Planar algebras, quantum information theory and subfactors

After choosing the external $*$-arc appropriately, the relations on top in Figure 8 are equivalent to the unitarity of $u$ while the relations on the bottom are equivalent to the unitarity of $Z_{R(k,\epsilon,\ell)}(u)$. $\Box$

1) Let $\displaystyle{u=\sum_{i,j}a^{i}_{j}~{}~{}e^{i}_{j}\in P_{(2,+)}}$. Then,

and

Let $H=((\sqrt{n}a^{i}_{j}))$. Then Equation 3.1 is equivalent to $HH^{*}=n~{}I$ and Equation 3.2 is equivalent to $|\sqrt{n}a^{i}_{j}|=1$. Thus the association of $u$ with $H$ is a 1-1 correspondence between $\{0,1\}$-biunitary elements of $P_{(2,+)}$ and Hadamard matrices of size $n\times n$.

2) Now let $\displaystyle{u=\sum_{i,j,k}a_{ij}^{k}~{}~{}e^{i}_{k}(j]\in P_{(3,+)}}$. Then,

and

Let $Q=((Q^{i}_{j}))$ where $Q^{i}_{j}=(a_{1j}^{i},a_{2j}^{i},\ldots,a_{nj}^{i})\in{\mathbb{C}}^{n}$. Then, Equation 3.3 is equivalent to the column vectors of $Q$ forming an orthonormal basis for $\mathbb{C}^{n}$ and Equation 3.4 is equivalent to the row vectors of $Q$ forming an orthonormal basis for $\mathbb{C}^{n}$. Thus the association of $u$ with $Q$ is a 1-1 correspondence between $\{0,1\}$-biunitary elements of $P_{(3,+)}$ and quantum Latin squares of size $n\times n$.

3) Let $\displaystyle{u=\sum_{i,j,k,\ell}a^{ij}_{k\ell}e^{ij}_{\ell k}\in P_{(4,+)}}$. Then,

and

Here let $U=((a^{ij}_{k\ell}))$. Then, Equation 3.5 is equivalent to $U$ being unitary and Equation 3.6 is equivalent to the block transpose of $U$ being unitary. Thus the association of $u$ with $U$ is a 1-1 correspondence between $\{0,2\}$-biunitary elements of $P_{(4,+)}$ and biunitary matrices of size $n^{2}\times n^{2}$. $\Box$

Let $\displaystyle{u=\sum_{i,j,k,\ell}~{}~{}a^{ij}_{k\ell}~{}~{}e^{ij}_{\ell k}~{}~{}}\in P_{(4,+)}$. Then,

by Lemma 16 and the observation above. Hence

and

Let $B(j,\ell)$ be the $n\times n$ matrix given by $B(j,\ell)=((\sqrt{n}a^{ij}_{k\ell}))$. Then Equation 3.8 is equivalent to $B(j,\ell)$ being a unitary matrix for all $j,\ell$ and Equation 3.7 is equivalent to the collection $\{B(j,\ell)\}_{j,\ell}$ forming an orthonormal basis for $M_{n}({\mathbb{C}})$. Thus the association of $u$ with $\{B(j,\ell)\}_{j,\ell}$ is a 1-1 correspondence between $\{A,R(4,+)\}$-biunitary elements of $P_{(4,+)}$ and unitary error bases of $M_{n}({\mathbb{C}})$. $\Box$

Consider the maps $\sigma$ and $\sigma^{*}$ defined by the left and right side pictures in Figure 15 respectively. It is clear that they are equivariant for the $*$-operations on $P_{(n\ell,\epsilon\eta^{\ell})}$ and $P_{(n\ell+k-\ell,\epsilon\eta^{\ell})}$. To show that they are actually adjoints of each other, it suffices to verify the equality of Figure 16 for arbitrary $x\in P_{(n\ell,\epsilon\eta^{\ell})}$ and $z\in P_{(n\ell+k-\ell,\epsilon\eta^{\ell})}$.

This is clear by isotopy. Finally, that $\sigma^{*}\sigma=id$ is a simple pictorial verification using the equalities of Figure 8. $\Box$

(1) $\Rightarrow$ (2) Using the relations in Figure 8, it is easy to see that the equation in Figure 12 implies those of Figure 13.
(2) $\Rightarrow$ (3) This is clear.
(3) $\Rightarrow$ (1) We need to see that the double circle relation of Figure 14 implies the existence of a $y$ satisfying the equation in Figure 12.

Observe that the picture on the left in Figure 14 is given by $\sigma^{*}F\sigma(x)$. Thus the double circle relation implies that $\sigma^{*}F\sigma(x)=x$ and hence (applying $\sigma$ on both sides and using the definition of $E$) that $EF\sigma(x)=\sigma(x)$. Since $E$ and $F$ are projections, norm considerations imply that $||EF\sigma(x)||\leq||F\sigma(x)||\leq||\sigma(x)||$. Therefore equality holds throughout and so $F\sigma(x)=\sigma(x)$. Now define $y$ by the first equality in Figure 13. The equation $F\sigma(x)=\sigma(x)$ then implies that the equation on the left of Figure 18 holds and therefore also the equation on the right.

Finally, using the relations of Figure 8, this completes the proof. $\Box$

In order to prove that $Q$ is a $C^{*}$-planar subalgebra of ${}^{(\ell,\epsilon)}P$ it is enough to prove that it is a planar subalgebra of ${}^{(\ell,\epsilon)}P$ and that it is closed under $*$. Closure under $*$ is clear from the double circle condition of Figure 14.

To verify that $Q$ is a planar subalgebra of ${}^{(\ell,\epsilon)}P$, it suffices to see that it is closed under the action of any set of ‘generating tangles’. A set of such generating tangles, albeit for the class of ‘restricted tangles’ - see [DeKdy2018] - was given in Theorem 3.5 of [KdySnd2004]. It follows easily from that result that a set of generating tangles for all tangles is given by $\{1^{(0,\pm)}\}\cup\{M_{(n,\eta),(n,\eta)}^{(n,\eta)},I_{(n,\eta)}^{(n+1,\eta)},E_{(n+1,\eta)}^{(n,\eta)}:n\geq 0\}\cup\{R^{(n,-\eta)}_{(n,\eta)}:n\geq 1\}$. Here $E$, $M$ and $I$ are the tangles in Figure 19 below. We will show, case by case, that $Q$ is closed under the action of each of these tangles.

Closure under $1^{(0,\pm)}$: We need to check that $Z_{1^{(0,\pm)}}^{(^{(\ell,\epsilon)}P)}(1)=1_{(0,\epsilon(\pm)^{\ell})}\in Q_{(0,\pm)}$. This follows directly from the double circle relation of Figure 14.

Closure under $M=M_{(n,\eta),(n,\eta)}^{(n,\eta)}$: We need to check that if $x_{1},x_{2}\in Q_{(n,\eta)}$, then $Z_{M}^{(^{(\ell,\epsilon)}P)}(x_{1}\otimes x_{2})=Z_{M^{(\ell,\epsilon)}}^{P}(x_{1}\otimes x_{2})\in Q_{(n,\eta)}$. Observe that $M^{(\ell,\epsilon)}$ is the multiplication tangle of colour $(n\ell,\epsilon\eta^{\ell})$. Now suppose that $y_{1},y_{2}\in P_{(n\ell,\epsilon\eta^{\ell}(-)^{k-\ell})}$ are such that the equation in Figure 12 holds for $x_{1},y_{1}$ and for $x_{2},y_{2}$. It is easy to see that then the same equation also holds for $x_{1}x_{2},y_{1}y_{2}$.

Closure under $I=I_{(n,\eta)}^{(n+1,\eta)}$: We need to see that if $x\in Q_{(n,\eta)}$, then $Z_{I}^{(^{(\ell,\epsilon)}P)}(x)=Z^{P}_{I^{(\ell,\epsilon)}}\in Q_{(n+1,\eta)}$. Observe that the tangle $I^{(\ell,\epsilon)}$ is the $\ell$-fold iterated inclusion tangle from $P_{(n\ell,\epsilon\eta^{\ell})}$ to $P_{((n+1)\ell,\epsilon\eta^{\ell})}$. Suppose that $y\in P_{(n\ell,\epsilon\eta^{\ell}(-)^{k-\ell})}$ is such that the equation in Figure 12 holds for $x,y$. Again, an easy verification shows that the equation in Figure 12 also holds for $Z_{I^{(\ell,\epsilon)}}(x),Z_{\tilde{I}^{(\ell,\epsilon)}}(y)$, where $\tilde{I}^{(\ell,\epsilon)}$ is the $\ell$-fold iterated inclusion tangle from $P_{(n\ell,\epsilon\eta^{\ell}(-)^{k-\ell})}$ to $P_{((n+1)\ell,\epsilon\eta^{\ell}(-)^{k-\ell}))}$.

Closure under $E=E_{(n+1,\eta)}^{(n,\eta)}$: We need to see that if $x\in Q_{(n+1,\eta)}$, then $Z_{E}^{(^{(\ell,\epsilon)}P)}(x)$ $=Z^{P}_{E^{(\ell,\epsilon)}}(x)\in Q_{(n,\eta)}$. Observe that the tangle $E^{(\ell,\epsilon)}$ is the $\ell$-fold iterated conditional expectation tangle from $P_{((n+1)\ell,\epsilon\eta^{\ell})}$ to $P_{(n\ell,\epsilon\eta^{\ell})}$. Take $y\in P_{((n+1)\ell,\epsilon\eta^{\ell}(-)^{k-\ell})}$ such that the equation in Figure 12 holds for $x,y$. We will verify that then, the equation in Figure 12 also holds for $Z_{E^{(\ell,\epsilon)}}(x),Z_{\tilde{E}^{(\ell,\epsilon)}}(y)$, where $\tilde{E}^{(\ell,\epsilon)}$ is the $\ell$-fold iterated conditional expectation tangle from $P_{((n+1)\ell,\epsilon\eta^{\ell}(-)^{k-\ell})}$ to $P_{(n\ell,\epsilon\eta^{\ell}(-)^{k-\ell}))}$.

First note that, it is an easy consequence of the relations in Figure 8 that the relations of Figure 20 hold for all $(n,\eta)$.

Now, these relations, in turn, imply the equations in Figure 21. In this figure, the first and the third equalities follow from Figure 20 while the second equality is a consequence of the proof of closure under $I$.

Closure under $R=R^{(n,-\eta)}_{(n,\eta)}$: We need to see that if $x\in Q_{(n,\eta)}$, then $Z_{R}^{(^{(\ell,\epsilon)}P)}(x)=Z^{P}_{R^{(\ell,\epsilon)}}(x)\in Q_{(n,-\eta)}$. We illustrate how this is done when $(n,\eta)=(3,+)$. It should be clear that the proof of the general case is similar. Begin by observing that since $x\in Q_{(3,+)}\subseteq P_{(3\ell,\epsilon)}$, it satisfies the double circle relation of Figure 22.

Now, moving the external $*$ $\ell$ steps counterclockwise and redrawing yields the equation in Figure 23.

A little thought now shows that this is precisely the double circle relation for $Z^{P}_{R^{(\ell,\epsilon)}}(x)\in P_{(3\ell,\epsilon(-)^{\ell})}$, establishing that $Z^{P}_{R^{(\ell,\epsilon)}}(x)$ indeed belongs to $Q_{(3,-)}$ as desired. $\Box$

Given Theorem 25, what remains to be seen is that $Q$ is connected, has modulus $(\sqrt{n})^{\ell}$ and is spherical with positive definite picture trace. Since $P$ has modulus $\sqrt{n}$, the cabling ${}^{(\ell,\epsilon)}P$ has modulus $(\sqrt{n})^{\ell}$ and so does $Q$. The other assertions need a little work.

Note that $Q_{(0,\eta)}$ is a subspace of $(^{(\ell,\epsilon)}P)_{(0,\eta)}=P_{(0,\epsilon\eta^{\ell})}$ and so if $\epsilon\eta^{\ell}=+$, then $Q_{(0,\eta)}$ is 1-dimensional since $P_{(0,+)}$ is so. If $\epsilon\eta^{\ell}=-$, then, $Q_{(0,\eta)}$ is the subspace of all $x\in P_{(0,-)}$ such that the double circle relation of Figure 14 holds for $x$. From Theorem 2, a basis of $P_{(0,-)}$ is given by all $S(i)$ for $i=1,\cdots,n$ and by the black and white modulus relations, a double circle around these gives a scalar multiple of $1^{(0,-)}$. It follows that $x$ is necessarily a scalar multiple of $1^{(0,-)}$ so that $Q_{(0,\eta)}$ is 1-dimensional, in this case as well. Hence $Q$ is connected.

To see that $Q$ is spherical, observe first that on any $P_{(n,\eta)}$ the composites of the left and right picture traces with the traces $\tau_{\pm}$ on $P_{(0,\pm)}$ (which specify its $C^{*}$-planar algebra structure - see Definition 12 of [KdyMrlSniSnd2019]) are equal. This is seen by explicit computation with the bases of $P_{(n,\eta)}$ and can be regarded as a version of sphericality for $P$. It is clear that this property descends to $Q$.

Finally observe that the picture trace on $Q_{(n,\eta)}$ is exactly the composite of $\tau_{\pm}$ with the picture trace on $P_{(n\ell,\epsilon\eta^{\ell})}$ and is consequently positive definite. $\Box$

Only the irreducibility of $Q$ needs to be seen and we will show using explicit bases computations that $dim(Q_{(1,+)})=1$. We only consider the $\epsilon=+$ case, the proof in the other case being similar. Thus $Q_{(1,+)}\subseteq P_{(1,+)}$. Begin with $x=\sum_{i}\lambda_{i}e(i]\in Q_{(1,+)}$. The double circle relation for $x$ implies that the equation of Figure 4 holds.