Oeding (Auburn) tensors of rank 5 December 15, / 24

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1 Oeding (Auburn) tensors of rank 5 December 15, / 24

2 Recall Peter Burgisser s overview lecture (Jan Draisma s SIAM News article). Big Goal: Bound the computational complexity of det n, perm n, M n. Algorithmic Complexity is often governed by the rank of a tensor. Need: tools for bounding the rank (or border rank) of tensors. How can we find such lower bounds? Polynomials! Common theme: Exploit all available symmetry to aid in computations. Take the n n matrix multiplication tensor M n C n2 C n2 C n2. In 2013 Ikenmeyer, Hauenstein and Landsberg provided a new computational proof that the border rank of 2 2 matrix multiplication M 2 is 7 by providing a degree 20 invariant F for border rank 7 that did not vanish on M 2. Driasma highlighted the dramatic power of symmetry: Straightforward combinatorics shows that the space of degree-20 polynomials on the 64-dimensional space is C( , 20) = 8, 179, 808, 679, 272, 664, 720-dimensional it is striking how representation theory helps us to find F and evaluate it at M 2! Our work echoes this theme. Oeding (Auburn) tensors of rank 5 December 15, / 24

3 First questions for tensors Consider tensors of format n 1 n d. 1 What notion of rank are you using? (tensor rank / border rank via secant varieties) 2 What is the expected (generic) tensor rank? (defectivity) 3 How can you find a minimal decomposition of a given tensor? (Find effective algorithms) 4 How can you detect the rank of a given tensor? (Provide certificates) 5 How many decompositions does a given tensor have? (identifiability) Knowing equations of secant varieties can help with all of these questions, especially if they re determinantal. Oeding (Auburn) tensors of rank 5 December 15, / 24

4 Secant varieties and tensors Let V 1,..., V d, be C-vector spaces. The tensor product V 1 V d is the vector space with elements (T i1,...,i d ) considered as hyper-matrices or tensors. Segre variety (rank 1 tensors): Defined by Seg : PV 1 PV d P ( V 1 V d ) ([v 1 ],..., [v d ]) [v 1 v d ]. In coordinates: T i1,...,i d = v 1,i1 v 2,i2 v d,id. The r th secant variety of a variety X P N : σ r (X) := P(span{x 1,..., x r }) P N. x 1,...,x r X General points of σ r (Seg(PV 1 PV d )) have rank r: [ r ] v1 v s 2 s... vd s, or in coordinates: T i1,...,i n s=1 = r s=1 vs 1,i 1 v s 2,i 2 v s d,i d. Oeding (Auburn) tensors of rank 5 December 15, / 24

5 A first case: matrices Suppose k m n. If M Mat m n (C) has rank k then (full rank) A Mat m k (C) and (full rank) B Mat k n (C) such that for any (full rank) U Mat k k (C). M = AB but also M = (AU)(U 1 B), So dim(σ k (P m 1 P n 1 )) has dimension m k + n k k 2 1. Expected dimension: ExpDim(σ k (P m 1 P n 1 )) = min{n m 1, k (n + m 1) 1}. So the defect is k (n + m 1) 1 (m k + n k k 2 1) = k 2 k. Rank k matrices are defective (and thus not identifiable) for k 0, 1. Oeding (Auburn) tensors of rank 5 December 15, / 24

6 First examples of equations of secant varieties: Flattenings Given T V 1 V d, take index sets I J = [d]. Tensor product is associative, so V 1 V d = ( i I V i) ( j J V j), And we can view T as a (flattening) matrix: Facts: If Rank(T ) = 1, then rank F I (T ) = 1. F (T + T ) = F (T ) + F (T ). F I (T ): V I V J Sub-additivity of matrix rank implies if Rank T = r then Rank F ((T )) r. So, (r + 1) (r + 1) minors of flattenings (if non-trivial) are equations for tensors of rank r. Oeding (Auburn) tensors of rank 5 December 15, / 24

7 tensors Suppose V i = C 2 for 1 i 4. Four different 1-flattenings (up to transpose): F 1 (T ): V 1 V 2 V 3 V 4 F 2 (T ): V 2 V 1 V 3 V 4 F 3 (T ): V 3 V 1 V 2 V 4 F 4 (T ): V 4 V 1 V 2 V 3 All 8 2 matrices, max rank 2. If all have rank 1, then the tensor T actually has rank 1 (and vice versa). Three different 2-flattenings (up to transpose) F 1,2 (T ): (V 1 V 2 ) V 3 V 4 F 1,3 (T ): (V 1 V 3 ) V 2 V 3 F 1,4 (T ): (V 1 V 4 ) V 2 V 3 All 4 4. So determinants vanish for tensors of rank 3. Oeding (Auburn) tensors of rank 5 December 15, / 24

8 A defective secant variety Suppose V i = C 2 for 1 i 4. Counting parameters, expect σ 3 (PV 1 PV 2 PV 3 PV 4 ) to be 3 ( ) + 2 = 14, so expect codimension 1 in P 15 = P(V 1 V 2 V 3 V 4 ). On the other hand, F 1,2 (T ): (V 1 V 2 ) V 3 V 4 F 1,3 (T ): (V 1 V 3 ) V 2 V 3 F 1,4 (T ): (V 1 V 4 ) V 2 V 3 Check: any two of det(f 1,2 (T )), det(f 1,3 (T )), det(f 1,4 (T )) are algebraically independent, so dim zeros(det(f 1,2 (T )), det(f 1,3 (T ))) = 13 In fact, σ 3 (PV 1 PV 2 PV 3 PV 4 ) = zeros(det(f 1,2 (T )), det(f 1,3 (T ))) The expected dimension is not the actual dimension, so the variety is defective, and thus rank 3 tensors of format are not identifiable. Theorem (Catalisano-Geramita-Gimigliano) Suppose V i = C 2 for 1 i d. For all d 5 and all k, σ k (PV 1 V d ) has the expected dimension (non-defective). Oeding (Auburn) tensors of rank 5 December 15, / 24

9 Identifiability for Binary Tensors dim σ 3 (PV 1 PV 2 PV 3 PV 4 ) = 13 < 14 also implies that the generic tensor of rank 3 and format has infinitely many decompositions. Definition If the general tensor of format n 1 n d and rank k has finitely many decompositions the variety is not k-defective. Definition A tensor format n 1 n 2... n d is called k-identifiable if the generic tensor of that format and rank k has a unique (up to trivial re-ordering) decomposition as the sum of k rank-1 tensors. Theorem (Bocci-Chiantini 2014) tensors are not identifiable in rank 5, but the generic tensor of that format has exactly 2 decompositions. Theorem (Bocci-Chiantini-Ottaviani 2014) For 6 factors, the Segre is almost always k-identifiable. Oeding (Auburn) tensors of rank 5 December 15, / 24

10 Identifiability for Perfect Tensors A tensor of format n 1 n d is of perfect format if [ i=1 n i]/[1 + i (n i + 1)] is an integer (generically have finitely many decompositions). Theorem (Hauenstein-Oeding-Ottaviani-Sommese 14) The general tensor has a unique decomposition of rank 6. The general tensor has a unique decomposition of rank 4. Conjecture (Hauenstein-Oeding-Ottaviani-Sommese 14) The only perfect formats (n 1,..., n d ) where a general tensor has a unique decomposition are: 1 (2, k, k) for some k matrix pencils, classical Kronecker normal form, 2 (3, 4, 5), and 3 (2, 2, 2, 3). (see arxiv: ) Oeding (Auburn) tensors of rank 5 December 15, / 24

11 Out of Bernd Sturmfels s Algebraic Fitness Session Find the equations of σ 5 (Seg(P 1 P 1 P 1 P 1 P 1 )): Theorem (Oeding-Sam [Exp. Math 2015]) The affine cone of σ 5 (Seg(P 1 5 )) is a complete intersection of two equations: one of degree 6, and one of degree 16. The star refers to the careful numerical, sometimes probablistic computations used in our proofs, which took around two weeks of human/computer time. Note this result implies that: σ 5 (Seg(P 1 5 )) is arithmetically Cohen-Macaulay. Oeding (Auburn) tensors of rank 5 December 15, / 24

12 Find equations in the ideal Consider σ 5 (P 1 P 1 P 1 P 1 P 1 ) P 31. Look for equations in R = C[x 00000,..., x 11111] of low degree. The Hilbert function of R starts like this: d = HF R (d) = HF R/I (d) = ?... dim(i d ) = ?... Naively, to find the space of sextics in the ideal: compute points on the variety, evaluate them on the monomials of degree 6 and compute the kernel of the resulting matrix. The space of degree 16 equations has dimension , so it seems hopeless to work here. Oeding (Auburn) tensors of rank 5 December 15, / 24

13 The equation f 6 Choose a basis e 0, e 1 for V i so that we can identify the coordinates of P(V) with x I where I {0, 1} 5. Given a monomial in the x I, define its skew-symmetrization to be c 1 σ Σ 5 sgn(σ)x σ(i) where c is the coefficient of x I in the sum. The polynomial f 6 has 864 monomials and is the sum of the skew-symmetrizations of the following 15 monomials: x 00000x 01010x 01101x 10011x 10100x 11111, x 00000x 01100x 01111x 10010x 10111x 11001, x 00000x 01100x 01111x 10011x 10110x 11001, x 00000x 01101x 01110x 10011x 10110x 11001, x 00110x 01000x 01101x 10000x 10011x 11111, x 00100x 01010x 01111x 10000x 10111x 11001, x 00100x 01000x 01111x 10011x 10110x 11001, x 00110x 01000x 01101x 10001x 10010x 11111, x 00100x 01010x 01111x 10001x 10111x 11000, x 00100x 01010x 01111x 10011x 10101x 11000, x 00101x 01010x 01111x 10000x 10110x 11001, x 00100x 01011x 01110x 10011x 10101x 11000, x 00110x 01001x 01100x 10001x 10010x 11111, x 00110x 01001x 01111x 10011x 10100x 11000, x 00111x 01010x 01101x 10011x 10100x Oeding (Auburn) tensors of rank 5 December 15, / 24

14 Alternative description in terms of Young symmetrizers, see [Bates-Oeding 10]. The Young Symmetrizer algorithm takes as input a set of fillings of five Young diagrams, performs a series of skew-symmetrizations and symmetrizations, and produces as output a polynomial in the associated Schur module. There are 5 standard tableaux of shape (3, 3) and content {1, 2,..., 6}. The following Schur module, which uses one of each of the 5 standard fillings, realizes the non-trivial copy of 5 i=1 (S 3,3V i ) inside of Sym 6 (V 1 V 2 V 3 V 4 V 5 ) S V 1 S V 2 S V 3 S V 4 S V Can show that the image of the Young symmetrizer vanishes on an open subset of points of X. Oeding (Auburn) tensors of rank 5 December 15, / 24

15 From Ikenmeyer s intro lecture: Polynomials and graphs Each tableau gets a color. Each column becomes a directed colored arrow. where color 1 corresponds to, color 2 to, color 3 to, color 4 to. color 5 to, The degree 6 invariant is described by S V 1 S V 2 S V 3 S V 4 S V Combinatorics of these graphs played a strong role in the resolution of the Garcia-Stillman-Sturmfels Conjecture on ideals of secant line varieties [Raicu 12], and the resolution of the Landsberg-Weyman Conjecture on ideals of tangential varieties [Oeding-Raicu 13]. Oeding (Auburn) tensors of rank 5 December 15, / 24

16 The Young symmetrizer algorithm for evaluations 1 input: Point: Z C 2 C 2 C 2 C 2 C 2, Multi-partition: ((3, 3), (3, 3), (3, 3), (3, 3), (3, 3)) a c e a c d a b e a b d a b c Filling: F = b d f b e f c d f c e f d e f 2 Construct a product of determinants: p = a1 1 a1 2 c1 b 1 1 c1 2 e1 1 b1 2 d 1 1 e1 2 1 d1 2 f 1 1 f a2 1 1 a2 2 c2 2 b 2 1 c2 2 d2 1 b2 2 e 2 1 d2 2 1 e2 2 f 2 1 f a3 2 1 a3 2 2 c 3 1 c3 2 a4 1 a4 2 b4 1 b4 2 d4 1 d4 2 a5 1 a5 2 b5 1 b5 2 c5 1 c5 2 b3 1 b3 2 d 3 1 d3 2 e3 1 e3 2 f 3 1 f 3 2 c 4 1 c4 2 e 4 1 e4 2 f 4 1 f 4 2 d 5 1 d5 2 e 5 1 e5 2 f 5 1 f Start with p(a, b, c, d, e, f, x) of multi-degree (5, 5, 5, 5, 5, 5, 0). 4 Substitutions: a 1 i a2 j a3 k a4 l a5 ma 6 n x i,j,k,l,m,n and x i,j,k,l,m,n Z i,j,k,l,m,n. Produce a polynomial of multi-degree (0, 5, 5, 5, 5, 5, 0). 5 Substitutions: b 1 i b2 j b3 k b4 l b5 mb 6 n x i,j,k,l,m,n and x i,j,k,l,m,n Z i,j,k,l,m,n. Produce a polynomial of multi-degree (0, 0, 5, 5, 5, 5, 0). 6 Repeat for c, d, e, f 7 output: the value of p(z). Producing p(z) takes much less time and memory than p(x). = 2 15 terms (don t expand!). Oeding (Auburn) tensors of rank 5 December 15, / 24

17 Evaluate the highest-weight space of S π V M π on X. 1 Use characters to compute m := dim M (d,d),(d,d),(d,d),(d,d),(d,d). 2 Make m random points y i of ambient space, and m + 5 points x i of X. 3 Repeat the following for k = 1..m: 3.0 Start with linearly independent fillings F = {F 1,..., F k 1 }. 3.1 At step k take a random filling F k of shape (d, d) Evaluate p k (y 1) using the Young symmetrizer algorithm for F k. If p k (y 1 ) is non-zero, continue. Otherwise return to (3.1). 3.3 Populate the k k matrix (one processor core per entry) Q k (y) := (p j(y i)) (i,j) 3.4 Compute Rank(Q k (y) If Q k has rank k, add F k to F increment k, and return to (3.0). Otherwise return to (3.1). 4 Take linearly independent fillings F = {F 1,..., F m } 5 Populate the (m + 5) m matrix (one processor core per entry) Q m (x) := (p j (x i )) i,j 6 The kernel of Q m (x) is the subspace of M π vanishing on the x i 7 The extra points increases likelihood that ker(q m (x)) also vanishes on X. Oeding (Auburn) tensors of rank 5 December 15, / 24

18 From Jon Hauenstein s Intro Lectures: Numerical Algebraic Geometry & Bertini input: An irreducible variety H. Output: deg H 1 Choose a random linear space L with dim L = codim H. 2 Generate a point x H L. Initialize W := {x}. 3 Perform a random monodromy loop starting at the points in W: (a) Pick a random loop M(t) in the grassmannian of linear spaces so that M(0) = M(1) = L. (b) Trace the curves H M(t) starting at the points in W at t = 0 to compute the endpoints E at t = 1. (Hence, E H L). (c) Update W := W E. 4 Repeat (2) until #W stabilizes. 5 Use the trace test to verify that W = H L. 6 Return deg H = #H( L). Proposition The degree of σ 5 ( (P 1 ) 5) is 96. Oeding (Auburn) tensors of rank 5 December 15, / 24

19 More Symmetry! Let V := V 1 V 2 V 3 V 4 V 5. Let U d = Sym d (V) SL 5 2 (the superscript denotes taking invariants). This has an action of Σ 5. Here are the dimensions of U d, and the spaces of Σ 5 -invariants and Σ 5 skew-invariants in U d. Degree d dim U d dim U Σ5 d This follows from standard character theory calculations. dim U Σ5,sgn d Use linear algebra to compute bases of each space of invariants, and compute the subspaces of invariants vanishing on rank 5 tensors. Oeding (Auburn) tensors of rank 5 December 15, / 24

20 Invariant Invariants How to compute with U Σ5 d and U Σ5,sgn d : Apply same Young symmetrizer method to detect non-zero fillings F. Compute for y in ambient space, σ.p F (y) or σ.p F (y) sgn(σ) σ Σ 5 σ Σ 5 by evaluating σ.p F (y) = p σ.f (y) on a different processor for each σ. Then sum the results (with or without signs). Repeat for new fillings and evaluating on y in ambient space until finding enough linearly independent fillings. Evaluate again on x j X and compute the kernel of the associated matrix ( ) Q(x) = p σfj (x i ). σ Σ Each evaluation took between 500 and 23,000 seconds and up to approximately 10GB of RAM on our servers: 24 cores 2.8 GHz Intel Xeon processors, 144 GB RAM 40 cores 2.8 GHz Intel Xeon processors, 256GB of RAM Oeding (Auburn) tensors of rank 5 December 15, / 24

21 Check invariants of degrees 8, 10, 12, 14, 16. In degree 16, dim U 16 = Compute evaluations in parallel, and then sum results with / without signs. Try S 5 -skew-invariants, U S5,sgn 16 : Find a basis of the 10-dimensional space U S 5,sgn 16 of skew invariants. Evaluate the basis on 10 random points of σ5 Store results in a matrix and compute its rank discover U S 5,sgn 16 I(X) is full-dimensional, so no new equations. Try S 5 -invariants, U S5 16 : Find a basis of the 39-dimensional space U S 5 16 of invariants. Evaluate the basis on 10 random points of σ5 Store results in a matrix and compute its rank discover U S 5 16 I(X) has dimension 36 (random). f6 U S 5,sgn 10 is 2-dimensional, so there is 1 minimal generator of degree 16 in I(X). Oeding (Auburn) tensors of rank 5 December 15, / 24

22 Theorem (Oeding-Sam 2015) The affine cone of σ 5 (Seg(P 1 5 )) is a complete intersection of two equations: one of degree 6, and one of degree 16. Use Bertini (with Hauenstein s help) to find deg σ 5 (Seg(P 1 5 )) = 96 Known codim 2, so we suspect complete intersection of two polynomials. Compute the only degree 6 invariant f 6, and show that it vanishes on an open subset of X (and thus on all of X). Check invariants of degree 8,10,12,14, 16. In degree 16, discover one new generator, f 16 vanishes on any number of random points of X. Y = V (f 6, f 16 ), a complete intersection. Also X Y and deg X deg Y = 96. Since X is irreducible of codimension 2, and Y is equidimensional, Y is also irreducible (otherwise the degree inequality would be violated). So X is the reduced subscheme of Y. Also, this implies that deg(x) = deg(y ), so Y is generically reduced. Since Y is Cohen Macaulay, generically reduced is equivalent to reduced. Hence X = Y is a complete intersection. Oeding (Auburn) tensors of rank 5 December 15, / 24

23 Consequences of main result Consider X := σ 5 (PV 1 PV 5 ) for any V i. Let G = 5 i=1 GL(V i). Functoriality implies that the equations G.f 6 and G.f 16 vanish on X, and are the only minimal generators of I( X) coming from modules where all partitions have at most 2 parts. (Use Sam and Snowden s -module and twisted commutative algebra theory.) These are new equations from secant varieties, which, as far as we know, don t come from flattenings. A new example of a secant variety of a segre product that is a complete intersection and hence arithmetically Cohen-Macaulay. Oeding (Auburn) tensors of rank 5 December 15, / 24

24 The End Oeding (Auburn) tensors of rank 5 December 15, / 24