Generators for finite depth subfactor planar algebras

Abstract

We show that a subfactor planar algebra of finite depth k is generated by a single s-box, for $s \leq \min \{k+4,2k\}$.

The main result of Kodiyalam and Tupurani [3] shows that a subfactor planar algebra of finite depth is singly generated with a finite presentation. If P is a subfactor planar algebra of depth k, it is shown there that a single 2k-box generates P. It is natural to ask what the smallest s is such that a single s-box generates P. While we do not resolve this question completely, we show in this note that $s \leq \min \{k+4,2k\}$ and that k does not suffice in general. All terminology and unexplained notation will be as in [3].

For the rest of the paper fix a subfactor planar algebra P of finite depth k. Let 2t be such that it is the even number of k+3 and k+4. We will show that some s-box generates P as a planar algebra, where $s = \min \{2k,2t\}$. The main observation is the following result about involutive algebra anti-automorphisms of finite-dimensional complex semisimple algebras. We mention as a matter of terminology that we always deal with $\mathbb {C}$-algebra anti-automorphisms and automorphisms (as opposed to those that might induce a non-identity involution on the base field $\mathbb {C}$). Also, as is common in Hopf algebra literature, we will use S a instead of S(a) to demote the image of a under a map S of algebras.

FormalPara Theorem 1.

Let A be a finite-dimensional complex semisimple algebra and let $S:A \rightarrow A$ be an involutive algebra anti-automorphism. Suppose that A has no 2×2 matrix summand. Then, there exists a∈A such that a and Sa generate A as an algebra.

Before beginning the proof of this theorem, we observe that the somewhat peculiar restriction on A not having an $M_{2}(\mathbb {C})$ summand is really necessary.

FormalPara Remark 2.

The map S : M ₂($\mathbb {C}$) → M ₂($\mathbb {C}$) defined by S a = a d j(a) is easily verified to be an involutive algebra anti-automorphism, while there exists no $a \in M_{2}(\mathbb {C})$ that together with S a generates $M_{2}(\mathbb {C})$ since these generate only a commutative subalgebra.

We pave the way for a proof of Theorem 1 by studying the two special cases when $A = M_{n}(\mathbb {C})$ and $A = M_{n}(\mathbb {C}) \oplus M_{n}(\mathbb {C})$. In these, n is a fixed positive integer. We will need the following lemmas that specify a ‘standard form’ for each of these two special cases.

FormalPara Lemma 3.

Let S be an involutive algebra anti-automorphism of $M_{n}(\mathbb {C})$ . There is an algebra automorphism of $M_{n}(\mathbb {C})$ under which S is identified with either (i) the transpose map or (ii) the transpose map followed by conjugation by the matrix

$$\begin{array}{@{}rcl@{}} J = \left[ \begin{array}{cc} 0 & I_{k}\\ -I_{k} & 0 \end{array} \right] (= -J^{T} = -J^{-1}). \end{array} $$

The second case may arise only when n=2k is even (and I _k denotes, of course, the identity matrix of size k).

FormalPara Proof.

Let T denote the transpose map on $M_{n}(\mathbb {C})$. The composite map TS is then an algebra automorphism of $M_{n}(\mathbb {C})$ and is consequently given by conjugation with an invertible matrix, say u. Thus S x=(u x u ⁻¹)^T. Involutivity of S implies that u is either symmetric or skew-symmetric. By Takagi’s factorization (see p. 204 and p. 217 of [1]), u is of the form v ^T v if it is symmetric and of the form v ^T J v if it is skew-symmetric, for some invertible v. For the algebra automorphism of $M_{n}(\mathbb {C})$ given by conjugation with v, S gets identified in the symmetric case with the transpose map and in the skew-symmetric case with the transpose map followed by conjugation by J. □

FormalPara Lemma 4.

Let S be an involutive algebra anti-automorphism of $M_{n}(\mathbb {C}) \oplus M_{n}(\mathbb {C})$ that interchanges the two minimal central projections. There is an algebra automorphism of $M_{n}(\mathbb {C}) \oplus M_{n}(\mathbb {C})$ fixing the minimal central projections under which S is identified with the map x⊕y↦y ^T ⊕x ^T.

FormalPara Proof.

The map x⊕y↦S(y ^T⊕x ^T) is an algebra automorphism of $M_{n}(\mathbb {C}) \oplus M_{n}(\mathbb {C})$ fixing the minimal central projections and is therefore given by x⊕y↦u x u ⁻¹⊕v y v ⁻¹ for invertible u,v. Hence S(x⊕y) = u y ^T u ⁻¹⊕v x ^T v ⁻¹.

Thus, S ²(x⊕y) = u(v ⁻¹)^T x v ^T u ⁻¹⊕v(u ⁻¹)^T y u ^T v ⁻¹. Involutivity of S now implies that v ^T u ⁻¹ and u ^T v ⁻¹ are both scalar matrices, or equivalently, v ^T = λ u and u ^T = μ v for non-zero scalars λ,μ. Taking transposes shows that λ μ=1 and finally, by replacing u by λ u, we may assume that v = u ^T.

The commutativity of the following diagram:

now implies that under the algebra automorphism of $M_{n}(\mathbb {C}) \oplus M_{n}(\mathbb {C})$ given by x⊕y↦u ⁻¹ x u⊕y, S is identified with x⊕y↦y ^T⊕x ^T. □

The proof of Theorem 1 in the case $A = M_{n}(\mathbb {C})$ (for n ≠ 2) needs some preparation. For a subset $S \subseteq M_{n}(\mathbb {C})$ we use the notation $S^{\prime }$, as usual, to denote its commutant in $M_{n}(\mathbb {C})$.

FormalPara Lemma 5.

If $U \subseteq \mathbb {C}^{2N}$ is non-empty and Zariski open, then

$$\begin{array}{@{}rcl@{}} U \cap \{(z_{1},\ldots,z_{N},\overline{z_{1}},\ldots,\overline{z_{N}}) : z_{i} \in \mathbb{C}\} \neq \emptyset. \end{array} $$

FormalPara Proof.

It suffices to show that $S = \{(z_{1},\ldots ,z_{N},\overline {z_{1}},\ldots ,\overline {z_{N}}) : z_{i} \in \mathbb {C}\}$ is Zariski dense in $\mathbb {C}^{2N}$. If a polynomial f in 2N variables vanishes on S, then the polynomial p(u ₁,…,u _N ,v ₁,…v _N ) = f(u ₁ + i v ₁,…,u _N + i v _N ,u ₁−i v ₁,…,u _N −i v _N ) vanishes on $\mathbb {R}^{2N}$. It is then easily seen by induction on the number of variables that p identically vanishes and then, so does f. □

FormalPara PROPOSITION 6

For n>1, the set

$$\begin{array}{@{}rcl@{}} U = \left\{\vphantom{\frac{A^{A}}{A^{A}_A}} (P,Q) \in M_{n}(\mathbb{C}) \times M_{n}(\mathbb{C}): P,Q {\text{~invertible and~}} \left. \left. \begin{array}{cc} & \\ & \end{array} \right. \right.^{} \right. \\ \left. \left\{ \left[ \begin{array}{cc} 0 & P\\ Q & 0 \end{array} \right], \left[ \begin{array}{cc} 0 & P^{T}\\ Q^{T} & 0 \end{array} \right] \right\}^{\prime} = \mathbb{C}I_{2n} \right\}. \end{array} $$

is a non-empty Zariski open subset of $M_{n}(\mathbb {C}) \times M_{n}(\mathbb {C}).$

FormalPara Proof.

For an arbitrary matrix $\left [ \begin {array}{cc} X & Y\\ Z & W \end {array} \right ] \in M_{2n}(\mathbb {C})$, the condition that it commute with both $\left [ \begin {array}{cc} 0 & P\\ Q & 0 \end {array} \right ]$ and $\left [ \begin {array}{cc} 0 & P^{T}\\ Q^{T} & 0 \end {array} \right ]$ is given by a set of 8n ² homogeneous linear equations in the 4n ² entries of X,Y,Z,W with coefficient (linear) polynomials in the entries of P and Q.

The solution space for this system is at least one dimensional (since it certainly contains the identity matrix) and thus the coefficient matrix has rank at most 4n ²−1. The condition that the solution space is exactly one dimensional is hence equivalent to the condition that the coefficient matrix has rank at least 4n ²−1, which is clearly Zariski open condition in the entries of P and Q. It follows that U is Zariski open.

To show non-emptiness of U, choose an invertible $Q \in M_{n}(\mathbb {C})$ such that Q and Q ^T generate $M_{n}(\mathbb {C})$ as an algebra. For instance, Q could be I _n + N _n , where N _n is the n×n nilpotent matrix with super-diagonal entries, all 1 and 0 entries elsewhere. The condition that $\left [ \begin {array}{cc} X & Y\\ Z & W \end {array} \right ] \in M_{2n}(\mathbb {C})$commutes with both $\left [ \begin {array}{cc} 0 & I\\ Q & 0 \end {array} \right ]$ and $\left [ \begin {array}{cc} 0 & I\\ Q^{T} & 0 \end {array} \right ]$ is equivalent to the set of equations:

$$\begin{array}{@{}rcl@{}} &YQ=QY=Z=YQ^{T}=Q^{T}Y,\\ &WQ=QX,~X=W,~WQ^{T}=Q^{T}X. \end{array} $$

Since Y commutes with Q and Q ^T (which generate $M_{n}(\mathbb {C})$), Y = λ I _n for a scalar $\lambda \in \mathbb {C}$. Thus Z = λ Q = λ Q ^T. Now (and this is the crucial point where n>1 is needed), since Q and Q ^T generate $M_{n}(\mathbb {C})$ which is not commutative, they cannot be equal and so λ=0. Since X = W and hence commutes with both Q and Q ^T, X = W = μ I for some scalar $\mu \in \mathbb {C}$. Thus (I,Q)∈U. □

FormalPara PROPOSITION 7

Let S be an involutive algebra anti-automorphism of $M_{m}(\mathbb {C})$ with m ≠ 2. There exists invertible $x \in M_{m}(\mathbb {C})$ which, together with S x, generates $M_{m}(\mathbb {C})$ as an algebra.

FormalPara Proof.

First, we may assume by Lemma 3 that S is either (i) the transpose map or (ii) the transpose map followed by conjugation by J. In Case (i), as in the proof of Proposition 6, x = I _m + N _m is invertible and such that x and S x generate $M_{m}(\mathbb {C})$ as an algebra.

In Case (ii), m=2n is necessarily even. It then follows from Proposition 6 and Lemma 5 that there is an invertible $P \in M_{n}(\mathbb {C})$ such that

$$\begin{array}{@{}rcl@{}} \left\{ \left[ \begin{array}{cc} 0 & P\\ \bar{P} & 0 \end{array} \right], \left[ \begin{array}{cc} 0 & P^{T}\\ \bar{P}^{T} & 0 \end{array} \right] \right\}^{\prime} = \mathbb{C}I_{2n} \end{array} $$

The commutant of these two matrices is the same as that of the algebra they generate which is a ∗-subalgebra of $M_{m}(\mathbb {C})$ since they are adjoints of each other. By the double commutant theorem, it follows that the algebra generated by these is the whole of $M_{m}(\mathbb {C})$. Now take $x = \left [ \begin {array}{cc} 0 & P\\ \bar {P} & 0 \end {array} \right ]$. □

In proving Theorem 1 for $A = M_{n}(\mathbb {C}) \oplus M_{n}(\mathbb {C})$, we will need the following lemma.

FormalPara Lemma 8.

Let A and B be finite dimensional complex unital algebras and let a∈A and b∈B be invertible. Then, for all but finitely many $\lambda \in \mathbb {C}$ , the algebra generated by a⊕λb∈A⊕B contains both a (=a⊕0) and b (=0⊕b).

FormalPara Proof.

We may assume that λ ≠ 0 and then it suffices to see that a is expressible as a polynomial in a⊕λ b. Note that since a⊕λ b is invertible and A⊕B is finite dimensional, the algebra generated by a⊕λ b is actually unital. In particular, it makes sense to evaluate any complex univariate polynomial on a⊕λ b.

Let p(X) and q(X) be the minimal polynomials of a and b respectively. By invertibility of a and b, neither p nor q has 0 as a root. The minimal polynomial of λ b is $q(\frac {X}{\lambda })$. Unless λ is the quotient of a root of p by a root of q, p(X) and $q(\frac {X}{\lambda })$ will have no common roots and therefore be coprime. So there will exist a polynomial r(X) that is divisible by $q(\frac {X}{\lambda })$ but is X modulo p(X). Thus r(a⊕λ b) = a, as desired. □

FormalPara PROPOSITION 9

Let S be an involutive algebra anti-automorphism of $M_{n}(\mathbb {C}) \oplus M_{n}(\mathbb {C})$ that interchanges the two minimal central projections. There exists invertible $x \oplus y \in M_{n}(\mathbb {C}) \oplus M_{n}(\mathbb {C})$ which together with S(x⊕y) generates $M_{n}(\mathbb {C}) \oplus M_{n}(\mathbb {C})$ as an algebra.

FormalPara Proof.

First, by Lemma 4, we may assume that S is the map x⊕y↦y ^T⊕x ^T. Now, as in the proof of Proposition 7, there is an invertible $x \in M_{n}(\mathbb {C})$ such that x and x ^T generate $M_{n}(\mathbb {C})$. By Lemma 8, for all but finitely many $\lambda \in \mathbb {C}$, the algebra generated by x⊕λ x contains x⊕0 and 0⊕x and similarly the algebra generated by λ x ^T⊕x ^T contains x ^T⊕0 and 0⊕x ^T. Thus the algebra generated by x⊕λ x and λ x ^T⊕x ^T is the whole of $M_{n}(\mathbb {C}) \oplus M_{n}(\mathbb {C})$. □

FormalPara Proof Proof of Theorem 1.

Let $\hat {A}$ denote the (finite) set of all inequivalent irreducible representations of A and for $\pi \in \hat {A}$, let d _π denote its dimension. Since S is an involutive anti-automorphism, it acts as an involution on the set of minimal central projections of A. It is then easy to see that there exist subsets $\hat {A}_{1}$ and $\hat {A}_{2}$ of $\hat {A}$ and an identification

$$\begin{array}{@{}rcl@{}} A \rightarrow \bigoplus_{\pi \in \hat{A}_{1}} M_{d_{\pi}}(\mathbb{C}) \oplus \bigoplus_{\pi \in \hat{A}_{2}} (M_{d_{\pi}}(\mathbb{C}) \oplus M_{d_{\pi}}(\mathbb{C})) \end{array} $$

such that each summand is S-stable.

Now, by Propositions 7 and 9, in each summand of the above decomposition, either $M_{d_{\pi }}(\mathbb {C})$ or $M_{d_{\pi }}(\mathbb {C}) \oplus M_{d_{\pi }}(\mathbb {C})$, there is an invertible element which together with its image under S generates that summand.

Finally, an inductive application of Lemma 8 shows that if a is a general linear combination of these generators, then a and S a generate A as an algebra. □

Before we prove our main result, we will need a result about connected pointed bipartite graphs. Recall that a bipartite graph has its vertex set partitioned into ‘even’ and ‘odd’ vertices and all its edges connect an even and an odd vertex. It is pointed if a certain even vertex, normally denoted by ∗, is distinguished. Its depth is the largest distance of a vertex from ∗.

FormalPara PROPOSITION 10

Let Γ be a connected pointed bipartite graph of depth k≥3. For any vertex v of Γ, let t be the one of k+3,k+4 with the same parity as v. The number of paths of length t from ∗ to v is at least 3.

FormalPara Proof.

We analyse three cases depending on the distance of v from ∗.

Case I: If v=∗, note that t≥6 is even. To show that there are at least 3 paths of length t from ∗ to ∗, it suffices to show that there are at least 3 paths of length 6 from ∗ to ∗. Since k≥3, choose any vertex at distance 2 from ∗ and a path from ∗ to the chosen vertex. It is easy to see that there are at least 3 paths of length 6 from ∗ to ∗ supported on the edges of this path.

Case II: If v is at distance 1 from ∗, then t≥7 is odd. As observed in Case I, there are at least 3 paths of length 6 from ∗ to ∗ and consequently at least 3 paths of length 7 from ∗ to v.

Case III: Suppose v is at a distance n from ∗, where n>1. Observe that if n and k have the same parity, then n≤k while in the other case, n≤k−1. Choose a path ξ ₁ ξ ₂ ξ ₃⋯ξ _n from ∗ to v. Then $\xi _{2} \neq \overline {\xi _{1}}.$ Then we have three paths $\xi _{1} \overline {\xi _{1}} \xi _{1} \overline {\xi _{1}} \xi _{1} \xi _{2} {\cdots } \xi _{n}$, $\xi _{1} \xi _{2} \overline {\xi _{2}} \xi _{2} \overline {\xi _{2}} \xi _{2} {\cdots } \xi _{n}$, and $\xi _{1} \overline {\xi _{1}} \xi _{1} \xi _{2} \overline {\xi _{2}} \xi _{2} {\cdots } \xi _{n}$ of length n+4 from ∗ to v. Thus if n and k have the same parity, so that t = k+4, then there exist at least 3 paths of length t from ∗ to v. If n and k have opposite parity then t = k+3 and since n≤k−1 in this case, since there exist at least 3 distinct paths of length n+4 from ∗ to v, there also exist 3 distinct paths of length t from ∗ to v.

□

We now prove the main result.

FormalPara Theorem 11.

Let P be a subfactor planar algebra of finite depth k. Let 2t be the even number in {k+3,k+4}. Let $s=\min \{2k,2t\}$ . Then P is generated by a single s-box.

FormalPara Proof.

Case I: If k≤3, s=2k. Then by Proposition 5.1 of [3], P is generated by a single s box.

Case II: If k>3, so that s = 2t, let Γ be the principal graph of the subfactor planar algebra P. Then from Proposition 10, the number of paths of length s from the ∗-vertex to any even vertex v in Γ is at least 3. So P _s does not have an $M_{2}(\mathbb {C})$ summand. Consider the t-th power, say X, of the s-rotation tangle. This tangle changes the position of ∗ on an s-box from the top left to the bottom right position. Clearly ${Z^{P}_{X}}$ : P _s → P _s is an involutive algebra anti-automorphism. From Theorem 1, there exists an element a∈P _s such that a and ${Z^{P}_{X}}(a)$ generate P _s as an algebra. Since s≥k, the planar algebra generated by P _s contains P _k and thus is the whole of P. Hence the single s-box containing a generates the planar algebra P.

□

We finish by showing that k+1 might actually be needed.

FormalPara Example 12.

Let P =P(V) be the tensor planar algebra (see [ 2 ]) for details) of a vector space V of dimension greater than 1. It is easy to see that depth(P) = 1. However, given any a ∈ P ₁ = End(V), if Q is the planar subalgebra of P generated by a, a little thought shows that Q ₁ is just the algebra generated by a and is hence abelian while P ₁ is not.