Abstract
We show that a subfactor planar algebra of finite depth k is generated by a single s-box, for \(s \leq \min \{k+4,2k\}\).
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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.
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.
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.
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
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 \(M_{n}(\mathbb {C})\).
If \(U \subseteq \mathbb {C}^{2N}\) is non-empty and Zariski open, then
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 6For n>1, the set
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:
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 7Let 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
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.
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 9Let 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
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 ∗.
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.
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.
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.
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Kodiyalam V and Tupurani S, Universal skein theory for finite depth subfactor planar algebras, (English) Zbl 1252.46064, Quantum Topol. 2 (2) (2011) 157–172
Acknowledgement
The authors are grateful to Prof. T Y Lam for his remarks.
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Communicating Editor: B V Rajarama Bhat
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KODIYALAM, V., TUPURANI, S. Generators for finite depth subfactor planar algebras. Proc Math Sci 126, 235–240 (2016). https://doi.org/10.1007/s12044-016-0273-8
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DOI: https://doi.org/10.1007/s12044-016-0273-8