $u_z{\mathrm{\mid }}_{V_c}$ for $z\in Z_G(c)$ form a finite-dimensional group representation. If it had a nonzero proper invariant subspace $W$, choose for each $h\in C$ an element $t_h$ with $t_{h}c{t}_h^{-1}=h$. For $h^{\mathrm{\prime }}=gh{g}^{-1}$ the element $z=t_{h^{\mathrm{\prime }}}^{-1}g{t}_h$ satisfies $zc{z}^{-1}=t_{h^{\mathrm{\prime }}}^{-1}gh{g}^{-1}{t}_{h^{\mathrm{\prime }}}=c$, so $z\in Z_G(c)$. Therefore $u_g{u}_{t_h}W=u_{t_{h^{\mathrm{\prime }}}}{u}_{z}W\subseteq u_{t_{h^{\mathrm{\prime }}}}W$. The direct sum ${⨁}_{h\in C}{u}_{t_h}W$ is invariant under every $u_g$ and $p_a$; it is nonzero, and it is proper because each $u_{t_h}$ is an isomorphism from $V_c$ to $V_h$ while $W⊊V_c$. It is thus a nonzero proper $D(G)$-submodule, contradicting irreducibility. Hence $V_c$ is irreducible as a $Z_G(c)$-representation.

The constructed module has one fiber $W_h\cong W$ for each $h\in C$, so its dimension is $\mathrm{\mid }C\mathrm{\mid }\dim \pi$. The map $G\mathrm{/}{Z}_G(c)\to C$, $g{Z}_G(c)\mapsto gc{g}^{-1}$, is well-defined and onto; equality of two conjugates implies the two representatives differ by an element of $Z_G(c)$, so it is injective. Therefore $\mathrm{\mid }C\mathrm{\mid }\mathrm{\mid }{Z}_G(c)\mathrm{\mid }=\mathrm{\mid }G\mathrm{\mid }$.

For an arbitrary irreducible $V$ with support $C$, set $W=V_c$ and let $\pi$ be its centralizer action from step 3.1. The maps $u_{t_h}$ then identify each fiber $V_h$ with the constructed copy $W_h$, and the relation $u_g{u}_{t_h}=u_{t_{gh{g}^{-1}}}{u}_{t_{gh{g}^{-1}}^{-1}g{t}_h}$ shows that this fiberwise identification intertwines the full $D(G)$-action. Distinct classes have distinct supports; for a fixed $C$, an isomorphism restricts on the $p_c$ fiber to an isomorphism of centralizer representations. Hence the pairs $(C,[\pi ])$ classify the irreducibles without duplication.

If $c^{\mathrm{\prime }}=rc{r}^{-1}$ is another representative of $C$, then $Z_G(c^{\mathrm{\prime }})=r{Z}_G(c)r^{-1}$. Sending $\pi$ to ${\pi }^{\mathrm{\prime }}(z^{\mathrm{\prime }})=\pi (r^{-1}{z}^{\mathrm{\prime }}r)$ gives a bijection of the irreducible representation classes of the two centralizers. The fiber map $u_r:V_c\to V_{c^{\mathrm{\prime }}}$ implements this relabeling because $u_{z^{\mathrm{\prime }}}{u}_r=u_r{u}_{r^{-1}{z}^{\mathrm{\prime }}r}$, by $u_g{u}_l=u_{gl}$ from step 1.1. Thus the parametrization is independent of the representative up to the stated conjugation.

The finite centralizer $Z_G(c)$ satisfies the finite-group and characteristic hypotheses of [F5], and [F6] supplies the resulting squared-dimension sum ${\sum }_{[\pi ]\in {\mathrm{Irr}}_{\mathbb{C}}(Z_G(c))}(\dim \pi {)}^2=\mathrm{\mid }{Z}_G(c)\mathrm{\mid }$. Summing the squared dimensions from step 3.2 over conjugacy classes gives ${\sum }_C\mathrm{\mid }C{\mathrm{\mid }}^2\mathrm{\mid }{Z}_G(c)\mathrm{\mid }={\sum }_C\mathrm{\mid }C\mathrm{\mid }\mathrm{\mid }G\mathrm{\mid }=\mathrm{\mid }G{\mathrm{\mid }}^2$, since the conjugacy classes partition $G$. Finally, [F1] gives $\dim D(G)=\mathrm{\mid }G{\mathrm{\mid }}^2$.