aa r X i v : . [ m a t h . C O ] F e b An Optimal Inverse Theorem
Alex Cohen ∗ Guy Moshkovitz † Abstract
We prove that the partition rank and the analytic rank of tensors are equal upto a constant, over any large enough finite field. The proof constructs rational mapscomputing a partition rank decomposition for successive derivatives of the tensor, onan open subset of the kernel variety associated with the tensor. This largely settlesthe main question in the “bias implies low rank” line of work in higher-order Fourieranalysis, which was reiterated by Kazhdan and Ziegler, Lovett, and others.
The interplay between the structure and randomness of polynomials is a recurrent theme incombinatorics, analysis, computer science, and other fields. A basic question in this line ofresearch asks: if a polynomial is biased—in the sense that its output distribution deviatessignificantly from uniform—must it be the case that it is structured, in the sense that itis a function of a small number of lower-degree polynomials? Quantifying the trade-offbetween “biased” and “structured” for polynomials has been the topic of many works, andhas applications to central questions in higher-order Fourier analysis, additive combinatorics,effective algebraic geometry, number theory, coding theory, and more.Let us state the question formally. As is often done, we will only consider polynomials thatare multilinear forms ; it is known that the original question for degree- k polynomials reducesto the case of k -linear forms, by symmetrization, with little loss in the parameters and overany field F of characteristic char( F ) > k (see, e.g., [6, 9]). A k -linear form, or a k -tensor , overa field F is a function T : V × · · · × V k → F , with V i finite-dimensional linear spaces over F ,that is separately linear in each of its k arguments; equivalently, T is a degree- k homogeneouspolynomial of the form T ( x ( ) , . . . , x ( k ) ) = P I =( i ,...,i k ) T I Q kj =1 x ( j ) i j with T I ∈ F . One mayalso identify a k -tensor T with the k -dimensional matrix ( T I ) I ∈ F n ×···× n k , or with anelement of the tensor space V ⊗ · · · ⊗ V k . Structure and randomness for tensors are definedusing the following two notions: ∗ Yale University, New Haven, CT 06511. Email: [email protected] . † Department of Mathematics, City University of New York (Baruch College), New York, NY 10010.Email: [email protected] . Part of this research was conducted during the 2020 CUNY Discrete MathREU. The partition rank
PR( T ) of T over a field F is the smallest r such that T is a sum of r tensors of the form pq , where p and q are tensors over F in complementary, nonemptysubsets of the variables x ( ) , . . . , x ( k ) (i.e., p : Q i ∈ S V i → F and q : Q i ∈ T V i → F forsome partition S ∪ T = [ k ] with S, T = ∅ ). • The analytic rank
AR( P ) of T over a finite field F isAR( P ) := − log | F | E x ∈ V ×···× V k χ ( T ( x ))where χ is any nontrivial additive character (the definition is independent of the choiceof χ ).Partition rank was introduced by Naslund [18] (a similar notion for polynomials was con-sidered earlier by Schmidt [19], and more recently by Ananyan and Hochster [2]). Analyticrank was introduced by Gowers and Wolf [6] in the context of higher-order Fourier analysis.It is known that AR( T ) ≤ PR( T ) (see Kazhdan and Ziegler [10] and Lovett [16]). Thestructure-vs-randomness question is the “inverse” problem, asking to bound PR( T ) fromabove in terms of AR( T ).Following Green and Tao [7], Lovett and Bhowmick [3] proved that PR( T ) ≤ f k (AR( T ))with f k ( x ) of Ackermann-type growth. Much better bounds were obtained for specific valuesof k by Haramaty and Shpilka [8]: PR( T ) ≤ AR( T ) for k = 3, and PR( T ) ≤ O (AR( T )) for k = 4; Lampert [14] improved the bound for k = 4 to a polynomial bound. Back to k = 3, a recent result of the authors [4] gives PR( T ) ≤ o | F | (1)) AR( T ) if F = F ;Adiprasito, Kazhdan, and Zieglar [1] independently obtained a similar result using relatedideas. However, the problem of obtaining a reasonable bound for k -tensors remained elusive,until the groundbreaking works of Mili´cevi´c [17], and of Janzer [9] for small fields, whoobtained polynomial bounds: PR( T ) ≤ AR( T ) D with D = 2 poly ( k ) , thus proving a conjectureof Kazhdan and Ziegler [11].Multiple authors (e.g., [1, 8, 12, 16]) have asked whether—or conjectured that— PR( T )and AR( T ) are in fact equal up to a constant: that structure and randomness of polynomialsare two sides of the same coin. Conjecture 1.
For every k ≥ there is C = C ( k ) such that the following holds for everyfinite field F . For every k -tensor T over F , PR( T ) ≤ C · AR( T ) . In this paper we prove the conjecture, over any large enough field.
Theorem 1.1.
For every k ≥ and n there is F = F ( n, k ) such that the following holds forevery finite field F with | F | > F . For every k -tensor T ∈ F n ×···× n k with n , . . . , n k ≤ n , PR( T ) ≤ (2 k − −
1) AR( T ) + 1 . We note that Theorem 1.1 applies in arbitrary characteristic. Moreover, the bound onthe field size F ( n, k ) is quite effective (no more than an i -fold exponential with a fixed i ;possibly double exponential). 2 .1 Proof overview We begin our proof, as mathematicians often do, by moving beyond the given setting inConjecture 1 of a finite field F , and instead work in the algebraic closure F . This enablesus to use tools from algebraic geometry. The main drawback of this approach, as onemight assume, is that finding a partition rank decomposition over F —that is, one where thesummands are tensors over F —need not yield a partition rank decomposition over F , whichis what Conjecture 1 calls for.The analogue of analytic rank over an algebraically closed field F is the geometric rank . Todefine it, we view a k -tensor T : V ×· · ·× V k → F ( V i are finite-dimensional vector spaces over F ) as a ( k − T : V × · · · × V k − → V k ; this is done by considering the “slices” of T , say along the V k axis, as ( k − k = 2 this is the familiar correspondencebetween matrices and linear maps). The geometric rank GR( T ) is the codimension of thekernel variety of T , ker T = n x ∈ V × · · · × V k − (cid:12)(cid:12)(cid:12) T ( x ) = o . Geometric rank was introduced in [13], where in particular it was shown to be independentof the choice of axis V k to slice T along; for k = 2 this is column rank equals row rank.(A similar definition over the complex numbers was used by Schmidt [19] in the context ofnumber theory.) That GR( T ) is the analogue of AR( T ) over F is suggested by the identityAR( T ) = ( P k − i =1 dim V i ) − log | F | | ker T ( F ) | , where ker T ( F ) is the set of F -rational points inker T . Thus, moving to the algebraic closure recasts Conjecture 1 as asking to bound analgebraic notion of complexity (partition rank) by a geometric notion of complexity (thecodimension of the kernel variety).In order to circumvent the main drawback of moving to F mentioned above, an importantidea in our proof is that we not only find a partition rank decomposition for T but also for awhole family of related tensors: the total derivatives of T of all orders, D T , D T , . . . , D k − T ,at all points in an open subset of ker T . Specifically, for each derivative D i T we constructa rational map that takes each point from that open subset to some partition rank decom-position. To find such a rational (partition rank) decomposition , we consider the tangentspace of the variety at each of the points, parameterize it using a rational map, and take aderivative along the tangent vectors.We note that the total derivative D T is given by the Jacobian matrix, which makes it a polynomial matrix , meaning a matrix whose entries are polynomials; more generally, D d T isa polynomial d -tensor . At the core of our result is a proof that if a polynomial d -tensor f has a rational decomposition on an open subset of a variety V , then the total derivative D f has a rational decomposition, on an open subset of V , of size bounded by the codimensionof V . Since the total derivative of f is a polynomial ( d + 1)-tensor, we are back in thesame setting as with f , enabling an iterative process. We apply this process starting withthe polynomial matrix D T and the kernel variety ker T . To begin the process, however,we must first obtain a rational decomposition of D T by other means. This turns out torequire a rational map computing a rank factorization of the polynomial matrix D T , whichwe explicitly construct. We then apply the iterative process so that, by the end of it, we have3 rational decomposition, on an open subset of ker T , for the polynomial k -tensor D k − T .Since the total derivative retains all the partial derivatives throughout the iterations, we canreconstruct T from D k − T by evaluating the latter at any point. Now, one can obtain apartition rank decomposition over F by simply evaluating at any point from the open subsetof ker T on which the rational decomposition is defined. This gives the following bound. Theorem 1.2.
For every k -tensor T over any algebraically closed field, PR( T ) ≤ (2 k − −
1) GR( T ) . Crucially, our proof method is flexible enough to yield a partition rank decomposition overthe finite field F —as long as we can guarantee that our open subset of ker T has an F -rationalpoint. In the last part of the proof of Theorem 1.1, we first observe that in order to boundthe analytic rank of T it suffices to apply our iterative process on any subvariety of ker T that contains all its F -rational points. We then show, using results from effective algebraicgeometry, that one can indeed find such a subvariety that, moreover, can be guaranteed tohave at least one F -rational point in the domain of our rational decomposition. Evaluatingour rational decomposition at that point yields a small partition rank decomposition over F .As is evident from the overview above, our proof is self-contained and does not applyresults from additive combinatorics, nor any “regularity lemma” for polynomials or notionsof quasi-randomness. This makes our proof quite different from arguments previously appliedin this line of research.Finally, we note that, among other applications, Theorem 1.1 gives the best knownbounds, over large enough finite fields, for the inverse theorem for Gowers norms in thespecial case of polynomial phase functions (see, e.g., Green and Tao [7] and Janzer [9]). In Section 2 we specify the space in which our decompositions live, and relate them topartition rank decompositions. In Section 3 we define the total derivative of rational maps,and prove that a total derivative of a tensor of a sufficiently high order contains enoughinformation to reconstruct the original tensor. In Section 4 we give a rational map for a rankfactorization of a matrix (used in the base case of our inductive proof) and a rational mapparameterizing the kernel of a matrix (used in the induction step). In Section 5 we definetangent spaces, and obtain a rational map parameterizing the tangent space of a variety atnon-singular points. In Section 6 we give our inductive proof, which yields a partition rankdecomposition assuming a non-singular F -rational point on the kernel variety of the tensor.Finally, in Section 7 we complete the proof of Theorem 1.1 by showing that over large enoughfields, a non-singular F -rational point can always be found. As we later learned through [1], Theorem 1.2 improves a previous upper bound of Schimdt [19] of roughly k !(log 2) − k GR( T ). Schmidt’s proof is over the complex numbers, and it is unclear whether it can be usedto obtain a partition rank decomposition over finite fields. .3 Notation An F -polynomial is a polynomial with coefficients in F (i.e., an element of F [ x ]). A ra-tional map f : F n F m is an m -tuple of rational functions over F (i.e., quotients of F -polynomials). Note that a rational map can be viewed as a function whose domain Dom( f )is the set of inputs for which all m rational functions are defined. We will use the conventionthat f can also be evaluated at points x ∈ F n , in which case f ( x ) ∈ F m if x ∈ Dom( f )( ⊆ F n ).We will need some basic algebro-geometric terminology. We say that a variety V ⊆ F n is defined over F if it can be cut out by some F -polynomials f , . . . , f m , meaning V = { x ∈ F n | f ( x ) = · · · = f m ( x ) = 0 } . The ideal of V is I( V ) = { f ∈ F [ x ] | ∀ x ∈ V : f ( x ) = 0 } . Any variety V can be uniquelywritten as the union of irreducible varieties, where a variety is said to be irreducible if itcannot be written as the union of strictly contained varieties. The dimension of a variety V ,denoted dim V , is the maximal length d of a chain of irreducible varieties ∅ 6 = V ( · · · ( V d ( V . The codimension of V ⊆ F n is simply codim V = n − dim V . A point x ∈ V issaid to be F -rational if x ∈ F n ; we denote by V ( F ) the set of F -rational points in V .Crucial to our proof is the simultaneous use of different aspects of a k -tensor. Henceforth,we fix bases for our vector spaces, so that we will freely pass between a vector space V andits dual V → F , between matrices in V ⊗ V and linear maps V → V , and more generally,between k -tensors in V ⊗ · · · ⊗ V k and k -linear forms V × · · · × V k → F or ( k − V × · · · × V k − → V k . Let V = N ki =1 V i be a space of k -tensors over a field F . The r -constructing space over V isthe vector space C r ( V ) = M ∅6 = S ⊆ [ k − (cid:0) V ⊗ S × V ⊗ S (cid:1) r where V ⊗ X := N i ∈ X V i for X ⊆ [ k ]. Note that the direct summands correspond to un-ordered, non-trivial partitions { S, T } of [ k ] (i.e., S ∪ T = [ k ] with S, T = ∅ ). For example,the order-2 and order-3 r -constructing spaces are C r ( V ⊗ V ) = (cid:0) V × V (cid:1) r , C r ( V ⊗ V ⊗ V ) = (cid:0) V × ( V ⊗ V ) (cid:1) r ⊕ (cid:0) V × ( V ⊗ V ) (cid:1) r ⊕ (cid:0) ( V ⊗ V ) × V (cid:1) r . We will frequently identify C r ( V ) with the isomorphic linear space C r ( V ) × C r ( V ) = (cid:18) M ∅6 = S ⊆ [ k − V ⊗ S (cid:19) r × (cid:18) M ∅6 = S ⊆ [ k − V ⊗ S (cid:19) r . We take the tensor product in order; that is, if X = { i < i < · · · } then V ⊗ X = V i ⊗ V i ⊗ · · · . k = 3, C r ( V ) × C r ( V ) = (cid:16) V ⊕ V ⊕ ( V ⊗ V ) (cid:17) r × (cid:16) ( V ⊗ V ) ⊕ ( V ⊗ V ) ⊕ V (cid:17) r . Let IP : U n × V n → U ⊗ V be the inner product function ,IP( u , . . . , u n , v , . . . , v n ) = X i ∈ [ n ] u i ⊗ v i . Observe that applying IP on the constructing space C r ( V ) = C r ( V ) × C r ( V ) gives a map intothe tensor space V :IP : ( u ( i , S ) , v ( i , S ) ) i ∈ [ r ] , ∅6 = S ⊆ [ k − X i ∈ [ r ] X ∅6 = S ⊆ [ k − u ( i , S ) ⊗ v ( i , S ) ∈ V , where ( u ( i , S ) ) ∈ C r ( V ) and ( v ( i , S ) ) ∈ C r ( V ). Note that, if we identify each vector space V i of V with some F n i , then IP is a polynomial (in fact bilinear) map into F n ×···× n k ; explicitly,IP : ( x ( i,S ) I , y ( i,S ) J ) i ∈ [ r ] , ∅6 = S ⊆ [ k − ,I ∈ n S ,J ∈ n S (cid:16) X i ∈ [ r ] X ∅6 = S ⊆ [ k − x ( i,S ) I S y ( i,S ) I S (cid:17) I ∈ n [ k ] (2.1)where n X = Q i ∈ X [ n i ] for X ⊆ [ k ], and I X denotes the projection of I = ( i , . . . , i k ) ∈ n [ k ] onto X , meaning I X = ( i x ) x ∈ X ∈ n X .Since the number of sets ∅ 6 = S ⊆ [ k −
1] is 2 k − −
1, we have, immediately from definition,that if a k -tensor lies in the image of IP on the r -constructing space then its partition rankis at most (2 k − − r . Fact 2.1.
Let T ∈ V = N ki =1 V i with V i vector spaces over F . If T ∈ IP( C r ( V )) then PR( T ) ≤ (2 k − − r . The total derivative of a rational map f = ( f , . . . , f m ) : F n F m is the rational mapD f : F n F m ⊗ F n given by D f ( x ) = ( ∂ j f i ( x )) i ∈ [ m ] ,j ∈ [ n ] . That is, the total derivative maps each point to the Jacobian matrix of the rational functions f , . . . , f m evaluated at that point. In particular, D f is a matrix whose every entry is arational function. The directional derivative of f along v ∈ F n is thus given by(D f ( x )) v = ( ∂ f i ( x ) v + · · · + ∂ n f i ( x ) v n ) i ∈ [ m ] ∈ F m . a ∈ N , theorder- a total derivative of a rational map f = ( f , . . . , f m ) : F n F m is the rational mapD a f : F n F m ⊗ ( F n ) ⊗ a given byD a f ( x ) = ( ∂ j ,...,j a f i ( x )) i ∈ [ m ] ,j ,...,j a ∈ [ n ] . In particular, D a f is an ( a + 1)-dimensional matrix, or ( a + 1)-tensor, whose every entry isa rational map. We will need the following properties of the total derivative: • Sum rule: D( f + g ) = D f + D g . • Product rule: D( f g ) = g D f + f D g . • Quotient rule: D( f /g ) = (1 /g )( g D f − f D g ). • Chain rule: D( f ◦ g )( x ) = (D f )( g ( x )) D g ( x ). • The rational maps f and D a f have the same domain.Note that if f is a single polynomial of degree d then D a f is an a -tensor whose everycomponent is a polynomial of degree d − a . In particular, D d f is a constant d -tensor.Henceforth, a slicing of a k -tensor T ∈ N ki =1 V i is the tuple of ( k − T = ( T , T , . . . ) ∈ V j ⊗ N i = j V i of T along one of the axes V j , which we view as a ( k − T : Q i = j V i → V j . Recall that for any a ∈ N and x , the order- a total derivativeD a T ( x ) ∈ V j ⊗ ( Q i = j V i ) ⊗ a is an ( a + 1)-tensor. Claim 3.1.
Let T be a k -tensor over F , and let T be a slicing of T . Then PR( T ) ≤ PR(D k − T ( x )) for every x .Proof. If T is the slicing of T along V j then by identifying V j with F m and Q i = j V i with F n , we see that it suffices to prove the following claim: For every d -linear map f : F n → F m and x ∈ F n we have that f (cid:22) D d f ( x ), by which we mean that f can be obtained from the( d + 1)-tensor D d f ( x ) ∈ F m ⊗ ( F n ) d by viewing it as a d -linear map D d f ( x ) : ( F n ) d → F m and evaluating it at some point y ∈ ( F n ) d . Clearly, it suffices to prove, for each of the md -linear forms p : F n → F of f : F n → F m , that p (cid:22) D d p ( x ) for every x ∈ F n . Write p ( x ) = p ( x , . . . , x d ) = X J =( j ,...,j d ) c J x ,j · · · x d,j d with c J ∈ F . The d -tensor D d p ( x ) ∈ ( F n ) ⊗ d is given byD d p ( x ) = ( ∂ ( i ,j ) ,..., ( i d ,j d ) p ( x )) ( i ,j ) ,..., ( i d ,j d ) = ( c j ,...,j d i , . . . , i d ∈ [ d ] distinct0 otherwise . d -linear form D d p ( x ) : ( F n ) d → F is given byD d p ( x ) = X j ,...,j d X i ,...,i d distinct c j ,...,j d y (1) i ,j · · · y ( d ) i d ,j d and thus evaluates (at y ( t ) i,j = [ t = i ] · y i,j ) to(D d p ( x ))(( y , . . . , , (0 , y , , . . . , , . . . , (0 , . . . , , y d )) = p ( y ) . Therefore, p (cid:22) D d p ( x ) for every x . As explained above, this completes the proof. For a matrix A ∈ F m × n , we denote by A x × y = ( A i,j ) i ∈ [ x ] ,j ∈ [ y ] the submatrix of A consistingof the first x rows and the first y columns. A rank factorization of a matrix A ∈ F m × n ofrank r is a pair of matrices ( A , A ) ∈ F m × r × F r × n such that A = A A . We next show thatrank factorization can be done via a rational map, mapping (an open subset of) matrices ofrank exactly r to a rank- r factorization thereof. Lemma 4.1.
Let m, n, r ∈ N and F any field. There is a rational map F : F m × n F m × r × F r × n such that for any matrix A of rank r with A r × r invertible, F ( A ) = ( A , A ) isa rank factorization of A . Moreover, A has I r as a submatrix.Proof. We first show that if rank( A ) = r and A r × r is invertible then we have the matrixidentity A = A m × r ( A r × r ) − A r × n . (4.1)Since the column rank of A equals rank( A ) = r , the first r columns of A are not just linearlyindependent but also a basis for the column space. Therefore, there is a matrix X ∈ F r × n such that A m × r X = A. Restricting this equality to the first r rows, we have A r × r X = A r × n . Since A r × r is invertible,we obtain X = ( A r × r ) − A r × n . Combining the above gives the identity (4.1).Let F : F m × n F m × r × F r × n be the rational map F ( A ) = ( A m × r , det( A r × r ) − adj( A r × r ) A r × n ) , where we recall that the adjugate matrix adj( A r × r ) has the property that each entry is apolynomial in the entries of A r × r (a cofactor). We have (via Cramer’s rule) that ( A r × r ) − =det( A r × r ) − adj( A r × r ). Thus, by (4.1), F ( A ) is a rank factorization for any rank- r matrix A with det A r × r = 0. Finally, for the “moreover” part, note that ( A r × r ) − A r × n contains as itsfirst r columns the submatrix ( A r × r ) − A r × r = I r .8ext, we show that projecting onto the kernel of a matrix can be done by a rational map.Recall that a matrix P is a projection matrix onto a subspace U if P = P and im P = U . Lemma 4.2.
Let m, n, r ∈ N and F any field. There is a rational map P : F m × n F n × n such that for any matrix A of rank r with A r × r invertible, P ( A ) is a projection matrix onto ker A ; moreover, in the projection matrix I n − P ( A ) all but the first r rows are zero.Proof. Let F : F m × n F m × r × F r × n be the rank-factorization rational map given byLemma 4.1. Let A be a matrix of rank r with A r × r invertible, and write F ( A ) = ( A , A ).Then A = A A and A = (cid:0) I r X (cid:1) r × n for some X ∈ F r × ( n − r ) . Let P = (cid:18) − X I n − r (cid:19) n × n . (4.2)Note that P = P . Furthermore, for d = n − r , we have that A P = (cid:0) I r X (cid:1) r × n (cid:18) r × r − X d × r I d (cid:19) n × n = (cid:0) (cid:1) r × n . It follows that AP = A A P = ( ) m × n . Thus, the d nonzero columns of P lie in ker A , andsince they are linearly independent and dim ker A = n − rank A = d , we deduce that thecolumn space of P is ker A , that is, im P = ker A . This means P is a projection matrix ontoker A . Now, observe that the function that maps any matrix A —provided A is of rank r with A r × r invertible—to the matrix in (4.2) is given by a rational map P : F m × n F n × n .This completes the proof. The tangent space of a variety V ⊆ F n at x ∈ V is the linear space T x V = n v ∈ F n (cid:12)(cid:12)(cid:12) ∀ g ∈ I( V ) : (D g ( x )) v = 0 o . Equivalently, if { g , . . . , g s } ⊆ F [ x , . . . , x n ] is a generating set for the ideal I( V ) then thetangent space to V at x ∈ V is the kernel T x V = ker J( x ), where J( x ) is the Jacobianmatrix ( ∂ x g )( x ) · · · ( ∂ x n g )( x )... . . . ...( ∂ x g s )( x ) · · · ( ∂ x g s )( x ) s × n . A point x on an irreducible variety V is non-singular if dim T x V = dim V . We have thefollowing basic fact about tangent spaces (see, e.g., Theorem 2.3 in [20]). Fact 5.1.
For every irreducible variety V and x ∈ V , dim T x V ≥ dim V . perfect if each element has a root of order char( F ) or char( F ) = 0. Inparticular, algebraically closed fields and finite fields are perfect. We will use the followingknown fact (which can be obtained from Lemma 9.5.19 in [22]). Fact 5.2.
If an irreducible variety V is defined over a perfect field F then I( V ) has a gen-erating set of F -polynomials. We next show that over a perfect field, the tangent spaces—at non-singular points in anontrivial open subset—can be parameterized by a rational map.
Lemma 5.3.
Let V ⊆ F n be an irreducible variety defined over a perfect field F , and x a non-singular point on V . There is a rational map P : F n F n × n that is defined at x ,such that for every x ∈ V ∩ Dom( P ) , P ( x ) is a projection matrix onto T x V ; moreover, inthe projection matrix I n − P ( x ) all but the first codim V rows are zeroProof. Since F is a perfect field, Lemma 5.2 implies that the ideal I( V ) has a generating setof F -polynomials, { g , . . . , g m } ∈ F [ x ]. Let J : x ( ∂ j g i ( x )) i ∈ [ m ] ,j ∈ [ n ] be the correspondingJacobian matrix, viewed as a polynomial map J : F n → F m × n . By definition, ker J( x ) = T x V for every x ∈ V .Put r = codim V . Since x is a non-singular point on V , we have that rank(J( x )) =codim V = r . In particular, the submatrix J( x ) I × J is invertible for some I ⊆ [ m ] and J ⊆ [ n ] with | I | = | J | = r . Apply Lemma 4.2 with m, n, r , together with appropriate rowand column permutations, to obtain a rational map P : F m × n F n × n such that for anymatrix A , with rank A = r and A I × J invertible, P ( A ) is a projection matrix onto ker A .By construction, P is defined at J( x ). Let P : F n F n × n be the rational map given bythe composition P = P ◦ J. Then P is defined at x ; moreover, for every x ∈ V , whererank J( x ) = r and J( x ) I × J invertible, we have that P ( x ) = P (J( x )) is a projection matrixonto ker J( x ) = T x V . This completes the proof. Recall that a polynomial matrix is polynomial map into matrices (or a matrix whose everyentry is a polynomial). For a polynomial matrix f : F n → F m × n , we say that a point x on avariety V ⊆ F n is f -singular if rank f ( x ) < max x ∈ V rank f ( x ). Note that this extends thenotion of a singular point, which is obtained by taking f to be the Jacobian of a generatingset of I( V ). In this section we prove the following bound on the partition rank—over anyperfect field. Theorem 6.1.
Let T be a k -tensor over a perfect field F . Let X ⊆ ker T be a variety definedover F which has an F -rational point that is neither singular nor D T -singular. Then PR( T ) ≤ (2 k − −
1) codim X . F in Theorem 6.1 is algebraically closed, taking X = ker T and a genericpoint on it immediately gives the bound PR( T ) ≤ (2 k − −
1) GR( T ), thus proving Theo-rem 1.2.The proof of Theorem 6.1 starts with a slicing T of T , obtains a rational decomposition forthe polynomial matrix D T using the rank-factorization map from Lemma 4.1, then proceedsto iteratively construct a rational decomposition for D T , D T , . . . , D k − T by parameterizingthe tangent spaces to X and taking derivatives along tangent vectors, and finally evaluates therational decomposition at the guaranteed non-singular F -rational point to obtain a partitionrank decomposition over F for T itself.The following theorem gives the inductive step in the proof of Theorem 6.1. For rationalmaps f and g we write f | V = g | V if f ( x ) = g ( x ) for every x ∈ V for which f ( x ) and g ( x )are defined. Theorem 6.2.
Let V ⊆ F n be an irreducible variety defined over a perfect field F , x anon-singular point on V , and f : F n → N ki =1 V i =: V a polynomial map. If there existsa rational map H : F n C r ( V ) defined at x such that f | V = IP ◦ H | V then there existsa rational map H ′ : F n C s ( V ⊗ F n ) defined at x such that (D f ) | V = IP ◦ H ′ | V with s = max { r, codim V } . Theorem 6.2 can be illustrated using the commutative diagrams: V C r ( V ) F n V H IP f = ⇒ V C s ( V ⊗ F n ) F n V ⊗ F n H ′ IPD f The proof of Theorem 6.2 proceeds by splitting the derivative D f into two parts: a mainterm which is mapped into the portion of the constructing space V ⊗ F n that only involvestensors on V , . . . , V n , and a remainder term—controlled by the codimension of V —which ismapped into the complementary portion of V ⊗ F n .Before going into the proof, we remind the reader that if g : F n V is a rational mapinto a space V of k -tensors, then the total derivative D g ( x ) ∈ V ⊗ F n is a ( k + 1)-tensor, whilethe directional derivative of g along any vector v ∈ F n is again a k -tensor, (D g ( x )) v ∈ V . Proof.
For h ∈ C r ( V ) we henceforth write h = ( h , h ) ∈ C r ( V ) × C r ( V ). Recall from (2.1)that IP on the constructing space C r ( V ) can be viewed as a bilinear map. We claim that forevery x = ( x , x ) , y = ( y , y ) ∈ C r ( V ) we have(D IP( x , x ))( y , y ) = IP( x , y ) + IP( x , y ) . To see this, note that each component of the map IP is of the form b ( x , x ) = P ℓi =1 x ,i x ,i (where the sum is over a subset of the variables on C r ( V )). Using the product rule for the11otal derivative, we have(D b ( x , x ))( y , y ) = ( x , , . . . , x ,ℓ , x , , . . . , x ,ℓ ) · ( y , , . . . , y ,ℓ , y , , . . . , y ,ℓ )= ℓ X i =1 x ,i y ,i + ℓ X i =1 x ,i y ,i = b ( x , y ) + b ( x , y ) , which proves our claim. Using the chain rule for the total derivative,D(IP ◦ H )( x ) = (D IP)( H ( x )) D H ( x )= (D IP)( H ( x ) , H ( x ))(D H ( x ) , D H ( x ))= IP( H ( x ) , D H ( x )) + IP( H ( x ) , D H ( x )) . (6.1)Put O = V ∩ Dom( H ). Since x ∈ O , we have that O is a non-empty open subset of theirreducible variety V , and thus its Zariski closure is O = V . Taking the total derivative atany x ∈ O along a tangent vector v ∈ T x V , we claim that(D f ( x )) v = (D IP ◦ H ( x )) v . (6.2)Indeed, if p/q ( p, q ∈ F [ x ]) is any one of the components of f − IP ◦ H then, since ( f − IP ◦ H )( x ) = 0, we have p ( x ) = 0. Thus, p ( y ) = 0 for every y ∈ O = V , and so p ∈ I( V ).Since (D p ( x )) v = 0 by the definition of a tangent space, we haveD pq ( x ) = 1 q ( x ) ( q ( x ) D p ( x ) − p ( x ) D q ( x )) = 1 q ( x ) D p ( x ) , which means (D( p/q )( x )) v = 0, and so (D( f − IP ◦ H )( x )) v = 0, verifying (6.2).Apply Lemma 5.3, with V and the non-singular x , to obtain a rational map P : F n F n × n that is defined at x , such that P ( x ) is a projection matrix onto T x V for every x ∈ V ∩ Dom( P ). Let Q ( x ) = I n − P ( x ). Put O ′ = V ∩ Dom( H ) ∩ Dom( P ), and note that x ∈ O ′ . We deduce that for every x ∈ O ′ we haveD f ( x ) = (D f ( x ))( P ( x ) + Q ( x )) = (D IP ◦ H ( x )) P ( x ) + (D f ( x )) Q ( x )= IP( H ( x ) , (D H ( x )) P ( x )) + IP( H ( x ) , (D H ( x )) P ( x ))+ codim V X i =1 (D f ( x )) Q ( x ) i ⊗ ( Q t ( x ) i ) t , (6.3)where the second equality follows from (6.2), and the last equality applies (6.1), and more-over, uses the decomposition of the projection matrix Q = Q ( x ) as Q = Q = codim V X i =1 Q i (( Q t ) i ) t = codim V X i =1 Q i ⊗ ( Q t ) i where A i denotes column i of a matrix A . 12ut r ′ = codim V . Let H ′ ( x ) = ( H ′ ( x ) , H ′ ( x )) be given by H ′ ( x ) = H ( x ) ⊕ H ( x ) ⊕ (D( f ( x )) Q ( x ) i ) r ′ i =1 ,H ′ ( x ) = (D H ( x )) P ( x ) ⊕ (D H ( x )) P ( x ) ⊕ ( Q t ( x ) i ) r ′ i =1 . Note that for every x ∈ F n we have H ′ ( x ) ∈ (cid:16) M ∅6 = S ⊆ [ k − ( V ⊗ S ) r ⊕ M ∅6 = S ⊆ [ k − ( V ⊗ [ k ] \ S ) r ⊕ ( V ⊗ [ k ] ) r ′ (cid:17) ,H ′ ( x ) ∈ (cid:16) M ∅6 = S ⊆ [ k − ( V ⊗ [ k ] \ S ⊗ V k +1 ) r ⊕ M ∅6 = S ⊆ [ k − ( V ⊗ S ⊗ V k +1 ) r ⊕ ( V k +1 ) r ′ (cid:17) . Let U = N k +1 i =1 V i = V ⊗ V k +1 with V k +1 a vector space isomorphic to F n . Observe that H ′ ( x ) ∈ (cid:18) M ∅6 = S ⊆ [ k ] (cid:0) U ⊗ S (cid:1) max { r,r ′ } (cid:19) × (cid:18) M ∅6 = S ⊆ [ k ] (cid:0) U ⊗ [ k +1] \ S (cid:1) max { r,r ′ } (cid:19) = C s ( U ) × C s ( U ) . Thus, H ′ : F n C s ( U ) is a rational map with Dom( H ′ ) = Dom( H ) ∩ Dom( P ). We deducethat H ′ is defined at x and, by (6.3), satisfies for every x ∈ V ∩ Dom( H ′ ) thatD f ( x ) = IP( H ′ ( x )) . This completes the proof.
Corollary 6.3.
Let V ⊆ F n be an irreducible variety defined over a perfect field F , x a non-singular point on V , and f : F n A ⊗ B a polynomial matrix. If there exists arational map H : F n C r ( A ⊗ B ) defined at x such that f | V = IP ◦ H | V then, for every a ∈ N , there exists a rational map H ′ : F n C s ( A ⊗ B ⊗ ( F n ) ⊗ a ) defined at x such that (D a f ) | V = IP ◦ H ′ | V with s = max { r, codim V } .Proof. Let H : F n C r ( A ⊗ B ) be a rational map defined at x such that f | V = IP ◦ H | V .We proceed by induction on a ≥
0. The induction base a = 0 clearly holds with H ′ = H ,since (D f ) | V = f | V = IP ◦ H ′ | V , hence we move to the induction step. By the inductionhypothesis, there exists a rational map H ′ : F n C s ( A ⊗ B ⊗ ( F n ) ⊗ a ) defined at x suchthat (D a f ) | V = IP ◦ H ′ | V . Apply Theorem 6.2 with the rational map f ′ = D a f : F n A ⊗ B ⊗ ( F n ) ⊗ a =: V . As H ′ : F n C s ( V ) satisfies f ′ | V = (D a f ) | V = IP ◦ H ′ | V , it follows from this applicationthat there exists a rational map H ′′ : F n C s ′ ( V ⊗ F n ) defined at x such that(D a +1 f ) | V = (D f ′ ) | V = IP ◦ H ′′ | V with s ′ = max { s, codim V } = s . This completes the induction step and the proof.13 roof of Theorem 6.1. Let T ∈ N i ∈ [ k ] V i be a k -tensor over a perfect field F . Identify V ×· · · × V k − with F n , and V k with F m . Let T : F n → F m be a slicing of T , and consider thepolynomial matrix D T : F n → F m ⊗ F n . First, we claim that for every x ∈ X ,rank(D T ( x )) ≤ codim X . (6.4)Write T = ( T , . . . , T m ), so that the ( k − T , . . . , T m cut out ker T . This meansthat { T , . . . , T m } ⊆ I(ker T ) ⊆ I( X ), using the statement’s assumption X ⊆ ker T . We claimthat T x ( X ) ⊆ ker(D T ( x )); indeed, for every x ∈ X , if v ∈ T x ( X ) then (D g ( x )) v = 0 forevery g ∈ I( X ), and in particular for every T i , so(D T ( x )) v = (D T ( x ) , . . . , D T m ( x )) v = , meaning v ∈ ker(D T ( x )). We obtain (6.4) sincerank(D T ( x )) = n − dim ker(D T ( x )) ≤ n − dim T x ( X ) ≤ codim X where the last inequality applies Fact 5.1 using the irreducibility of X .Put f = D T : F n → F m ⊗ F n . By the statement’s assumption, X has an F -rationalpoint x which is neither singular nor f -singular. The latter means that r := rank f ( x )equals max x ∈ X rank f ( x ); or, put differently, f ( X ) ⊆ M r where M r ⊆ F m ⊗ F n is the setof matrices of rank at most r . Apply Lemma 4.1 with m , n , and r , and use appropriaterow and column permutations, to obtain a rational map F : F m × n F m × r × F r × n that isdefined at f ( x ) (using rank f ( x ) = r ), such that for any matrix A ∈ M r ∩ Dom( F ) we havethat F ( A ) = ( F ( A ) , F ( A )) is a rank factorization of A . Let F : F n F m × r × F r × n be therational map given by the composition F = F ◦ f . Note that Dom( F ) = Dom( F ). Thus, F is defined at x , and for any x ∈ f − ( M r ) ∩ Dom( F ) we have that F ( x ) = ( F ( x ) , F ( x ))satisfy f ( x ) = F ( x ) F ( x ) = r X i =1 F ( x ) i ( F t ( x ) i ) t = r X i =1 F ( x ) i ⊗ F t ( x ) i , where A i denotes column i of a matrix A . Let H ( x ) := ( F ( x ) i , F t ( x ) i ) i ∈ [ r ] . Observe thatfor every x ∈ F n we have that H ( x ) ∈ ( F m × F n ) r = C r ( F m ⊗ F n ). Note that Dom( H ) =Dom( F ). Thus, H : F n C r ( F m ⊗ F n ) is a rational map defined at x such that for every x ∈ f − ( M r ) ∩ Dom( H ) we have f ( x ) = IP( H ( x )). Since X ⊆ f − ( M r ), this means that f | X = IP ◦ H | X . Now, apply Corollary 6.3 with the irreducible X ⊆ F n , the non-singular x ∈ X , thepolynomial matrix f : F n → F m ⊗ F n , the rational map H : F n C r ( F m ⊗ F n ), and a = k − H ′ : F n C s ( F m ⊗ ( F n ) ⊗ k − ) that is defined at x such thatD k − f | X = IP ◦ H ′ | X for s = max { r, codim X } = codim X by (6.4). Evaluating at x ∈ X ,we obtain D k − T ( x ) = D k − f ( x ) = IP( H ′ ( x )) . x is F -rational, this implies that D k − T ( x ) ∈ IP( C codim X ( F m ⊗ ( F n ) ⊗ k − )). Therefore,PR( T ) ≤ D k − T ( x ) ≤ (2 k − −
1) codim X , where the first inequality follows from Claim 3.1 and the second inequality from Fact 2.1.This completes the proof. In this section we show how to construct from any variety a subvariety that captures all F -rational points, is defined over F and so are its irreducible components, and is of boundedcomplexity if the original variety is. We use this subvariety to finish the proof of Theorem 1.1 We will need several known results. For a variety V ⊆ F qn , where F q is the finite field of size q , we denote the image of the Frobenius automorphism on V by Frob V = { ( x q , . . . , x qn ) | ( x , . . . , x n ) ∈ V } . The Frobenius automorphism characterizes definability over F q (see, e.g.,the proof of Corollary 4 in [21]). Fact 7.1.
For V ⊆ F qn an irreducible variety, V is defined over F q if and only if Frob( V ) = V . We say that the complexity of variety V ⊆ F n is at most M if n ≤ M , and V can be cutout by at most M polynomials each of degree at most M . Similarly, the complexity of anideal h f , . . . , f m i ⊂ F [ x , . . . , x n ] is at most M if n, m ≤ M .To show that the algebro-geometric operations we will perform have a bounded effect onthe complexity, we state the following (see, e.g., [5]). Fact 7.2.
Let I ⊂ F [ x , . . . , x n ] be an ideal of complexity at most M , and F a perfect field.The primary decomposition of I can be computed in O M (1) steps assuming char( F ) = 0 orthat char( F ) is sufficiently large relative to M . As a consequence, the primary decompositionfor I consists of O M (1) ideals of complexity O M (1) . Remark 7.3.
The best algorithms for computing a primary decomposition take doublyexponential time in M , and they require the characteristic to be sufficiently large (or zero).Although we have not inspected these algorithms in detail, it seems likely that a doublyexponential bound on the characteristic suffices as well.As a consequence, we have the following bounds. Fact 7.4.
There is C : N → N such that for all varieties X , Y ⊆ F n of complexity at most M , each of the following is at most C ( M ) : • The number of irreducible components of X , The complexity of any irreducible components of X , • The complexity of the union X ∪ Y , • The complexity of the intersection X ∩ Y , • The complexity of the singular locus
Sing X . Fact 7.4 follows from Fact 7.2 since each of the items can be easily obtained from theprimary decomposition of the ideal of the variety—for instance, the irreducible componentscorrespond to the minimal primes in the decomposition. The first four bounds are similar.The bound on the complexity of the singular locus Sing X is obtained from the followingobservation. If X has complexity at most M , and so is cut out by an ideal I generated byat most M polynomials, then the ideal I( X ) = √ I has complexity O M (1) since it can beobtained from the primary decomposition of I . Since Sing X is cut out by minors of theJacobian matrix of any generating set, its complexity is O M (1).We will use the following version of the Lang-Weil bound (see [21]). Theorem 7.5 (Lang-Weil bound [15]) . Let F be a finite field. For any variety V definedover F and of complexity at most M , | V ( F ) | = | F | dim V ( c ( V ) ± O M ( | F | − / )) where c ( V ) is the number of irreducible components of dimension dim V defined over F . Lemma 7.6.
Let F be a finite field. Any variety V defined over F of complexity at most M has a subvariety V F ⊆ V satisfying:1. Every irreducible components of V F is defined over F V F has complexity at most some O M (1) .3. V ( F ) ⊆ V F .Proof. For an irreducible variety Y ⊆ F n , let F ( Y ) = Y ∩ Frob( Y ). Observe that, byLemma 7.1, if Y is not defined over F then F ( Y ) ( Y , and thus, by the irreducibilityof Y , dim F ( Y ) < dim Y . Moreover, Y ( F ) ⊆ F ( Y ). For an arbitrary variety X ⊆ F n ,let F ( X ) = S Y F ( Y ) where the union is over the irreducible components Y of X . Thendim( S Y F ( Y )) < dim( S Y Y ) where both unions are over the irreducible components Y of X that are not defined over F . Moreover, X ( F ) ⊆ F ( X ).For a variety V ⊆ F n , let V F = F (dim V +1) ( V ). By the above discussion, V F is a fixedpoint of F , that is, F ( V F ) = V F . Thus, every irreducible component of V F ⊆ V is definedover F . Moreover, V ( F ) ⊆ V F . Finally, it follows from Fact 7.4 that if V has complexityat most M then F ( V ), and therefore V F , has complexity bounded by some function of M .This completes the proof. 16e are now ready to prove the main theorem. Proof of Theorem 1.1.
Let T be a slicing of T . Since the variety ker T ⊆ F N , where N :=( k − n , is cut out by at most n polynomials of degree k − N . Let X be the subvariety of ker T given by Lemma 7.6. Then X satisfies the followingproperties:1. Every irreducible component of X is defined over F ,2. X has complexity O N (1),3. ker T ( F ) ⊆ X .Let Y be a top-dimensional irreducible component of X . Consider the singular locus,Sing Y = { x ∈ Y | rank J( x ) < codim Y } , where J : F n → F m × n is the Jacobian ma-trix of some m polynomials generating I( Y ). Similarly, consider the D T -singular locus,Sing D T Y = { x ∈ Y | rank D T ( x ) < max x ∈ Y rank D T } . Note that both subvarietiesare strict subsets of Y and thus have dimension smaller than dim Y . Moreover, bothhave complexity O N (1) by (2) and Fact 7.4, as they are cut out by minors of J and ofD T : F N → F n , respectively. We apply the Lang-Weil bound of Theorem 7.5 on Y , Sing Y ,and Sing D T Y . Crucially, Y is defined over F by (1), which implies that c ( Y ) = 1. Moreover, c (Sing Y ) , c (Sing D T Y ) ≤ O N (1) using Fact 7.4. We have | Y ( F ) | − | Sing Y ( F ) ∪ Sing D T Y ( F ) | ≥ | F | dim Y (1 − O N ( | F | − / )) − | F | dim Y − O N (1)= | F | dim Y (1 − O N ( | F | − / )) . Therefore, provided | F | > O N (1), we have that | Y | > | Sing Y ∪ Sing D T Y | , which meansthat Y has an F -rational point that is neither singular nor D T -singular. Apply Theorem 6.1on Y ⊆ ker T to obtain the boundPR( T ) ≤ (2 k − −
1) codim Y = (2 k − −
1) codim X . On the other hand, using (3) we obtainAR( T ) = N − log | F | | ker T ( F ) | = N − log | F | | X ( F ) | = codim X − log | F | | X ( F ) || F | dim X = codim X − log | F | O N (1) , where the last step again uses Theorem 7.5 together with (2). From the above two inequalitieswe deduce thatPR( T ) ≤ (2 k − −
1) codim X = (2 k − − T ) + log | F | O N (1)) ≤ (2 k − −
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