Explicit polynomial sequences with maximal spaces of partial derivatives and a question of K. Mulmuley

We answer a question of K. Mulmuley: In 2016arXiv160902103E it was shown that the method of shifted partial derivatives cannot be used to separate the padded permanent from the determinant. Mulmuley asked if this"no-go"result could be extended to a model without padding. We prove this is indeed the case using the iterated matrix multiplication polynomial. We also provide several examples of polynomials with maximal space of partial derivatives, including the complete symmetric polynomials. We apply Koszul flattenings to these polynomials to have the first explicit sequence of polynomials with symmetric border rank lower bounds higher than the bounds attainable via partial derivatives.


Introduction
Let S d C N denote the space of homogeneous polynomials of degree d in N variables and let p ∈ S d C N . Let S e C N * denote the space of homogenous differential operators of order e with constant coefficients, which acts on S d C N when e ≤ d. The e-th partial derivative map of p (or e-th flattening of p) is We call the image of p e,d−e the e-th space of partial derivatives of p; it is straightforward to verify that rank(p e,d−e ) = rank(p d−e,e ) and that given e ′ ≤ e ≤ d/2, if p e,d−e has full rank then p e ′ ,d−e ′ has full rank.
Let M ≥ N . A polynomial p ∈ S d C N is a degeneration of q ∈ S d C M , if p ∈ GL M · q ⊂ S d C M , where here we have chosen a linear inclusion C N ⊂ C M and we consider p as a polynomial in M variables that just happens to only use N of them. Similarly, p is a specialization of q if p ∈ End M ·q ⊆ S d C M . Notice that if p is a specialization of q then it is a degeneration of q. In complexity theory, one is interested in finding obstructions to specialization of a polynomial p to a polynomial q.
The method of partial derivatives and other flattening methods (see, e.g., [4]) exploit semicontinuity of matrix rank to prove that a polynomial is not a degeneration of another. Indeed, if p is a degeneration of q, then rank(p e,d−e ) ≤ rank(q e,d−e ) for all e, therefore, comparing the ranks of the partial derivatives maps of p and q for various (or all) e, one can prove that p is not a degeneration of q (and thus nor is p a specialization of q).
The method of partial derivatives dates back to Sylvester in 1852 [22], who called the maps (1) catalecticants. These maps have been used to obtain lower bounds on the Waring rank, Waring border rank and cactus border rank of polynomials (see, e.g., [3,12]). The symmetric or Waring rank of a polynomial p ∈ S d C N is the smallest r such that p = r j=1 ℓ d j where ℓ j ∈ C N are linear forms. One writes R S (p) = r. The symmetric or Waring border rank of p is the smallest r such that p is a limit of polynomials of Waring rank r, and one writes R S (p) = r. The ranks of the partial derivatives maps give lower bounds for the symmetric border rank of p: R S (p) ≤ min e {rank(p e,d−e )}.
In [9,11], it was shown that for a general polynomial p all the maps p e,d−e are of maximal rank. When the second author was preparing [15], he asked several experts if they knew of an explicit sequence of polynomials (e.g., in the complexity class VNP) with partial derivatives of maximal rank, as the standard references [12] in mathematics and [4] in computer science did not have one. Those asked did not furnish any example, so we wrote down several, see below. One example we found surprised us: the polynomial (p n,2 ) k = (x 2 1 + · · · + x 2 n ) k , because it is in the complexity class VP e of sequences of polynomials admitting polynomial size formulas. It turns out this example had been discovered by Reznick in 1991 [20, Thm. 8.15], and in the same memoir he describes an explicit sequence that essentially dates back to Bierman [2] (the proof, if not the statement appeared in 1903), see below.
Note that h n,d ∈ VP s , the complexity class determined by the determinant, because the complete symmetric functions can be expressed as a determinant of a matrix whose entries are power sum functions. While it is obvious to the experts, since it is important, we present a proof of the following: A ΣΛΣ circuit is a depth three arithmetic circuit whose first and third layers consist of addition gates and whose middle layer consists of powering gates, sending z → z d for some d. Our results show this model is quite weak.
The method of shifted partial derivatives is a variant of the method of partial derivatives. It was introduced in [13] and exploited in [10] to prove super-polynomial lower complexity bounds for depth four circuits for the permanent (and determinant). In the same paper the authors ask if the method could be used to separate VP from VNP.
For p ∈ S d C N the method of shifted partials is based on the study of the following maps (for judiciously chosen e and τ ): Notice that if τ = 0, then p (e,d−e)[τ ] is the partial derivative map defined in (1). Let Again, semicontinuity of matrix rank guarantees that if p is a degeneration of q, then dim ∂ =e q =τ ≥ dim ∂ =e p =τ and the method of shifted partials can be used to prove that p is not a degeneration of q by showing that dim There is a geometric interpretation of the image of the shifted partial derivative map: given p ∈ S d C n , let V (p) ⊆ P(C n ) * be the hypersurface of degree d cut out by p. The image of p e,d−e generates an ideal in S d C n that we denote by J e (p); it cuts out a subvariety of V (p) that is called the e-th Jacobian locus of p. The image of the shifted partials map p (e,d−e)[τ ] is the component of degree d + τ of J e (p). In particular, the study of the ranks of the shifted partials maps of p is equivalent to the study of the growth of the ideals J e (p).
..,n is an n × n matrix of indeterminates.
By [18], both IMM d n and Pow d n are VP s -complete, the same complexity class for which the determinant polynomial det n is complete. By the homogenization result of [18], the VP s = VNP conjecture can be rephrased by stating that there is no polynomially bounded function n(m) such that the permanent polynomial perm m is a specialization of IMM m n(m) (or of Pow m n(m) ). Note that in the cases of Pow m n and IMM m n , we are comparing directly with the permanent polynomial rather than its padded version as in the classical setting of the determinant versus permanent conjecture. Note further that Pow m n is a specialization of IMM m n . We prove that the method of shifted partials cannot be used to achieve a superpolynomial separation between perm m and IMM m n : Theorem 3. If n > m 6 , then perm m cannot be separated from IMM m n by the method of shifted partial derivatives. More precisely, given any linear inclusion C m 2 ⊂ C mn 2 considering perm m ∈ S m C mn 2 as a polynomial that just involves m 2 of the mn 2 variables, then for all choices of e,τ , dim ∂ =e (perm m ∈ S m C mn 2 ) =τ ≤ dim ∂ =e IMM m n =τ .
Additional results. We give a priori upper bounds for the utility of Koszul flattenings (Proposition 16), another variant of the partial derivatives map. We show that these bounds are sharp for the first Koszul flattenings in low dimensions and degree (Proposition 21). We obtain explicit (but not sharp) lower bounds for the Koszul flattenings of f k,n , showing one obtains better Waring border rank lower bounds with this method than by the method of partial derivatives (Proposition 19). Ironically, now the simple polynomial f k,n has the highest Waring border rank lower bound of all explicit polynomials of odd degree.
Related work. Let e n,d = 1≤i 1 <i 2 <···<i d ≤n x i 1 · · · x i d ∈ S d C n denote the elementary symmetric polynomial. The complexity of e n,d has been well studied: its symmetric border rank is bounded below by n ⌊ d 2 ⌋ because its space of partial derivatives of order e is spanned by the square free monomials of degree d − e. It symmetric rank is bounded above by n i when d is odd and there is a similar formula for even d [16]. Its padded version can be computed by a homogeneous depth three (ΣΠΣ) circuit of size n 2 [1], and when log(n) ≤ d ≤ 2n 3 , one has the lower bound of max(Ω( n 2 d ), Ω(nd)) from [21] for its depth three circuit size. The lower bounds appear to translate to complete symmetric functions, however the upper bound relies on the generating function for the elementary symmetric functions being a product of linear forms, whereas the complete symmetric functions have generating function The gap between the padded and unpadded depth three circuit complexity may have led researchers to think the results of [5] might fail in a model without padding, motivating Mulmuley's question (although Mulmuley himself anticipated our affirmative answer).
The shifted partial derivative complexity of elementary symmetric polynomials is studied in [6], where strong lower bounds are proved, which in turn give complexity lower bounds for depth four circuits.
Remark 5. The same argument as Proposition 4 allows us to prove a more general result. Use q n to identify C n with (C n ) * and let W be a subspace of C n such that q n | W is nondegenerate. Then, via the identification q n : C n → C n * induced by q n , we may consider W * as a subspace of (C n ) * . Then The argument is the same as above, considering that Theorem 6 (Reznick, [20]). For every n, k and e ≤ k, we have (f n,k ) e,2k−e (S e (C n ) * ) = q k−e n S e C n . In particular, for all n, k, e, the flattening (f n,k ) e,2k−e has full rank.
Proof. We proceed by induction on e. For e = 0 there is nothing to prove. Let e ≥ 1.
By the induction hypothesis, the (e − 1)-st flattening surjects onto q k−e+1 n S e−1 C n . It suffices to show that Up to rescaling q n and the differential operators, the term in parenthesis is ∂ ∂x i (x α q n ). These terms span S e C n by Proposition 4.
Theorem 7. For every n, k, e, the flattening ( f n,k ) e,2k+1−e has full rank.
Proof. Write C n = ℓ n ⊕ W as representation of the symmetric group S n , where ℓ n spans a S n -invariant and W is isomorphic to the Specht modules [n−1, 1]. S 2 W contains a unique S n -invariant up to scale, that we denote by g n . Notice that q n = 1 n ℓ n + g n (after possibly rescaling g n ).
Let e ≤ k. We show that the image of ( f n,k ) e,2k+1−e has dimension e+n−1 e . The decomposition C n = W ⊕ ℓ n is orthogonal with respect to the S n -invariant nondegenerate inner-product q n , and q n is non-degenerate restricted to each space. Write ℓ * n = 1 n ∂ ∂x 1 + · · · + ∂ ∂xn , so that (C n ) * = ℓ * n ⊕ W * with W * · ℓ n = 0 and ℓ * n · W = 0.
The last equality follows from Remark 5: for every j, S e−j W * · q k n = q k−e+j n S e−j W . Now, since ℓ * n · g n = 0, we obtain In particular, the highest power of This shows that all the summands in the last line of (2) are linearly independent. Therefore, their span has dimension

Two auxiliary results
Proof of Proposition 1. Consider the circuits in Figure 1 and Figure 2. Together, they give a formula of size k(4n − 2) + 2k − 2 that computes f n,k = q k n . Adding n − 1 addition gates to compute ℓ n and one multiplication gate to multiply ℓ with q k n gives a formula of size k(4n − 2) + 2k − 2 + n for f n,k . Proposition 8. If n = 2m (resp. n = 2m+1) then the polynomial f n,k is a specialization of the matrix powering polynomial Pow 2k m+1 (resp. Pow 2k m+2 ). If n = 2m (resp. n = 2m + 1) then the polynomial f n,k is a specialization of the iterated matrix multiplication polynomial IMM 2k+1 m+2 (resp. IMM 2k+1 m+3 ).
x n x n * * · · · * + + + Figure 1. A circuit to compute q n of size 4n − 2 q n q n q n · · · · · · q n * * * * Proof. Let n = 2m + 1 and set y ± j = x 2j−1 ± √ −1x 2j for j = 1, . . . , m. Consider the specialization of Pow 2k m+2 to the matrix Thus, the nonzero eigenvalues of Q m (as functions of x 1 , ..., x n ), are ± √ q n . In particular, If n = 2m is even, apply the same argument to the matrix obtained from Q m by removing the last row and the last column.
Similarly, f n,k is a specialization of IMM 2k+1 m+2 or IMM 2k+1 m+3 (depending on the parity of n) by making the first matrix ℓ n Id and specializing the remaining matrices to the matrix above.

Proof of Theorem 3
Choose a linear inclusion C m 2 ⊂ C mn 2 and regard perm m ∈ S m C mn 2 . Our goal is to show that for every s, τ We split the proof into three cases. In the first and in the second case, we degenerate IMM m n to f n,k if m = 2k is even and to f n,k if m = 2k + 1 is odd. This is possible by Proposition 8. Write F m,n for either f n,k or f n,k in what follows. Since IMM m n degenerates to F m,n , we have dim ∂ =s IMM m n =τ ≥ dim ∂ =s F m,n =τ . In the third case, we specialize IM M m n to the power sum polynomial of degree m in m 2 variables y m 1 + · · · + y m m 2 by specializing every argument of IM M m n to the diagonal matrix of size n × n with y 1 , . . . , y m 2 in the first m 2 diagonal entries and 0 elsewhere.
Up to the action of GL mn 2 , assume C m ⊆ C n ⊆ C mn 2 , where C m is the space spanned by the variables of perm m and C n is the space spanned by the variables of F m,n . It will suffice to prove dim ∂ =s (perm m ∈ S m C n ) =τ ≤ dim ∂ =s F m,n ∈ S m C n =τ because the remaining mn 2 − n variables will contribute the same growth to the ideals J s (perm m ) and J s (F m,n ). Since s ≥ ⌈ m 2 ⌉, (F m,n ) s,m−s surjects onto S m−s C n , and thus, for every shift τ , the shifted partial derivative map surjects onto S m−s+τ C n for all τ . This shows ∂ =s F m,n ∈ S m C n =τ = S m−s+τ C n and proves this case.
Taking the logarithm of the right hand side of (4), we obtain ln m s 2 ≤ s ln(m 2 ).
Case 3: s < m 2 and τ > m 3 . Here set all matrices (X 1 , ..., X m ) equal to a matrix that is zero except for the first m 2 entries on the diagonal, call them y 1 , ..., y m 2 . The resulting degeneration of IMM m n is y m 1 + · · · + y m m 2 . We compare the shifted partials of both polynomials in m 2 variables. The space of partial derivatives of order s is y m−s

Complete symmetric functions
Recall that h n,d is the complete symmetric function of degree d in n variables: where the summation is over all multi-indices α = (α 1 , . . . , α n ) with α 1 + · · · + α n = d. Proof. We proceed by induction on e. If e = 1, suppose x β = x n and write h n,d = d j=0 x j n h n−1,d−j (x 1 , . . . , x n−1 ), so that where the first summation in the last line contains one term from each summand in the previous line, and the second summation contains the remaining terms (with shifted indices). The first summation adds up to h n,d−1 (x 1 , . . . , x n ); by repeating this on the second summation we obtain and iterating this process we obtain d j=0 x j n h n,d−1−j (x 1 , . . . , x n ) = h n+1,d−1 (x 1 , . . . , x n , x n ) proving the base case.

By chain rule and by symmetry
where we used the case e = 1 again. Since γ! · β 1 = β!, we conclude.
Proof. Write [f ] γ for the coefficient of x γ in the polynomial f .
We use induction on p. If p = 0, then h n+p,e (x 1 , . . . , x n , x (β) ) = h n,p and for every γ we Let p ≥ 1 and suppose β 1 ≥ 1. Write h n+p,e (y) = e j=0 y j 1 h n+p−1,e−j (y 2 , . . . , y n+p ). Let η 1 = (1, 0, . . . , 0) ∈ Z n ≥0 . We have Apply the inductive hypothesis to the summands of the right hand side to get Let a 1 , . . . , a N ∈ Z ≥0 be nonnegative integers and let G(a 1 , . . . , a N ) to be the N × N symmetric matrix whose (i, j)-th entry is a i + a j a i . The Lindström-Gessel-Viennot Lemma (see [8, §2]) guarantees that G(a 1 , . . . , a N ) is a totally nonnegative matrix (in the sense that every minor is nonnegative), and its rank is equal to the number of distinct Our goal is to prove that G(a 1 , . . . , a N ) is positive definite if the a i are distinct.
We will need the following two technical results. Given a matrix A we denote by A i Proof. Without loss of generality, suppose I = {1, . . . , r} and let R = A I I . We want to prove that R is full rank, namely that Ru = 0 for some u ∈ R r implies u = 0.
In particular Bv = 0, therefore Av = 0; since the first r columns of A are linearly independent, we deduce v = 0, so that u = 0 and R is nonsingular.  are equal, we show that the corresponding m-tuples a i and a j are equal. Consider the principal 2 × 2 submatrix obtained from these two columns: then they are linearly independent by induction hypothesis, and by Lemma 11 the submatrix A ij ij is positive definite. The submatrix B ij ij is positive semidefinite and has strictly positive diagonal entries, therefore by Lemma 12 C ij ij is positive definite, in contradiction with the assumption. This shows that if In particular a i = a j . Therefore, we may assume that C has distinct columns and our goal is to show that C has full rank. Suppose by contradiction that C does not have full rank and let be a vanishing linear combination of the columns of C.
Up to conjugation by a permutation matrix, suppose there exist 0 = k 0 < k 1 < · · · < k r = N such that a ′ i = a ′ j if k s < i, j ≤ k s+1 and a ′ i = a ′ j otherwise. Notice that if the a i 's are distinct, then A has full rank and so does C from Lemma 12 because B is positive semidefinite with strictly positive entries. Therefore, suppose k 1 ≥ 2 and up to reducing to a principal submatrix suppose that α k 1 = 0. Since the first k 1 columns (and rows) of A are equal, the first k 1 columns (and rows) of B are linearly independent, otherwise two of them would be equal, providing that two m-tuples a i and a j for i, j ≤ k 1 would be equal. By Lemma 11, the principal submatrix B 1,...,k 1 1,...,k 1 is positive definite. The linear combination (5) can be written as Define A to be the matrix obtained from A by removing the first k 1 −1 rows and columns, that is A = G(a ′ k 1 , . . . , a ′ N ). A has the same rank as A. Let B ′ = P T BP , for notice that B ′ is obtained from B by performing row and column operations. In particular B ′ has the same signature as B; moreover, from the block structure of P , we deduce that the submatrix B ′ 1,...,k 1 1,...,k 1 has the same signature as B 1,...,k 1 1,...,k 1 , namely it is positive definite. This shows that the k 1 -th diagonal entry of B ′ is strictly positive. Define B to be the submatrix obtained from B ′ by removing the first k 1 − 1 rows and columns. Define C = A⊙ B. The linear combination of (5) induces a vanishing linear combination among the columns of C.
By repeating this procedure at most r times, we find a singular r × r matrix C = A ⊙ B with A positive semidefinite and of full rank (so positive definite) and B positive semidefinite with strictly positive diagonal entries. By Lemma 12, we obtain a contradiction. This concludes the proof.
Using these results, we can finally prove: Theorem 14. For every n, d, e, the flattening (h n,d ) e,d−e : S e (C n ) * → S d−e C n of the complete symmetric function h n,d has full rank.
Proof. First, we consider the case d = 2k even.
It suffices to prove the result for e = k.

Define two column vectors
From Proposition 9 and Proposition 10, we have h = Ab where the (β, γ)-th entry of A is Since the multi-indices β are all distinct, by applying Proposition 13 we deduce that A is nonsingular and therefore the entries of b are linear combinations of the entries of h. This shows that (h n,d ) k,k is full rank.
Now consider d = 2k + 1 odd. It suffices to prove the result for e = k + 1. Let The image of the flattening g k,k : S k (C n ) * → S k C n is contained in the image of (h n,d ) k+1,k . To conclude, we will show that g k,k is full rank.
For readers familiar with cactus rank and border rank, by [12,Thm. 5.3D], we obtain the same lower bounds for cactus rank and border rank.

Koszul flattenings
We recall the Koszul flattenings introduced in [14]: Tensor the s-th partial derivative map (1) with the identity map on Λ q C N for some q, and then compose with the exterior derivative map Let p ∧q s,d−s : Λ q C N ⊗S s C N * → Λ q+1 C N ⊗S d−s−1 C N denote the composition of the partial derivative map and the exterior derivative.
Then, by [14,Prop 4 When p s,d−s is of maximal rank for all s, Koszul flattenings can only give a better Waring border rank lower bound than the partial derivative maps when d = 2k + 1 is odd and s = k. For example, if d = 2k is even, then the rank of the Koszul flattening is bounded above by dim Λ q C N ⊗ S k−1 C N so the Waring border rank lower bound from Koszul flattenings is bounded above by N +k−2 k−1 N N −q whereas from flattenings alone, one already gets the larger lower bound of N +k−1 k . Similarly, if d is odd but N = 2q + 1, the bound from Koszul flattenings is lower than the one from standard flattenings.
Suppose d = 2k + 1 and p k,k+1 is of maximal rank. Viewed naïvely, one might think Koszul flattenings could prove Waring border rank lower bounds of up to but this is not possible because the exterior derivative map is a GL N -module map that is not surjective unless q = N, N − 1. Indeed, we have the decompositions so, by Schur's Lemma, S k+2,1 q−1 C N = ker(δ) and image(δ) = S k+1,1 q C N .
The following result gives an a priori upper bound for the rank of the Koszul flattening, by determining a lower bound for the dimension of ker(p ∧q k,k+1 ).
Proposition 16. Let p ∈ S 2k+1 C N . Then for every q Proof. We will prove that dim image(p ∧(q−1) e−1,d−e+1 ) ≤ dim ker(p ∧q e,d−e ) and conclude that the estimate holds for every q via an induction argument. Indeed, image(p and we conclude by induction. Remark 17. This is still not the end of the story: when N = 2q + 1 with q odd, then the linear map p ∧q k,k+1 : Λ q C N ⊗ S k C N * → Λ q+1 C N S k C N ≃ Λ q C N * S k C N was observed in [14], at least in certain cases, to be skew-symmetric. In particular, in this case, if the bound in (7) is odd, it cannot be attained.
Remark 18. Since the border rank bound is obtained by dividing rank(p ∧q k,k+1 ) by , we see asymptotically, the best potential lower bound is obtained when N = 2q+1 and there the limit of the method is twice the bounds obtained via flattenings minus lower order terms. This improvement is irrelevant for complexity. It is known more generally that the improvement in best possible lower bounds beyond the best possible bounds of partial derivatives are limited for any determinantal method. This was observed independently by Efremenko, Garg, Oliviera and Wigderson (personal communication), and Galcazka [7] for completely different reasons.
We now show Koszul flattenings can indeed give border rank lower bounds beyond the best lower bound attainable via the method of partial derivatives.

Proposition 19.
The Koszul flattening ( f k,n ) ∧q k,k+1 , when n > 2, and all q < n 2 , has rank at least In particular, it implies R S ( f k,n ) ≥ n+k−1 k + q − 1, which is greater than the lower bound obtainable by flattenings.
Proof. For fixed n, recall the unique S n -invariant g := g n ∈ S 2 W from the proof of Thm. 7, where W is the subspace of C N isomorphic to the Specht module [n − 1, 1] under the action of S n . Let L be the span of ℓ := ℓ n . For every s, write p s := (ℓ * ) s−1 ·(ℓq s−1 ), which from Eqn. (3) is a polynomial of degree s with non-zero projection onto L s . From Eqn. 2, the image of the k-th flattening map of f n,k is image( f n,k ) k,k+1 = k s=0 p s+1 S k−s W ⊆ S k+1 C n .
We will give a lower bound for the dimension of the image of Λ q C n ⊗ image( f n,k ) k,k+1 under the exterior derivative map. We have Λ q C n = Λ q W ⊕ L ∧ Λ q−1 W as a S nmodule.
Consider the image of (L ∧ Λ q−1 W ) ⊗ (p s+1 S k−s W ) under the exterior derivative. The S n -equivariant projection of this space onto (L ∧ Λ q−1 W ) ⊗ (L s+1 ⊗ S k−s W ) commutes with the exterior derivative, which is GL n -equivariant and therefore S n -equivariant. The image of (L ∧ Λ q−1 W ) ⊗ (L s+1 ⊗ S k−s W ) under the exterior derivative is (when s ≤ k − 1 and after reordering the factors) the subspace S (k−s,1 q−1 ) W ⊗L⊗L s+1 ⊆ L ∧ Λ q W ⊗L s+1 S k−s−1 W . Now consider the image of Λ q W ⊗ p s+1 S k−s W . Again, consider its projection onto Λ q W ⊗(L s+1 ⊗S k−s W ). By applying the exterior derivative map, we obtain a subspace of (L ∧ Λ q W ) ⊗ (L s ⊗ S k−s W ) ⊕ Λ q+1 W ⊗(L s+1 ⊗ S k−s−1 W ) ; consider its projection to the second summand Λ q+1 W ⊗L s+1 S k−s−1 W when s ≤ k − 1. For the same reason as above, the image of this projection is S (k−s,1 q ) W ⊗L s+1 up to reordering the factors.
Note that S (k−t+1,1 q−1 ) W ⊕ S (k−t,1 q ) W = S k−t W ⊗Λ q W as a GL(W )-module. Consider the summands for s ranging from 0 to k − 2 in the first case and 1 to k − 1 in the second, we obtain components S k−t W ⊗Λ q W for t from 1 to k − 1. We obtain a subspace in the image of the Koszul flattening that is isomorphic as a GL(W )-module to The first factor of the space above has the same dimension as S k−1 C n minus dim(S 0 W ) = 1.
So far we have a contribution to the rank of n − 1 q ( n + k − 2 k − 1 − 1).
Next consider the s = 0 contribution that one obtains by applying the exterior derivative to the component with the factor Λ q W . The exterior derivative map is Λ q W ⊗ (L ⊗ S k W ) → Λ q W ∧ L ⊗ S k W ⊕ Λ q W ⊗ (L ⊗ S k−1 W ). This projects isomorphically onto the first term in the target and (Λ q W ∧ L) ⊗ S k W does not intersect the image of any other term that we considered so far, so we obtain an additional contribution to the image of dimension n−1+k−1 k n−1 q . We now have a contribution of to the rank.
Finally consider the s = k and s = k − 1 terms in the term with a factor L ∧ Λ q−1 W : the sources are respectively L ∧ Λ q−1 W ⊗ p k+1 and (L ∧ Λ q−1 W ) ⊗ p k W . Consider the projections respectively to (L ∧ Λ q−1 W ) ⊗ ℓ k−1 S 2 W (the second factor is ℓ k−1 g) and (L ∧ Λ q−1 W ) ⊗ ℓ k−2 S 3 W (the second factor is ℓ k−2 gW ). Now, applying the exterior derivative, these spaces map injectively to (L∧Λ q W )⊗L k−1 W and (L∧Λ q W )⊗ℓ k−2 S 2 W respectively. These targets do not appear in other terms analyzed above, so we pick up Collecting all the contributions together, we conclude.
Remark 20. A more careful analysis of the q = 1 case shows it also improves the flattening lower bound.
For n = 3, . . . , 6 and k = 1, . . . , 6, let p = h n,2k+1 or p = f k,n . The Koszul flattenings of p ∧1 k,k+1 have rank equal to the bound in (6) if k is odd and one less than the bound in (6) if k is even.
Notice that in the cases n = 4, 5, 6 with k even, the Koszul flattenings give a border rank lower bound for h n,2k+1 and f k,n that is one less than the bound for a generic polynomial.