Construction 3.8. We now construct an isomorphism of nets of algebras from the fusion categorical net $\mathfrak {F}$ for $\mathcal {C}=\mathsf {Hilb_{fd}}(\mathbb {Z}/2)$ from Example 2.5 with $X = 1 \oplus g$ , where $1, g \in \mathbb {Z}_2$ , to either $\mathfrak {C}'$ or $\mathfrak {D}'$ from Remark 3.7 above. This isomorphism is essentially a planar algebra embedding from the $\mathcal {C}$ planar algebra with generator X to the $\mathbb {C}^2$ -spin model planar algebra from [Reference JonesJon21, Ex. 2.8].

The ‘box space’ $\mathfrak {F}(J)=\operatorname {\mathrm {End}}_{\mathcal {C}}(X^{\#J})$ where $\#J=n$ , is spanned by Temperley-Lieb string diagrams with n top boundary points and n bottom boundary points with three types of strands, $X,1_{\mathcal {C}},g$ subject to the following relations (in addition to $1_{\mathcal {C}}$ being the empty strand):

The first diagram denotes an orthogonal direct sum, which suppresses distinguished isometries $\iota _1:1_{\mathcal {C}} \to X$ and $\iota _g: g\to X$ satisfying $\iota _1\iota _1^\dagger + \iota _g\iota _g^\dagger = \operatorname {\mathrm {id}}_X$ . The rotations of $\iota _1$ and $\iota _g$ are their adjoints. We may always expand every diagram with X strands, so we see that $\mathfrak {F}(J)$ is spanned by diagrams with only $1$ and g strands; we only work with these string diagrams. Multiplication is stacking of boxes, where we get zero if the string types of $1_{\mathcal {C}}$ and g do not match.

Observe that $X^{n+1} \cong 2^n \cdot 1_{\mathcal {C}} \oplus 2^n\cdot g$ , so $\operatorname {\mathrm {End}}_{\mathcal {C}}(X^{n+1}) \cong M_{2^n}(\mathbb {C})\oplus M_{2^n}(\mathbb {C})$ . Let $p_1,p_g\in \operatorname {\mathrm {End}}_{\mathcal {C}}(X)$ be the orthogonal projections onto the first and second copy of $M_1(\mathbb {C}) \cong \mathbb {C}$ for $n=0$ . The operator $u:=p_1 - p_g\in \operatorname {\mathrm {End}}_{\mathcal {C}}(X)$ is a self-adjoint unitary generating $\operatorname {\mathrm {End}}_{\mathcal {C}}(X)$ . Now consider the morphism

Here, the crossings mean mapping between the two orthogonal copies of g in $X^2\cong 2\cdot 1_{\mathcal {C}} \oplus 2\cdot g$ . It is easily verified that v is a self-adjoint unitary which anticommutes with $u\otimes \operatorname {\mathrm {id}}_X$ and $\operatorname {\mathrm {id}}_X\otimes u$ . For the algebra $\mathfrak {F}(J)=\operatorname {\mathrm {End}}_{\mathcal {C}}(X^{\#J})$ where $\#J=n+1$ , we write $u_i$ for the copy of u on the i-th strand, and we write $v_j$ for the copy of v on the j-th and $(j+1)$ -th strands, where $i = 1, \dots , n+1$ and $j = 1, \dots , n$ . Observe that the $u_i, v_j$ give $2n+1$ self-adjoint unitaries satisfying the relations of Lemma 3.5. This gives an abstract isomorphism $\mathfrak {F}\cong \mathfrak {C}'\cong \mathfrak {D}'$ , where $\mathfrak {C}'$ and $\mathfrak {D}'$ are the concrete realizations in terms of Pauli matrices of the abstract presentations $\mathfrak {C}$ and $\mathfrak {D}$ , as defined above. We now give a concrete isomorphism.

The $\mathbb {C}^2$ -spin model planar algebra is a diagrammatic representation of the 1D spin chain with $\mathbb {C}^2$ -spins at each site of $\mathbb {Z}$ , with a local $M_2(\mathbb {C})$ acting at each site. We represent a distinguished ONB of $\mathbb {C}^2$ by an unshaded and a red node . (For the isomorphism $\mathfrak {F}\to \mathfrak {C}'$ , and , while for the isomorphism $\mathfrak {F}\to \mathfrak {D}'$ , and in the notation of Remark 3.7). Product tensors in $\bigotimes ^n \mathbb {C}^2$ are represented by drawing n nodes on a line, for example:

Matrix units for this computational basis $|\eta \rangle \langle \xi |$ are represented by rectangles where $|\eta \rangle $ is represented by nodes on the top of the rectangle and $|\xi \rangle $ is represented by nodes on the bottom of the rectangle, e.g.:

Composition is the bilinear extension of stacking boxes, where we get zero unless all nodes match along the middle.

When $\#J=n+1$ , we get an injection $\mathfrak {F}(J)\hookrightarrow \bigotimes ^{n+1} M_2(\mathbb {C})=M_{2^{n+1}}(\mathbb {C})$ by mapping a string diagram in the $1_{\mathcal {C}}$ and g strings to the matrix unit which only remembers the shadings at the end points, e.g.:

This map is well-defined and injective as all relations in $\mathcal {C}$ lie in its kernel. Indeed, observe that recabling the red strands has no effect on the location of the endpoints. Denote by $J+1$ the interval obtained from J by including one site to the right, and denote by $1+J$ the interval obtained from J by adding one site to the left. The following squares commute:

Moreover, these squares fit into a larger commutative cube with $\mathfrak {F}(1+J+1)$ , as adding strings/tensoring on the left and right commute.

We have thus constructed an embedding of nets of algebras from $\mathfrak {F}$ into the 1D spin chain. It remains to identify the image of $\mathfrak {F}$ under this map. By inspection, the image is exactly spanned by those diagrams with an even number of $ =|-\rangle $ boundary nodes, which is exactly the subalgebra of $M_{2^{n+1}}(\mathbb {C})$ which preserves the subspaces spanned by product tensors in the computational ONB which have an even or odd number of $=|-\rangle $ nodes. Hence if and $=|-\rangle $ , the image is exactly $\mathfrak {C}'$ , where the image of $u_i$ is $\sigma ^X_i$ and the image of $v_j$ is $\sigma ^Z_j\sigma _{j+1}^Z$ . If and $=|-\rangle $ , the image is exactly $\mathfrak {D}'$ , where the image of $u_i$ is $\sigma ^Z_i$ and the image of $v_j$ is $\sigma _j^X \sigma _{j+1}^X$ .

Proof. We prove the first case, and the second is similar and left to the reader. Without loss of generality, we may assume that $\partial \Delta $ is rough on all sides of $\Delta $ ; if this is not the case, we can replace $\Delta $ with a larger region satisfying this property. Observe that $xp_\Delta $ preserves the space of ground states for $H_\Delta $ , which is isomorphic to $\mathbb {C}^{2^{2i+2j-1}}$ where $\Delta $ has dimensions $i\times j$ (with $n+1 \leq i+4$ as $\Lambda \Subset _2\Delta $ ). Indeed, the space of ground states for $H_\Delta $ can be identified with the space of states along $\partial \Delta $ spanned by the simple tensors with an even number of $|-\rangle $ ’s in the $|\pm \rangle $ computational basis. Comparing with the faithful action of $\mathfrak {C}(I)$ on $\mathbb {C}^{2^{n+1}}$ from Remark 3.7 above, we can view the even-parity subspace of $\mathbb {C}^{2^{n+1}}$ as a subspace of the $\partial \Delta $ subspace by extending by all $|+\rangle $ outside sites in I, and we can view the odd-parity subspace of $\mathbb {C}^{2^{n+1}}$ as a subspace of the $\partial \Delta $ subspace by extending by all $|+\rangle $ outside sites in I except for a single site $j\in \partial \Delta \setminus I$ which is always $|-\rangle $ . These two subspaces witness a faithful action of $\mathfrak {C}(I)$ on the ground state subspace, and thus the map $x\mapsto xp_\Delta $ is injective.

Algorithm 3.10. Suppose we have rectangles Λ ≪₂Δ or Λ ⋐₂Δ, and set J := ∂Λ ∩ ∂Δ. We assume either J = ∅ or J is rough, and the case when J is smooth is entirely similar. The following algorithm expresses a local operator a ∈ 𝔄(Λ) which is a monomial in the Pauli operators that commutes with all A _s , B _p for s, p ⊂Δ as a product of the A _s , B _p for s, p ⊂Λ times an operator in ℭ(J).

• $\underline {\text{Step 1:}}$ We only apply this step if a is supported entirely on two adjacent columns or two adjacent rows of sites, e.g.,

  (3.11)

  

  We call these two rows or columns H. If a is supported on a larger region, go to Step 2.

  First, if H ∩ J = ∅, we claim a = 1. The Pauli operator for a on the outermost edge ℓ commutes with the A _s for the outermost vertex s, and thus must be either 1 _ℓ or σ _ℓ ^X , where 1 _ℓ is the unit of the algebra at edge ℓ. But it also commutes with the B _p next to it, so it must be either 1 _ℓ or σ _ℓ ^Z . We conclude it is 1 _ℓ . Working from the outside in, the result follows.

  Otherwise, H ∩ J≠∅, and arguing as in the previous paragraph, we may assume H ⊂J~. We claim a ∈ ℭ(J). We assume J~ is oriented similar to the left hand side of (3.11) as in the left hand side of (3.4); the other cases are similar. First, consider an extremal vertical edge ℓ of J~ beyond any rough horizontal edges as in (3.4). These extremal edges must always be 1 _ℓ , as they commute with the B _p to the left and the A _s above or below.

  Now, consider a non-extremal vertical edge ℓ of J~. By considering the plaquette term on the left side of ℓ, we know that a must be 1 _ℓ or σ _ℓ ^Z on this edge, as this plaquette only intersects J~ at ℓ. If a is σ _ℓ ^Z here, we can multiply a by a D _p ∈ ℭ(J) operator to the right of ℓ and thus assume that a acts as the identity on this edge. Hence, we may assume that the support of a is contained in J, the horizontal rough edges. Now, considering the star terms to the left of these edges, we know that a must be 1 _ℓ or σ _ℓ ^X = C _ℓ ∈ ℭ(J) for each rough edge ℓ, as the star term to the right of ℓ only intersects J at ℓ. Thus a ∈ ℭ(J).

• $\underline {\text{Step 2:}}$ Now suppose a is supported on a larger region. We pick a distinguished side of ∂Λ which is necessarily either rough or smooth. If J = ∅, any side of ∂Λ works. If J≠∅, we pick the distinguished side of ∂Λ which is opposite J. If the distinguished side is rough, go to Step 3; if it is smooth, go to Step 4.

• $\underline {\text{Step 3:}}$ Since the distinguished edge is rough, the Pauli operators on the rough edges must commute with A _s terms for vertices s on the outside of the rough edges, which are necessarily in Δ ∖Λ. This means these Pauli operators must be either 1 _ℓ or σ _ℓ ^X . For each σ _ℓ ^X that appears, multiply by an A _s for the star s ⊂Λ containing ℓ to ‘cancel’ the σ _ℓ ^X . This will work except possibly at the two edges of the distinguished edge, where the canceling star s may not be contained in Λ. In this case, we see the edge in question also commutes with a plaquette operator in Δ, forcing the Pauli operator to be 1 _ℓ .

  We may now view a′ as a monomial supported on a smaller rectangle Λ′⊂Λ with a smooth edge opposite J. Go back to Step 1 with a′ supported on Λ′.

• $\underline {\text{Step 4:}}$ Since the distinguished edge is smooth, the Pauli operators on the smooth edges must commute with B _p terms for plaquettes p on the outside of the smooth edges, which are necessarily in Δ ∖Λ. This means these Pauli operators must be either 1 _ℓ or σ _ℓ ^Z . For each σ _ℓ ^Z that appears, multiply by a B _p for the plaquette p ⊂Λ containing ℓ to ‘cancel’ the σ _ℓ ^Z . This will work except possibly at the two edges of the distinguished edge, where the canceling plaquette p may not be contained in Λ. In this case, we see the edge in question also commutes with a star operator in Δ, forcing the Pauli operator to be 1 _ℓ .

  We may now view a′ as a monomial supported on a smaller rectangle Λ′⊂Λ with a rough edge opposite J. Go back to Step 1 with a′ supported on Λ′.

Observe that the product obtained from this algorithm is independent of the choice of Δ beyond that Λ ≪₂Δ or Λ ⋐₂Δ with J = ∂Λ ∩ ∂Δ.

Proof. By Lemma 4.5, $p_\Delta \bigotimes _{v\in \Delta } \mathcal {H}_v$ is unitarily isomorphic to $\mathcal {S}_{\mathcal {C}}(\#\partial \Delta )$ , and $\Gamma _\varphi $ acts by gluing $\varphi $ onto the sites in $I\subset \partial \Delta $ . Clearly $\Gamma _\varphi $ preserves $p_\Delta \bigotimes _{v\in \Delta } \mathcal {H}_v$ , so $[\Gamma _\varphi , p_\Delta ]=0$ . That $\Gamma _\varphi p_\Delta =0$ implies $\varphi =0$ is easily verified using the positive definite skein module inner product. Finally, if $\Gamma _\varphi p_\Lambda =0$ , then $\Gamma _\varphi p_\Delta =0$ , so $\varphi =0$ .

Proof.

• $\underline {(\text {LTO1}):}$ Suppose $\Lambda \ll _1 \Delta $ . We define the $\partial \Delta $ -tube algebra $\operatorname {Tube}_{\mathcal {C}}(\partial \Delta )$ with internal and external boundaries given by $\partial \Delta $ :

  

  Observe that $p^A_\Delta \bigotimes _{v\in \Delta } \mathcal {H}_v$ carries a $\operatorname {Tube}_{\mathcal {C}}(\partial \Delta )$ -action where we resolve the $\mathcal {C}$ -morphism from the annulus into the outer-most vertex spaces. (This action is similar to [Reference Green, Huston, Kawagoe, Penneys, Poudel and SanfordGHK⁺24, Eq. (10)].) Since $\Lambda \ll \Delta $ , every $p_\Delta ^A x p_\Delta ^A$ commutes with this $\operatorname {Tube}_{\mathcal {C}}(\partial \Delta )$ -action for $x\in \mathfrak {A}(\Lambda )$ .

  Now consider the evaluation map into the skein module $\operatorname {\mathrm {eval}}: p^A_\Delta \bigotimes _{v\in \Delta }\mathcal {H}_v\to \mathcal {S}(\#\partial \Delta )$ . Observe that $\mathcal {S}(\#\partial \Delta )$ also carries a $\operatorname {Tube}_{\mathcal {C}}(\partial \Delta )$ -action, and the evaluation map clearly intertwines the actions. We also observe that $\mathcal {S}(\#\partial \Delta )$ is an irreducible $\operatorname {Tube}_{\mathcal {C}}(\partial \Delta )$ -module, as all rank-one operators on $\mathcal {S}(\#\partial \Delta )$ can be realized by operators in an annulus with no $\mathcal {C}$ -strings going around the annulus. By Lemma 4.5, for any $x\in \mathfrak {A}(\Lambda )$ ,

  $$ \begin{align*}p_\Delta x p_\Delta = p^A_\Delta p^B_\Delta p^A_\Delta x p^A_\Delta p^B_\Delta p^A_\Delta = D_{\mathcal{C}}^{-2\#p\in \Delta} p^A_\Delta \operatorname{\mathrm{eval}}^\dagger \underbrace{ \operatorname{\mathrm{eval}} (p^A_\Delta x p^A_\Delta) \operatorname{\mathrm{eval}}^\dagger }_{\in \operatorname{\mathrm{End}}_{\operatorname{Tube}_{\mathcal{C}}(\partial \Delta)}(\mathcal{S}(\#\partial \Delta))\cong \mathbb{C}} \operatorname{\mathrm{eval}} p^A_\Delta. \end{align*} $$

  This means that the compression of x above lies in

  $$ \begin{align*}\mathbb{C} (\operatorname{\mathrm{eval}}^\dagger \operatorname{\mathrm{eval}}) \underset{\text{(Lem.~4.5)}}{=} \mathbb{C} p^B_\Delta \end{align*} $$
  acting on $p^A_\Delta \bigotimes _{v\in \Delta } \mathcal {H}_v$ . Thus $p_\Delta \mathfrak {A}(\Lambda )p_\Delta = \mathbb {C} p_\Delta $ .

• $\underline {(\text {LTO2}):}$ Suppose $\Lambda \Subset _1 \Delta $ with $\partial \Lambda \cap \partial \Delta =I\neq \emptyset $ . Without loss of generality, we may assume I is at the top or right of $\Lambda $ . We claim that $p_\Delta \mathfrak {A}(\Lambda )p_\Delta =\mathfrak {F}(I)p_\Delta $ . This will prove (LTO2) as

  $$ \begin{align*}\mathfrak{F}(I)p_\Delta \underset{\text{(Lem.~4.7)}}{\subseteq} \mathfrak{B}(\Lambda\Subset_1 \Delta) \subseteq p_\Delta \mathfrak{A}(\Lambda)p_\Delta = \mathfrak{F}(I)p_\Delta. \end{align*} $$

  It remains to prove $p_\Delta \mathfrak {A}(\Lambda )p_\Delta \subseteq \mathfrak {F}(I)p_\Delta $ . By Lemma 4.5, $p_\Delta \bigotimes _{v\in \Delta } \mathcal {H}_v \cong \mathcal {S}(\#\partial \Delta )$ , which can be identified with

  $$ \begin{align*}\bigoplus_{\vec{c}_{\partial\Delta\setminus I},\vec{c}_{ I}}\mathcal{C}(\vec{c}_{\partial\Delta\setminus I} \to \vec{c}_I) \cong \mathcal{C}(X^{\#\partial\Delta\setminus I} \to X^{\# I}) \end{align*} $$
  with the skein module inner product, where $\vec {c}_{\partial \Delta \setminus I}= c_{i_1}\cdots c_{i_m}$ is a tensor product of simples in $\mathcal {C}$ over the sites of $\partial \Delta \setminus I$ , and $\vec {c}_I = c_{j_1}\cdots c_{j_n}$ is a tensor product of simples over the sites of I. Observe that this space is canonically an invertible $\operatorname {\mathrm {End}}_{\mathcal {C}}(X^{\# I}) - \operatorname {\mathrm {End}}_{\mathcal {C}}(X^{\# \partial \Delta \setminus I})$ bimodule, i.e., these algebras are each other’s commutants by the Yoneda Lemma.

  Now given $x\in \mathfrak {A}(\Lambda )$ , $p_\Delta xp_\Delta $ acts as the identity on all tensorands $\mathcal {H}_v$ with $v\in \partial \Delta \setminus I$ . Thus the action of $p_\Delta ap_\Delta $ commutes with the right action of $\operatorname {\mathrm {End}}_{\mathcal {C}}(X^{\# \partial \Delta \setminus I})$ transported back to $p_\Delta \bigotimes _{v\in \Delta } \mathcal {H}_v$ from the skein module. We conclude that there is a $\varphi \in \operatorname {\mathrm {End}}_{\mathcal {C}}(X^{\# I})$ such that $p_\Delta xp_\Delta = \Gamma _{\varphi }p_\Delta $ .

• $\underline {\text {(LTO3)}:}$ Suppose $\Lambda _1 \subset \Lambda _2 \Subset _1 \Delta $ with $\partial \Lambda _1 \cap \partial \Delta = \partial \Lambda _2 \cap \partial \Delta =I\neq \emptyset $ . Without loss of generality, we may assume I is at the top or right of $\Lambda _i$ for $i=1,2$ . By the proof of (LTO2) above,

  $$ \begin{align*}\mathfrak{B}(\Lambda_1 \Subset_1 \Delta) = \mathfrak{F}(I)p_{\Delta} = \mathfrak{B}(\Lambda_2 \Subset_1 \Delta). \end{align*} $$

• $\underline {\text {(LTO4)}:}$ Suppose $\Lambda \Subset _1 \Delta _1\subset \Delta _2$ with $\partial \Lambda \cap \partial \Delta _1 = \partial \Lambda \cap \partial \Delta _2 =I\neq \emptyset $ . Without loss of generality, we may assume I is at the top or right of $\Lambda $ . We saw in the proof of (LTO2) above that $\mathfrak {B}(\Lambda \Subset _1 \Delta _1) = \mathfrak {F}(I)p_{\Delta _1}$ . By Lemma 4.7 applied to $\Lambda \Subset _1 \Delta _2$ , $\Gamma _\varphi p_{\Delta _2}=0$ implies $\varphi =0$ , so $\Gamma _\varphi p_{\Delta _1}=0$ .