From Holant to Quantum Entanglement and Back
aa r X i v : . [ c s . CC ] A p r From Holant to Quantum Entanglement and Back
Jin-Yi Cai ∗ [email protected] Zhiguo Fu † [email protected] Shuai Shao ∗ [email protected] Abstract
Holant problems are intimately connected with quantum theory as tensor networks. We firstuse techniques from Holant theory to derive new and improved results for quantum entangle-ment theory. We discover two particular entangled states | Ψ i of 6 qubits and | Ψ i of 8 qubitsrespectively, that have extraordinary and unique closure properties in terms of the Bell prop-erty. Then we use entanglement properties of constraint functions to derive a new complexitydichotomy for all real-valued Holant problems containing an odd-arity signature. The signaturesneed not be symmetric, and no auxiliary signatures are assumed. ∗ Department of Computer Sciences, University of Wisconsin-Madison. Supported by NSF CCF-1714275. † School of Information Science and Technology and KLAS, Northeast Normal University, China. Supported byNSFC-61872076.
Introduction
Holant problems are a broad class of Sum-of-Products. It generalizes other frameworks such ascounting constraint satisfaction problems ( n > Z n → C . Let F be any fixed set ofsignatures. A signature grid Ω = ( G, π ) over F is a tuple, where G = ( V, E ) is a graph withoutisolated vertices, π labels each v ∈ V with a signature f v ∈ F of arity deg( v ), and labels theincident edges E ( v ) at v with input variables of f v . We consider all 0-1 edge assignments σ , andeach gives an evaluation Q v ∈ V f v ( σ | E ( v ) ), where σ | E ( v ) denotes the restriction of σ to E ( v ). Definition 1.1 (Holant problems) . The input to the problem
Holant( F ) is a signature grid Ω =(
G, π ) over F . The output is the partition function Holant Ω = X σ : E ( G ) →{ , } Y v ∈ V ( G ) f v ( σ | E ( v ) ) . Bipartite Holant problems
Holant(
F | G ) are Holant problems over bipartite graphs H = ( U, V, E ) ,where each vertex in U or V is labeled by a signature in F or G respectively. Weighted ∗ , Holant + and Holant c [19, 2, 21, 3]. Withoutassuming auxiliary signatures a Holant dichotomy is established only for non-negative real valuedsignatures [37]. Holant problems are in fact synonymous with tensor networks in quantum theory. The partitionfunction Holant Ω can be used in a (strong) simulation of quantum circuits [41]. A signature grid isjust a tensor network, where each signature is a tensor with its inputs associated with its incidentedges. In this sense, a signature of arity n represents a state of n qubits . In quantum theory,the basic component of a system is a qubit. The (pure) state of an n qubits | Ψ i is describedby a vector in C n . (The standard notion requires quantum states to have norm 1, but in thispaper, normalization by a nonzero scalar makes no difference for complexity, so we work with1tates having arbitrary nonzero norms.) A nonzero n -ary signature f is synonymous with an n -qubit state | f i = P x ∈{ , } n f ( x ) | x i . In this paper, we use them interchangeably. When f is a zerosignature (i.e., f ≡ | f i is a null state, denoted by N .A core concept in quantum theory is entanglement . It is perhaps the most distinguishingcharacteristic feature separating quantum and classical physics. Definition 1.2 (Quantum entanglement) . A state of n qubits ( n > , representing a multiplesystem) is entangled if it cannot be decomposed as a tensor product of single-qubit states (individualsystems). It is genuinely entangled if it cannot be decomposed as a tensor product of states ofproper subsystems. It exhibits multipartite entanglement if it involves a genuinely entangled stateof subsystem of more than two qubits (i.e., it cannot be decomposed as a tensor product of single-qubit states and 2-qubit states). Today, entanglement is recognized as an important resource in quantum computing and quan-tum information theory. It has been shown that quantum computing speedups essentially dependon unbounded entanglement [34]. While in quantum information theory, an entangled state isshared by several parties, one can perform operations on a subsyetem locally without access tothe other subsystems. This set-up is commonly used in quantum teleportation and quantum keydistribution [26, 4]. For different information-theoretic tasks, different types of entanglement can beused [39]. The classification of them under stochastic local operation with classical communication (SLOCC) equivalence was proposed in 2000 by D¨ur et al. [23], and is an area of active research[43, 38, 36, 35, 1, 29]. Yet so far, even the classification of entangled 4-qubit states is not completelysettled. For more about quantum entanglement theory, we refer to the survey [32].
There are many natural connections between Holant problems and quantum theory. The intro-duction of Holant problems is inspired by holographic transformations [42]. Such a holographictransformation applied separately on each qubit i with a matrix A i is just a SLOCC in quantumtheory. Also, many known P-time computable signature sets for Holant problems can be clearlydescribed in the quantum literature [18, 2] and they correspond directly to sets of states that areof independent interest in quantum theory [22, 31].Going beyond that, Backens recently applied knowledge from the theory of quantum entangle-ment, directly to the study of Holant problems and derived new dichotomy results [2, 3]. We givea short description for these results in this subsection. We use h Φ | to denote the Hermitian adjoint(complex conjugate) of | Φ i , and h Φ | Ψ i to denote the (complex) inner product of two n -qubit states. Definition 1.3 (Projection) . The projection of the i -th qubit of an n -qubit ( n > state | Ψ i ontoa single-qubit state | θ i = a | i + b | i is defined as h θ | i | Ψ i = ¯ a | Ψ i i + ¯ b | Ψ i i where ¯ a and ¯ b are complexconjugates of a and b , and | Ψ i i and | Ψ i i are states of the remaining n − qubits when the i-thqubit of | Ψ i is set to and respectively. Theorem 1.4 ([40],[28]) . Let | Ψ i be a genuinely entangled n -qubit ( n > state. For any twoqubits of | Ψ i , there exist projections of the other n − qubits onto n − many single-qubit statesthat result in an entangled 2-qubit state. This result was presented to show that any pure entangled multipartite quantum state violatessome Bell’s inequality [40]. The original proof [40] was flawed and was corrected recently [28].2heorem 1.4 shows that two particle entanglement can be realized via performing local projectionson a multiparticle state. It is observed in [2] that the theorem holds even when restricted to onlylocal projections onto computational or Hadamard basis states, i.e., | i , | i , | + i = | i + | i and |−i = | i − | i .Based on Theorem 1.4 and the inductive entanglement classification under SLOCC equivalence[36, 35, 1], Backens showed that beyond entangled 2-qubit states, genuinely entangled 3-qubit statescan be realized via local projections onto computational or Hadamard basis states (Theorem 12 in[2]). This theorem is equivalent to the following inductive statement. Theorem 1.5 ([2]) . Let | Ψ i be an n -qubit ( n > state exhibiting multipartite entanglement.Then, there exists some i and some | θ i ∈ {| i , | i , | + i , |−i} such that h θ | i | Ψ i exhibits multipartiteentanglement. Remark:
This result shows that multipartite entanglement of an n -qubit ( n >
4) state can be preserved under projections onto states | i , | i , | + i and |−i .The Holant + ( F ) problem is defined as Holant( F ∪ {| i , | i , | + i , |−i} ). According to Theorem1.5, we know that in the framework of Holant + problems, a genuinely entangled 3-qubit state canalways be realized from an n -qubit ( n >
4) state exhibiting multipartite entanglement. Then, usinga genuinely entangled 3-qubit state, a full dichotomy was proved for Holant + problems [2]. Later,it was generalized to Holant c problems [21, 3] where Holant c ( F ) is defined as Holant( F ∪ {| i , | i} ). In this paper, we consider whether multipartite entanglement can be preserved under projectionsonto only computational basis states, i.e., | i or | i . We have the following result. Theorem 1.6.
Let | Ψ i be an n -qubit ( n > state exhibiting multipartite entanglement and h n | Ψ i 6 = 0 . If | Ψ i is not of the form a | n i + b | n i when n > , or when n = 4 and | Ψ i isnot of the form a | i + b | i + c | i + d | i (up to a permutation of the four qubits) where a, b, c and d can possibly be zero, then there exists some i such that | Ψ i i or | Ψ i i exhibits multipartiteentanglement. That h n | Ψ i 6 = 0 is a normalization condition, and the other conditions are all necessary toensure the preservation of multipartite entanglement under projections to | i and | i . Thus The-orem 1.6 is a strengthening of Theorem 1.5. More importantly, our approach is in the oppositedirection to Backens. While Backens proved results in quantum entanglement theory to apply it tothe complexity classification of Holant problems, we prove new results in quantum entanglementtheory by employing the machinery from Holant problems. We prove Theorem 1.6 using a tech-nique developed for Holant problems called the interplay between the unique prime factorization ofsignatures and gadget constructions . This technique is at the heart of a standard approach (arityreduction) to build inductive arguments for Holant problems [14]. The new result in quantumentanglement theory sheds light on the classification of entanglement under SLOCC equivalence.Going one step further, we ask whether we can restrict projections onto only one state | i ,while multipartite entanglement is still preserved. The answer is no. Then, one way to salvage thesituation is to consider the self-loop gadget using one of the Bell states, | φ + i = | i + | i togetherwith projections onto | i . 3 efinition 1.7 (Self-loop) . The self-loop on the i -th and j -th qubits of a state | Ψ i by the Bell state | φ + i = | i + | i is defined as h φ + | ij | Ψ i = | Ψ ij i + | Ψ ij i , where | Ψ ij i and | Ψ ij i are states of n − qubits when setting the i-th and j-th qubits of | Ψ i to and respectively. Lemma 1.8.
Let | Ψ i be an n -qubit ( n > state exhibiting multipartite entanglement. Thereexists some choice of three or four of the n qubits such that by performing self-loops by | φ + i andprojections onto | i of the other qubits, we get • a -qubit state exhibiting multipartite entanglement, or • a GHZ type -qubit state, i.e., | GHZ i = | i + | i , or • the state | i . Why do we consider | φ + i and | i ? The state | φ + i is synonymous with the binary equality signature = . It is always available in the Holant framework as it means merging two danglingedges in a graph. Moreover, we can show that | i is realizable from any state of odd number ofqubits under some mild assumptions. Then, we can apply Lemma 1.8 to get a new dichotomy forHolant problems where at least one signature of odd arity is present. Theorem 1.9.
Let F be a set of real-valued signatures containing at least one signature of oddarity. Then Holant( F ) is either P-time computable or Remark:
Theorem 1.6 and Lemma 1.8 hold for complex-valued n -qubit states. However, Theorem1.9 is restricted to real-valued signatures, in which the Hermitian conjugate and the complex innerproduct can be represented by a mating gadget in the Holant framework. What about signature sets containing only even arity signatures, in which | i cannot be realized.Since | φ + i is always available, we consider whether multipartite entanglement is preserved underself-loops by | φ + i alone. Given an n -qubit ( n > | Ψ i exhibiting multipartiteentanglement, are there some i and j such that performing a self-loop by | φ + i on the i -th and j -th qubits of | Ψ i results in an ( n − n it must be n ≥ exist genuinely entangled6-qubit and 8-qubit states such that multipartite entanglement is not preserved under self-loops.Furthermore, it is not preserved under self-loops not only by | φ + i , but also by all four Bell states, | φ + i , | ψ + i = | i + | i , | φ − i = | i − | i , and | ψ − i = | i − | i . The self loop of the i -th and j -thqubits of | Ψ i by | ψ + i is defined as h ψ + | ij | Ψ i = | Ψ ij i + | Ψ ij i . Similarly, we can define h φ − | ij | Ψ i and h ψ − | ij | Ψ i . Definition 1.10 (Bell property) . Let | Ψ i be a genuinely entangled state. We say that it satisfiesthe Bell property if for any two qubits i and j of | Ψ i and any Bell state | φ i , h φ | ij | Ψ i is a tensorproduct of Bell states. It satisfies the strong Bell property if for any two i and j and any Bell state | φ i , h φ | ij | Ψ i is a tensor product of the Bell state | φ i , i.e., h φ | ij | Ψ i = | φ i ⊗ · · · ⊗ | φ i . Theorem 1.11.
There exist genuinely entangled -qubit states that satisfy the Bell property, andgenuinely entangled -qubit states that satisfy the strong Bell property.
4e first give an 8-qubit state | Ψ i that satisfies the strong Bell property. | Ψ i = | i + | i + | i + | i + | i + | i + | i + | i + | i + | i + | i + | i + | i + | i + | i + | i . | Ψ i can be represented by an 8-ary signature Ψ . The support of Ψ has the following structure:the sums of the first four variables, and the last four variables are both even; the assignment ofthe first four variables are either identical to, or complement of the assignment of the last fourvariables. While it is not obvious from this description that the support set is an affine subspace of Z , but it is . Another interesting description of Ψ is as follows: Take 4 bits x , x , x , x , (theseare not the first 4 bits in the description above), then on the support the remaining 4 bits are mod2 sums of (cid:0) (cid:1) subsets of { x , x , x , x } .The 6-qubit state | Ψ i satisfying the Bell property has 32 nonzero coefficients. We give it inthe signature form. Ψ ( x , . . . , x ) = ( − x x + x x + x x + x x + x x + x x , where the support of Ψ is P i =1 x i = 0 mod 2 (even parity). We can write Ψ as the following8-by-8 matrix where the assignment of the first three variables in lexicographic order (from 000 to111) is the row index and the assignment of the last three variables in lexicographic order is thecolumn index. M , (Ψ ) = − −
10 1 − − − − − − − − − − . We can use Pauli operations to generate more states satisfying the Bell property. Consider thefollowing four Pauli operators I = (cid:20) (cid:21) , X = (cid:20) (cid:21) , Y = (cid:20) − ii (cid:21) and Z = (cid:20) − (cid:21) . A Pauli operation on an n -qubit state | Ψ i is defined as P ⊗ P ⊗ . . . ⊗ P n | Ψ i (which producesanother n -qubit) where each P i is a Pauli operator. Let | Ψ i and | Ψ i be states described above.Let P and P denote the sets of states realized by performing Pauli operations on | Ψ i and | Ψ i respectively. All states in P and P satisfy the Bell property.Due to the existence of these 6-qubit and 8-qubit states with such extraordinary properties, itremains as a difficult task to achieve a full dichotomy for real-valued Holant problems. On the otherhand, we hope such states can be further investigated and perhaps applied to quantum computingor quantum information theory.The paper is organized as follows. In Section 2, we give a proof of our main quantum entan-glement result (Theorem 1.6) by using the theory of signatures. We give some preliminaries forHolant problems in Section 3. Then, we give a proof outline for our dichotomy result (Theorem1.9) in Section 4 and the full proof in Section 5. 5 Preservation of Multipartite Entanglement under Projections
We use the theory of signatures to prove Theorem 1.6. Recall that by our definition, a signaturealways has arity at least one. A nonzero signature g divides f denoted by g | f , if there is asignature h such that f = g ⊗ h (with possibly a permutation of variables) or there is a constant λ such that f = λ · g . In the latter case, if λ = 0, then we also have f | g since g = λ · f . Fornonzero signatures, if both g | f and f | g , then they are nonzero constant multiples of each other,and we say g is an associate of f , denoted by g ∼ f . In terms of this division relation, irreducible signatures and prime signatures are defined. It is proved that they are equivalent, which gives usthe unique prime factorization (UPF) of signatures [14]. Definition 2.1 (Irreducible signature) . A nonzero signature f is irreducible if g | f implies g ∼ f . Definition 2.2 (Prime signature) . A nonzero signature f is a prime signature, if for any nonzerosignatures g and h , f | g ⊗ h implies that f | g or f | h . Lemma 2.3.
The notions of irreducible signatures and prime signatures are equivalent.
A prime factorization of a signature f is f = g ⊗ . . . ⊗ g k up to a permutation of variables,where each g i is a prime (irreducible) signature. Lemma 2.4 (Unique prime factorization) . Every nonzero signature f has a prime factorization.If f has prime factorizations f = g ⊗ . . . ⊗ g k and f = h ⊗ . . . ⊗ h ℓ , both up to a permutation ofvariables, then k = ℓ and after reordering the factors we have g i ∼ h i for all i . A nonzero signature f is irreducible if f cannot be written as g ⊗ h for some signatures g and h . This is equivalent to saying that | f i is a genuinely entangled state of multiple qubits or | f i isa single-qubit state. Let T denote the set of tensor products of unary signatures and T denotethe set of tensor products of unary and binary signatures. Then a state | f i of multiple qubits isentangled iff f / ∈ T , and | f i exhibits multipartite entanglement iff f / ∈ T . The following result isa direct corollary of Lemma 2.4. Corollary 2.5.
Let f be a nonzero n -ary signature. Suppose that there are two irreducible signa-tures g on a variable set A and h on a variable set B such that g | f and h | f . Then, either A isdisjoint with B , or A = B and g ∼ h . We use f i and f i to denote the signature forms of quantum states | f i i and | f i i realized byprojections onto | i and | i . In the Holant framework, these signatures are realized by a pinning gadget, i.e., connecting the variable x i of f with unary signatures ∆ = (1 ,
0) and ∆ = (0 , and ∆ are signature forms of | i and | i . We may further pick a variable x j of f ci and pin it to the value d ( c, d ∈ { , } ). Obviously, the pinning gadgets on different variables x i and x j commute. Thus, we have ( f ci ) dj = ( f dj ) ci . We denote it by f cdij .The following lemma is easy to check. Lemma 2.6.
Let f has arity n . If f i ≡ for all i ∈ [ n ] and f j ≡ for some j ∈ [ n ] , then f ≡ . Suppose that g | f where g is on a variable set A . Then for any variable x i of f that is not in A , we have g | f i and g | f i (By definition, the division relation holds even if f i or f i is a zerosignature). Thus, the division relation is unchanged under pinning gadgets on variables out of A .The following lemma shows that a stronger converse is also true.6 emma 2.7. Let f be an n -ary ( n > signature. If there exists a signature g on a variable set A such that g | f i for all x i / ∈ A , and furthermore g | f j for some x j / ∈ A , then g | f .Proof. We may assume f is nonzero, for otherwise the conclusion trivially holds. We now provethis for a unary signature g = ( a, b ). We assume g is on the variable x u . Consider the signature f ′ = bf u − af u . Clearly, f ′ has arity at least 1. For every i = u , we have f i = ( a, b ) ⊗ h for some h . Then, ( f i ) u = a · h , ( f i ) u = b · h , and hence( f ′ ) i = ( bf u − af u ) i = bf ui − af ui = b ( f i ) u − a ( f i ) u = ba · h − ab · h ≡ . Moreover, there is an index j = u such that g | f j , i.e., f j = ( a, b ) ⊗ h ′ for some h ′ . Then,( f j ) u = a · h ′ , ( f j ) u = b · h ′ , and hence ( f ′ ) j = ( bf u − af u ) j = b ( f j ) u − a ( f j ) u = ba · h ′ − ab · h ′ ≡ . By Lemma 2.6, we have f ′ ≡
0. Thus, we have f u : f u = a : b , and hence g | f .For a signature g of arity > n −
2, the proof is essentially the same, which we omit here.
We restate Theorem 1.6 in terms of signatures. We use S ( f ) to denote the support of f , i.e., S ( f ) = { α | f ( α ) = 0 } . We use 0 n and 1 n to denote the n -bit all-0 and all-1 strings. Theorem 2.8.
Let f be an n -ary ( n > signature, f / ∈ T and f (0 n ) = 0 . If S ( f )
6⊆ { n , n } when n > , or S ( f )
6⊆ { , , , } up to any permutation of four variables when n = 4 , then there exists some i such that f i or f i is not in T .Proof. Since f (0 n ) = 0, we have f i f ij i and j . Also,since the support S ( f )
6⊆ { n , n } , there exist some s and t such that f st
0. For a contradiction,we assume f i , f i ∈ T for all i . We consider the following two possible cases. Case 1 . For all indices i , f i ∈ T (i.e., tensor product of unary signatures).We will show that in this case, there is a unary signature a ( x u ) on some variable x u , such that a ( x u ) | f . This will lead us to a contradiction.Recall that there exist variables x s and x t such that f st
0. We consider f t . Clearly, f t f t ∈ T , in the UPF of f t , the variable x s may appear in a unary signature or an irreduciblebinary signature. In both cases, since f has arity at least 4, we can pick a variable x u such that x u and x s appear in two distinct irreducible signatures in the UPF of f t (i.e., x u and x s are notentangled in f t ). Then, we show that x u must appear in a unary signature in the UPF of f t .Otherwise, there is an irreducible binary signature b ( x u , x v ) such that b ( x u , x v ) | f t . Since x u isnot entangled with x s in f t , we have v = s . Then, b ( x u , x v ) | ( f t ) s . On the other hand, weconsider f s . By our assumption, f s ∈ T and hence there exists some unary signature a ′ ( x u ) suchthat a ′ ( x u ) | f s . Then, we have a ′ ( x u ) | ( f s ) t . Recall that the pinning gadgets on different variablescommute. Thus, ( f t ) s = ( f s ) t = f st , and we know that it is a nonzero signature. By Corollary2.5, we have b ( x u , x v ) ∼ a ′ ( x u ). This is a contradiction. Thus, there exists some unary signature a ( x u ) such that a ( x u ) | f t .Now we show that a ( x u ) | f i for all indices i = u . First, we show that a ( x u ) | f s . Since f s ∈ T , there exists some unary signature a ′ ( x u ) such that a ′ ( x u ) | f s , and then a ′ ( x u ) | f st .Also, we have a ( x u ) | f st since a ( x u ) | f t . Since f st
0, by Corollary 2.5, we have a ( x u ) ∼ a ′ ( x u ).Thus, a ( x u ) | f s . Then, we consider f i for all indices i / ∈ { u, s } . Since f i ∈ T , there exists aunary signature a ′′ ( x u ) such that a ′′ ( x u ) | f i , and then a ′′ ( x u ) | f is . Also, we have a ( x u ) | f is since7 ( x u ) | f s . Recall that f is i . Then, by Corollary 2.5, we have a ( x u ) ∼ a ′′ ( x u ). Thus, a ( x u ) | f i .Since a ( x u ) | f i for all i = u and a ( x u ) | f t for some t = u , by Lemma 2.7, we have a ( x u ) | f .In other words, f = a ( x u ) ⊗ g where g is a nonzero signature of arity n − x u . Since f / ∈ T , we have g / ∈ T . Consider f u . We know that it is a nonzero signature and hence f u ∼ g . Thus, f u / ∈ T . We have reached a contradiction. Case 2.
There exists some index k and an irreducible binary signature b ( x v , x w ) such that b ( x v , x w ) | f k .We will show that in this case, b ( x v , x w ) | f . First, we show that b ( x v , x w ) | f i for all i / ∈ { v, w } .We already have b ( x v , x w ) | f k . Consider f i for all indices i / ∈ { v, w, k } . Since f i ∈ T and f i a ( x v ) or an irreducible binary signature b ′ ( x v , x w ′ ) that appearsin the UPF of f i , i.e., a ( x v ) | f i or b ′ ( x v , x w ′ ) | f i . In the former case, we have a ( x v ) | f ik . Inthe latter case and if w ′ = k , we have b ′ ( x v , x w ′ ) | f ik . In the latter case and if w ′ = k , then let a ′ ( x v ) be the unary signature realized from b ′ ( x v , x w ′ ) by pinning x w ′ = x k to 0, we get a ′ ( x v ) | f ik .On the other hand, since b ( x v , x w ) | f k , we have b ( x v , x w ) | f ik . Since f ik
0, by Corollary 2.5,we know that the two cases that a ( x v ) | f ik and a ′ ( x v ) | f ik cannot occur. Thus, w ′ = k and b ′ ( x v , x w ′ ) | f ik . By Corollary 2.5, b ( x v , x w ) ∼ b ′ ( x v , x w ′ ). Thus, we have that b ( x v , x w ) | f i for all i / ∈ { v, w } .Then we want to show that there exists some j / ∈ { v, w } such that b ( x v , x w ) | f j . • We first consider the case that there exist some indices i and j where { i, j } is disjoint with { v, w } such that f ij
0. We show that b ( x v , x w ) | f j . Since b ( x v , x w ) | f i , we have b ( x v , x w ) | f ij . By assumption f ij
0, and then clearly f j
0. Recall that f j ∈ T .Again, there is either a unary signature a ( x v ) or an irreducible binary signature b ′ ( x v , x w ′ )that appears in the UPF of f j , i.e., a ( x v ) | f j or b ′ ( x v , x w ′ ) | f j . In the first case since i = v , we can pin x i of f j to 0, and we get a ( x v ) | f ij . In the second case and if w ′ = i ,again we can get a ′ ( x v ) | f ij , where a ′ ( x v ) = b ′ ( x v , x i to 0. But f ij b ( x v , x w ) | f ij . Then, in the UPF of f ij , it does not have a unary signature on x v as a factor. Thus, it must be the case that b ′ ( x v , x w ′ ) | f j where w ′ = i . Then, we have b ′ ( x v , x w ′ ) | f ij . Since b ( x v , x w ) | f ij and f ij
0, by Corollary 2.5, b ′ ( x v , x w ′ ) ∼ b ( x v , x w ),and thus b ( x v , x w ) | f j . Then, by Lemma 2.7, we have b ( x v , x w ) | f . In other words, f = b ( x v , x w ) ⊗ h where h is a nonzero signature of arity n − x v and x w . Since f / ∈ T , we have h / ∈ T . Then consider f v . We know that it is a nonzero signatureand h | f v . Thus, f v / ∈ T . Contradiction. • Then we consider the case that f ij ≡ { i, j } that are disjoint with { v, w } .Consider an n -bit input α of f . We write α as α v α w β where α v is the input on variable x v , α w is the input on variable x w , and β is the input on the other n − f ( α ) = 0 if β is not the all-0 or all-1 bit string in { , } n − . It follows that f has at mosteight nonzero entries. We list all its entries by the following 4-by-2 n − matrix M vw ( f ) withvariables ( x v , x w ) ∈ { , } as the row index (in the order 00, 01, 10, 11) and the assignmentof the other variables in lexicographic order as the column index. M vw ( f ) = c . . . . . . c c . . . . . . c c . . . . . . c c . . . . . . c . c = f (0 n ) = 0. Consider signatures f v and f v . They have the following matrix formswith the variable x w ∈ { , } as the row index. M w ( f v ) = (cid:20) c . . . . . . c c . . . . . . c (cid:21) and M w ( f v ) = (cid:20) c . . . . . . c c . . . . . . c (cid:21) . Also consider signatures f w and f w . They have the following matrix forms with the variable x v ∈ { , } as the row index. M v ( f w ) = (cid:20) c . . . . . . c c . . . . . . c (cid:21) and M v ( f w ) = (cid:20) c . . . . . . c c . . . . . . c (cid:21) . Consider f v . Since f has arity at least 4, f v has arity at least 3. Since f v ∈ T , the variable x w either appears in a unary factor a ( x w ) of f v or an irreducible binary factor b ( x w , x w ′ ) of f v .In the latter case, we can pick another variable x r of f v where r = w or w ′ , and we consider f vr . We know that f vr c = 0 and b ( x w , x w ′ ) | f vr since b ( x w , x w ′ ) | f v . Notice thatthe column with c and c does not appear in M w ( f vr ). Thus, the signature f vr is of the form( c , c ) ⊗ (1 , ⊗ ( n − which is a tensor product of unary signatures. Contradiction. Thus,there is a unary signature a ( x w ) such that a ( x w ) | f v . Then, we have c c = c c . Similarlyby considering f w , we have c c = c c . Now, we consider f v , and prove c c = c c . If c = c = 0, then clearly we have c c = c c = 0. Otherwise, for any r = w or v , wehave f vr = ( c , c ) ⊗ (1 , ⊗ ( n − b ( x w , x w ′ ) such that b ( x w , x w ′ ) | f v , then we can find some r = w, w ′ such that b ( x w , x w ′ ) | f vr . Contradiction. Thus, there is a unary signature a ( x w ) such that a ( x w ) | f v . Then, we have c c = c c . Similarly by considering f w , we have c c = c c . – Suppose n >
5. Then f v has arity at least 4. We first show that c = 0. We consider f vw = [ c , , . . . . . . , , c ] . Since f v ∈ T , we have f vw ∈ T . Note that f vw has arity atleast 3. Since c = 0, the only possible value of c to make f vw ∈ T is 0. Thus, c = 0.Since c c = c c = 0 and c = 0, we have c = 0. Also, since c c = c c = 0 and c = 0,we have c = 0. If c = 0, then f = b ( x v , x w ) ⊗ (1 , ⊗ ( n − ∈ T . A contradiction with f / ∈ T . Thus, we have c = 0. Since c c = c c = 0 and c = 0, we have c = 0. Alsosince c c = c c = 0 and c = 0, we have c = 0. Consider f vw = [ c , , . . . . . . , , c ].Since f vw ∈ T and it has arity at least 3, and c = 0, we have c = 0. Thus, f has onlytwo nonzero entries that are on the all-0 input and the all-1 input. A contradiction withour assumption that S ( f )
6⊆ { n , n } . – Suppose n = 4. If c = 0, then with the same proof as in the case that n >
5, wehave c = c = 0, c = 0 and then c = c = 0. Thus, S ( f ) ⊆ { , , } .Contradiction. Otherwise, c = 0. Suppose that c = kc . Then c = kc since c c = c c and c = kc since c c = c c . If c and c are not zero, then c = kc since c c = c c . Then, f = b ( x v , x w ) ⊗ (1 , , , k ) ∈ T . Contradiction. Thus, c = c = 0.Similarly, if c and c are not zero, then we still have c = kc since c c = c c . Then, wehave f ∈ T . Contradiction. Thus, c = c = 0. Then, S ( f ) ⊆ { , , , } .Contradiction.Therefore, there exists some i such that f i or f i is not in T .Our result can be used in the classification of entanglement under SLOCC equivalence. An n -qubit state | Ψ i is equivalent to another n -qubit state | Φ i under SLOCC if there exist some9nvertible 2-by-2 matrices M , M , . . . , M n such that | Ψ i = M ⊗ M ⊗ . . . ⊗ M n | Φ i . Physicistsare interested in the classification of SLOCC equivalence classes. For 2-qubit states there are twoSLOCC classes, and for 3-qubit states there are six SLOCC classes [23]. However, for states of4 or more qubits there are infinitely many SLOCC classes [23]. Then, the goal is to categorizethese classes into some finitely many families with common physical or mathematical properties.Depending on which properties are used, there are different approaches. One powerful approachthat can possibly handle states of a high number of qubits is by induction [36, 35, 1, 29]. In thisapproach, the classification of n -qubit states relies on the classification of ( n − n -qubit state | Ψ i . We can pick some index i and write | Ψ i as | Ψ i = | i| Ψ i i + | i| Ψ i i .Families of entanglement classes of | Ψ i can be defined according to the types of entanglements foundin the linear span {| Ψ i i , | Ψ i i} which is related to the entanglement types of | Ψ i i and | Ψ i i them-selves. Theorem 1.6 gives a direct relation between the entanglement types of | Ψ i and {| Ψ i i , | Ψ i i} .For example, consider a 5-qubit state exhibiting multipartite entanglement. First, by performingSLOCC using the matrix N = [ ] on this state, we can always get a state | Ψ i where the coefficientof | i is nonzero. If | Ψ i has the form a | i + b | i , then it is equivalent to | GHZ i = | i + | i .Otherwise, we can apply Theorem 1.6. There exists some i such that | Ψ i i or | Ψ i i exhibits multi-partite entanglement. Then, in order to classify the state | Ψ i , we only need to consider possibleentanglement types of {| Ψ i i , | Ψ i i} where at least one state exhibits multipartite entanglement.This eliminates many cases compared to considering all entanglement types of {| Ψ i i , | Ψ i i} . A constraint function f , or a signature, of arity n > Z n → C . We use f α to denote f ( α ). If f α = f α for all α where f α denotes the complex conjugation of f α and α denotes thebit-wise complement of α , we say f satisfies the ars . We use wt( α ) to denote the Hamming weightof α . The support S ( f ) of a signature is { α ∈ Z n | f α = 0 } . A signature f of arity n has even(odd) parity if S ( f ) ⊆ { α ∈ Z n | wt( α ) is even (odd) } . Let F be any fixed set of signatures. Asignature grid Ω = ( G, π ) over F is a tuple, where G = ( V, E ) is a graph without isolated vertices, π labels each v ∈ V with a signature f v ∈ F of arity deg( v ), and labels the incident edges E ( v )at v with input variables of f v . We consider all 0-1 edge assignments σ , each gives an evaluation Q v ∈ V f v ( σ | E ( v ) ), where σ | E ( v ) denotes the restriction of σ to E ( v ). Definition 3.1 (Holant problems) . The input to the problem
Holant( F ) is a signature grid Ω =(
G, π ) over F . The output is the partition function Holant Ω = X σ : E ( G ) →{ , } Y v ∈ V ( G ) f v ( σ | E ( v ) ) . The bipartite Holant problems
Holant(
F | G ) are Holant problems over bipartite graphs H =( U, V, E ) , where each vertex in U or V is labeled by a signature in F or G respectively. Counting constraint satisfaction problems ( n to denote the Equality signature of arity n , which takes value 1 on the all-0or all-1 input and 0 elsewhere. (We may also denote the n-bits all-0 and all-1 strings by ~ n and ~ n n when it is clear from the context.) Let EQ = { = , = , . . . , = n , . . . } de-note the set of all Equality signatures. We use ( = ) to denote the binary Disequality signaturewith truth table (0 , , , Lemma 3.2. [8] F ) ≡ T Holant(
EQ | F ) . A signature f of arity n > k × n − k matrix M [ k ] , [ n − k ] ( f ), which lists the2 n many values of f with the [ k ] many variables as row index and the other [ n − k ] many variablesas column index. In particular, f can be expressed as a 2 × n − matrix M i ( f ) which lists the 2 n values of f with variable x i ∈ { , } as row index and the assignments of the other n − M i ( f ) = (cid:20) f , ... f , ... . . . f , ... f , ... f , ... . . . f , ... (cid:21) = (cid:20) f i f i (cid:21) , where f ai denotes the row vector indexed by x i = a in M i ( f ). We may omit the subscript when themeaning is clear from context. To introduce the idea of holographic transformation, it is convenient to consider bipartite graphs.For a general graph, we can always transform it into a bipartite graph while preserving the Holantvalue, as follows. For each edge in the graph, we replace it by a path of length two. (This operationis called the of the graph and yields the edge-vertex incidence graph.) Each new vertexis assigned the binary
Equality signature (= ). Thus, we have Holant(= | F ) ≡ T Holant( F ).For an invertible 2-by-2 matrix T ∈ GL ( C ) and a signature f of arity n , written as a columnvector f ∈ C n , we denote by T f = T ⊗ n f the transformed signature. For a signature set F , define T F = { T f | f ∈ F } the set of transformed signatures. For signatures written as row vectorswe define f T − and F T − similarly. Whenever we write T f or T F , we view the signatures ascolumn vectors; similarly for f T − or F T − as row vectors. We can also represent T f as thematrix M [ k ] , [ n − k ] ( T f ) with [ k ] variables as row index and the other [ n − k ] variables as columnindex. Then, we have M [ k ] , [ n − k ] ( T f ) = T ⊗ k M [ k ] , [ n − k ] ( f )( T T ) ⊗ n − k . Similarly, M [ k ] , [ n − k ] ( f T − ) =( T − T ) ⊗ k M [ k ] , [ n − k ] ( f )( T − ) ⊗ n − k .Let T ∈ GL ( C ). The holographic transformation defined by T is the following operation:given a signature grid Ω = ( H, π ) of Holant(
F | G ), for the same bipartite graph H , we get a newsignature grid Ω ′ = ( H, π ′ ) of Holant( F T − | T G ) by replacing each signature in F or G with thecorresponding signature in F T − or T G . Theorem 3.3 (Valiant’s Holant Theorem [42]) . For any T ∈ GL ( C ) , Holant Ω ( F | G ) = Holant Ω ′ ( F T − | T G ) . Therefore, a holographic transformation does not change the complexity of the Holant problemin the bipartite setting. Let Q ∈ R × be a 2-by-2 orthogonal matrix. Note that (= ) Q − = (= ).We have Holant(= | F ) ≡ T Holant(= | Q F ) . A particular holographic transformation that will be commonly used in this paper is the trans-formation defined by Z − = √ (cid:2) − i i (cid:3) . Recall that = denotes the binary Disequality signature11ith truth table (0 , , , N = [ ] with onevariable indexing rows and the other indexing columns respectively. Note that (= ) Z = ( = ).Therefore, we have Holant(= | F ) ≡ T Holant( = | Z − F ) . We denote Z − F by b F and Z − f by b f . We know f and b f have the following relation [14]. Lemma 3.4. f is a real valued signature iff b f satisfies ars . The following fact is easy to check.
Lemma 3.5.
Let Q ∈ R × . Q is orthogonal up to a scalar iff b Q = Z − Q ( Z − ) T is diagonal oranti-diagonal.Proof. For Q = (cid:2) a b − b a (cid:3) , we have b Q = (cid:2) a + b i a − b i (cid:3) ; for Q = (cid:2) a bb − a (cid:3) we have b Q = (cid:2) a − b i a + b i (cid:3) . One basic tool used throughout the paper is gadget construction. An F -gate is similar to a signaturegrid ( G, π ) for Holant( F ) except that G = ( V, E, D ) is a graph with internal edges E and danglingedges D . The dangling edges D define input variables for the F -gate. We denote the regular edgesin E by 1 , , . . . , m and the dangling edges in D by m + 1 , . . . , m + n . Then the F -gate defines afunction f f ( y , . . . , y n ) = X σ : E →{ , } Y v ∈ V f v (ˆ σ | E ( v ) )where ( y , . . . , y n ) ∈ { , } n is an assignment on the dangling edges, ˆ σ is the extension of σ on E bythe assignment ( y , . . . , y m ), and f v is the signature assigned at each vertex v ∈ V . This function f is called the signature of the F -gate. There may be no internal edges in an F -gate at all. Inthis case, f is simply a tensor product of these signatures f v , i.e., f = N v ∈ V f v (with possibly apermutation of variables). We say a signature f is realizable from a signature set F by gadgetconstruction if f is the signature of an F -gate. If f is realizable from a set F , then we can freelyadd f into F while preserving the complexity (Lemma 1.3 in [8]). Lemma 3.6. [8] If f is realizable from a set F , then Holant( f, F ) ≡ T Holant( F ) . Note that, in the setting of Holant(= | F ), every edge is labeled by = , while in the setting ofHolant( = | b F ), every edge in a gadget is labeled by = .A basic gadget construction is merging . In the setting of Holant(= | F ), given a signature f ∈ F of arity n , we can connect two variables x i and x j of f using = , and this operation gives a signatureof arity n −
2. We use ∂ ij f to denote this signature and ∂ ij f = f ij + f ij , where f abij denotes thesignature obtained by setting ( x i , x j ) = ( a, b ) ∈ { , } . (We use f abij to denote a function, and f abij to denote a vector that lists the truth table of f abij in a given order.) While in the setting ofHolant( = | b F ), the above merging operation is equivalent to connecting two variables x i and x j of b f using = . We denote the resulting signature by b ∂ ij b f , and we have d ∂ ij f = b ∂ ij b f = b f ij + b f ij . Notethat, the merging operation using = on a signature is synonymous with the self-loop of an n -qubitby bell state | φ + i , and the merging operation using = is synonymous with the self-loop by bellstate | ψ + i .If by merging any two variables of b f in the setting of Holant( = | b F ), we can only realize thezero signature, then the following result shows that b f itself is “almost” a zero signature [14].12 emma 3.7. [14] Let b f be a signature of arity n > . If for any indices { i, j } , by merging variables x i and x j of b f using = , we have b ∂ ij b f ≡ , then b f α = 0 for any α with < wt( α ) < n .Proof. Suppose there exists some α , where 0 < wt( α ) < n , such that f α = 0. Since α is neitherall-0 or all-1, and α has length at least 3, we can find three bits in some order such that on thesethree bits, α takes value 001 or 110. Without loss of generality, we assume they are the first threebits of α and we denote α by 001 δ or 110 δ ( δ maybe empty). We first consider the case that α = 001 δ . Consider another two strings β = 010 δ and γ = 100 δ . Note that if we merge variables x and x of f , we get ∂ f , its entry ( ∂ f ) δ on the input 0 δ (for bit positions 3 to n ) is the sumof f δ and f δ . Since ∂ f ≡
0, we have f δ + f δ = 0 . Similarly, by merging variables x and x , we have f δ + f δ = 0 , and by merging variables x and x , we have f δ + f δ = 0 . These three equations have only a trivial solution, f δ = f δ = f δ = 0. A contradiction. If α = 110 δ , the proof is symmetric. Remark:
The above proof actually gives a stronger result.
Lemma 3.8.
Let b f be a signature of arity n > . If b f α = 0 for some α with < wt( α ) < n , thenthere is a pair of indices { i, j } such that ( b ∂ ij b f ) β = 0 for some β , with wt( β ) = wt( α ) − . Recall that we can construct the pinning gadget when ∆ = (1 ,
0) is available.
Lemma 3.9.
Let f be a signature of arity n > . If for any index i , by pinning the variable x i of f to , we have f i ≡ , then f α = 0 for any wt( α ) = n . If furthermore, there is a pair of indices { j, k } such that ∂ jk f ≡ , then f ≡ .Proof. For any wt( α ) = n , there is an index i such that α i = 0. By pinning x i to 0, we get thesignature f i . We know f α is an entry in f i , and then f α = 0 since f i ≡ { j, k } such that ∂ jk f ≡
0. Let β denote the string of n bitswhere β j = β k = 0 and β ℓ = 1 elsewhere, and γ denote the string of n bits 1s. Consider the signature ∂ jk f . We know f β + f γ is an entry in ∂ jk f (when ∂ jk f is a constant, we have f β + f γ = ∂ jk f ). Weknow f β + f γ = 0 since ∂ jk f ≡
0. Clearly, wt( β ) = n and we have f β = 0. Thus, we have f γ = 0.Thus, we have f ≡ mating . Given a real valued signature f ofarity n >
2, we connect two copies of f in the following manner: For any m < n , fix a set S of n − m variables among all n variables of f . For each x k ∈ S , connect x k of one copy of f with x k of theother copy using = . The variables that are not in S are called dangling variables. In this paper,we only consider the case that m = 1 or 2. For m = 1, there is one dangling variable x i . Then,13he mating construction realizes a signature of arity 2, denoted by m i f . It can be represented bymatrix multiplication. We have M ( m i f ) = M i ( f ) I ⊗ ( n − M T i ( f ) = (cid:20) f i f i (cid:21) h f i T f i T i = (cid:20) | f i | h f i , f i ih f i , f i i | f i | (cid:21) , (3.1)where h· , ·i denotes the (complex) inner product and | · | denotes the norm defined by this innerproduct. Note that |h f i , f i i| | f i | | f i | by Cauchy-Schwarz inequality. Similarly, in the settingof Holant( = | b F ), the above mating operation is equivalent to connecting variables in S using = .We denote the resulting signature by b m i b f = d m i f and we have M ( b m i b f ) = M i ( b f ) N ⊗ n − M T i ( b f ) = "b f i b f i (cid:21) ⊗ ( n − h b f i T b f i T i . (3.2)Note that b f satisfies the ars since f is real, we have N ⊗ ( n − b f i T = ( b f , ... , b f , ... , . . . , b f , ... ) T = ( b f , ... , b f , ... , . . . , b f , ... ) = b f i T . Thus, we have M ( b m i b f ) = "b f i b f i (cid:21) ⊗ ( n − h b f i T b f i T i = "b f i b f i f i T b f i T i = " h b f i , b f i i | b f i | | b f i | h b f i , b f i i . (3.3) In this subsection, we give more results for the factorization of real-valued signatures and signaturessatisfying ars . We say a signature f is reducible if f = g ⊗ h , up to a permutation of variables, forsome signatures g and h . All zero signatures (of arity greater than 1) are reducible.The following lemma is an equivalent restatement of Lemma 2.15 in [14]. Lemma 3.10.
Let b f be a reducible signature satisfying ars , then there exists a factorization b f = b g ⊗ b h such that b g and b h both satisfies ars .Let f be a real valued reducible signature, then there exists a factorization f = g ⊗ h such that g and h are both real-valued signatures. If a vertex v in a signature grid is labeled by a reducible signature f = g ⊗ h , we can replacethe vertex v by two vertices v and v and label v with g and v with h , respectively. The incidentedges of v become incident edges of v and v respectively according to the partition of variablesof f in the tensor product of g and h . This does not change the Holant value. Clearly, f = g ⊗ h is realizable from { g, h } . On the other hand, Lin and Wang proved in [37] (Lemma 3.3) that,from a reducible signature f = g ⊗ h f by g and h while preserving thecomplexity of a Holant problem. Lemma 3.11.
If a nonzero real-valued signature f has a real factorization g ⊗ h , then Holant( g, h, F ) ≡ T Holant( f, F ) and Holant( = | b g, b h, b F ) ≡ T Holant( = | b f , b F ) for any signature set F ( b F ) . In this case, we say g ( b g ) and h ( b h ) are realizable from f ( b f ) byfactorization. .5 Polynomial interpolation Polynomial interpolation is a powerful technique to prove
Lemma 3.12.
Let g and g be two nonzero binary signatures with M ( g ) = P − [ ] P and M ( g ) = P − h λ λ i P for some invertible matrix P . If λ λ = 0 and | λ λ | 6 = 1 , then Holant( g , F ) T Holant( g, F ) for any signature set F . Lemma 3.13.
Let g be a nonzero binary signature with M ( g ) = P − h λ λ i P for some invertiblematrix P , and h be a nonzero unary signature. If λ λ = 0 , | λ λ | 6 = 1 , and h (as a column vector)is not an eigenvector of M ( g ) , then Holant( h ′ , g, F ) T Holant( h, g, F ) for any unary signature h ′ and any signature set F . We give some known signature sets that define polynomial time computable (tractable) countingproblems.
Definition 3.14.
Let T denote the set of tensor products of unary and binary signatures. Definition 3.15.
A signature on a set of variables X is of product type if it can be expressed as aproduct of unary functions, binary equality functions (= ) , and binary disequality functions ( = ) ,each on one or two variables of X . We use P to denote the set of product-type functions. Definition 3.16.
A signature f ( x , . . . , x n ) of arity n is affine if it has the form λ · χ AX =0 · i Q ( X ) , where λ ∈ C , X = ( x , x , . . . , x n , , A is a matrix over Z , Q ( x , x , . . . , x n ) ∈ Z [ x , x , . . . , x n ] is a quadratic (total degree at most 2) multilinear polynomial with the additional requirement thatthe coefficients of all cross terms are even, i.e., Q has the form Q ( x , x , . . . , x n ) = a + n X k =1 a k x k + X ≤ i We say a signature set F is C -transformable if there exists a T ∈ GL ( C ) suchthat (= )( T − ) ⊗ ∈ C and T F ⊆ C . This definition is important because if Holant( C ) is tractable, then Holant( F ) is tractable forany C -transformable set F . Then, the following tractable result is known. Theorem 3.19. For any set of complex valued signatures F , Holant( F ) is P-time computable if • F ⊆ T , or • F is P -transformable, or • F is A -transformable, or • F is L -transformable.We denote the above tractable conditions by conditions (3.19). Let H = √ (cid:2) − (cid:3) be the 2-by-2 Hadamard matrix. We have the following result. Lemma 3.20. Let F be a set of real valued signatures and F does not satisfy conditions (3.19).Then for any orthogonal matrix Q ∈ R × , Q F also does not satisfy conditions (3.19). Moreover,we have H F 6⊆ P , b F 6⊆ P , b F 6⊆ A and T α F 6⊆ A . Let F by a set of real valued signatures. The following three results are known [20, 21, 3]. Theorem 3.21. F ) is F ⊆ A or P . Theorem 3.22. ( F ) is F ⊆ A , P , L , or T α F ⊆ A Theorem 3.23. Holant c ( F ) is F ⊆ T , A , P , L , H F ⊆ P , b F ⊆ P or T α F ⊆ A . Based on the above three theorems and Lemma 3.20, we have the following hardness result. Theorem 3.24. Let F be a set of real valued signatures and F does not satisfy conditions (3.19).Then for any orthogonal matrix Q ∈ R × , we have Q F ) , ( Q F ) and Holant c ( Q F ) are The following two reductions are known [8]. One states that we can realize all = k once we have= . The other states that we can realize all = k once we have = . Lemma 3.25. F ) T Holant(= , F ) . Lemma 3.26. ( F ) T Holant(= , F ) . In the following, without other specifications, we use f to denote a real-valued signature and F to denote a set of real-valued signatures. While b f denotes a signature satisfying ars and b F denotes a set of such signatures. We use Q ∈ R × to denote an orthogonal matrix and b Q denotes Z − Q ( Z − ) T . 16 Proof Outline for Theorem 1.9 We first give a more concrete restatement of Theorem 1.9. Theorem 4.1. Let F be a set of real-valued signatures containing a (nonzero) signature of oddarity. If F satisfies the tractability conditions stated in Theorem 3.19 then Holant( F ) is P-timecomputable. Otherwise, Holant( F ) is P-hard. By Theorem 3.19, the tractability part is known. We prove F does notsatisfy these conditions. First, we show that under some holographic transformations, either onecan use a signature of odd arity in F to realize the unary signature ∆ = (1 , k ) ( k > = | = k , b F )(Corollary 5.4) and Holant(∆ , F ) separately (Theorem 5.15).It is easy to verify that for any k > 3, k ( = , b F ) T Holant( = | = k , b F ). Thus, we provethe = | = k , b F ) by giving a dichotomy of k ( = , G ) for any set G ofcomplex-valued signatures (Theorem 5.3). This result should be of independent interest.Next, we focus on the , F ). This is where our entanglement result canbe applied. If ∆ is realizable from Holant(∆ , F ), then we reduce Holant(∆ , F ) from Holant c ( F )and we are done by the dichotomy of Holant c ( F ). By using ∆ , we first give two conditions that ∆ can be easily realized by pinning (Lemma 5.5) or interpolation (Lemma 5.6). As a corollary (Corol-lary 5.7) of Lemma 5.6, we show that either Holant c ( F ) T Holant(∆ , F ), or every irreducible f ∈ F satisfies the following important first order orthogonality condition. Definition 4.2 (First order orthogonality, c.f. Definition 5.8) . Let f be a complex-valued signatureof arity n > , we say that it satisfies the first order orthogonality condition if there exists some µ = 0 such that for all indices i ∈ [ n ] , the entries of f satisfy the following equations | f i | = | f i | = µ, and h f i , f i i = 0 . To restate it in the quantum terminology, let | Ψ i be a normalized n -qubit ( n > state, i.e., h Ψ | Ψ i = 1 . Then it satisfies the first order orthogonality if for every i -th qubit of | Ψ i , h Ψ i | Ψ i i = h Ψ i | Ψ i i = 1 / and h Ψ i | Ψ i i = 0 . Remark: A real-valued signature f satisfies the first order orthogonality precisely when there issome µ = 0 such that for all indices i , M ( m i f ) = µI . In this case, the complex inner product isrepresented by mating real-valued signatures using (= ). Also, by (3.2), when b f is a signature with ars , the complex inner product can also be represented by mating using ( = ). We can show that f satisfies first order orthogonality if and only if b f satisfies it. Although first order orthogonality iswell-defined for any complex-valued signature, when f is not real-valued ( b f does not satisfy ars ),we can not say anything about m i f and b m i b f since now they can not represent the complex innerproduct. The properties of m i f and b m i b f crucially depend on f being real (equivalently b f satisfying ars ).Back to the proof of the , F ). Since F does not satisfy conditions(3.19), F 6⊆ T . Hence, there is a signature f ∈ F of arity n > f / ∈ T . In other words, F contains an n -qubit state exhibiting multipartite entanglement. We will prove the n . We first consider the base case that n = 3. We show that an irreducible ternarysignature (a genuinely entangled 3-qubit state) satisfying first order orthogonality has some special17orms, from which one can realize (= ) or (= ) after some holographic transformations. Then,we can reduce the problem from F ), or ( F ), or Holant( = | = , b F ), to Holant(∆ , F )(Lemma 5.9). This allows us to finish the proof by citing existing dichotomy results for F ),or ( F ), or the result we showed above for Holant( = | = k , b F ).Then, we consider the inductive step. The general strategy is that we start with a signature f ∈ F of arity n > T , and realize a signature g of arity n − n − T , by pinning or merging. (By the definition of T , when n = 4, this g must have arity 3.) Bya sequence of reductions (that is constant in length independent of the problem instance size), wecan realize a signature h of arity 3 that is not in T . Then we are done. In other words, given an n -qubit state with multipartite entanglement, we want to show that multipartite entanglement ispreserved under projections onto | i and self-loops by | φ + i . Lemma 1.8 says that the preservationholds, or | i or | GHZ i is realizable (Lemmas 5.13 and 5.14). We give an inductive restatement ofLemma 1.8 in the Holant framework. Lemma 4.3. Let f ∈ F be a signature of arity n > and f / ∈ T . Then one of the followingalternatives must hold: • ∆ is realizable: Holant(∆ , ∆ , F ) T Holant(∆ , F ) , or • (= ) is realizable: Holant(= , F ) T Holant(∆ , F ) , or • a signature g / ∈ T of arity n − or n − can be realized: Holant(∆ , g, F ) T Holant(∆ , F ) .Proof Sketch. For all indices i and all pairs of indices { j, k } , consider f i and ∂ jk f . If there exists i or { j, k } such that f i or ∂ jk f / ∈ T , then we can realize g = f i or ∂ jk f which has arity n − n − 2, and we are done. Otherwise, f i and ∂ jk f ∈ T for all i and all { j, k } . Under this assumption,our goal is to show that we can realize ∆ , or there is a unary signature a ( x u ) or a binary signature b ( x v , x w ) such that a ( x u ) | f or b ( x v , x w ) | f . Then, we have f = a ( x u ) ⊗ g or f = b ( x v , x w ) ⊗ g forsome g of arity n − n − 2. We know g can be realized from f by factorization. By the definitionof T , we have g / ∈ T since f / ∈ T , and we are done. When n > 5, the above induction proof can beachieved by the interplay of the unique prime factorization, and the commutivity of f i (pinning)and ∂ jk f (merging) gadgets on disjoint indices. (Lemmas 5.11 and 5.12). For n = 4, there is theadditional case that (= ) can be realized. Thus for n = 4, it requires more work (Lemma 5.14); weneed to combine the induction proof and first order orthogonality to handle it.We know that Holant c ( F ) (which is just Holant(∆ , ∆ , F )) and ( F ) are both F does not satisfy the tractability conditions (3.19). Thus, we have shown that when F doesnot satisfy the tractability conditions (3.19), Holant(∆ , F ) is We give the full proof of Theorem 4.1 in this section. Lemma 5.1. Let F be a set of real-valued signatures containing a signature of odd arity. Then,there exists some orthogonal matrix Q such that • Holant(∆ , Q F ) T Holant( F ) or • Holant( = | = k +1 , b Q b F ) T Holant( F ) , for some k > .Proof. Suppose f has arity n . We prove our lemma by induction on n .18f n = 1, let f = ( a, b ) where a, b are not both zero. Let Q = √ a + b (cid:2) a b − b a (cid:3) , an orthogonalmatrix. Note that Holant( F ) is just Holant(= | F ), and = is invariant under an orthogonalholographic transformation (= )( Q − ) ⊗ = (= ), and Q ( a, b ) T = √ a + b (1 , T . Thus,Holant(= | ( a, b ) , F ) ≡ T Holant(= | ∆ , Q F ) . The base case is proved.We assume our claim is true for n = 2 k − 1. Now, we consider n = 2 k + 1 > 3. If there is apair of indices { i, j } such that ∂ ( ij ) f 0, then we can realize a signature of arity 2 k − f .By induction hypothesis, we haveHolant(∆ , Q F ) T Holant( ∂ ij f, F ) T Holant( F ) . Otherwise, ∂ ij f ≡ { i, j } . Thus, we also have b ∂ ij b f ≡ { i, j } . Then,by Lemma 3.7, we have b f = a (1 , ⊗ n + ¯ a (0 , ⊗ n for some a = 0. Let c Q = h n √ ¯ a n √ a i . Q isorthogonal (up to a scalar) by Lemma 3.5. We have ( c Q ) ⊗ n b f = | a | (cid:0) (1 , ⊗ n + (0 , ⊗ n (cid:1) . Thus, aholographic transformation by c Q and Z − yieldsHolant( F ) = Holant(= | f, F ) ≡ T Holant( = | b f , b F ) ≡ T Holant( = | = k +1 , c Q b F ) . Thus, Holant( = | = k +1 , b Q b F ) T Holant( F ) where k > = | = k , b F ) and Holant(∆ , F ) in the followingsubsections. ( = | = k , b F ) Recall that EQ k denotes the set of equalities of arity nk for all n > 1, i.e., EQ k = { = k , = k , . . . , = nk , . . . } . The problem k ( = , F ) is defined as Holant( EQ k |6 = , F ). This is equivalentto Holant( EQ k , = | EQ k , = , F ). First, we have the following reduction. Lemma 5.2. For any k > , k ( = , F ) T Holant( = | = k , F ) .Proof. Connecting one input each of k copies of ( = ) on the LHS with the k variables of (= k ) onthe RHS produces one (= k ) on the LHS. Once we have (= ( n − k ) on the LHS, we take one copyof (= ( n − k ) and two copies of (= k ) on the LHS, and one copy of (= k ) on the RHS with variables x , . . . , x k . Connect x and x to one variable each of the two copies of (= k ) on the LHS, and x , . . . , x k all to variables of (= ( n − k ). This produces one copy of (= nk ) on the LHS.Then, we give a dichotomy of k ( = , F ) for any set F of complex-valued signatures. Let ρ be a 4 k -th primitive root of unity, T k = (cid:2) ρ (cid:3) , and A dk = { f | T dk f ∈ A } , for d ∈ [ k ]. Theorem 5.3. k ( F , = ) is -hard except for the following cases • F ⊆ P ; • there exists k ∈ [ d ] such that F ⊆ A dk ,which can be computed in polynomial time.Proof. We will give a proof in Section 5.5. This result should be of independent interest.19 emark: Let F be a set of real-valued signatures. It is easy to see that when F does not satisfy theconditions in Theorem 3.19, b F does not satisfy the tractable conditions of Theorem 5.3, therefore k ( = , b F ) is b Q b F for any orthogonal matrix Q ∈ R × .Let b f ∈ b F be a signature of arity n . By definition, we have b Q b f = [ Z − Q ( Z − ) T ] ⊗ n [( Z − ) ⊗ n f ] = ( Z − ) ⊗ n [ Q ( Z − ) T Z − ] ⊗ n f. Note that 2( Z − ) T Z − = (cid:2) − (cid:3) . Thus, 2 Q ( Z − ) T Z − is also a real orthogonal matrix, denoted by Q − . Therefore, up to a scalar b Q b f = ( Z − ) ⊗ n ( Q ⊗ n − f ) = \ Q − f . Thus, b Q b F = \ Q − F . By Lemma3.20, Q − F does not satisfy conditions (3.19) when F does not satisfy conditions (3.19), hence k ( = , \ Q − F ) is k ( = , b Q b F ) is Corollary 5.4. Let F be a set of real-valued signatures. For any k > and any orthogonal matrix Q , Holant( = | = k , b Q b F ) is F does not satisfy the conditions in Theorem 3.19. ∆ , F ): the base case In the following two subsections, we will show that Holant(∆ , F ) is F does not satisfyconditions (3.19). Since F 6⊆ T , there is a signature f ∈ F of arity n > T . Recallthat f / ∈ T means that the state of | f i has multipartite entanglement. We will prove our claim byinduction on the arity n .In this subsection, we focus on the base case that f has arity 3. In other words, there is amultipartite entangled 3-qubit. We first give two conditions that ∆ = (0 , 1) can be easily realizedfrom a signature of arity not necessarily 3 by pinning (Lemma 5.5) or interpolation (Lemma 5.6).Then, we have Holant c ( F ) T Holant(∆ , F ), and we are done by the hardness of Holant c ( F ).For cases in which ∆ cannot be realized from a ternary signature f , we show that f has specialforms. Then, we can use this f to realize = or = , and reduce the problem from Q F ), ( Q F ) or ( = , b Q b F ) for some orthogonal matrix Q . Based on the known hardnessresults, we will finish the proof for the base case that f has arity 3 (Lemma 5.9). Lemma 5.5. Let f ∈ F be a nonzero signature and f ~ = 0 . Then Holant c ( F ) T Holant(∆ , F ) .Proof. We prove this by induction on the arity n of f .If n = 1, we have f = (0 , λ ) for some λ = 0 since f 0. Clearly, ∆ is realizable from f .Assuming our claim is true when n = k , we consider the case that n = k + 1. For all indices i ∈ [ n ], consider signatures f i realized from f by pinning variable x i to 0. We know f i is signatureof arity k and f i ( ~ k ) = f ( ~ k +1 ) = 0. • If there is an index i such that f i 0, then by induction hypothesis, we have Holant c ( F ) T Holant(∆ , f i , F ) T Holant(∆ , F ) . • Otherwise, f i ≡ i . Then, by Lemma 3.9, we have f = λ (0 , ⊗ n for some λ = 0 since f 0. Thus, ∆ is realizable from f by Lemma 3.11.Thus, we have Holant c ( F ) T Holant(∆ , F ) . Now for all indices i , we consider signatures m i f realized from f by mating. We give a conditionby which ∆ can be realized from m i f by interpolation.20 emma 5.6. Let f ∈ F be a nonzero signature of arity n > . If there is an index i such that M ( m i f ) (as a 2-by-2 matrix) is not the identity matrix up to a scalar ( M ( m i f ) = µ i I ), then • there is an unary signature a ( x i ) on variable x i such a ( x i ) | f , or • Holant c ( F ) T Holant(∆ , F ) .Proof. We denote M ( m i f ) = (cid:20) | f i | h f i , f i ih f i , f i i | f i | (cid:21) by (cid:20) a bb c (cid:21) . Since f is real, M ( m i f ) is real symmetric, and thus diagonalizable with real eigenvalues. We firstconsider the case that M ( m i f ) is degenerate. Then, we have |h f i , f i i| = | f i | | f i | , so f i and f i are linearly dependent by Cauchy-Schwarz. Since f 0, either f i and f i is nonzero. Assume f i isnonzero (the other case is similar). Then, we have f i = c · f i for some constant c . It follows that f = a ( x i ) ⊗ f i , for a unary signature a ( x i ) = (1 , c ).Now we assume M ( m i f ) has rank 2, then we have a, c > 0. We consider the value of b . • If b = 0, then M ( m i f ) = [ a c ]. Clearly, a = c since M ( m i f ) is not I up to a scalar. Given a = c and ac > 0, we have | ac | 6 = 1. By Lemma 3.12, we can realize (0 , , , 1) = (0 , ⊗ from m i f by interpolation. Then, by Lemma 3.11, we can realize ∆ = (0 , 1) by factorization. • Otherwise, b = 0. Clearly, we know (1 , T is not an eigenvector of M ( m i f ). Suppose M ( m i f ) = P − h λ λ i P , where λ and λ are two real eigenvalues of M ( m i f ). Since M ( m i f )has rank 2 and M ( m i f ) is not I up to a scalar, we have λ λ = 0 and λ = λ . Also, by thetrace formula, λ + λ = a + c > 0. Thus λ λ = − 1. Then we have | λ λ | 6 = 1 . By Lemma 3.13,we can realize ∆ = (0 , 1) by interpolation.Thus, we have Holant c ( F ) T Holant(∆ , F ).Moreover, if for all indices i , M ( m i f ) = µ i I , then we show all µ i have the same value. Corollary 5.7. Let f ∈ F be an irreducible signature of arity n > . Then we have • Holant c ( F ) T Holant(∆ , F ) , or • there exists some µ = 0 such that for all indices i , M ( m i f ) = µI i.e., m i f = µ · (= ) .Proof. In the above proof, if f is irreducible, either Holant c ( F ) T Holant(∆ , F ), or M ( m i f ) = µ i [ ] for every index i , where µ i = | f i | = | f i | = 0 . If we further connect the two danglingvariables x i of m i f , which totally connects the corresponding pairs of variables in two copies of f ,we get a value 2 µ i . This value does not depend on the particular index i . Thus, all µ i have thesame value for i ∈ [ n ]. We denote this value by µ .We introduce the following key property, which we call first order orthogonality . Definition 5.8 (First order orthogonality) . Let f be a complex-valued signature of arity n > , wesay it satisfies first order orthogonality if there exists some µ = 0 such that for all indices i ∈ [ n ] ,the entries of f satisfy the following equations | f i | = | f i | = µ, and h f i , f i i = 0 . (5.1) Remark: When f is a real-valued signature, it satisfies first order orthogonality is precisely thatthere is µ = 0 such that for all indices i , M ( m i f ) = µI . Now, we consider b f . Since d (= ) = Z − (= 21 = ( = ), we know that for every index i , b m i b f = d m i f = \ µ (= ) = µ ( = ). Thus, M ( b m i b f ) = " h b f i , b f i i | b f i | | b f i | h b f i , b f i i = µ (cid:20) (cid:21) . This implies that b f also satisfies the first order orthogonality. The reversal is also true. Therefore, f satisfies the first order orthogonality is equivalent to b f satisfying it. Later, based on first orderorthogonality of b f , we will carve out second order orthogonality. For a real valued binary signature,it satisfies first order orthogonality if and only if its (2-by-2) signature matrix is an orthogonalmatrix. Thus, a binary signature with ars satisfies first order orthogonality if and only if it hasparity.Although the first order orthogonality is well-defined for any complex valued signature, when f is not real-valued ( b f does not satisfy ars ), we can not say anything about m i f and b m i b f . Theproperties of m i f and b m i b f crucially depend on f being real ( b f satisfying ars ).First order orthogonality implies some non-trivial properties. Consider the vector f i . We canpick a second variable x j and separate f i into two vectors f ij and f ij according to x j = 0 or 1.Then | f i | = | f ij | + | f ij | = µ. Similarly, we have | f j | = | f ij | + | f ij | = µ. Comparing the above two equations, we have | f ij | = | f ij | . (5.2)This is ture for all pairs of indices { i, j } . Similarly, by considering | f j | = | f ij | + | f ij | = µ, we have | f ij | = | f ij | , (5.3)for all pairs { i, j } .Now, we are ready to finish the proof of the base case n = 3 using the above equations. Abinary signature satisfies the first order orthogonality iff its signature matrix is orthogonal, and inthis case, we say the binary signature is orthogonal. Lemma 5.9 (Base case n = 3) . Let f ∈ F be a signature of arity and f / ∈ T . Then Holant(∆ , F ) is P-hard unless F satisfies conditions (3.19).Proof. Since f is a ternary signature and f / ∈ T , we know f is irreducible. If f = 0 or f does not satisfy the first order orthogonality, then by Lemma 5.5 or Corollary 5.7, we haveHolant c ( F ) T Holant(∆ , F ). By Theorem 3.24, Holant c ( F ) is F does not satisfyconditions (3.19), and hence Holant(∆ , F ) is f = 1 bynormalization and f satisfies first order orthogonality and thus equations (5.1), (5.2) and (5.3).We consider binary signatures f , f and f realized by pinning. If there is an index i suchthat the binary signature f i is irreducible and not orthogonal, then by Corollary 5.7 we are done.Otherwise, f , f and f are all either reducible or orthogonal. Let N be the number of orthogonalsignatures among f , f and f . According to N = 0 , , N = 0. Then f , f and f are all reducible. So, f is of the form (1 , a, b, ab ), and so are f and f . Thus f has the following matrix M , ( f ) = (cid:20) a b abc ac bc d (cid:21) . By the equation | f | = | f | from (5.3), we have b + a b = c + a c . Then, (1 + a )( b − c ) = 0. Being real, we have 1 + a > 0, and thus b = c . Similarly bysymmetry, we have a = b = c . By the equation | f | = | f | from (5.2), we have1 + a = b c + d . Then, d = 1 + a − a . By the equation h f , f i = 0 from (5.1), we have c + a c + b c + abd = 0 . Then, c (1 + 2 a ) = − abd . Taking squares of both sides, we have a (1 + 4 a + 4 a ) = a d . Plug in d = 1 + a − a , and we have a (1 + 4 a + 4 a − a − a + a ) = a (1 + a ) = 0 . Since 1 + a > 0, we have a = 0, and hence b = c = 0 and d = 1. – If d = 1, then f has the signature matrix [ ], which is (= ). Then, by Lemma 3.25,we can realize all equality signatures (= k ). Thus, we have F ) T Holant(= , F ) T Holant(∆ , F ) . By Theorem 3.24, we know F ) is F does not satisfy conditions(3.19), and hence Holant(∆ , F ) is – Otherwise, d = − 1. We perform a holographic transformation by the orthogonal matrix Q = (cid:2) − (cid:3) . Note that(= )( Q − ) ⊗ = (= ) and Q ⊗ f = (= ) . Thus, the holographic transformation by Q yieldsHolant(= | f, F ) ≡ T Holant(= | = , Q F ) . Again by Lemma 3.25, we have Q F ) T Holant(∆ , F ) . By Theorem 3.24, weknow that Q F ) is F does not satisfy conditions (3.19), and henceHolant(∆ , F ) is N = 1. Without loss of generality, we may assume f is orthogonal and f and f arereducible. Then f has the form (1 , a, ǫa, − ǫ ), f has the form (1 , a, b, ab ) and f has theform (1 , ǫa, b, ǫab ), for some ǫ = ± 1. Therefore, for some value x , f has the signature matrix, M ( f ) = (cid:20) a ǫa − ǫb ab ǫab x (cid:21) . By the equation | f | = | f | from (5.3), we have( ǫa ) + ( − ǫ ) = b + ( ab ) . Thus (1 + a )(1 − b ) = 0. So b = 1. By the equation | f | = | f | from (5.2), we have1 + a = ( ǫab ) + x = a + x . Then, x = 1. By the equation h f , f i = 0 from (5.1), we have b + a b + ǫ a b − ǫx = 0 . (5.4)Then, ǫx = b (1+2 a ). Taking squares of both sides, we have 1 = (1+2 a ) , which implies that a = 0. So by (5.4), we have b − ǫx = 0, and thus x = bǫ = ǫb . It follows that M ( f ) = (cid:2) − ǫb ǫb (cid:3) ,with b = ǫ = 1.Mating variable x of one copy of f with variable x of another copy of f (with x and x asdangling variables), we get a 4-ary signature m f with the signature matrix M ( m f ) = M x x ,x ( f ) M x ,x x ( f ) = b − ǫ ǫb (cid:20) − ǫb ǫb (cid:21) = = 2 M (= ) . Therefore, we can realize (= ), and then by Lemma 3.26 we can realize all equality signatures(= k ) of even arity. Thus, ( F ) T Holant(= , F ) T Holant(∆ , F ) . By Theorem 3.24, we know ( F ) is F does not satisfy conditions 3.19,and hence Holant(∆ , F ) is • N = 2. Without loss of generality, we may assume f and f are orthogonal, and f isreducible. Then, f has the form (1 , ǫ a, a, − ǫ ) where ǫ = ± f has the form (1 , ǫ a, a, − ǫ )where ǫ = ± 1, and f has the form (1 , ǫ a, ǫ a, ǫ ǫ a ). Then for some x , f has the form M ( f ) = (cid:20) ǫ a ǫ a ǫ ǫ a a − ǫ − ǫ x (cid:21) . By the equation | f | = | f | , we have( ǫ a ) + ( ǫ ǫ a ) = a + ( − ǫ ) . So we get a = 1. Since a is real, we have a = ± 1. By the equation h f , f i = 0, we have a − ǫ a − ǫ a + ǫ ǫ x = − a + ǫ ǫ x = 0 . x = aǫ ǫ = ǫ ǫ a . By mating we get m f , and we have M ( m f ) = aǫ a − ǫ ǫ a − ǫ ǫ ǫ ǫ ǫ a (cid:20) ǫ a ǫ a ǫ ǫ a − ǫ − ǫ ǫ ǫ a (cid:21) = 2 ǫ ǫ ǫ ǫ ǫ ǫ ǫ ǫ . – If ǫ ǫ = 1, then m f is 2 times the Is-Even signature, which takes value 1 on all inputsof even weight, and 0 otherwise. Note that, for H = √ (cid:2) − (cid:3) we have(= )( H − ) ⊗ = (= ) and H ⊗ ( m f ) = 4(= ) . Thus, a holographic transformation by H yieldsHolant(= | m f, F ) ≡ T Holant(= | = , H F ) . By Lemma 3.26, we have ( H F ) T Holant(= | = , H F ) T Holant(∆ , F ) . By Theorem 3.24, ( H F ) is F does not satisfy conditions (3.19),and hence Holant(∆ , F ) is – Otherwise, ǫ ǫ = − 1. Then g ( y , y , y , y ) = m f can be normalized as (cid:20) − 10 1 − − − (cid:21) ,where the row index is y y and column index is y y ∈ { , } , both listed lexicograph-ically. After a permutation of variables, we have M y y ,y y ( m f ) = (cid:20) − − − − (cid:21) . Con-necting variables y , y of a copy of m f with variables y , y of another copy of m f respectively, we get a signature with the signature matrix M y y ,y y ( m f ) M y y ,y y ( m f ) = (cid:20) − − − − (cid:21) (cid:20) − − − − (cid:21) = 2 (cid:20) (cid:21) . Now perform a holographic transformation by H , and we get (= ), which implies thatHolant(∆ , F ) is F does not satisfy conditions (3.19). • N = 3. Then for some values a , x and ǫ , ǫ = ± 1, the signature f has the signature matrix M ( f ) = (cid:20) ǫ a ǫ a − ǫ ǫ a − ǫ − ǫ x (cid:21) . By the equation h f , f i = 0, we have a − ǫ a − ǫ a − ǫ ǫ x = − a − ǫ ǫ x = 0 . Hence, x = − ǫ ǫ a . A holographic transformation by Z − yieldsHolant(= | f, ∆ , F ) ≡ T Holant( = | b f , c ∆ , b F ) . Note that c ∆ = Z − (1 , T = (1 , T , and a simple calculation shows M ( b f ) = Z − M ( f )(( Z − ) T ) ⊗ = (cid:2) − a i 00 1+ a i (cid:3) h (1+ ǫ )(1+ ǫ ) (1 − ǫ )(1+ ǫ ) (1+ ǫ )(1 − ǫ ) (1 − ǫ )(1 − ǫ )(1 − ǫ )(1 − ǫ ) (1+ ǫ )(1 − ǫ ) (1 − ǫ )(1+ ǫ ) (1+ ǫ )(1+ ǫ ) i . If ǫ = ǫ = 1, then up to a constant, M ( b f ) = (cid:2) − a i a i (cid:3) . Let c Q = h √ a i √ − a i i .Then ( c Q ) ⊗ b f has the signature matrix (1 + a ) [ ]. Thus, a holographic transfor-mation by c Q yieldsHolant( = | b f , c ∆ , b F ) ≡ T Holant( = | = , c Q c ∆ , c Q b F ) . Thus, we have Holant( = | = , c Q c ∆ , c Q b F ) T Holant(∆ , F ) . By Corollary 5.4, we know Holant( = | = , c Q c ∆ , c Q ) is F does not satisfyconditions (3.19), and hence Holant(∆ , F ) is – Otherwise, M ( b f ) has the signature matrix (cid:2) − a i a i (cid:3) up to a permutation of vari-ables. Connecting b f with c ∆ = (1 , 1) using = , we get a binary signature b g with matrix M ( b g ) = [1 , (cid:20) (cid:21) (cid:20) − a i a i (cid:21) = (1 + a i , , , − a i ) . Connecting one variable of b f with one variable of b g using = , we get a signature b h withthe signature matrix M ( b h ) = (cid:20) a i 00 1 − a i (cid:21) (cid:20) (cid:21) (cid:20) − a i a i (cid:21) = (cid:20) (1 + a i ) − a i ) (cid:21) . Then, a holographic transformation by c Q = (cid:20) √ (1 − a i ) √ (1+ a i ) (cid:21) yieldsHolant( = | b h, c ∆ , b F ) ≡ T Holant( = | = , c Q c ∆ , c Q b F ) . Then similarly by Corollary 5.4, we have Holant(∆ , F ) is , F ) is F satisfies conditions (3.19). ∆ , F ): the induction step Now, we consider the inductive step. The general strategy is that we start with a signature f ∈ F of arity n > T , and realize a signature g of arity n − n − T . By a sequence of reductions (that is constant in length independentof the problem instance size), we can realize a signature h of arity 3 that is not in T . Then weare done. In other words, we want to show that the multipartite entanglement is preserved underprojections onto | i or forming self-loops by | φ + i .For all indices i and all pairs of indices { j, k } , consider f i and ∂ jk f . If there exist i or { j, k } such that f i or ∂ jk f / ∈ T , then we can realize g = f i or ∂ jk f which has arity n − n − f i and ∂ jk f ∈ T for all i and all { j, k } . We denote this propertyby f ∈ R T . Under the assumption that f ∈ R T , our goal is to show that we can realize ∆ and hence we are done by the hardness of Holant c ( F ), or there is an unary signature a ( x u ) orbinary signature b ( x v , x w ) such that a ( x u ) | f or b ( x v , x w ) | f. Then, we have f = a ( x u ) ⊗ g or f = b ( x v , x w ) ⊗ g for some g of arity n − n − 2. By the definition of T , we know g / ∈ T since f / ∈ T . By Lemma 3.11, we can realize g by factorization, and we are done. When n > f i (pinning) and ∂ jk f (merging) operations on disjoint indices (Lemmas 5.11 and5.12). For n = 4, the proof requires more work (Lemma 5.14); we need to combine the inductionproof and first order orthogonality to handle it.We denote the property that f i ∈ T for all i by R T . We carry out our induction proof bythe following lemmas. Lemma 5.10. Let f be a signature of arity n > . If there exists a nonzero signature g , the scopeof which is a subset of the scope of f , such that g | f i for all indices i disjoint with the scope of g and furthermore, g | ∂ jk f for some pair of indices { j, k } disjoint with the scope of g , then g | f .(Note that if ∂ jk f ≡ then g | ∂ jk f is satisfied.)Proof. We may assume f is nonzero, for otherwise the conclusion trivially holds. We now provethis for a unary signature g = ( a, b ). We assume g is on the variable x u . Consider the signature f ′ = bf u − af u . Clearly, x j and x k are in the scope of f ′ . Thus, f ′ has arity at least 2. For every i = u , we have f i = ( a, b ) ⊗ h for some h . Then, ( f i ) u = a · h , ( f i ) u = b · h , and hence( f ′ ) i = ( bf u − af u ) i = bf ui − af ui = b ( f i ) u − a ( f i ) u = ba · h − ab · h ≡ . Moreover, there are indices j, k = u such that g ( x u ) | ∂ jk f . Then, ∂ jk f = ( a, b ) ⊗ h ′ , for some h ′ .Then, we have ( ∂ jk f ) u = a · h ′ , ( ∂ jk f ) u = b · h ′ , and hence ∂ jk ( f ′ ) = ∂ jk ( bf u − af u ) = b ( ∂ jk f ) u − a ( ∂ jk f ) u = ba · h ′ − ab · h ′ ≡ . By Lemma 3.9, we have f ′ ≡ 0. Thus, we have f u : f u = a : b , and hence g ( x u ) | f .For a signature g of arity > n − 2, the proof is essentially the same, which we omit here. Lemma 5.11. Let f be a signature of arity n > , f / ∈ T , f ∈ R T and f ∈ R T . Then there isa unary signature a ( x u ) such that a ( x u ) | f , or ∆ is realizable from f .Proof. As f / ∈ T , f is nonzero. We may further assume f i i . Otherwise, wehave f ~ = 0. Then, by Lemma 5.5, we can realize ∆ . By the same reason, we may further assume f ij { i, j } .For some arbitrary index r , we consider f r . Since f r ∈ T , there exists some unary signature a ( x u ) such that a ( x u ) | f r . We show a ( x u ) | f . Consider f i for all indices i = u, r . Since f i ∈ T ,there is a unary signature a ′ ( x u ) such that a ′ ( x u ) | f i , and hence we have a ′ ( x u ) | ( f i ) r . On theother hand, since a ( x u ) | f r , we also have a ( x u ) | ( f r ) i . Note that the pinning operations ondifferent variables commute. Thus, we have ( f r ) i = ( f i ) r , and we know it is a nonzero signature.Then, by UPF (Lemma 2.4), we have a ( x u ) ∼ a ′ ( x u ). Thus, a ( x u ) | f i for all indices i = u .Then, we show a ( x u ) | ∂ jk f for some arbitrary pair of indices j, k = u . If ∂ jk f ≡ 0, then wehave a ( x u ) | ∂ jk f and hence a ( x u ) | f by Lemma 5.10. Next, we assume ∂ jk f 0. Similarly, iffor some index i = j, k , we have ( ∂ jk f ) i ≡ 0, then we have ∂ jk f ( ~ 0) = 0 and hence by Lemma 5.6,we can realize ∆ . Otherwise, ( ∂ jk f ) i i / ∈ { j, k } . Recall that ∂ jk f ∈ T . We show thevariable x u must appear in a unary signature in the UPF of ∂ jk f ∈ T . • For a contradiction, suppose there is an irreducible binary signature b ( x u , x v ) such that b ( x u , x v ) | ∂ jk f . Since f has arity n > 5, we can pick some index i / ∈ { u, v, j, k } suchthat b ( x u , x v ) | ( ∂ jk f ) i . Note that ( ∂ jk f ) i = ∂ jk ( f i ) ∂ jk ( f i ) has an irreducible binary tensor divisor b ( x u , x v ). However, f i ∈ T and so is ∂ jk ( f i ). By UPF, we get a contradiction.27 Thus, there is a unary signature a ′′ ( x u ) such that a ′′ ( x u ) | ∂ jk f . Pick some index i / ∈ { u, j, k } ,and we have a ′′ ( x u ) | ( ∂ jk f ) i . We also have a ( x u ) | ∂ jk ( f i ) since a ( x u ) | f i . Again, ( ∂ jk f ) i = ∂ jk ( f i ) a ′′ ( x u ) ∼ a ( x u ). Thus, a ( x u ) | f byLemma 5.10 and we are done. Lemma 5.12. Let f be a signature of arity n > , f / ∈ T , f ∈ R T and f / ∈ R T . Then, there isan irreducible binary signature b ( x v , x w ) such that b ( x v , x w ) | f , or ∆ is realizable from f .Proof. Since f / ∈ R T , but f ∈ R T , there is some index r such that f r is nonzero and has anirreducible binary signature factor b ( x v , x w ). We will show this b ( x v , x w ) divides f . Again, we mayassume f i f ij i and all { i, j } . Otherwise, we can realize ∆ by Lemma 5.5.Consider f i for all indices i / ∈ { v, w, r } . Since f i ∈ T and f i 0, there is either a unarysignature a ( x v ) or an irreducible binary signature b ′ ( x v , x w ′ ) such that a ( x v ) | f i or b ′ ( x v , x w ′ ) | f i .We also have b ( x v , x w ) | ( f r ) i since b ( x v , x w ) | f r . Again, we have ( f r ) i = ( f i ) r 0. Then byUPF, we know that the unary signature a ( x v ) does not exist, and it must be b ′ ( x v , x w ′ ) | f i and b ( x v , x w ) = b ′ ( x v , x w ′ ). Thus, we have b ( x v , x w ) | f i for all i / ∈ { v, w } .Then, for an arbitrary pair of indices { j, k } disjoint with { v, w } , we show b ( x v , x w ) | ∂ jk f . Again,we may assume ∂ jk f b ( x v , x w ) | ∂ jk f is proved) and furthermore ( ∂ jk f ) i i disjoint with { j, k } , for otherwise, we can realize ∆ . Since f has arity n > 5, we can picksome index i / ∈ { u, v, j, k } such that b ( x v , x w ) | ∂ jk ( f i ) due to b ( x v , x w ) | f i . Recall that ∂ jk f ∈ T ,we consider the UPF of ∂ jk f . Using a similar argument as in the previous paragraph, we have b ( x v , x w ) | ∂ jk f by UPF.Combining the above two lemmas, we have the following result. Lemma 5.13 (Inductive step for n > . If f ∈ F is a signature of arity n > and f / ∈ T , then • Holant c ( F ) T Holant(∆ , F ) or • there is a signature g / ∈ T of arity n − or n − such that Holant(∆ , g, F ) T Holant(∆ , F ) . Now, the only case left for the induction proof is when f is a signature of arity 4. We deal withit by using the first order orthogonality condition. Lemma 5.14 (Inductive step n = 4) . Let f ∈ F be a signature of arity and f / ∈ T . Then • Holant c ( F ) T Holant(= | ∆ , F ) , or • ( F ) T Holant(= | = , F ) T Holant(= | ∆ , F ) , or • there is a signature g / ∈ T of arity such that Holant(∆ , g, F ) T Holant(∆ , F ) .Proof. First, we may assume f is irreducible. Otherwise, we consider its irreducible factors. Since f / ∈ T , it has an irreducible factor g of arity 3 such that g / ∈ T . By Lemma 3.11, g is realizable from f by factorization, and the lemma is proved. Also we may assume f = 1 after normalizationand f satisfies first order orthogonality; otherwise, by Lemma 5.5 and Corollary 5.7, we are done.We consider signatures f i realized by pinning x i to 0 in f , for all i . If there is i such that theternary signature f i / ∈ T , then we are done, since f i has arity 3. Also, since f has arity 4, ∂ ij f isa binary signature for any pair of indices { i, j } . Hence ∂ ij f ∈ T . Thus, we may assume f ∈ R T . • If f ∈ R T , then there are three unary signatures such that f = a ( x ) ⊗ a ( x ) ⊗ a ( x ). Bythe same proof in Lemma 5.11, we have a ( x ) | f and a ( x ) | f . Thus, a ( x ) ⊗ a ( x ) | f .28 Otherwise, there is an index i such that f i has an irreducible binary factor. Without loss ofgenerality, we assume that f = a ( x ) ⊗ b ( x , x ) where b ( x , x ) is irreducible. By thesame proof as in Lemma 5.12, we have b ( x , x ) | f .Therefore, in both cases, there is a binary signature b ( x , x ), which may be reducible, i.e., b ( x , x ) = a ( x ) ⊗ a ( x ), such that b ( x , x ) | f and b ( x , x ) | f . Thus, we have f = a ( x ) ⊗ b ( x , x ) and f = a ′ ( x ) ⊗ b ( x , x ) . By a normalization we may let b ( x , x ) = (1 , a, b, c ), a ( x ) = (1 , x ) and a ′ ( x ) = (1 , y ). Then, f has the signature matrix, for some z, z , z , z M , ( f ) = a b cx ax bx cxy ay by cyz z z z . Then, we consider the signature f . It has the signature matrix M , ( f ) = ax axy ayz z . We have f ∈ T , and nonzero. In its unique factorization, if x and x belong to an irreduciblebinary signature, then ( f ) , which has the signature matrix M , ( f ) = [ ax ax ], would have beenan irreducible binary signature, a contradiction. Similarly x and x do not belong to an irreduciblebinary signature in the unique factorization of f . Therefore x appears in a unary signature inthe factorization of f . It follows that z = az . Similarly from f ∈ T , we can prove z = bz . Wealso write z as cz + w . Thus, we have M , ( f ) = (1 , x, y, z ) T ⊗ (1 , a, b, c ) + w ((0 , T ) ⊗ ⊗ (0 , ⊗ . We know w = 0 since f / ∈ T . By pinning any 3 of the 4 variables to 0, we can realize four unarysignatures (1 , a ) , (1 , b ) , (1 , x ) and (1 , y ). For example, (1 , x ) can be realized from f by pinningvariables x , x and x to 0. • Suppose a, b, x, y are not all zero, say x = 0. We connect the unary (1 , x ) with the variable x of f , and we get a signature g with the signature matrix M , ( g ) = (cid:20) x a (1 + x ) b (1 + x ) c (1 + x ) y + xz a ( y + xz ) b ( y + xz ) c ( y + xz ) + xw (cid:21) . Clearly, 1 + x = 0. By normalization, we have M , ( g ) = (cid:20) a b cx ′ ax ′ bx ′ cx ′ + w ′ (cid:21) , where x ′ = y + xz x and w ′ = xw x , and w ′ = 0 since xw = 0. Thus g = (1 , x ′ ) x ⊗ (1 , a, b, c ) x ,x + w ′ (0 , ⊗ . (5.5)29e claim that g / ∈ T . Otherwise consider the unique factorization of g in T . By the sameproof above for f ∈ T , we can see that x of g cannot appear in an irreducible binarysignature, either with x or with x , as a tensor factor in the unique prime factorization of g . Hence x must appear in a unary signature in this factorization. This would imply that w ′ = 0, by the form of M , ( g ), a contradiction.It follows that g / ∈ T , and we are done. • Otherwise, a = b = x = y = 0. Then, we know M , ( f ) = c z z . Here, we write z as cz + w . By equation (5.2), we have 1 + c = z + z and 1 + z = c + z .Thus, we have c = z and z = 1. By pinning variables x and x to 0, we can realize thebinary signature (1 , , , c ). If it is not reducible or orthogonal, then by Lemma 5.6 we canrealize ∆ . Otherwise, we have c = 0 or c = ± 1. Similarly, we have z = 0 or z = ± 1. Aswe already have c = z , we get c = z = 0 or c = z = 1. We consider the signature m f realized by mating variables x , x of f . We have M ( m f ) = M , ( f )( M , ( f )) T = c z + cz z + cz z + z . If c = z = 0, then we have M ( m f ) = (cid:20) (cid:21) = M (= ) . Otherwise, c = z = 1. Also,we know z = cz since f / ∈ T . Note that z = 1 and ( cz ) = 1. This implies that z = − cz .Then, we have z + cz = z − c z = z − z = 0. Thus, we have M ( m f ) = (cid:20) (cid:21) = 2 M (= ) . Therefore, we can realize (= ) from f , and then by Lemma 3.26 we can realize all equalitysignatures = k of even arity. Thus, we have ( F ) T Holant(= , F ) T Holant(∆ , F ) . This completes the proof of the lemma. Theorem 5.15. Holant(∆ , F ) is P-hard unless F satisfies the tractable conditions (3.19).Proof. Assume F does not satisfy conditions (3.19). Then F 6⊆ T . There is a signature f ∈ F ofarity n > T . If n = 3, then by Lemma 5.9, we are done.Suppose our statement is true for 3 n k . Consider n = k + 1 > 4. By Lemmas 5.13and 5.14, we have Holant c ( F ), or ( F ), or Holant(= | ∆ , g, F ) T Holant(= | ∆ , F ) forsome g / ∈ T of arity k − k at least 3. By Theorem 3.24 and the induction hypothesis, weknow Holant c ( F ), ( F ) and Holant(= | ∆ , g, F ) are all F does not satisfyconditions (3.19), and hence Holant(= | ∆ , F ) is .4 The proof of Theorem 4.1 Proof. The tractability is known by Theorem 3.19.By Lemma 5.1, for some orthogonal matrix Q , we have Holant(= | ∆ , Q F ) T Holant(= | F )or Holant( = | = k +1 , b Q b F ) T Holant(= | F ). Since F does not satisfy conditions (5.1), we know Q F also does not satisfy conditions (5.1). Then, by Theorem 5.15 and Corollary 5.4, we haveHolant(= | ∆ , Q F ) and Holant( = | = k +1 , b Q b F ) are both | F ) is In this subsection, we will prove Theorem 5.3. For k = 1, Theorem 5.3 follows Theorem 3.21. Notethat ( = ) / ∈ L . Thus for k = 2, Theorem 5.3 follows Theorem 3.22. This implies that Theorem 5.3has been proved for k = 1 , i = − , α = i and β = α in this section. And we use [1 , , · · · , , a ] r to denote a general equality signature of arity r in the following.Note that k ( = , F ) ≡ T Holant( E Q k | 6 = , F ) . Moreover, by the following two gadgets, we have ( = ) , E Q k on both sides i.e., k ( = , F ) ≡ Holant( E Q k , = |E Q k , = , F ) . (5.6) ✟✟✟✟✟✟❍❍❍❍❍❍ (cid:4)(cid:4)(cid:4) • ... • (cid:4)(cid:4)(cid:4) ... • Fig 1. In an instance of Holant( E Q k | 6 = , F ), the first gadget realizes (= nk ) on the RHS.The squares are labeled by ( = ) and the round vertices are labeled by (= nk ). The secondgadget realizes ( = ) on the LHS, where the two round vertices are labeled by (= k ) on the RHS.The following two lemmas show that in k ( = , F ) the pinning signatures ∆ = [1 , 0] and∆ = [0 , 1] are freely available. The first lemma is from [33]. It shows that we have [1 , ⊗ k , [0 , ⊗ k freely in k ( F ). Lemma 5.16. k ( F , [1 , ⊗ k , [0 , ⊗ k ) ≤ T k ( F ) . The second lemma is from [37]. It shows that we can remove the tensor powers of [1 , ⊗ k , [0 , ⊗ k . Lemma 5.17. k ( F , [1 , , [0 , ≤ T k ( F , [1 , ⊗ k , [0 , ⊗ k ) . Firstly, we prove Theorem 5.3 for the case that there exists a general equality of arity less than k in F . 31 emma 5.18. Let f = [1 , , · · · , , a ] r with r < k and a = 0 , and F be a signature set. Then k ( F , = , f ) is -hard except for the following cases • F ⊆ P ; • there exists d ∈ [ k ] such that F ∪ { f } ⊆ A dk ,which can be computed in polynomial time.Proof. Note that the lemma has been proved for the cases k = 1 , k in the following. Let k = tr + r with0 < r ≤ r . Note that k > r , so t ≥ 1. In Holant( EQ k , = |EQ k , = , f, F ), connecting ℓt copies of[1 , , · · · , , a ] r to (= ℓk ) we get [1 , , · · · , , a tℓ ] ℓr for ℓ = 1 , , · · · , i.e.,Holant( EQ ar , = |EQ k , = , f, F ) ≤ T Holant( EQ k , = |EQ k , = , f, F ) , where EQ ar = { [1 , , · · · , , a t ] r , [1 , , · · · , , a t ] r , · · · , [1 , , · · · , , a ℓt ] ℓr , · · · } . Let T = h a tr i ,then T − ( EQ ar ) = EQ r . Thus after the holographic transformation using T , we haveHolant( EQ r | T EQ k , = , T ⊗ r f, T F ) ≤ T Holant( EQ k , = |EQ k , = , f, F ) , i.e., r ( T EQ k , = , T ⊗ r f, T F ) ≤ T k ( = , f, F ) . By induction, if { T EQ k , T F , T ⊗ r f } * P and { T EQ k , T F , T ⊗ r f } * A d ′ r for any d ′ ∈ [ r ], then r ( T EQ k , = , T ⊗ r f, T F ) is k ( = , f, F ) is { T EQ k , T F , T ⊗ r f } ⊆ P , then F ⊆ P since T is a diagonal matrix. Moreover,if { T EQ k S T F S T ⊗ r f } ⊆ A d ′ r for some d ′ ∈ [ r ], let T ′ = h γ d ′ i , where γ is a 4 r -th primitiveroot of unary, so, γ r = 1, then T ′⊗ k T ⊗ k (= k ) ∈ A , T ′⊗ r T ⊗ r f ∈ A and T ′ T F ⊆ A . Firstly, by T ′⊗ k T ⊗ k (= k ) ∈ A , we have ( γ d ′ k a ktr ) = 1. This implies that γ d ′ a tr is a 4 k -th root of unity, i.e.,there exists d ∈ [ k ] such that γ d ′ a tr = ρ d , then T ′ T = h ρ d i . Thus F ⊆ A dk and f ∈ A dk . Thisfinishes the proof. Definition 5.19. Let f = ( f i i ··· i n ) , g = ( g i i ··· i n ) be two n -ary signatures, then f g is an n -ary signature and ( f g ) i i ··· i n = f i i ··· i n g i i ··· i n for any i i · · · i n ∈ { , } n . In particular, f k =( f ki i ··· i n ) , and for a signature set F , F k = { f k | f ∈ F } . Definition 5.20. If f has affine support of rank r , and X = { x j , x j , · · · , x j r } is a set offree variables, then f X is the compressed signature of f for X such that f X ( x j , x j , · · · , x j r ) = f ( x , x , · · · , x n ) , where ( x x · · · x n ) is in the support of f . When it is clear from the context, weomit X and use f to denote f X . Note that if f has affine support, then f ∈ A iff f ∈ A .Theorem 5.3 has been proved for k = 1 , 2. We will prove it by induction on k . The followinglemma is the main tool to do the induction. More precisely, the following lemma shows that if f = f f · · · f k ′ for some k ′ | k , then we can simulate kk ′ ( = , f ) by k ( = , f , f , · · · , f k ′ ).If f / ∈ P and f / ∈ A d ′ kk ′ for any d ′ ∈ [ kk ′ ], then by induction, kk ′ ( = , f ) is k ( = , f , f , · · · , f k ′ ) is emma 5.21. Let F be a signature set. Signatures f , f , · · · , f k ′ has the same arity and f = f f · · · f k ′ , where k ′ | k , then kk ′ ( f, F k ′ ) ≤ T k ( f , f , · · · , f k ′ , F ) . Proof. In an instance of kk ′ ( f, F k ′ ), by expanding each variable x to k ′ copies of x , replac-ing each h k ′ ∈ F k ′ by k ′ copies of h , and replacing f by f , f , · · · , f k ′ , we get an instance of k ( f , f , · · · , f k ′ , F ) and its value is same as the value of the instance of dd ′ ( f, F k ′ ).This finishes the proof.If the support of f is not affine, then the support of f k is not affine. By Lemma 5.21, f k ) ≤ T k ( f ) . Moreover, by Theorem 3.21, f k ) is k ( f ) is P , every signature f ∈ P has a decomposition as a product of signaturesover disjoint of variables, where each factor has support contained in a pair of antipodal points:There exists a partition X = { x , x , · · · , x n } = S ℓj =1 X j , and a signature f j on X j such that f ( X ) = Q ℓj =1 f j ( X j ), and for all 1 ≤ j ≤ ℓ, the support of f j is contained in { α j , ¯ α j } for some α j ∈ { , } | X j | .The proof of the following lemma totally follows the the proof of Lemma 4.8 and Lemma 4.9 of[21]. We just generalize it from k = 2 to general k . Lemma 5.22. If f * P but f k ∈ P , then for any d , in k ( F ) , we can construct h , h , · · · , h d such that h = h h · · · h d / ∈ P , i.e., k ( h, F ) ≤ T k ( F ) . Proof. Just replacing 2 by k in the proof of Lemma 4.8 and Lemma 4.9 of [21], we can constructa rank-2 signature g from f in k ( F ) such that g has the support ( x ) k ( x ) k and g =(1 , a, b, − ab ) up to a nonzero scalar, where ab = 0. By pinning all the x = 0, we get a signature u has the form ( x ) k and u = (1 , a ), and by pinning all the x = 0, we get a signature u hasthe form ( x ) k and u = (1 , b ). Let u = u ⊗ u . For any d , let h = g, h = · · · = h k = u , then h = h h · · · h d has the compressed signature (1 , a k , b k , − a k b k ) that is not in P .Since F * P , by Lemma 5.22 and Lemma 5.21, we have h, = , F k ) ≤ k ( = , F ) , where h / ∈ P . If F k * A , then h, = , F k ) is k ( = , F ) is F k ⊆ A . Lemma 5.23. [13] If a signature f has affine support and f ( x , x , · · · , x n ) is a power of ρ forany ( x x · · · x n ) in the support of f , then there exists multilinear polynomial Q ( x i , x i , · · · , x i r ) such that f ( x , x , · · · , x n ) = ρ Q ( x i ,x i , ··· ,x ir ) , where { x i , x i , · · · , x i r } is a set of free variables. F k ⊆ A , for any signature f ∈ F of rank r , without loss of generality, assume that { x , x , · · · , x r } is a set of free variables of f , by Lemma 5.23, there exists a multilinear polynomial Q ( x , x , · · · , x r ) ∈ Z [ X ] such that f ( x , x , · · · , x r ) = ρ Q ( x ,x , ··· ,x r ) up to a scalar. Definition 5.24. Suppose f has affine support of rank r with { x , x , · · · , x r } as a set of freevariables. We use all non-empty combinations P rj =1 a j x j ( a j ∈ Z , not all zero ) of x , x , · · · , x r asthe names of bundles of f . The type of each bundle is a possibly empty multiset of “ + ” and “ − ” ,and is defined as follows: For every input variable x k (1 ≤ k ≤ n ) of f there is a unique bundlenamed P rj =1 a j x j such that on the support of f , x k is either always equal to P rj =1 a j x j (mod 2) or always equal to P rj =1 a j x j + 1 (mod 2) . In the former case we add a “ + ” , and in the lattercase we add a “ − ” to the bundle type for the bundle named P rj =1 a j x j , and we say the variable x k belong to this bundle.All input variables are partitioned into bundles. If the number of variables in each bundle ismultiple ℓ for some integer ℓ , then we say f has ℓ -type support. We can list a signature’s input variables, by listing all its non-empty bundles followed by thebundle type. For example, f ( x (++) , x (++) , ( x + x )( −− )) has rank 2, arity 6 and its supportis 2-type.Connecting one variable x i of a signature to (= k ) using ( = ), is equivalent to replace x i by( k − 1) copies of ¯ x i . We call this operation is ( k − k , by connecting d copies of them to = k , we make these k variables disappear. Wecall this operation is collation. Definition 5.25. Let S be a subset of [ k ] . If f ∈ A dk iff k ∈ S , then we call f is an S -affinesignature. In particular, if S is exactly all the even (odd) numbers of [ k ] , then we call f is aneven-affine (odd-affine) signature. Lemma 5.26. Let f be a signature of affine support. By doing ( k − -multiple and collation to f we get a new signature f ′ , then f ∈ P iff g ∈ P ; f ∈ A dk iff g ∈ A dk for any d ∈ [ k ] . ✟✟✟✟✟✟❍❍❍❍❍❍ (cid:4) x ¯ x ¯ x ¯ x ¯ x N ❍❍❍❍❍❍ ✟✟✟✟✟✟ • ...... Fig. 2 Transforming the variable x to k − x by connecting (= k ) using ( = ).The triangle, square and bullet are labeled by f , ( = ) and (= k ) respectively. Proof. We prove the lemma for ( k − f ∈ A dk . Other cases aresimilar and we omit them here. Note that ( = ) ∈ A dk and (= k ) ∈ A dk for each d ∈ [ k ]. Thus if f ∈ A dk for some d , then g ∈ A dk by the closure of A . Conversely, note that the variable x of f which is connected to (= k ) using ( = ) is flipped into ( k − 1) copies of ¯ x in g . By connecting these( k − 1) copies of ¯ x to (= k ) using ( k − 1) copies of ( = ), we get f ′ that is same as f . Thus if g ∈ A dk ,then f ′ ∈ A dk and so does f .By Lemma 5.26, we can flip the variable ¯ x to x in the support of f at the same time. Thus wecan assume that there is no ¯ x . 34 efinition 5.27. Let f be an n -ary signature which has support of rank r . If there exists a set offree variables, without loss of generality, assume that it is { x , x , · · · , x r } , such that all the bundleshave type “+”, i.e., f ( x (+ + · · · + | {z } n ) x (+ + · · · + | {z } n ) · · · ( x + x + · · · + x r )(+ + · · · + | {z } n ··· r )) , where n + n + · · · + n ··· r = n , then we call f is monotone. And we denote its support by ( x ) n ( x ) n · · · ( x r ) n r ( x + x ) n · · · ( x + x + · · · + x r ) n ··· r . Lemma 5.28. If f / ∈ A dk for some d ∈ [ k ] , then we have k ( = , g, F ) ≤ T k ( = , f, F ) , where g is monotone, g * A dk and has the same rank as f .Proof. Without loss of generality, assume that { x , x , · · · , x r } is a set of free variables of f . Ifthere exists variable x i such that x i = a x + a x + · · · + a r x r + 1 in the support of f , then a( k − x i , we get a new signature g which is not in A dk by Lemma 5.26. Moreover, g has affine support and { x , x , · · · , x r } is a set of free variables and the variable x i is flipped into k − x i = a x + a x + · · · + a r x r .By Lemma 5.28, in the following we can assume that all the signatures are monotone. Thefollowing lemma will simplify the signatures further by reducing ranks. Lemma 5.29. Let f be a signature and F be a signature set. If f / ∈ A dk for some d ∈ [ k ] , then wehave k ( g, F , [1 , , [0 , ≤ T k ( f, F , [1 , , [0 , , where g / ∈ A dk and g has rank at most 3.Proof. Without loss of generality, assume { x , x , · · · , x r } is a set of free variables of f , and f ( x , x , · · · , x r ) = ρ Q ( x ,x , ··· ,x r ) by Lemma 5.23 up to a scalar, where Q ( x , x , · · · , x r ) is a mul-tilinear polynomial. By the holographic transformation using T dk = h ρ k i , we haveHolant( E dk | b f , b F , [1 , , [0 , ≡ T k ( f, F , [1 , , [0 , , where b f = ( T dk ) ⊗ arity ( f ) f , b F = T dk F , and E dk = E Q k ( T dk ) − . Note that b f has the same support as f and the ratio of f ( x , x , · · · , x n ) and b f ( x , x , · · · , x n ) is a power of ρ for any ( x x · · · x n )in the support of f . Thus there exists a multilinear polynomial b Q ( x , x , · · · , x r ) such that b f ( x , x , · · · , x r ) = ρ b Q ( x ,x , ··· ,x r ) . Assume that Q ( x , x , · · · , x r ) = X ≤ i ≤ r a i x i + X ≤ j,ℓ ≤ r a jℓ x j x ℓ + P ( x , x , · · · , x r ) , (5.7)where P ( x , x , · · · , x r ) is a polynomial and all the terms have power at least 3. In (5.7), • If there exists a i k ), we pin all free variables to 0 by [1 , 0] except x i , then we get arank-1 signature that is not in A . 35 If there exists a jℓ k ), we pin all free variables to 0 by [1 , 0] except x j , x ℓ , then weget a rank-2 signature that is not in A . • Finally, if P ( x , x , · · · , x r ) k ), suppose the monomial M has the minimum degree,among all monomials in P whose coefficient that is nonzero modulo 4 k . We pin all freevariables which are not in M to 0 by [1 , 0] and pin the variables in M to 1 by [0 , 1] except 3of them, then we get a rank-3 function that is not in A .If a i ≡ k ), a jℓ ≡ k ) for all a i and a jℓ , and P ( x , x , · · · , x r ) ≡ k ), then b f ∈ A . This is a contradiction. In total, we always can get a non-affine signature of degree at most3 in Holant( E dk | b f , b F , [1 , , [0 , . This means we can get a signature g / ∈ A dk of degree at most 3 in k ( f, F , [1 , , [0 , . This finishes the proof. Remark: Let f be a rank-3 signature with f ( x , x , x ) = i a x + a x + a x + a x x + a x x + a x x + a x x x and the support ( x ) k ( x ) k ( x ) k ( x + x ) k ( x + x ) k ( x + x ) k ( x + x + x ) k , and g bea signature with g ( y , y ) = i b y + b y + a x + b y y and the support ( y ) k ′ ( y ) k ′ ( y + y ) k ′ . Assume that k + k ′ < ℓk, k + k ′ < ℓk, k + k ′ < ℓk for some integer ℓ . By connecting the variable bundles ( y ) , ( y ) , ( y + y ) of g to the variablesbundles ( x + x ) , ( x + x ) , ( x + x ) of f by (= ℓk ) to get a new signature f ′ . Note that f ′ ( x , x , x ) = f ( x , x , x ) g ( x + x , x + x , x + x ) , i.e., f ′ ( x , x , x ) = i a x + a x + a x + a x x + a x x + a x x + a x x x + b ( x + x )+ b ( x + x )+ b ( x + x )( x + x ) . Let f ′ ( x , x , x ) = i c x + c x + c x + c x x + c x x + c x x + c x x x . We emphasis that a = c . In the following, we will connect the variables of g to f using (= ℓk ) in different ways. We emphasisthat the coefficient of x x x will not be changed.The proof of the following lemma follows the proof of Lemma 4.17 of [21]. We just generalizeit from to k . Lemma 5.30. Let f be a rank-3 signature which has the support ( x ) k ( x ) k ( x ) k ( x + x ) k ( x + x ) k ( x + x ) k ( x + x + x ) k , and the compressed signature f ( x , x , x ) = i a x + a x + a x +2 a x x +2 a x x +2 a x x +2 a , where a is odd, and g be a rank-2 signature that has the support ( y ) k ( y ) k ( y + y ) k , nd the compressed signature g ( y , y ) = i b y + b y +2 b y y . Then k ( f ′ ) ≤ k ( f, g ) , where f ′ / ∈ A is a rank-3 signature and has k -type support.Proof. We take one copy of f (( x ) k ( x ) k ( x ) k ( x + x ) k ( x + x ) k ( x + x ) k ( x + x + x ) k ) andthree copies of g : g (( u ) k ( u ) k ( u + u ) k ), g (( v ) k ( v ) k ( v + v ) k ), g (( w ) k ( w ) k ( w + w ) k ).We connect the variables x x · · · x | {z } k of f and u u · · · u | {z } k of g by (= k ) as the following figuresshows. (cid:4) ❅❅❅ ❍❍❍ (cid:0)(cid:0)(cid:0) ... x x N ❅❅❅ ❍❍❍ (cid:0)(cid:0)(cid:0) ... u u • (cid:0)(cid:0)(cid:0)✟✟✟❅❅❅ ......... Fig. 3 The square, triangle and bullet is labeled by f , g and (= k ) respectively.Similarly, we connect the variables x x · · · x | {z } k and v v · · · v | {z } k , x x · · · x | {z } k and w w · · · w | {z } k ,( x + x )( x + x ) · · · ( x + x ) | {z } k and u u · · · u | {z } k ,( x + x )( x + x ) · · · ( x + x ) | {z } k and v v · · · v | {z } k ,( x + x )( x + x ) · · · ( x + x ) | {z } k and w w · · · w | {z } k by (= k ) respectively. Then we get a signature f ′ which has the support( x ) k ( x ) k ( x ) k ( x + x ) k ( x + x ) k ( x + x ) k ( x + x + x ) k and f ′ ( x , x , x ) = i c x + c x + c x +2 c x x +2 c x x +2 c x x +2 c x x x for some c , c , c , c , c , c . But c = a by the remark. Since a is odd, f ′ is not affine andwe finish the proof. 37n the following proof, we often use the following obvious fact: If f has k -type support and f / ∈ A d k for some d ∈ [ k ], then f / ∈ A dk for any d ∈ [ k ]. Lemma 5.31. If F * A dk for any d ∈ [ k ] , then in k ( = , F ) we can construct [1 , , · · · , , a ] r ,where a = 0 and < r < k , or for some ≤ k ′ < k and k ′ | k , we can construct h such that h / ∈ A d ′ kk ′ for any d ′ ∈ [ kk ′ ] , and h = f f · · · f k ′ , where f , f , · · · , f k ′ can be constructed in k ( = , F ) .Proof. By Lemma 5.28, we can assume all the signatures in F are monotone. Moreover, we canassume that all the signature in F has rank less than 4 by Lemma 5.29. Then by Lemma 5.16 andLemma 5.17, the pinning signatures [1 , , [0 , 1] are available freely in k ( = , F ).Assume that f ∈ F and f / ∈ A dk .If f has rank-1, we can assume that f = [ x, , · · · , , y ] since f is monotone. If xy = 0,then f ∈ A dk for any d ∈ [ k ], this is a contradiction. Thus up to a scalar, we can assume that f = [1 , , · · · , , a ] with a = 0. Let r be the arity of f . If r k ), then we can assume that r = r + ℓ k , where 0 < r < k . After collations we get [1 , , · · · , , a ] of arity r and finish theproof. If r ≡ k ), we can assume that f = [1 , , · · · , , a ] ℓk . Since f has k -type support and f / ∈ A dk for some d , we have f / ∈ A . Then we are done by letting h = f .If f has rank 2, we can assume that f has the support ( x ) k ( x ) k ( x + x ) k since f is mono-tone. By pinning x = 0, we have a rank-1 signature f whose support has the form ( x ) k + k , i.e., f = [ f (0 , , · · · , , , · · · , , f (1 , , · · · , f (0 , , · · · , 0) = f (0 , 0) and f (1 , , · · · , 1) = f (0 , 1) are both nonzero. Thus up to a scalar, we can assume that f = [1 , , · · · , , a ] with a = 0.Moreover, by the proof for the rank-1 case, if k + k k ), then we are done. Thus k + k ≡ k ) . (5.8)Then by pinning x = 0 and x + x = 0, we have k + k ≡ k ) ,k + k ≡ k ) . (5.9)By (5.8) and the first equation of (5.9), we have k ≡ k (mod k ). Then by the second equation,we have 2 k ≡ k ) . This implies that k ≡ k ) or k ≡ k (mod k ). So we have k ≡ k ≡ k ≡ k ) , or k ≡ k ≡ k ≡ k k ) . If k ≡ k ≡ k ≡ k ), then we are done since the support of f has k -type and f / ∈ A .Now we can assume that the support of f has the form ( x ) k ( x ) k ( x + x ) k after collations. If f / ∈ A , then we are done by letting h = f . Otherwise, we can assume that f ( x , x ) = α b x + b x +2 b x x up to a scalar. By pinning x = 0 to f , we get [1 , , · · · , , α b ] k . If b , , · · · , , α b ] / ∈ A and has k -type. Thus we finish the proof. So we have b ≡ x = 0 , x + x = 0, we have b ≡ b ≡ f ( x , x ) = i b x + b x + b x x . b ≡ f is even-type affine, and b ≡ f is odd-typeaffine. Claim : There is no rank-1 signature in F which is not in A dk for some d ∈ [ k ]. If thereexists a rank-2 signature f ∈ F such that f / ∈ A dk for some d ∈ [ k ], then f has the support( x ) k ( x ) k ( x + x ) k with k ≡ k ≡ k ≡ k (mod k ) and f ( x , x ) = α b x + b x +2 b x x , where b ≡ b ≡ . If f has rank-3, then f has the support( x ) k ( x ) k ( x ) k ( x + x ) k ( x + x ) k ( x + x ) k ( x + x + x ) k since f is monotone. By pinning x = 0 , x = 0 , x = 0 to f , we get three signatures which havethe support ( x ) k + k ( x ) k + k ( x + x ) k + k , ( x ) k + k ( x ) k + k ( x + x ) k + k , ( x ) k + k ( x ) k + k ( x + x ) k + k respectively. By the Claim , we have k + k ≡ k + k ≡ k + k ≡ k ,k + k ≡ k + k ≡ k + k ≡ k ,k + k ≡ k + k ≡ k + k ≡ k . (5.10)and k + k ≡ k + k ≡ k + k (mod k ) ,k + k ≡ k + k ≡ k + k (mod k ) ,k + k ≡ k + k ≡ k + k (mod k ) . (5.11)By (5.10) we have k ≡ k ≡ k ≡ k ≡ − k ≡ − k ≡ − k (mod k . Moreover, by pinning x + x = 0, we get the signature which has the support( x ) k + k ( x ) k + k ( x + x ) k + k . By the Claim , k + k ≡ k + k ≡ k + k ≡ k . This implies that 2 k ≡ k ≡ k . 39o after collations, by (5.11), the support of f has to be one of the following forms:( x ) ǫ k ( x ) ǫ k ( x ) ǫ k ( x + x ) ǫ k ( x + x ) ǫ k ( x + x ) ǫ k ( x + x + x ) ǫ k , (5.12)( x ) k ( x ) k ( x ) k ( x + x ) k ( x + x ) k ( x + x ) k ( x + x + x ) k , (5.13)( x ) ǫ k ( x ) ǫ k ( x ) ǫ k ( x + x ) k ( x + x ) k ( x + x ) k ( x + x + x ) ǫ k , (5.14)( x ) k ( x ) k ( x ) k ( x + x ) ǫ k ( x + x ) ǫ k ( x + x ) ǫ k ( x + x + x ) k , (5.15)( x ) k ( x ) k ( x ) k ( x + x ) k ( x + x ) k ( x + x ) k ( x + x + x ) k , (5.16)( x ) k ( x ) k ( x ) k ( x + x ) k ( x + x ) k ( x + x ) k ( x + x + x ) k , (5.17)( x ) k ( x ) k ( x ) k ( x + x ) k ( x + x ) k ( x + x ) k ( x + x + x ) k , (5.18)( x ) k ( x ) k ( x ) k ( x + x ) k ( x + x ) k ( x + x ) k ( x + x + x ) k , (5.19)where ǫ i = 0 or 1.Note that except for the cases (5.12) and (5.13), we always have a rank-2 signature g by pinning x = 0 to f which has support ( y ) k ( y ) k ( y + y ) k after collations. If f has the support as (5.12),then we are done by letting h = f since it is k -type.In the following, we firstly assume all the signatures in F have the form (5.13). Otherwise,by the Claim , we can assume that there is a rank-2 signature g ∈ F which has the support( y ) k ( y ) k ( y + y ) k .Firstly, assume that all the signatures in F have the support (5.13). If f has the support (5.13)and f / ∈ A , then we are done by letting h = f . Otherwise, we can assume that f ( x , x , x ) = α a x + a x + a x +2 a x x +2 a x x +2 a x x +4 a x x x . By pinning x = 0, we have a rank-2 signature g with the support ( x ) k ( x ) k ( x + x ) k and g ( x , x ) = α a x + a x + a x +2 a x x . If g / ∈ A , then we are done by letting h = g . Otherwise, a ≡ a ≡ a ≡ x = 0 , x = 0, we have a ≡ a ≡ a ≡ a ≡ a ≡ a ≡ f ( x , x , x ) = i a x + a x + a x +2 a x x +2 a x x +2 a x x +2 a x x x . Let b f = h ρ d i ⊗ k f , then b f ( x , x , x ) = ρ dk (4 x +4 x +4 x +4 x x +4 x x +4 x x +4 x x x ) f ( x , x , x ) , i.e., b f ( x , x , x ) = i (2 d + a ) x +(2 d + a ) x +(2 d + a ) x +2( d + a ) x x +2( d + a ) x x +2( d + a ) x x +2( d + a ) x x x . So f is even-type if a is even and f is odd type if a is odd. Since F * A dk for any k ∈ [ k ],there at least two signatures f , f ∈ F and f i ( x , x , x ) = i a ( i )1 x + a ( i )2 x + a ( i )3 x +2 a ( i )12 x x +2 a ( i )13 x x +2 a ( i )23 x x +2 a ( i )123 x x x i = 1 , 2, where a (1)123 is odd and a (2)123 is even. Let h = f f , then h is k -type and h ( x , x , x ) = i ( a (1)1 + a (2)1 ) x +( a (1)2 + a (2)2 ) x +( a (1)3 + a (2)3 ) x +2( a (1)12 + a (2)12 ) x x +2( a (1)13 + a (2)13 ) x x +2( a (1)23 + a (2)23 ) x x +2( a (1)123 + a (2)123 ) x x x . Since a (1)123 + a (2)123 is odd, h / ∈ A . Moreover, h has k -type and we are done.Now we assume that there exists a rank-2 signature g ∈ F . By the Claim , g is k -type and g ( y , y ) = i b x + b x + b x x , where b ≡ b ≡ . Depending on whether b is even or odd, g is even-affine or odd-affine.We assume that b is even. After the holographic transformation using (cid:2) ρ (cid:3) , the following proofcan work for the case that b is odd and we omit it here. Since g is even-affine, for each even d ∈ [ k ], there exists f ∈ F such that f / ∈ A dk since F * A dk for any d ∈ [ k ]. Let F even = { f ∈ F | f / ∈ A dk for some even d ∈ [ k ] } .If there exists another rank-2 signature g ′ ∈ F even , by the Claim , g ′ is k -type and g ′ = i b ′ x + b ′ x + b ′ x x , where b ′ is odd. Then we are done by letting h = gg ′ . • If there exists a rank-3 signature f ∈ F even of the type (5.12), then we are done by letting h = f . If there exists a rank-3 signature f ∈ F even of the type (5.13), then we are done byLemma 5.30 since we have f and g in hand. • Assume that there is a signature f ∈ F even that has the support (5.14). If f / ∈ A , then weare done by letting h = f . Otherwise, we can assume that f ( x , x , x ) = α a x + a x + a x +2 a x x +2 a x x +2 a x x +4 a x x x . By pinning x = 0, we get a rank-2 signature g ′′ . After collations, g ′′ has the support( x ) k ( x ) k ( x + x ) k and g ′′ ( x , x ) = α a x + a x +2 a x x . If g ′′ / ∈ A , then we are done by letting h = gg ′ . Otherwise, we have a ≡ a ≡ a ≡ x = 0 , x = 0, we have a ≡ a ≡ a ≡ a ≡ a ≡ a ≡ f ( x , x , x ) = i a x + a x + a x +2 a x x +2 a x x +2 a x x +2 a x x x . Let b f = h ρ d i ⊗ k f , then b f = ρ dk (6 x +6 x +6 x +6 x x +6 x x +6 x x +8 x x x ) f ( x , x , x ) , i.e., b f = i (3 d + a ) x +(3 d + a ) x +(3 d + a ) x +(3 d +2 a ) x x +(3 d +2 a ) x x +(3 d +2 a ) x x +(4 d +2 a ) x x x . f / ∈ A dk for some even d , a is odd. By connecting the k copies of variables y , y , y + y of g to the k copies of variables of x + x , x + x , x + x of f using (= k ), we get a k -typesignature f ′ and f ′ = i a ′ x + a ′ x + a ′ x +2 a ′ x x +2 a ′ x x +2 a ′ x x +2 a x x x . Since a is odd, we have f ′ / ∈ A . Then we are done by letting h = f ′ . • Assume that there is a signature f ∈ F even which has the support (5.15). By the samestatement as the case (5.14), we can assume that f ( x , x , x ) = i a x + a x + a x +2 a x x +2 a x x +2 a x x +2 a x x x . Let b f = h ρ d i ⊗ k f , then b f = ρ dk (6 x +6 x +6 x +6 x x +6 x x +6 x x +4 x x x ) f ( x , x , x ) , i.e., b f = i (3 d + a ) x +(3 d + a ) x +(3 d + a ) x +(3 d +2 a ) x x +(3 d +2 a ) x x +(3 d +2 a ) x x +(2 d +2 a ) x x x . Since f / ∈ A dk for some even k , a is odd. By connecting the k copies of variables y + , y , y + y of g to the k copies of variables of x + x , x + x , x + x of f using (= k ), we get a k -type signature f ′′ which has the support( x ) k ( x ) k ( x ) k ( x + x ) k ( x + x ) k ( x + x ) k ( x + x + x ) k after collations and f ′′ = i a ′ x + a ′ x + a ′ x +2 a ′ x x +2 a ′ x x +2 a ′ x x +2 a x x x . Note that a is odd. Then we are done by Lemma 5.30 since we have g and f ′′ in hand..Now we can assume that there are no signatures that have the support (5.12), (5.13), (5.15) or(5.14) in F even , i.e., all the signatures in F even have the supports (5.16), (5.18), (5.17) or (5.19).For f ∈ F even , if f / ∈ A , then we are done by letting h = f . Otherwise, we have f ( x , x , x ) = β a x + a x + a x +2 a x x +2 a x x +2 a x x x +4 a x x x . Moreover, by pinning x = 0, we get a rank-2 signature u that has the support( x ) k ( x ) k ( x + x ) k and u = β a x + a x +2 a x x . If one of a , a , a is nonzero modulo 4, then we are done by letting h = gg ′ . Otherwise, we have a ≡ a ≡ a ≡ x = 0 , x = 0, we have a ≡ a ≡ a ≡ a ≡ a ≡ a ≡ f ( x , x , x ) = i a x + a x + a x +2 a x x +2 a x x +2 a x x x + a x x x . Note that 42 if a is odd, then f / ∈ A dk for any even d ∈ [ k ]; • if a ≡ f / ∈ A dk for d ≡ f ∈ A dk for d ≡ • if a ≡ f / ∈ A dk for d ≡ f ∈ A dk for d ≡ a is odd. Take two copies of f with free variable set { x , x , x } and { y , y , y } . By connecting the bundles ( x ) and ( y ), ( x ) and ( y ), ( x ) and ( y ), ( x + x ) and( y + y ), ( x + x ) and ( y + y ), ( x + x ) and ( y + y ), ( x + x + x ) and ( y + y + y ) by(= k ) respectively, we get a k -type signature f ′′′ and f ′′′ ( x , x , x ) = i a x +2 a x +2 a x +4 a x x +4 a x x +4 a x x x +2 a x x x . Note that a is odd. Then by Lemma 5.30, we are done since we have g and f ′′′ in hand.If a ≡ F * A dk for any k ∈ [ k ], there exists another signature f ∗ ∈ F even such that f ∗ is k -type and f ∗ ( x , x , x ) = i a ∗ x + a ∗ x + a ∗ x +2 a ∗ x x +2 a ∗ x x +2 a ∗ x x x + a ∗ x x x , where a ∗ ≡ f with free variables { x , x , x } and one copy of f ∗ with free variables { y , y , y } . By connecting the bundles ( x ) and ( y ), ( x ) and ( y ), ( x ) and( y ), ( x + x ) and ( y + y ), ( x + x ) and ( y + y ), ( x + x ) and ( y + y ), ( x + x + x ) and( y + y + y ) by (= k ) respectively, we get a k -type signature f ′′′′ and f ′′′′ ( x , x , x ) = i c x + c x + c x +2 c x x +2 c x x +2 c x x x + c x x x , where c i = a i + a ∗ i , c jk = a jk + a ∗ jk for 1 ≤ i ≤ , ≤ j < k ≤ c = a + a ∗ . Note that c ≡ g and f ′′′′ in hand.Now we can prove Theorem 5.3. Proof. We will prove the theorem by induction on k . Note that the theorem has been proved forthe cases k = 1 , k ≥ . If F ⊆ P or F ⊆ A dk for some k ∈ [ k ], the tractability is obvious. Then we assume that F * P and F * A dk for any d ∈ [ k ].For any d ′ ∈ [ k ], by Lemma 5.22 and F * P , we can construct g / ∈ P and there exist g , g , · · · , g k ′ such that g = g g · · · g k ′ , where the signatures g , g , · · · , g k ′ can be constructed in k ( = , F ), i.e., k ( = , g ) ≤ T k ( = , F ) . (5.20)By F * A dk and Lemma 5.31, we can construct general equality [1 , , · · · , , a ] of arity r < k ,or we can construct h such that for some k ′ | k , k ′ > h / ∈ A d ′ kk ′ for any d ′ ∈ [ kk ′ ], and there exist f , f , · · · , f k ′ such that h = f , f , · · · , f k ′ .If we have [1 , , · · · , , a ] of arity r < k , then we are done by Lemma 5.18. 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