GEOMETRY OF FEASIBLE SPACES OF TENSORS. A Dissertation YANG QI

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1 GEOMETRY OF FEASIBLE SPACES OF TENSORS A Dissertation by YANG QI Submitted to the Office of Graduate Studies of Texas A&M University in partial fulfillment of the requirements for the degree of DOCTOR OF PHILOSOPHY Approved by: Chair of Committee, Committee Members, Department Head, J.M. Landsberg Frank Sottile Paulo Lima-Filho Colleen Robles Katrin Becker Emil Straube August 2013 Major Subject: Mathematics Copyright 2013 Yang Qi

2 ABSTRACT Due to the exponential growth of the dimension of the space of tensors V 1 V n, any naive method of representing these tensors is intractable on a computer. In practice, we consider feasible subspaces (subvarieties) which are defined to reduce the storage cost and the computational complexity. In this thesis, we study two such types of subvarieties: the third secant variety of the product of n projective spaces, and tensor network states. For the third secant variety of the product of n projective spaces, we determine set-theoretic defining equations, and give an upper bound of the degrees of these equations. For tensor network states, we answer a question of L. Grasedyck that arose in quantum information theory, showing that the limit of tensors in a space of tensor network states need not be a tensor network state. We also give geometric descriptions of spaces of tensor networks states corresponding to trees and loops. ii

3 ACKNOWLEDGEMENTS I would like to thank my thesis advisor, J.M. Landsberg, for bringing these problems to my attention and very inspiring conversations. I am indebted for his constant support and guidance, both from a mathematical and human perspective. I would like to thank my committee members, Dr. Colleen Robles, Dr. Frank Sottile, Dr. Paulo Lima-Filho, and Dr. Katrin Becker for the guidance and support throughout the course of this research. I would like to thank the department faculty and staff, as well as my colleagues and friends for their support and help. iii

4 TABLE OF CONTENTS Page ABSTRACT ii ACKNOWLEDGEMENTS iii TABLE OF CONTENTS iv 1. INTRODUCTION AND BACKGROUND Motivation Equations for the secant varieties of Segre varieties Tensor network states PRELIMINARIES Dimensions of Secant Varieties of Segre Varieties Subspace Varieties Strassen s Equations Inheritance and Prolongation Normal forms of points in σ 3 (Seg(PA 1 PA n )) EQUATIONS FOR THE THIRD SECANT VARIETY OF AN N-FACTOR SEGRE VARIETY Outline of the proof of the main result Proof of the main theorem Case 1: T σ 3 (X 2 ) \ σ 2 (X 2 ), T / Sub 3,2,...,2 (A 1 A n ) Case 2: T σ 3 (X 2 ) \ σ 2 (X 2 ), T Sub 3,2,...,2 (A 1 A n ) \ Sub 2,2,...,2 (A 1 A n ) Case 3: T σ 3 (X 2 ) \ σ 2 (X 2 ), T Sub 2,2,...,2 (A 1 A n ) Case 4: T σ 2 (X 2 ) ON THE GEOMETRY OF TENSOR NETWORK STATES Definitions iv

5 4.2 Grasedyck s question Connections to the GCT program Critical loops Zariski closure Algebraic geometry perspective Reduction from the supercritical case to the critical case with the same graph Reduction of cases with subcritical vertices of valence one Trees SUMMARY REFERENCES v

6 1. INTRODUCTION AND BACKGROUND 1.1 Motivation Tensors are ubiquitous in mathematics and the sciences, and are especially important in algebraic statistics, biology, signal processing, and complexity theory [15, 16, 24, 29, 35, 37]. For example, in scientific computation the problem of determining the complexity of matrix multiplication can be viewed as decomposing a particular tensor (the matrix multiplication operator) according to its rank [28, 29]; in statistics, the problem of recovering the mixing matrix and source vector from the observation vector can be viewed as the symmetric tensor decomposition of the associated cumulants [25, 29, 36, 42]; in signal processing, CP decomposition, block term decomposition and other tensor decompositions are important [23, 27, 29]. In the study of tensors, the rank and border rank of a tensor are the standard measures of its complexity. Due to the geometric interpretations of rank and border rank, it is natural to study the secant varieties of Segre varieties since equations for these varieties produce tests for the border rank of a tensor. In practice, small secant varieties of Segre varieties play an important role as they correspond to tensors of low complexity. Another model defined to reduce the complexity of the spaces involved is tensor network states in quantum information theory. In this thesis we study both these models. 1.2 Equations for the secant varieties of Segre varieties The study of equations for secant varieties of Segre varieties is a classical problem in algebraic geometry, but these equations are still far from being understood [29]. Before exploring the known results of these equations, let us review the basic definitions of rank, border rank and secant varieties of Segre varieties. 1

7 Definition 1. A function f : A 1 A n C is multilinear if it is linear in each factor A l. The space of such multilinear functions is denoted by A 1 A n and called the tensor product of the vector spaces A 1,..., A n. Elements T A 1 A n are called tensors. Definition 2. Given β A 1,..., β n A n, define an element β 1 β n A 1 A n by β 1 β n (u 1,..., u n ) = β 1 (u 1 ) β n (u n ) for any u i A i. An element of A 1 A n is said to have rank one if it is of the form β 1 β n for some β i A i. The rank of a tensor T A 1 A n, denoted by R(T ), is the r minimum number r such that T = Z u with each Z u of rank one. u=1 Definition 3. A tensor T has border rank r, denoted by R(T ), if it is a limit of tensors of rank r but is not a limit of tensors of rank s for any s < r. Remark 1. Note that R(T ) R(T ). If T A 1 A 2 is a matrix, then R(T ) = R(T ). But this is not always true for T A 1 A n when n 3. For example, let T = a 1 b 1 c 1 + a 1 b 1 c 2 + a 1 b 2 c 1 + a 2 b 1 c 1 A B C. One can check 1 T has rank 3, but T = lim t 0 t [(t 1)a 1 b 1 c 1 + (a 1 + ta 2 ) (b 1 + tb 2 ) (c 1 + tc 2 )], hence R(T ) = 2. Definition 4. Define the n-factor Segre variety to be the image of the map Seg : PA 1 PA n P(A 1 A n ) ([v 1 ]..., [v n ]) [v 1 v n ] Remark 2. Seg(PA 1 PA 2 ) is the set of rank one matrices, and Seg(PA 1 PA n ) is the set of rank one tensors. 2

8 Definition 5. The join of two varieties Y, Z PV is J(Y, Z) = x Y,y Z,x y P 1 xy, where P 1 xy is the projective line through x and y. Definition 6. The join of k varieties X 1,..., X k PV is defined by induction to be J(X 1,..., X k ) = J(X 1, J(X 2,..., X k )), and the k-th secant variety of Y is defined to be the join of k copies of Y, σ k (Y ) = J(Y,..., Y ). Remark 3. σ k (Seg(PA 1 PA 2 )) is the set of matrices with rank at most k, and σ k (Seg(PA 1 PA n )) is the set of tensors with border rank at most k. It is clear that the ideal of Seg(PA 1 PA 2 ) is generated by all the 2 2 minors, denoted by 2 A 1 2 A 2, and the ideal of σ r (Seg(PA 1 PA 2 )) is generated by all the (r + 1) (r + 1) minors, denoted by r+1 A 1 r+1 A 2. Given W = A 1 A n, define a flattening A I A J of W to be a decomposition (A i1 A ip ) (A ip+1 A in ), where I = {i 1,..., i p } and J = {i p+1,..., i n }, I J = {1,..., n}, and I J =. Since Seg(PA 1 PA n ) can be embedded in Seg(PA I PA J ), then 2 A I 2 A J give equations for Seg(PA 1 PA n ). It turns out that Seg(PA 1 PA n ) is ideal theoretically defined by all the 2 2 minors of flattenings, i.e. all 2 A I 2 A J generate the ideal for Seg(PA 1 PA n ). Since σ r (Seg(PA 1 PA n )) can be embedded in σ r (Seg(PA I PA J )), r+1 A I r+1 A J give equations for σ r(seg(pa 1 PA n )). When studying Bayesian networks, Garcia, Stillman and Sturmfels conjectured that all the 3 3 minors of flattenings give all the equations for σ 2 (Seg(PA 1 PA n )) [22]. Landsberg and Manivel showed the set theoretic version of this conjecture is true [30], and Raicu proved the ideal theoretic version is true [44]. For more history, see [3, 30, 34, 44]. 3

9 It turns out that minors of flattenings are not enough to define higher secant varieties of Segre varieties. In 1983 Strassen discovered equations for σ 3 (Seg(PA 1 PA 2 PA 3 )) beyond 4 4 minors of flattenings [48]. Landsberg and Manivel proved σ 3 (Seg(PA 1 PA 2 PA 3 )) is set theoretically defined by Strassen s equations and 4 4 minors of flattenings [20, 31]. Landsberg and Weyman proved the ideal of σ 3 (Seg(PA 1 PA 2 PA 3 )) is generated in degree 4 by the module which arises from Strassen s commutation condition [34]. For the fourth secant varieties of Segre varieties, Friedland showed σ 4 (Seg(PA 1 PA 2 PA 3 )) is the zero set of certain equations of degree 5, 9 and 16 [20]. Bates and Oeding showed σ 4 (Seg(PA 1 PA 2 PA 3 )) is the zero set of certain equations of degree 5, 6 and 9 by numerical methods [4]. Friedland and Gross gave this result a computer-free proof [21]. For higher secant varieties of Segre varieties, for example σ 6 (P 3 P 3 P 3 ), there are no equations known. On the other hand, there are some qualitative descriptions of equations of secant varieties of Segre varieties. Draisma and Kuttler proved that for arbitrary fixed r, there is an uniform bound d(r) such that σ r (Seg(PA 1 PA n )) is set theoretically defined by equations of degree at most d(r) for any n [17]. In this thesis, we determine set theoretic equations for the third secant variety of the Segre product of n projective spaces, and from the proof of this statement we derive an upper bound for the degrees of these equations. Given any partition I J K = {1,..., n}, σ 3 (Seg(PA 1 PA n )) can be embedded in σ 3 (Seg(PA I PA J PA K )), thus Strassen s equations for all the partitions I J K = {1,..., n} and 4 4 minors for all the flattenings give us equations for σ 3 (Seg(PA 1 PA n )). Our main result is [43]: Theorem 1. σ 3 (Seg(PA 1 PA n )) is set theoretically defined by Strassen s 4

10 equations of all partitions I J K = {1,..., n} and all 4 4 minors of flattenings. Corollary 1. σ 3 (Seg(PA 1 PA n )) is set theoretically defined by Strassen s equations of degree 4 for the partitions {i} {j} {1,..., î,, ĵ,, n} and all 4 4 minors of flattenings. 1.3 Tensor network states Tensor network states are interesting models in physics defined to reduce the complexity of the spaces involved. In physics, tensors describe states of quantum mechanical systems. If a system has n particles, its state is an element of H 1 H n with H j Hilbert spaces. In numerical many-body physics, in particular solid state physics, one wants to simulate quantum states of thousands of particles, often arranged on a regular lattice (e.g., atoms in a crystal). Due to the exponential growth of the dimension of H 1 H n with n, any naive method of representing these tensors is intractable on a computer. Tensor network states were defined by restricting to a subset of tensors that is physically reasonable, in the sense that the corresponding spaces of tensors are only locally entangled because interactions (entanglement) in the physical world appear to just happen locally. These spaces are associated to graphs, i.e. for a fixed graph, we can associate complex vector spaces to each vertex and edge, and define a corresponding tensor network state. More precisely: Let V 1,..., V n be complex vector spaces, let v i = dim V i. Let Γ be a graph with n vertices v j, 1 j n, and m edges e s, 1 s m, and let e = (e 1,..., e m ) N m. Associate V j to the vertex v j and an auxiliary vector space E s of dimension e s to the edge e s. Make Γ into a directed graph. (The choice of directions will not effect the end result.) Let V = V 1 V n. For Γ, s e(j) means e s is incident to v j, s in(j) are the incoming edges and s out(j) the outgoing edges. 5

11 Define a tensor network state T NS(Γ, e, V) to be: T NS(Γ, e, V) := (1.1) {T V T j V j ( s in(j) E s ) ( t out(j) E t ), T = Con(T 1 T n )}, where Con is the contraction of all the E s s with all the Es s. Such spaces have been studied since the 1980 s, and go under different names: tensor network states, finitely correlated states (FCS), valencebond solids (VBS), matrix product states (MPS), projected entangled pairs states (PEPS), and multiscale entanglement renormalization ansatz states (MERA), see, e.g., [14, 18, 19, 26, 45, 49] and the references therein. We will use the term tensor network states. If Γ is a tree, then T NS(Γ, e, V) is closed [24]. Lars Grasedyck asked if every tensor network state is Zariski closed. In this thesis, we give a counterexample and show a tensor network state is not closed if the corresponding graph contains a cycle whose vertices have non-subcritical dimensions. We also give geometric descriptions of spaces of tensor networks states corresponding to trees and loops. Grasedyck s question has a surprising connection to the area of Geometric Complexity Theory, in that the result is equivalent to the statement that the boundary of the Mulmuley-Sohoni type variety associated to matrix multiplication is strictly larger than the projections of matrix multiplication (and re-expressions of matrix multiplication and its projections after changes of bases). Tensor Network States are also related to graphical models in algebraic statistics [29]. 6

12 2. PRELIMINARIES 2.1 Dimensions of Secant Varieties of Segre Varieties Terracini s lemma is a fundamental tool to compute the dimension of a join variety. Let Y, Z be projective varieties, and Ŷ, Ẑ be the cones over Y, Z. Lemma 1 (Terracini s lemma). Let (v, w) Ŷ Ẑ be a general point, and [u] = [v + w] J(Y, Z), then T [u] J(Y, Z) = T [v] Y + T [w] Z, where T [v] Y denotes the affine tangent space of Y at [v]. Definition 7. We call a variety X P n nondegenerate if it spans P n, i.e. is not contained in any hyperplane. If X P n is an irreducible nondegenerate variety whose r-th secant variety σ r (X) has dimension strictly less than min{r dim X +r 1, n}, we say that X is defective, and define the defect δ r (X) = r dim X + r 1 dim σ r (X). Here we list some known results on the dimensions of secant varieties of Segre varieties, for more results see [1, 8 11, 13]. Theorem 2 ( [12]). Consider σ r (Seg(P a 1 1 P an 1 )), and assume a n n 1 i=1 a i n 1 i=1 a i n If r n 1 i=1 a i n 1 i=1 a i n + 1, then σ r (Seg(P a 1 1 P an 1 )) has the expected dimension r(a a n n + 1) 1; 2. If a n > r n 1 i=1 a i n 1 i=1 a i n + 1, then σ r (Seg(P a 1 1 P an 1 )) has defect δ r = r 2 r( n 1 i=1 a i n 1 i=1 a i n + 1); 3. If r min{a 1,..., a n }, then σ r (Seg(P a 1 1 P an 1 )) = P n i=1 a i 1. 7

13 Theorem 3 ( [13]). The secant varieties of the Segre product of k copies of P 1, σ r (Seg(P 1 P 1 )), have the expected dimension except when k = 2, Subspace Varieties Subspace varieties are important auxiliary varieties in the study of equations for secant varieties. Definition 8. The subspace variety Sub b1,...,b n (A 1 A n ) is defined to be Sub b1,...,b n (A 1 A n ) := P{T A 1 A n dim(t (A j)) b j }. Proposition 1 ( [29]). The ideal of the subspace variety Sub b1,...,b n (A 1 A n ) is generated in degrees b j + 1 for 1 j n by the irreducible modules in b j+1 A j b j+1 (A 1 A j 1 A j+1 A n). The following Kempf-Weyman desingularization of Sub b1,...,b n (A 1 A n ) is useful for finding equations, minimal free resolutions, and establishing properties of singularities [29, 50]. Proposition 2 ( [50]). Consider the product of Grassmannians B = G(b 1, A 1 ) G(b n, A n ) and the bundle p : S 1 S n B, where S j is the tautological rank b j subspace bundle over G(b j, A j ). Assume that b 1 b n. Then the total space Z of S 1 S n maps to A 1 A n. The map Z A 1 A n gives a desingularization of Sub b1,...,b n (A 1 A n ). 8

14 2.3 Strassen s Equations In 1983 V. Strassen [48] discovered equations for tensors of bounded border rank beyond minors of flattenings. We present a version of Strassen s equations due to G. Ottaviani, which is easy to generalize to higher cases. Given T A B C, i.e. T : B A C, Id A T gives a linear map A B A A C, compose Id A T with the projection A A 2 A to define TBA : A B 2 A C. Theorem 4 ( [41]). Let T A B C, and assume 3 dim A dim B dim C. If [T ] σ r (Seg(PA PB PC)), then rank(tba ) r(dim A 1). Thus the size r(dim A 1) + 1 minors of T BA furnish equations for σ r(seg(pa PB PC)), which are called Strassen s equations. Proof. If T = a b c, then the image of TBA is a A c and thus rank(t BA ) = dim A 1 and the theorem follows because rank((t 1 + T 2 ) BA ) rank(t 1 BA)+rank(T 2 BA ) Theorem 5 ( [20,31]). σ 3 (Seg(PA PB PC)) is the zero set of the size 4 minors of flattenings and Strassen s equations. 2.4 Inheritance and Prolongation Inheritance is a general technique for studying equations of G-varieties. Proposition 3 ( [30]). For all vector spaces B j with dim B j = b j dim A j = a j r, a module S µ1 B1 S µn Bn such that l(µ j ) a j for all j, is in the ideal I d (σ r (Seg(PB 1 PB n ))) if and only if S µ1 A 1 S µn A n is in the ideal I d (σ r (Seg(PB 1 PB n ))). Corollary 2 ( [30]). Let dim A j r, 1 j n. The ideal of σ r (Seg(PA 1 PA n )) is generated by the modules inherited from the ideal of σ r (Seg(P r 1 9

15 P r 1 )) and the modules generating the ideal of Sub r,...,r (A 1 A n ). The analogous scheme and set theoretic results hold as well. According to this corollary, when studying these equations we only need consider the small dimensional cases. Prolongation is a general technique for finding equations of secant varieties. We list some basic facts about equations for secant varieties obtained by prolongation. Proposition 4 ( [29,47]). Let X, Y PV be subvarieties and assume that I δ (X) = 0 for δ < d 1 and I δ (Y ) = 0 for δ < d 2. Then I δ (J(X, Y )) = 0 for δ d 1 + d 2 2. Corollary 3 ( [29, 47]). Let X 1,..., X r PV be varieties such that I δ (X j ) = 0 for δ < d j. Then I δ (J(X 1,..., X r )) = 0 for δ d d r r. As a special case we have: Proposition 5 ( [29]). There are no nonzero degree d r homogeneous polynomials vanishing on σ r (Seg(PA 1 PA n )). 2.5 Normal forms of points in σ 3 (Seg(PA 1 PA n )) In this section we present how points of σ 3 (Seg(PA 1 PA n )) are explicitly parametrized. Proposition 6 ( [5]). Let X denote Seg(PA 1 PA n ) and p = [v] σ 2 (X), then v has one of the following normal forms: 1, p X; 2, v = x + y with [x], [y] X; 3, v = x with x T [x] X. Theorem 6 ( [5]). Let X denote Seg(PA 1 PA n ) and p = [v] σ 3 (X)\σ 2 (X), then v has one of the following normal forms: 10

16 1. v = x + y + z with [x], [y], [z] X; 2. v = x + x + y with [x], [y] X and x T [x] X; 3. v = x + x + x, where [x(t)] X is a curve and x = x (0), x = x (0); 4. v = x + y, where [x], [y] X are distinct points that lie on a line contained in X, x T [x] X, and y T [y] X. Normal forms for Theorem 6 are as follows: Theorem 7 ( [5]). Let X denote Seg(PA 1 PA n ) and p = [v] σ 3 (X)\σ 2 (X), then v has one of the following normal forms: 1. v = a 1 1 a n 1 + a 1 2 a n 2 + a 1 3 a n 3; n 2. v = a 1 1 a i 1 1 a i 2 a i+1 1 a n 1 + a 1 3 a n 3; i=1 3. v = a 1 1 a i 1 1 a i 2 a i+1 1 a j 1 1 a j 2 a j+1 1 a n 1 + i<j a i 1 1 a i 3 a i+1 1 a n 1; n n 4. v = a 1 2 a 2 1 a s 1 1 a s 2 a s+1 1 a n 1 + a i+1 s=2 1 a n 1, where a i j A i, and the vectors need not all be linearly independent. i=1 n a 1 1 i=1 a 1 1 a i 1 1 a i 3 11

17 3. EQUATIONS FOR THE THIRD SECANT VARIETY OF AN N-FACTOR SEGRE VARIETY 3.1 Outline of the proof of the main result Our main result on equations of the third secant varieties of Segre varieties is: Theorem 8. σ 3 (Seg(PA 1 PA n )) is set theoretically defined by Strassen s equations of all partitions I J K = {1,..., n} and all 4 4 minors of flattenings. Given T A 1 A n, for each A i we fix a basis {a i j} and its dual basis {α i j}. Let X k := Seg(PA 1 PA k P(A k+1 A n )), and X := Seg(PA 1 PA n ). Outline of the proof of the main result. If T A 1 A n satisfies all the equations given by 4 4 minors of flattenings, we may assume that 3 dim A 1 dim A n 2 [30]. If T satisfies Strassen s equations of the partition {1} {2} {3,..., n}, then T σ 3 (X 2 ). We split our discussion into 4 cases to show T σ 3 (X). Case 1: T σ 3 (X 2 ) \ σ 2 (X 2 ) and T / Sub 3,2,...,2 (A 1 A n ), then T has one of the four types of the normal forms in Theorem 7 for σ 3 (X 2 ). Because 4 4 minors of T : A 1 A 3 A 2 A 4 A n vanish, T has to have the same type of normal form for σ 3 (X 3 ). Similarly, by considering 4 4 minors of T : A 1 A k A 2 Âk A n we use induction to show that T has to maintain the same type of normal form for σ 3 (X). Case 2: T σ 3 (X 2 ) \ σ 2 (X 2 ) and T Sub 3,2,...,2 (A 1 A n ) \ Sub 2,2,...,2 (A 1 A n ), then T has one of the normal forms in Theorem 7 for σ 3 (X 2 ). Because dim A 2 = = dim A n = 2, the discussion of this case is more complicated than Case 1, and we split the argument into several subcases for each type of normal form. For each subcase, by considering 4 4 minors of T : A 1 A 3 A 2 A 4 A n 12

18 and T : A 2 A 3 A 1 A 4 A n, we show T has one of the normal forms for σ 3 (X 3 ). Note that the type of the normal form of T for σ 3 (X 2 ) could be different from the type of the normal form of T for σ 3 (X 3 ). By induction, we show that T has one of the normal forms of points in σ 3 (X). Case 3: T σ 3 (X 2 ) \ σ 2 (X 2 ) and T Sub 2,2,...,2 (A 1 A n ). In this case, T has two types of normal forms, T = (a 1 1 a 2 1 +a 1 2 a 2 2) b 3 1 +a 1 1 a 2 2 b 3 2 +a 1 2 a 2 1 b 3 3 or T = a 1 1 a 2 1 b a 1 1 a 2 2 b a 1 2 a 2 1 b 3 3 for some b 3 j A 3 A n. For the generic normal form T = (a 1 1 a a 1 2 a 2 2) b a 1 1 a 2 2 b a 1 2 a 2 1 b 3 3, we show that there is a rank 2 matrix φ 21 in the kernel of TA 2 A 1 : A 1 A 2 A 3 A n, and φ 21 (T ) S 2 A 1 (A 3 A n ). So if for each 2 i n, T has the generic type of normal form for σ 3 (Seg(PA 1 PA i P(A 2 Âi A n ))), then similarly we have a 2 2 matrix φ i1 Ker(TA i A 1 ) with full rank, and φ n1 φ 21 (T ) S n A 1. Since each φ i1 is nonsingular, T σ 3 (X) if and only if φ n1 φ 21 (T ) σ 3 (ν n (PA 1 )), where ν n is the n-th Veronese embedding. Since the equations for σ 3 (ν n (P 1 )) are known [33], we can check Strassen s equations and 4 4 minors of flattenings give equations for σ 3 (X) in this situation. If for some 2 i n, say i = 2, T does not have the generic normal form for σ 3 (X 2 ), T must have the other type of normal form T = a 1 1 a 2 1 b a 1 1 a 2 2 b a 1 2 a 2 1 b 3 3. By considering 4 4 minors of T : A 1 A 3 A 2 A 4 A n, T : A 2 A 3 A 1 A 4 A n, and T : A 1 A 2 A 3 A 4 A n, we deduce T σ 3 (X 3 ). Then we use induction to show T σ 3 (X) by checking each type of the normal forms in Theorem 7, under the assumption that T is not of the generic normal form for σ 3 (X 2 ). When proceeding by induction, because dim T (A 3 A n) 3 we can view T as a tensor in T (A 3 A n) A 3 A n and reduce most cases to Case 2. For the remaining cases, we show directly T σ 3 (X). Case 4: T σ 2 (X 2 ), then T has one of the three types of the normal forms 13

19 in Proposition 6 for σ 3 (X 2 ). We verify by induction that for each normal form T σ 3 (X). 3.2 Proof of the main theorem We only need to show that if T satisfies Strassen s equations of all partitions I J K = {1,..., n} and 4 4 minors of all flattenings I J = {1,..., n}, then T σ 3 (Seg(PA 1 PA n )). For each A i we fix a basis {a i j} and its dual basis {αj}. i Let X k := Seg(PA 1 PA k P(A k+1 A n )), and X := Seg(PA 1 PA n ). For any flattening I J = {1,..., n}, 4 4 minors of T : A I A J vanish if and only if dim T (A I ) 3. By Corollary 2, we can assume 3 dim A 1 dim A n 2. Since T satisfies Strassen s equations of the partition {1} {2} {3,..., n} and 4 4 minors of all flattenings, by Theorem 5 we have T σ 3 (X 2 ). We split our discussion into 4 cases to show T σ 3 (X) Case 1: T σ 3 (X 2 ) \ σ 2 (X 2 ), T / Sub 3,2,...,2 (A 1 A n ) Since T has one of the normal forms in Theorem 7, we use induction to show T σ 3 (X) by verifying each normal form. Type 1: Without loss of generality, let T = a 1 1 a 2 1 u 1 +a 1 2 a 2 2 u 2 +a 1 3 a 2 3 u 3, where u i A 3 A n. dim T (A 1 A 3) 3 implies that u i : A 3 A 4 A n has rank 1 for all i, say u i = b 3 i v i for some b 3 i A 3 and v i A 4 A n. Therefore T = a 1 1 a 2 1 b 3 1 v 1 + a 1 2 a 2 2 b 3 2 v 2 + a 1 3 a 2 3 b 3 3 v 3, i.e. T σ 3 (X 3 ). Now we use induction, assume T = a 1 1 a 2 1 b 3 1 b k 1 +a 1 2 a 2 2 b 3 2 b k 2 + a 1 3 a 2 3 b 3 3 b k 3, then dim T (A 1 A k ) 3 implies that bk i : A k A k+1 A n has rank 1 for all 1 i 3. Type 2: T = a 1 1 a 2 1 v2 3 + a 1 1 a 2 2 v1 3 + a 1 2 a 2 1 v1 3 + a 1 3 a 2 3 v3, 3 where vi 3 A 3 A n. Since T / σ 2 (X 2 ), v1 3 and v3 3 are non-zero. dim T (A 1 A 3) 3 14

20 implies v 3 1 and v 3 3 : A 3 A 4 A n have rank 1, say v 3 i = b 3 i v 4 i for i = 1, 3 and some b 3 i A 3, v 4 i A 4 A n, and for each j = 2, 3, a 2 1 v 3 2(α 3 j) + a 2 2 v 3 1(α 3 j) is a linear combination of a 2 1 v 3 2(α 3 1)+a 2 2 v 3 1(α 3 1) and a 2 1 v 3 1(α 3 1), then v 3 2 = b 3 1 v 4 2+b 3 2 v 4 1 for some b 3 2 A 3 and v 4 2 A 4 A n. Thus T = a 1 2 a 2 1 b 3 1 v a 1 1 a 2 1 b 3 1 v2 4 + a 1 1 a 2 1 b 3 2 v1 4 + a 1 1 a 2 2 b 3 1 v1 4 + a 1 3 a 2 3 b 3 3 v3. 4 k Now we use induction, and assume that T = b 1 1 b i 1 1 b i 2 b i+1 1 b k 1 + b 1 3 b k 3, where b i j = a i j for i = 1, 2 and 1 j 3. The induction argument is similar to the case k = 3 above. Type 3: T = a 1 1 a 2 2 v a 1 2 a 2 1 v a 1 2 a 2 2 v a 1 1 a 2 1 v a 1 1 a 2 3 v a 1 3 a 2 1 v 3 1, where v 3 i A 3 A n. If v 3 1 = 0, T has been discussed in Case 1 Type 1. If v 3 2 = 0, T has been discussed in Case 1 Type 2. So we assume v 3 1 and v 3 2 are non-zero. dim T (A 1 A 3) 3 implies v 3 1 = u 3 1 u 4 1, v 3 2 = u 3 1 u u 3 2 u 4 1 and v3 3 = u 3 1 u 4 3 +u 3 2 u 4 2 +u 3 3 u 4 1 for some u 3 1, u 3 2, u 3 3 A 3, and u 4 1, u 4 2, u 4 3 A 4 A n. Denote a i j by u i j when i = 1, 2, then T = u 1 1 u i 2 u j 2 u u 1 1 u i 3 u 4 1. i=1 1 i<j 4 The induction argument is similar the above argument. Type 4: T = a 1 2 a 2 1 v a 1 2 a 2 2 v a 1 1 a 2 1 v a 1 1 a 2 3 v a 1 3 a 2 1 v 3 1 for some v 3 j A 3 A n. Since T / σ 2 (X 2 ), v 3 1 0, then dim T (A 1 A 3) 3 implies v1 3 = u 3 1 u 4 1, v2 3 = u 3 1 u u 3 2 u 4 1, v3 3 = u 3 1 u u 3 3 u 4 1 for some u 3 j A 3, 4 u 4 j A 4 A n. Denote a i j by u i j for i = 1, 2, then T = u 1 2 u i 2 u u 1 1 u i 3 u 4 1. i=1 The induction argument is similar. i=1 i=2 15

21 3.2.2 Case 2: T σ 3 (X 2 ) \ σ 2 (X 2 ), T Sub 3,2,...,2 (A 1 A n ) \ Sub 2,2,...,2 (A 1 A n ) We show T σ 3 (X) by induction on each type of the normal forms. Type 1: T = a 1 1 b 2 1 b a 1 2 b 2 2 b a 1 3 b 2 3 b 3 3, where b 2 j A 2 and b 3 j A 3 A n. Without loss of generality, we can assume b 2 1 and b 2 2 are linearly independent, then b 2 3 = b 2 1 or b 2 3 = b b 2 2. If b 2 3 = a 2 1, since dim T (A 2 A 3) 3, then either b 3 2 : A 3 A 4 A n has rank 1, or both b 3 1 and b 3 3 have rank 1 as maps A 3 A 4 A n. When b 3 2 : A 3 A 4 A n has rank 1, let b 3 2 = a 3 2 b 4 2 for some b 4 2 A 4 A n. We only need to consider the case that at least one of b 3 1 and b 3 3 : A 3 A 4 A n has rank 2. Without loss of generality we can assume b 3 1 = u 3 1 b u 3 3 b 4 3 for some u 3 i A 3 and b 4 i A 4 A n where i = 1, 3, then dim T (A 1 A 3) 3 requires b 3 3(αj) 3 = x j b y j b 4 3 for some x j, y j, where j = 1, 2. Consider A 3 V 4, where V 4 is spanned by b 4 1 and b 4 3, after a change of basis, we can assume b 3 1 = u 3 1 b u 3 3 b 4 3 and b 3 3 = λu 3 1 b u 3 1 b λu 3 3 b 4 3, or b 3 3 = µu 3 1 b νu 3 3 b 4 3. Then T = T + a 1 2 b 2 2 a 3 2 b 4 2, where T = (a λa 1 3) b 2 1 u 3 1 b (a λa 1 3) b 2 1 u 3 3 b a 1 3 b 2 1 u 3 1 b 4 3 T (a 1 1 +λa 1 X 3, or 3 ) b2 1 u3 1 b4 3 T = (a µa 1 3) b 2 1 u 3 1 b (a νa 1 3) b 2 1 u 3 3 b a 1 2 b 2 2 a 3 2 b 4 2. When b 3 1 and b 3 3 : A 3 A 4 A n have rank 1, say b 3 1 = a 3 1 b 4 1 and b 3 3 = u 3 3 b 4 3 for some u 3 3 A 3 and b 4 i A 4 A n where i = 1, 3, and assume b 3 2 : A 3 A 4 A n has rank 2, dim T (A 2 A 3) 3 requires u 3 3 = a 3 1 up to a scalar, and dim T (A 1 A 3) 3 requires b 4 1 = b 4 3 up to a scalar, then T = (a a 1 3) b 2 1 a 3 1 b a 1 2 b 2 2 a 3 1 b 3 2(α1) 3 + a 1 2 b 2 2 a 3 2 b 3 2(α2). 3 If b 2 3 = b b 2 2, dim T (A 2 A 3) 3 implies b 3 1 or b 3 2 : A 3 A 4 A n has rank 1. If only one of them has rank 1, without loss of generality we assume that b 3 2 = a

22 u 4 1 +a 3 2 u 4 2, and b 3 1 = u 3 1 u 4 3. dim T (A 2 A 3) 3 implies b 3 3 = u 3 1 u 4 4 for some u 4 4 A 4 A n. dim T (A 1 A 3) 3 requires that u 4 3 and u 4 4 are linearly dependent, then we can assume u 4 4 = u 4 3. dim T (A 1 A 3) 3 also requires u 4 4 is a linear combination of u 4 1 and u 4 2. Consider A 3 V 4, where V 4 is the subspace of A 4 A n spanned by u 4 1 and u 4 2, after a change of basis, we can assume b 3 2 = a 3 1 u a 3 2 u 4 2 is still the identity matrix, and b 3 1 = b 3 3 = a 3 1 u 4 2 or a 3 1 u 4 1. Then T = (a 1 1+a 1 3) b 2 1 a 3 1 u 4 2+T, where T = a 1 2 b 2 2 a 3 1 u a 1 2 b 2 2 a 3 2 u a 1 3 b 2 2 a 3 1 u 4 2 T a 1 2 b 2 X 3, 2 a3 1 u4 2 or T = (a a 1 3) b 2 1 a 3 1 u (a a 1 3) b 2 2 a 3 1 u a 1 2 b 2 2 a 3 2 u 4 2. If both b 3 1 and b 3 2 have rank 1, let b 3 1 = a 3 1 u 4 1 and b 3 2 = u 3 2 u 4 2. If u 4 1 and u 4 2 are linearly independent, dim T (A 1 A 3) 3 implies b 3 3 : A 3 A 4 A n has rank 1. If u 4 1 and u 4 2 are dependent, say u 4 1 = u 4 2, and if u 3 2 = a 3 1 up to a scalar, since dim T (A 1 A 3) 3, then b 3 3(α1) 3 = xb 3 3(α2) 3 + yu 4 1 for some x, y. So T = (a 1 1+ya 1 3) b 2 1 a 3 1 u 4 1+(a 1 2+ya 1 3) b 2 2 a 3 1 u 4 1+a 1 3 (b 2 1+b 2 2) (xa 3 1+a 3 2) b 3 3(α2). 3 If u 3 2 and a 3 1 are independent, we can assume u 3 2 = a 3 2, since dim T (A 2 A 3) 3, then b 3 3 : A 3 A 4 A n has rank 1. Now we use induction. Assume T = a 1 1 b 2 1 b k 1 + a 1 2 b 2 2 b k 2 + a 1 3 b 2 3 b k 3, without loss of generality we can assume b 2 1 = a 2 1, b 2 2 = a 2 2, then b 2 3 = a 2 1 or b 2 3 = a a 2 2. The induction argument is similar to the case k = 3. Type 2: T = a 1 1 b 2 1 b a 1 1 b 2 2 b a 1 2 b 2 1 b a 1 3 b 2 3 b 3 3, without loss of generality we can assume b 2 1 = a 2 1 and b 2 2 = a 2 2, then b 2 3 = a 2 1, or b 2 3 = a λa 2 1 for some λ C. When b 2 3 = a 2 1, dim T (A 2 A 3) 3 forces b 3 1 : A 3 A 4 A n has rank 1, say b 3 1 = a 3 1 b 4 1. If b 3 3 : A 3 A 4 A n has rank 2, say b 3 3 = a 3 1 b a 3 2 b 4 3, then dim T (A 1 A 3) 3 requires that b 4 1 and b 3 2(α2) 3 are both in the subspace spanned by b 4 2 and b 4 3. After a change of basis, we can assume that b 3 3 = a 3 1 b a 3 2 b 4 3, and b 3 1 = a 3 1 b 4 2 or b 3 1 = a 3 1 b 4 3. We can assume b 3 2(α2) 3 = b λb 4 3 or b 3 2(α2) 3 = b 4 3. So we 17

23 have four cases: Case 1: If b 3 1 = a 3 1 b 4 3 and b 3 2(α2) 3 = b λb 4 3, T = a 1 1 a 2 1 a 3 1 b 3 2(α1) 3 + a 1 1 a 2 1 (λa 3 2) b 4 3+a 1 1 a 2 2 a 3 1 b 4 3+a 1 2 a 2 1 a 3 1 b 4 3+a 1 1 a 2 1 a 3 2 b 4 2+a 1 3 a 2 1 a 3 1 b 4 2+a 1 3 a 2 1 a 3 2 b 4 3. Let S(t) = (a 1 1+ta 1 3+t 2 a 1 2) (a 2 1+t 2 a 2 2) (a 3 1+ta 3 2+t 2 λa 3 2) (b 4 3+tb 4 2+t 2 b 3 2(α1)), 3 then T = S (0). Case 2: If b 3 1 = a 3 1 b 4 3 and b 3 2(α2) 3 = b 4 3, then T = T + T, where T = a 1 1 a 2 1 a 3 1 b 3 2(α1) 3 + a 1 1 a 2 1 a 3 2 b a 1 1 a 2 2 a 3 1 b a 1 2 a 2 1 a 3 1 b 4 3 T a 1 1 a 2 X 3, 1 a3 1 b4 3 and T = a 1 3 a 2 1 a 3 1 b a 1 3 a 2 1 a 3 2 b 4 3 T a 1 3 a 2 X 3. 1 a3 1 b4 3 Case 3: If b 3 1 = a 3 1 b 4 2 and b 3 2(α2) 3 = b 4 2+λb 4 3, T = T +(λa 1 1+a 1 3) a 2 1 a 3 2 b 4 3, where T = a 1 1 a 2 1 a 3 1 b 3 2(α1)+a a 2 1 a 3 2 b 4 2+a 1 1 a 2 2 a 3 1 b 4 2+(a 1 2+a 1 3) a 2 1 a 3 1 b 4 2 T a 1 1 a 2 X 3. 1 a3 1 b4 2 Case 4: If b 3 1 = a 3 1 b 4 2 and b 3 2(α2) 3 = b 4 3, then T = T +(a 1 1+a 1 3) a 2 1 a 3 2 b 4 3, where T = a 1 1 a 2 1 a 3 1 b 3 2(α1)+a a 2 2 a 3 1 b 4 2 +(a 1 2 +a 1 3) a 2 1 a 3 1 b 4 2 T a 1 1 a 2 X 3. 1 a3 1 b4 2 If b 3 3 : A 3 A 4 A n has rank 1, say b 3 3 = (xa ya 3 2) b 4 3, and b 4 1 and b 4 3 are linearly independent, dim T (A 1 A 3) 3 forces b 3 2(α2) 3 is a linear combination of b 4 1 and b 4 3. We can assume b 3 2(α2) 3 = b 4 1 or b 3 2(α2) 3 = b λb 4 1. If b 3 2(α2) 3 = b 4 1, T = T +a 1 3 a 2 1 (xa 3 1 +ya 3 2) b 4 3, where T = a 1 1 a 2 1 a 3 1 b 3 2(α1)+a a 2 1 a 3 2 b a 1 1 a 2 2 a 3 1 b 4 1+a 1 2 a 2 1 a 3 1 b 4 1 T a 1 1 a 2 X 3. If b 3 1 a3 1 b4 1 2(α2) 3 = b 4 3+λb 4 1, we can assume b 3 3 = a 3 2 b 4 3 or b 3 3 = (a 3 1+µa 3 2) b 4 3. If b 3 3 = a 3 2 b 4 3, then T = T +(a 1 1+a 1 3) a 2 1 a 3 2 b 4 3, where T = a 1 1 a 2 1 a 3 1 b 3 2(α1)+a a 2 1 (λa 3 2) b 4 1+a 1 1 a 2 2 a 3 1 b 4 1+a 1 2 a 2 1 a 3 1 b 4 1 T a 1 1 a 2 X 3. If b 3 1 a3 1 b4 1 3 = (a µa 3 2) b 4 3, and if µ 0, let ã3 2 = a µa 3 2, then T = T +(1/µa 1 1+a 1 3) a 2 1 ã3 2 b 4 3, where T = a 1 1 a 2 1 a 3 1 [b 3 2(α1) 1/µ(b λb 4 1)]+a 1 1 a 2 1 (λ/µã3 2) b 4 1+a 1 1 a 2 2 a 3 1 b 4 1+a 1 2 a 2 1 a 3 1 b 4 1 T a 1 1 a 2 X 3. If µ = 0, T = T +T, 1 a3 1 b4 1 where T = a 1 1 a 2 1 a 3 1 b 3 2(α1)+a a 2 1 (λa 3 2) b 4 1+a 1 1 a 2 2 a 3 1 b 4 1+a 1 2 a 2 1 a 3 1 b 4 1 T a 1 1 a 2 X 3, and T = a 1 1 a3 1 b4 1 1 a 2 1 a 3 2 b a 1 3 a 2 1 a 3 1 b 4 3 T a 1 1 a 2 X 3. 1 a3 1 b4 3 If b 4 1 and b 4 3 are linearly dependent, say b 4 1 = b 4 3, then T = T + T, where T = 18

24 a 1 1 a 2 1 a 3 1 b 3 2(α 3 1) + a 1 1 a 2 2 a 3 1 b (a xa 1 3) a 2 1 a 3 1 b 4 1 T a 1 1 a 2 1 a3 1 b4 1 X 3, and T = a 1 1 a 2 1 a 3 2 b 3 2(α 3 2) + (ya 1 3) a 2 1 a 3 2 b 4 1 T a 1 1 a 2 1 a3 2 b4 1 X 3. When b 2 3 = a 2 2 +λa 2 1, dim T (A 2 A 3) 3 implies three cases. Case 1: b 3 1 = a 3 1 b 4 1 and b 3 2 = a 3 1 b 4 2 for some b 4 1, b 4 2 A 4 A n ; Case 2: b 3 1 = a 3 1 b 4 1 and b 3 3 = a 3 1 b 4 3 for some b 4 1, b 4 3 A 4 A n ; Case 3: b 3 1 = a 3 1 b 4 1 and b 3 3 = a 3 2 b 4 3 for some b 4 1, b 4 3 A 4 A n. For case 1, if b 3 3 = u 3 3 u 4 3 for some u 3 3 A 3 and u 4 3 A 4 A n, then T = T +a 1 3 (a 2 2+λa 2 1) u 3 3 u 4 3, where T = a 1 1 a 2 1 a 3 1 b 4 2+a 1 1 a 2 2 a 3 1 b 4 1+a 1 2 a 2 1 a 3 1 b 4 1 T a 1 1 a 2 1 a3 1 b4 1 X 3. If b 3 3 : A 3 A 4 A n has rank 2, dim T (A 1 A 3) 3 requires b 4 1 = b 4 2 up to a scalar, and b 4 1 is a linear combination of b 3 3(α 3 1) and b 3 3(α 3 2), say b 3 3(α 3 1) = xb 3 3(α 3 2) + yb 4 1 or b 4 1 = b 3 3(α 3 2) up to a scalar, then T = (a a yλa 1 3) a 2 1 a 2 1 a 3 1 b 4 1+(a 1 1+ya 1 3) a 2 2 a 3 1 b 4 1+a 1 3 (a 2 2+λa 2 1) (xa 3 1+a 3 2) b 3 3(α 3 2), or T = T +T, where T = a 1 1 a 2 2 a 3 1 b 4 1+a 1 3 a 2 2 a 3 1 b 3 3(α 3 1)+a 1 3 a 2 2 a 3 2 b 4 1 T a 1 3 a 2 2 a3 1 b4 1 X 3, and T = (a 1 1+a 1 2) a 2 1 a 3 1 b 4 1+a 1 3 a 2 1 a 3 1 λb 3 3(α 3 1)+a 1 3 a 2 1 (λa 3 2) b 4 1 T a 1 3 a 2 1 a3 1 b4 1 X 3. For case 2, if b 4 3 = b 4 1 up to a scalar, then b 3 1 = b 3 3 up to a scalar, and T = a 1 1 a 2 1 b (a a 1 3) a 2 2 b (a λa 1 3) a 2 1 b 3 1, which is discussed in Case 2 Type 1. Hence we assume b 4 1 and b 4 3 are linearly independent. dim T (A 1 A 3) 3 implies b 3 2(α 3 2) = b 4 1 up to a scalar, then T = T + a 1 3 (a λa 2 1) a 3 1 b 4 3, where T = a 1 1 a 2 1 a 3 1 b 3 2(α 3 1) + a 1 1 a 2 1 a 3 2 b a 1 1 a 2 2 a 3 1 b a 1 2 a 2 1 a 3 1 b 4 1 T a 1 1 a 2 1 a3 1 b4 1 X 3. For case 3, dim T (A 2 A 3) 3 requires b 3 2(α 3 2) = b 4 1 up to a scalar. Then T = T + a 1 3 (a λa 2 1) a 3 2 b 4 3, where T = a 1 1 a 2 1 a 3 1 b 3 2(α 3 1) + a 1 1 a 2 1 a 3 2 b a 1 1 a 2 2 a 3 1 b a 1 2 a 2 1 a 3 1 b 4 1 T a 1 1 a 2 X 3. 1 a3 1 b4 1 k Now we assume T = b 1 1 b1 i 1 b i 2 b i+1 1 b k 1 + b 1 3 b k 3. The i=1 induction argument is similar to the case k = 3. 19

25 Type 3: T = a 1 1 b 2 2 b a 1 2 b 2 1 b a 1 2 b 2 2 b a 1 1 b 2 1 b a 1 1 b 2 3 b a 1 3 b 2 1 b 3 1. Without loss of generality, we can assume b 2 1 = a 2 1, b 2 2 = a 2 2, and b 2 3 = xa ya 2 2. dim T (A 2 A 3) 3 implies two cases. Case 1: b 3 1 = a 3 1 b 4 1 for some b 4 1 A 4 A n, b 3 2(α2) 3 = b 4 1 up to a scalar, and b 3 2(α1) 3 = b 3 3(α2) 3 + λb 4 1 for some λ C; Case 2: b 3 1 = a 3 1 b 4 1, and b 3 2 = a 3 1 b 4 2 for some b 4 1, b 4 2 A 4 A n. For case 1, T = a 1 1 a 2 2 a 3 1 b 3 3(α2) 3 + a 1 1 a 2 2 a 3 2 b a 1 2 a 2 1 a 3 1 b 3 3(α2) 3 + a 1 2 a 2 1 a 3 2 b a 1 2 a 2 2 a 3 1 b a 1 1 a 2 1 a 3 2 b 3 3(α2) 3 + a 1 1 a 2 1 a 3 1 b 3 3(α1) 3 + a 1 1 (y + λ)a 2 2 a 3 1 b (xa λa a 1 3) a 2 1 a 3 1 b 4 1. Let S(t) = [a 1 1+ta 1 2+t 2 (xa 1 1+λa 1 2+a 1 3)] [a 2 1+ta 2 2+t 2 (y+λ)a 2 2] (a 3 1+ta 3 2) [b 4 1+tb 3 3(α2)+t 3 2 b 3 3(α1)], 3 then T = S (0). For case 2, if b 4 2 = λb 4 1 for some λ C, then b 3 2 = λb 3 1, T = [(y + λ)a a 1 2] a 2 2 b (xa λa a 1 3) a 2 1 b a 1 1 a 2 1 b 3 3, which is discussed in Case 2 Type 1. Thus we assume b 4 1 and b 4 2 are independent. dim T (A 1 A 3) 3 implies b 3 3(α2) 3 = b 4 1 up to a scalar, so T = a 1 1 a 2 2 a 3 1 b a 1 2 a 2 1 a 3 1 b a 1 2 a 2 2 a 3 1 b a 1 1 a 2 1 a 3 1 b 3 3(α1) 3 + a 1 1 a 2 1 a 3 2 b a 1 1 (xa ya 2 2) a 3 1 b a 1 3 a 2 1 a 3 1 b 4 1. Let S(t) = [a ta t 2 a 1 3] [a ta t 2 (xa ya 2 2)] (a t 2 a 3 2) [b tb t 2 b 3 3(α1)], 3 then T = S (0). Now we assume T = i<j b 1 1 b i 1 1 b i 2 b i+1 1 b j 1 1 b j 2 b j+1 1 b k 1 + k i=1 b 1 1 b i 1 1 b i 3 b i+1 1 b k 1, and use induction to show T σ 3 (X). The induction argument is similar to the case k = 3. Type 4: T = a 1 2 b 2 1 b a 1 2 b 2 2 b a 1 1 b 2 1 b a 1 1 b 2 3 b a 1 3 b 2 1 b 3 1. If b 2 2 = b 2 1, T = a 1 2 b 2 1 b a 1 1 b 2 1 b a 1 1 b 2 3 b (a a 1 3) b 2 1 b 3 1, which is discussed in Case 2 Type 2. Hence we can assume b 2 i = a 2 i for 1 i 2. Assume b 2 3 = xa ya 2 2, then T = (ya a 1 2) a 2 1 b (ya a 1 2) a 2 2 b a 1 1 a 2 1 (b 3 3 yb 3 2) + (xa a 1 3) a 2 1 b 3 1. Therefore after a change of basis, we only need 20

26 to consider the case T = a 1 2 a 2 1 b a 1 2 a 2 2 b a 1 1 a 2 1 b a 1 3 a 2 1 b 3 1. dim T (A 2 A 3) 3 implies b 3 1 : A 3 A 4 A n has rank 1, say b 3 1 = a 3 1 b 4 1 for some b 4 1 A 4 A n. If b 3 3(α1) 3 and b 3 3(α2) 3 are linearly independent, dim T (A 1 A 3) 3 implies b 4 1, b 3 2(α2) 3 are in V 4, where V 4 is spanned by b 3 3(α1) 3 and b 3 3(α2). 3 For the subspace A 3 V 4, after a change of basis, we can assume a 3 1 and a 3 1 b 3 3(α1) 3 + a 3 2 b 3 3(α2) 3 are preserved, and b 4 1 = b 3 3(α1), 3 or b 4 1 = b 3 3(α2). 3 So we have two cases: Case 1: If b 4 1 = b 3 3(α1), 3 assume b 3 2(α2) 3 = xb 3 3(α1) 3 + yb 3 3(α2), 3 then T = T + (ya a 1 1) a 2 1 a 3 2 b 3 3(α2), 3 where T = a 1 2 a 2 1 a 3 1 b 3 2(α1) 3 + a 1 2 a 2 1 (xa 3 2) b 3 3(α1) 3 + a 1 2 a 2 2 a 3 1 b 3 3(α1) 3 + (a a 1 3) a 2 1 a 3 1 b 3 3(α1) 3 T a 1 2 a 2 1 a3 1 b3 3 (α3 1 ) X 3. Case 2: If b 4 1 = b 3 3(α2), 3 we can assume b 3 2(α2) 3 = b 3 3(α1) 3 + λb 3 3(α2) 3 for some λ C, or b 3 2(α2) 3 = λb 3 3(α2). 3 If b 3 2(α2) 3 = b 3 3(α1) 3 + λb 3 3(α2), 3 T = a 1 2 a 2 1 a 3 1 b 3 2(α1) 3 + a 1 2 a 2 1 (λa 3 2) b 3 3(α2) 3 + a 1 2 a 2 2 a 3 1 b 3 3(α2) 3 + a 1 3 a 2 1 a 3 1 b 3 3(α2) 3 + a 1 2 a 2 1 a 3 2 b 3 3(α1) 3 + a 1 1 a 2 1 a 3 1 b 3 3(α1) 3 + a 1 1 a 2 1 a 3 2 b 3 3(α2). 3 Let S(t) = (a ta t 2 a 1 3) (a t 2 a 2 2) (a ta t 2 λa 3 2) (b 3 3(α2) 3 + tb 3 3(α1) 3 + t 2 b 3 2(α1)), 3 then T = S (0). If b 3 2(α2) 3 = λb 3 3(α2), 3 T = T + T, where T = a 1 2 a 2 1 a 3 1 b 3 2(α1) 3 + a 1 2 a 2 1 λa 3 2 b 3 3(α2) 3 + a 1 2 a 2 2 a 3 1 b 3 3(α2) 3 + a 1 3 a 2 1 a 3 1 b 3 3(α2) 3 T a 1 2 a 2 1 a3 1 b3 3 (α3 2 ) X 3, and T = a 1 1 a 2 1 a 3 1 b 3 3(α1) 3 + a 1 1 a 2 1 a 3 2 b 3 3(α2) 3 T a 1 1 a 2 1 a3 1 b3 3 (α3 2 ) X 3. If b 3 3(α2) 3 = λb 3 3(α1) 3 for some λ C, then we can assume b 3 3 = a 3 1 b 3 3(α1) 3 or b 3 3 = a 3 2 b 3 3(α1). 3 Thus we have four cases: Case 1: If b 3 3 = a 3 1 b 3 3(α1), 3 b 3 3(α1) 3 and b 4 1 are linearly independent, we can assume b 3 2(α2) 3 = xb yb 3 3(α1) 3 for some x, y C due to dim T (A 1 A 3), then T = T + T, where T = a 1 2 a 2 1 a 3 1 b 3 2(α1) 3 + a 1 2 a 2 1 xa 3 2 b a 1 2 a 2 2 a 3 1 b a 1 3 a 2 1 a 3 1 b 4 1 T a 1 2 a 2 X 3, and T = a 1 1 a3 1 b4 1 2 a 2 1 ya 3 2 b 3 3(α1) 3 + a 1 1 a 2 1 a 3 1 b 3 3(α1) 3 T a 1 2 a 2 1 a3 1 b3 3 (α3 1 ) X 3. Case 2: If b 3 3 = a 3 1 b 3 3(α1) 3 and b 3 3(α1) 3 = µb 4 1 for some µ C, T = T + a 1 2 a

27 a 3 2 b 3 2(α 3 2), where T = a 1 2 a 2 1 a 3 1 b 3 2(α 3 1)+a 1 2 a 2 2 a 3 1 b 4 1+(µa 1 1+a 1 3) a 2 1 a 3 1 b 4 1 T a 1 2 a 2 1 a3 1 b4 1 X 3. Case 3: If b 3 3 = a 3 2 b 3 3(α 3 1), b 3 3(α 3 1) and b 4 1 are linearly independent, we can assume b 3 2(α 3 2) = xb 4 1+yb 3 3(α 3 1) due to dim T (A 1 A 3), then T = T +(ya 1 2+a 1 1) a 2 1 a 3 2 b 3 3(α 3 1), where T = a 1 2 a 2 1 a 3 1 b 3 2(α 3 1)+a 1 2 a 2 1 xa 3 2 b 4 1+a 1 2 a 2 2 a 3 1 b 4 1+a 1 3 a 2 1 a 3 1 b 4 1 T a 1 2 a 2 1 a3 1 b4 1 X 3. Case 4: If b 3 3 = a 3 2 b 3 3(α 3 1) and b 3 3(α 3 1) = µb 4 1 for some µ C, T = T + T, where T = a 1 2 a 2 1 a 3 1 b 3 2(α 3 1) + a 1 2 a 2 2 a 3 1 b a 1 3 a 2 1 a 3 1 b 4 1 T a 1 2 a 2 1 a3 1 b4 1 X 3, and T = a 1 1 a 2 1 a 3 2 µb a 1 2 a 2 1 a 3 2 b 3 2(α2) 3 T a 1 2 a 2 X 3. 1 a3 2 b4 1 k k Now assume T = b 1 2 b 2 1 b i 1 1 b i 2 b i+1 1 b k 1 + b 1 1 b i 1 1 i=2 b i 3 b i+1 1 b k 1, and use induction to show T σ 3 (X). The induction argument is similar to the case k = Case 3: T σ 3 (X 2 ) \ σ 2 (X 2 ), T Sub 2,2,...,2 (A 1 A n ) Since dim T (A 3 A n) 3, then after a change of basis we can assume T (A 3 A n) V, where V is spanned by {a 1 1 a a 1 2 a 2 2, a 1 1 a 2 2, a 1 2 a 2 1} or {a 1 1 a 2 1, a 1 1 a 2 2, a 1 2 a 2 1}. So T has 2 types of normal forms. Type 1: T = (a 1 1 a a 1 2 a 2 2) b a 1 1 a 2 2 b a 1 2 a 2 1 b 3 3, we reduce the problem to finding equations for σ 3 (ν n (P 1 )), which has been settled. i=1 Lemma 2. Let T A B C, where dim A = dim B. If there is an element φ Ker(T BA ) with full rank, then φ(t ) S2 A C. Proof of the lemma. Let {a i }, {b j }, {c k } be bases for A, B, C respectively, and {a i }, {b j }, {c k } their dual bases. Let T = α ijk a i b j c k, then T BA : a l b j i,k αijk (a l a i ) c k. Let φ = β l ja l b j Ker(T BA ), then β l jα ijk (a l a i ) c k = 0, which means j βl jα ijk = j βi jα ljk. Since φ(t ) = β l jα ijk a i a l c k, then φ(t ) S 2 A C. 22

28 Let V be a complex vector space. Given φ S d V, let φ a,d a S a V S d a V denote the (a, d a)-polarization of φ. As a linear map S a V S d a V, rank(φ a,d a ) r if [φ] σ r (ν d (PV )) [33]. Theorem 9 ( [33]). σ 3 (ν 3 (P n )) is ideal theoretically defined by Aronhold invariant and size 4 minors of φ 1,2. σ 3 (ν d (P n )) is scheme theoretically defined by size 4 minors of φ 2,2 and φ 1,3 when d 4. Now given any T A 1 A n, if there is some 1 i n, and for any j i, there is a φ ji Ker(TA j A i ) with full rank, then T = φ ni φ 1i (T ) S n A i has the same rank with T. If T satisfies 4 4 minors of flattenings, T satisfies size 4 minors of symmetric flattenings, by Theorem 9 T σ 3 (ν n (P 1 )), then T σ 3 (X). If T is of Type 1, we always have a 1 1 a a 1 2 a 2 1 Ker(TA 2 A 1 ) with full rank, hence if for any 2 i n, T is of Type 1 when viewed as a tensor in A 1 A i (A 2 A i 1 Âi A i+1 A n ), then T σ 3 (X). If T A 1 A 2 (A 3 A n ) is not of Type 1, then it must be of Type 2, and we will use induction to show that T σ 3 (X) in this situation. Type 2: T = a 1 1 a 2 1 b 3 1 +a 1 1 a 2 2 b 3 2 +a 1 2 a 2 1 b 3 3, the dimension of T (A 2 A 3) implies b 3 3 : A 3 A 4 A n has rank 1, or b 3 2 : A 3 A 4 A n has rank 1. If b 3 3 : A 3 A 4 A n has rank 1, say b 3 3 = a 3 1 b 4 3, and b 3 2 : A 3 A 4 A n has rank 2, say b 3 2 = a 3 1 b a 3 2 b 4 2, then dim T (A 2 A 3) 3 implies b 3 1(α2) 3 = λb µb 4 2 for some λ, µ C. If b 4 3, b 4 1 and b 4 2 are linearly independent, then dim T (A 1 A 2 A 3) 3 forces b 3 1(α1) 3 = xb yb zb 4 2 for some x, y, z C, thus T = a 1 1 a 2 1 (ya 3 1 b 4 1 +za 3 1 b 4 2 +λa 3 2 b 4 1 +µa 3 2 b 4 2)+a 1 1 a 2 2 (a 3 1 b 4 1 +a 3 2 b 4 2)+ (xa 1 1+a 1 2) a 2 1 a 3 1 b 4 3. For the subspace A 3 V 4, where V 4 A 4 A n is spanned by b 4 1 and b 4 2, after a change of basis we can assume a 3 1 b a 3 2 b 4 2 is preserved, a

29 is mapped to ua 3 1 +va 3 2 for some u, v C, and ya 3 1 b 4 1 +za 3 1 b 4 2 +λa 3 2 b 4 1 +µa 3 2 b 4 2 is of the Jordan canonical form, i.e. a 3 1 b a 3 2 b 4 2, or a 3 1 b 4 1, or a 3 1 b 4 2, or βa 3 1 b a 3 1 b βa 3 2 b 4 2 for some 0 β C. Hence we have: Subcase 1: T = a 1 1 (a a 2 2) a 3 1 b a 1 1 (a a 2 2) a 3 2 b (xa a 1 2) a 2 1 (ua va 3 2) b 4 3. Subcase 2: T = a 1 1 (a a 2 2) a 3 1 b a 1 1 a 2 2 a 3 2 b (xa a 1 2) a 2 1 (ua va 3 2) b 4 3. Subcase 3: T = T + (xa a 1 2) a 2 1 (ua va 3 2) b 4 3, where T = a 1 1 a 2 1 a 3 1 b a 1 1 a 2 2 a 3 1 b a 1 1 a 2 2 a 3 2 b 4 2 T a 1 1 a 2 X 3. 2 a3 1 b4 2 Subcase 4: T = T + (xa a 1 2) a 2 1 (ua va 3 2) b 4 3, where T = a 1 1 (βa a 2 2) a 3 1 b a 1 1 (βa a 2 2) a 3 2 b a 1 1 a 2 1 a 3 1 b 4 2 T a 1 1 (βa 2 X a2 2 ) a3 1 b4 2 If b 4 3 = pb qb 4 2 for some p, q C, for A 3 V 4, after a change of basis we can assume a 3 1 and a 3 1 b a 3 2 b 4 2 are preserved, b 4 3 = b 4 1 or b 4 2, and a 3 2 b 3 1(α2) 3 is of the form x 1 1a 3 1 b x 1 2a 3 1 b x 2 1a 3 2 b x 2 2a 3 2 b 4 2. If b 4 3 = b 4 1 we have: Subcase 5: T = T + a 1 1 (x 2 2a a 2 2) a 3 2 b 4 2, where T = a 1 1 a 2 1 a 3 1 [b 3 1(α1) 3 + x 1 1b 4 1+x 1 2b 4 2]+a 1 1 a 2 1 (x 2 1a 3 2) b 4 1+a 1 1 a 2 2 a 3 1 b 4 1+a 1 2 a 2 1 a 3 1 b 4 1 T a 1 1 a 2 X 3. 1 a3 1 b4 1 If b 4 3 = b 4 2, by changing a 3 2, b 4 2 and a 2 1, we can assume x 2 1 = 1 or 0. So we have: Subcase 6: T = a 1 1 a 2 1 a 3 1 [b 3 1(α1) 3 + x 1 1b x 1 2b 4 2] + a 1 1 a 2 1 (x 2 2a 3 2) b a 1 2 a 2 1 a 3 1 b a 1 1 a 2 1 a 3 2 b a 1 1 a 2 2 a 3 1 b a 1 1 a 2 2 a 3 2 b 4 2. Let S(t) = (a 1 1 +t 2 a 1 2) (a 2 1 +ta 2 2) (a 3 1 +ta 3 2 +t 2 x 2 2a 3 2) [b 4 2 +tb 4 1 +t 2 (b 3 1(α1)+x 3 1 1b 4 1 +x 1 2b 4 2)], so T = S (0). Subcase 7: T = T + T, where T = a 1 1 a 2 1 a 3 1 [b 3 1(α1) 3 + x 1 1b x 1 2b 4 2] + a 1 1 a 2 1 (x 2 2a 3 2) b a 1 2 a 2 1 a 3 1 b 4 2 T a 1 1 a 2 X 3, and T = a 1 1 a3 1 b4 2 1 a 2 2 a 3 1 b a 1 1 a 2 2 a 3 2 b 4 2 T a 1 1 a 2 X 3. 2 a3 1 b4 2 If b 3 2 : A 3 A 4 A n has rank 1, say b 3 2 = a 3 1 b 4 2 for some b 4 2 A 4 A n, and b 3 3 : A 3 A 4 A n has rank 2, say b 3 3 = a 3 1 b a 3 2 b 4 3 for some 24

30 b 4 1, b 4 3 A 4 A n, then dim T (A 1 A 3) 3 implies b 3 1(α2) 3 = λb µb 4 3 for some λ, µ C. If b 4 3, b 4 1 and b 4 2 are linearly independent, then dim T (A 1 A 2 A 3) 3 forces b 3 1(α1) 3 = xb yb zb 4 3 for some x, y, z C. For the subspace A 3 V 4, where V 4 A 4 A n is spanned by b 4 1 and b 4 3, after a change of basis we can assume a 3 1 b a 3 2 b 4 3 is preserved, a 3 1 is mapped to ua va 3 2 for some u, v C under the new basis, and xa 3 1 b za 3 1 b λa 3 2 b µa 3 2 b 4 3 is of the Jordan canonical form, i.e. a 3 1 b a 3 2 b 4 3, or a 3 1 b 4 1, or a 3 1 b 4 3, or βa 3 1 b a 3 1 b βa 3 2 b 4 3 for some 0 β C. Hence we have: Subcase 8: T = (a a 1 2) a 2 1 a 3 1 b (a a 1 2) a 2 1 a 3 2 b a 1 1 (ya a 2 2) (ua va 3 2) b 4 2. Subcase 9: T = (a a 1 2) a 2 1 a 3 1 b a 1 2 a 2 1 a 3 2 b a 1 1 (ya a 2 2) (ua va 3 2) b 4 2. Subcase 10: T = T + a 1 1 (ya a 2 2) (ua va 3 2) b 4 2, where T = a 1 1 a 2 1 a 3 1 b a 1 2 a 2 1 a 3 1 b a 1 2 a 2 1 a 3 2 b 4 3 T a 1 2 a 2 X 3. 1 a3 1 b4 3 Subcase 11: T = T + a 1 1 (ya a 2 2) (ua va 3 2) b 4 2, where T = (βa a 1 2) a 2 1 a 3 1 b (βa a 1 2) a 2 1 a 3 2 b a 1 1 a 2 1 a 3 1 b 4 3 T (βa 1 1 +a 1 X 3. 2 ) a2 1 a3 1 b4 3 If b 4 2 = pb qb 4 3 for some p, q C, for A 3 V 4, after a change of basis we can assume a 3 1 and a 3 1 b a 3 2 b 4 3 are preserved, b 4 2 = b 4 1 or b 4 3, and a 3 2 b 3 1(α2) 3 is of the form x 1 1a 3 1 b x 1 2a 3 1 b x 2 1a 3 2 b x 2 2a 3 2 b 4 3. If b 4 2 = b 4 1 we have: Subcase 12: T = T +(x 2 2a 1 1 +a 1 2) a 2 1 a 3 2 b 4 3, where T = a 1 1 a 2 1 a 3 1 [b 3 1(α1)+ 3 x 1 1b 4 1 +x 1 2b 4 3]+a 1 1 a 2 1 x 2 1a 3 2 b 4 1 +a 1 1 a 2 2 a 3 1 b 4 1 +a 1 2 a 2 1 a 3 1 b 4 1 T a 1 1 a 2 X 3. 1 a3 1 b4 1 If b 4 2 = b 4 3, by changing a 3 2, b 4 3 and a 2 2, we can assume x 2 1 = 1 or 0. So we have: Subcase 13: T = a 1 1 a 2 1 a 3 1 [b 3 1(α1) 3 + x 1 1b x 1 2b 4 3] + a 1 1 a 2 1 (x 2 2a 3 2) b a 1 1 a 2 2 a 3 1 b a 1 2 a 2 1 a 3 1 b a 1 1 a 2 1 a 3 2 b a 1 2 a 2 1 a 3 2 b 4 3. Let S(t) = (a 1 1 +ta 1 2) (a 2 1 +t 2 a 2 2) (a 3 1 +ta 3 2 +t 2 x 2 2a 3 2) [b 4 3 +tb 4 1 +t 2 (b 3 1(α1)+x 3 1 1b 4 1 +x 1 2b 4 3)], so T = S (0). 25