EQUATIONS FOR SECANT VARIETIES OF CHOW VARIETIES

EQUATIONS FOR SECANT VARIETIES OF CHOW VARIETIES

arXiv:1602.04275v1 [math.AG] 13 Feb 2016

YONGHUI GUAN Abstract. The Chow variety of polynomials that decompose as a product of linear forms has been studied for more than 100 years. Finding equations in the ideal of secant varieties of Chow varieties would enable one to measure the complexity the permanent polynomial. In this article, I use the method of prolongation to obtain equations for secant varieties of Chow varieties as GL(V )-modules.

1. Introduction 1.1. Motivation from geometry. Let V be a finite-dimensional complex vector space and let PS d V ∗ be the projective space of homogeneous polynomials of degree d on V . Define the Chow variety Chd (V ∗ ) ⊆ PS d V ∗ by (1)

Chd (V ∗ ) = {[z] ∈ PS d V ∗ ∣z = w1 ⋯wd for some wi ∈ V ∗ }.

ˆ ⊂ W ∗ denote the cone over X, define Let X ⊂ PW ∗ be an algebraic variety, and let X ˆ and i = 1, ⋯, r}, σr (X) = P{v ∈ W ∗ ∣v = x1 + ⋯ + xr for xi ∈ X where the overline denotes closure in the Zariski topology. The Chow variety Chd (V ∗ ) and σr (Chd (V ∗ )) are invariant under the action of GL(V ), therefore there ideals are GL(V )modules. The ideal of the Chow variety has been studied for over 100 years, dating back at least to Gordon and Hadamard. The Foulkes-Howe map, hδ,d ∶ S δ (S d V ) → S d (S δ V ) (see §2.3 for definition) was defined by Hermite [14] when dim V = 2, and Hermite proved the map is an isomorphism in his celebrated “Hermite reciprocity”. Hadamard [12] defined the map in general and observed that its kernel is Iδ (Chd (V ∗ )), the degree δ component of the ideal of the Chow variety , but we do not understand this map when d > 4 (see [13, 15, 6, 19, 1, 2]). Brill, Gordon [8] and others obtained set-theoretical equations for the Chow variety. I computed Brill’s equations as a GL(V )-module in [9]. 1.2. Motivation from complexity theory. The class VP is an algebraic analog of the class P, and the class VNP is an algebraic analog of the class VP. An algebraic analog of the famous P versus NP problem is Valiant’s Conjecture VP ≠ VNP [24] i.e. there does not exist polynomial size circuit that computes the permanent which is defined by permn = ∑σ∈Sn x1σ(1) x2σ(2) ⋯xnσ(n) . The variety σr (Chd (V )) is associated to depth 3 circuits called ∑ ⋀ ∑ circuits in [17]. Finding equations in the ideal of σr (Chd (V )) would enable one to prove lower complexity bounds for depth 3 circuits. The following theorem appeared in [17], it is a geometric rephrasing of results in [11]. √

Theorem 1.1. [11] If for all but a finite number of m, for all r, n with rn < 2 (2)

2

[ℓn−m permm ] /∈ σr (Chn (Cm

+1

m log(m)ω(1)

)),

Key words and phrases. Chow Variety, secant variety of Chow variety, prolongation, GL(V )-module. 1

,

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YONGHUI GUAN

then Valiant’s Conjecture VP ≠ VNP [24] holds. Theorem 1.1 motivated me to study the varieties σr (Chd (V )). Although the equations we obtain here cannot separate VP from VNP, the results come from a geometric perspective, and these are the first low degree equations for secant varieties of Chow varieties, in addition to the non-classical equations obtained by Koszul Young flattenings in [10]. My results include ● Equations for σ2 (Ch3 (C 6∗ )) (Theorems 1.2 and 1.3). ● Equations for σr (Ch4 (C4r∗ )) (Theorem 1.4). ● Properties related to plethysm coefficients (Theorems 6.3 and 7.2). ● Equations for σr (Chd (Cdr∗ )) when d is even (Theorem 1.5) 1.3. Results. Let X ⊂ W ∗ be an algebraic variety. Suppose we know the ideal of X, there is a systematic method called prolongation (or multiprolongation) to compute the ideal of σr (X), but this method is difficult to implement. This method was studied by J. Sidman and S. Sullivant [23], and J.M. Landsberg and L. Manivel [18]. For any partition λ, let Sλ V be the irreducible GL(V )-module determined by the partition λ. As a GL(V )-module, S k (S d V ) can be decomposed into a direct sum of irreducible GL(V )modules, the multiplicity of Sλ V in S k (S d V ) is the plethysm coeffcient pλ (k, d). To obtain equations for secant varieties, on one hand I compute prolongations directly via differential operators and representation theory. On the other hand, I rephrase prolongations and reduce computing prolongations to computing polarization maps via plethysm coefficients and generalized Littlewood-Richardson coefficients. This gives a path towards obtaining equations for secant varieties of Chow varieties and other varieties. For d = 3, Theorem 1.2. Let dim V ≤ 6, I7 (σ2 (Ch3 (V ∗ ))) = 0. Also Theorem 1.3. Let dim V ≥ 6, S(5,5,5,5,3,1) V ⊂ I8 (σ2 (Ch3 (V ∗ ))).

EQUATIONS FOR SECANT VARIETIES OF CHOW VARIETIES

3

For d = 4, Theorem 1.4. Consider dim V ≥ 4r, S(6,6,44r−2 ) V ⊂ I4r+1 (σr (Ch4 (V ∗ )). A partition is an even partition if all the components of the partition are even numbers. When d is even, any even partition with length no more than k has positive plethysm coefficients in S k (S d V ) [4]. Theorem 1.5. The isotypic component of S((2m+2)m ,(2m)2mr−m ) V is in I2mr+1 (σr (Ch2m (V ∗ ))). Moreover any module with even partition and smaller than ((2m + 2)m , 2) (with respect to the lexicographic order) is in I2mr+1 (σr (Ch2m (V ∗ ))). 1.4. Organization. In §2, I review semi-standard tableaux, how to write down highest weight vectors of a module via raising operators, and the Foulkes-Howe map related to the ideal of the Chow variety Chd (V ∗ ). In §3, I explain how to compute prolongations and multiprolongations of a GL(V )-module via differential operators and representation theory to obtain equations for σr (Chd (V ∗ )). In §4, I prove Theorems 1.2 and 1.3. In §5, I prove Theorem 1.4. In §6, I prove a theorem related to plethysm coefficients of S 2m (S 2m+1 V ) , and using this I prove Theorem 1.5. In §7, I prove a property about plethysm coefficients. 1.5. Acknowledgement. I thank my advisor J.M. Landsberg for discussing all the details throughout this article. I thank C. Ikenmeyer and M. Michalek for discussing the plethysm coefficients. Most of this work was done while the author was visiting the Simons Institute for the Theory of Computing, UC Berkeley for the Algorithms and Complexity in Algebraic Geometry program, I thank the Simons Institute for providing a good research environment. 2. Preliminaries 2.1. Semi-standard tableaux. I follow the notation in [7] and [16]. A partition λ of an integer n is λ = (λ1 , ⋯, λm ) with λ1 ≥ ⋯ ≥ λm and ∑m i=1 λi = n. We often denote this by λ ⊢ n. To a partition λ ⊢ n, we associate a Young diagram, which is a box diagram with λi boxes in row i. A filling of a Young diagram using the numbers {1, ⋯, d} is an assignment of one number to each box, with repetitions allowed. A filled Young diagram is called a Young tableau. A standard filling is one in which the entries are strictly increasing in the both the rows and columns, while a semi-standard filling is one in which the entries are strictly increasing in the columns and weakly increasing in the rows. Standard tableaux and semi-standard tableaux are similarly defined. 2.2. Highest weight vectors of modules in S k (S d V ) via raising operators. I follow the notation in [7]. The group GL(V ) has a natural action on V ⊗d such that g ⋅ (v1 ⊗ v2 ⋯ ⊗ vd ) = g ⋅ v1 ⊗ ⋯ ⊗ g ⋅ vd . Let dim V = n and let {e1 , e2 , ⋯, en } be a basis of V . Let B ⊂ GL(V ) be the subgroup of upper-triangular matrices (a Borel subgroup). For any partition λ = (λ1 , ⋯, λn ), let Sλ V be the irreducible GL(V )-module determined by the partition λ. For each Sλ V , there is a unique line that is preserved by B, called a highest weight line. Let gl(V ) be the Lie algebra of GL(V ), there is an induced action of gl(V ) on V ⊗d . For X ∈ gl(V ), X.(v1 ⊗ v2 ⋯ ⊗ vd ) = X.v1 ⊗ v2 ⋯ ⊗ vd + v1 ⊗ X.v2 ⊗ ⋯ ⊗ vd + ⋯ + v1 ⊗ v2 ⋯ ⊗ vd−1 ⊗ X.vd . Let Eji ∈ gl(V ) such that Eji (ej ) = ei and Eij (ek ) = 0 when k ≠ j. If i < j, Eji is called a raising operator; if i > j, Eji is called a lowering operator. A highest weight vector of a GL(V )-module is a weight vector that is killed by all raising operators. Each realization of the module Sλ V has a unique highest weight line. Let W be a

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YONGHUI GUAN

GL(V )-module, the multiplicity of Sλ V in W is equal to the dimension of the highest weight space with respect to the partition λ. Define the weight space W(a1 ,⋯,an ) ⊂ S k (S d V ) to be the set of all the weight vectors whose weights are (a1 , ⋯, an ). Note that S d V has a natural basis {eα1 1 ⋯eαnn }α1 +⋯+αn =d . Example 2.1. S(4,2) V ⊂ S 3 (S 2 V ) has multiplicity 1. Proof. Let v be a highest weight vector of S(4,2) V . The weight space W(4,2) has a basis {(e21 )2 (e22 ), (e21 )(e1 e2 )2 }. Write v = a(e21 )2 (e22 ) + b(e21 )(e1 e2 )2 , then E21 v = 0 implies (2a + 2b)(e21 )2 (e1 e2 ) = 0, therefore a = −b, so the multiplicity of S(4,2) V in S 3 (S 2 V ) is 1.  Proposition 2.2. The highest weight vector f of S(2k ) V ⊂ S k (S 2 V ) is determinant of the k × k matrix M with Mij = ei ej for 1 ≤ i, j ≤ k. Proof. Since S(2k ) V ⊂ S k (S 2 V ) is of multiplicity one, we only need to prove det M is killed by i all raising operators Ei+1 (i = 1, 2, ..., k − 1). By symmetry, we only need to prove det M is killed by the raising operator E21 . It is straightforward to verify det M is killed by the raising operator E21 .  Proposition 2.3. The highest weight vector f of S(7,3,2) V ⊂ S 4 (S 3 V ) is f

= (e31 )2 (e1 e22 )(e2 e23 ) − 2(e31 )2 (e1 e2 e3 )(e22 e3 ) + (e31 )2 (e1 e23 )(e32 ) − (e31 )(e21 e2 )2 (e2 e23 ) + 2(e31 )(e21 e2 )(e21 e3 )(e22 e3 ) − 4(e31 )(e21 e2 )(e1 e22 )(e1 e23 ) + 0(e31 )(e21 e3 )(e1 e22 )(e1 e2 e3 ) + 3(e21 e2 )3 (e1 e23 ) + 4(e1 e2 e3 )2 (e21 e2 )(e31 ) − (e31 )(e21 e3 )2 (e32 ) + 3(e21 e2 )(e1 e22 )(e21 e3 )2 − 6(e21 e2 )2 (e21 e3 )(e1 e2 e3 ).

Proof. Let f ∈ W(7,3,2) ⊂ S 4 (S 3 V ) be a weight vector. The weight space W(7,3,2) ⊂ S 4 (S 3 V ) has dimension 12. Write f as a linear combination of the basis vectors and apply E21 and E32 to f , we get two systems of linear equations. There is a unique solution up to scale.  Remark 2.4. S(7,3,2) V cuts out Ch3 (V ∗ ) set-theoretically [9]. Proposition 2.5. The highest weight vector f of S(5,4,2,1) V ⊂ S 4 (S 3 V ) is (3)

f = e22 e4 h1 + e1 e3 e4 h2 + e1 e2 e4 h3 + e21 e4 h4 .

Here h4 = (e21 e2 )(e32 )(e1 e23 ) − (e1 e22 )2 (e1 e23 ) − (e21 e2 )(e1 e2 e3 )(e22 e3 ) + (e21 e3 )(e1 e22 )(e22 e3 ) − (e1 e22 )(e1 e2 e3 )2 − (e21 e3 )(e1 e2 e3 )(e32 ), h3 = −E21 h4 , h1 = 12 E21 E21 h4 is a highest weight vector of S(5,2,2) V ⊂ S 3 (S 3 V ) and h2 = E32 E21 h4 is a highest weight vector of S(4,4,1) V ⊂ S 3 (S 3 V ). 2.3. Foulkes-Howe map and the ideal of Chow variety. I follow the notation in [16], §8.6. Define the Foulkes-Howe map F Hδ,d ∶ S δ (S d V ) → S d (S δ V ) as follows. First include S δ (S d V ) ⊂ V ⊗δd . Next, regroup and symmetrize the blocks to (S δ V )⊗d . Finally, thinking of S δ V as a single vector space, symmetrize again to land in S δ (S d V ). Example 2.6. Explicitly F H2,2 (x2 ⋅ y 2 ) = (xy)2 , and F H2,2 ((xy)2 ) = 21 [x2 ⋅ y 2 + (xy)2 ]. F Hδ,d is a GL(V )-module map and Hadamard [12] observed the following relationship between Foulkes-Howe map and ideal of Chow variety. Proposition 2.7. (Hadamard [12]) Ker F Hδ,d = Iδ (Chd (V ∗ )).

EQUATIONS FOR SECANT VARIETIES OF CHOW VARIETIES

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Corollary 2.8. When δ = d + 1, Ker F Hd+1,d = Id+1 (Chd (V ∗ )). Therefore as an abstract GL(V )-module, Id+1 (Chd (V ∗ )) ⊃ S d+1 (S d V ) − S d (S d+1 V ). Proposition 2.9. (Hermite [14], Hadamard [13], J.M¨ uler and M.Neunh¨ofer)[22]) When d = 2, 3, 4, F Hd,d are injective and hence surjective. Proposition 2.10. (T. McKay [21]) If F Hδ,d is surjective, then F Hδ+1,d is surjective. So when d = 2, 3, 4, F Hd+1,d are surjective, and Id+1 (Chd (V ∗ )) = S d+1 (S d V ) − S d (S d+1 V ) as GL(V )- modules. 3. Prolongations, multiprolongations and particle derivatives 3.1. Prolongations, multiprolongations and ideals of secant varieties. I study prolongations, multiprolongations and how they relate to ideals of secant varieties. Let W be a complex vector space with a basis {e1 , ⋯, en }. I follow the notation in [16]. Definition 3.1. For A ⊂ S d W , define A(p) = (A ⊗ S p W ) ∩ S p+d W. It is equivalent to saying that A(p) = {f ∈ S p+d W ∣

∂pf ∈ A any β ∈ Nn with ∣β∣ = p}. β ∂e

Here are some properties about prolongation. Proposition 3.2. For A ⊂ S d W , A(p) is the inverse image of A ⊗ S p W under the polarization map Pd,p ∶ S d+p W → S d W ⊗ S p W . Proof. For any f ∈ S (p+d) W , ∂pf ⊗ eα . α ∂e ∣α∣=p

Pd,p (f ) = ∑ Hence

∂pf ∂pf α p ⊗ e ∈ A ⊗ S W ⇔ ∈ A for any ∣α∣ = p ⇔ f ∈ A(p) . α α ∂e ∂e ∣α∣=p

Pd,p (f ) = ∑

 Theorem 3.3. (J. Sidman, S. Sullivant [23]) Let X ∈ PW ∗ be an algebraic variety and let d be the integer such that Id−1 (X) = 0 and Id (X) ≠ 0. Then Ir(d−1) (σr (X)) = 0 and Ir(d−1)+1 (σr (X)) = Id (X)(r−1)(d−1) . Remark 3.4. Theorem 3.3 bounds the lowest degree of an element in the ideal of σr (X) if we know generators of the ideal of X. Proposition 3.5. Let X ⊂ PW ∗ be an algebraic variety, then Id (X)(p) ⊂ Id+1 (X)(p−1) . Proof. Let f ∈ Id (X)(p) ⊂ S d+p W , consider

∂ p−1 f ∂eα

with ∣α∣ = p − 1,

n ∂pf ∂ p−1 f = ei ∈ Id+1 (X). ∑ α ∂eα i=1 ∂(e ei )



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YONGHUI GUAN

Example 3.6. Consider Ch3 (V ∗ ) with dim V ≥ 4, by Proposition 2.9 and Proposition 2.10, I3 (Ch3 (V ∗ )) = 0 and (4)

I4 (Ch3 (V ∗ )) = S 4 (S 3 V ) − S 3 (S 4 V ) = S(7,3,2) V + S(6,2,2,2) V + S(5,4,2,1) V.

Therefore by Theorem 3.3 I6 (σ2 (X)) = 0 and I7 (σ2 (X)) = I4 (X)(3) . The following proposition is about multiprolongations. Proposition 3.7. (Multiprolongation [16] ) Let X ⊂ P W ∗ be an algebraic variety, a polynomial P ∈ S δ W is in Iδ (σr (X)) if and only if for any nonnegative decreasing sequence (δ1 , δ2 , ⋯, δr ) with δ1 + δ2 + ⋯ + δr = δ, P¯ (v1 , ⋯, v1 , v2 , ⋯, v2 , ⋯, vr , ⋯, vr ) = 0 ˆ where the number of v ′ s appearing in the formula is mi . for all vi ∈ X, i

The following proposition rephrases multiprolongations. Proposition 3.8. Let X ∈ P W ∗ be an algebraic variety, for any positive integer δ and r, and for any decreasing sequence δ⃗ = (δ1 , δ2 , ⋯, δr ) with δ1 + δ2 + ⋯ + δr = δ, consider the following polarization maps Pδ1 ,δ2 ,⋯,δr ∶ S δ W → S δ1 W ⊗ S δ2 W ⊗ ⋯ ⊗ S δr W. δ1 δi−1 Let Aδ,i W ⊗ Iδi (X) ⊗ S δi+1 W ⊗ ⋯ ⊗ S δr W ⊂ S δ1 W ⊗ S δ2 W ⊗ ⋯ ⊗ S δr W , then ⃗ = S W ⊗⋯⊗S

Iδ (σr (X)) =

⋂

δ1 +δ2 +⋯+δr =δ

(Aδ,1 Pδ−1 ⃗ + ⋯ + Aδ,r ⃗ ) 1 ,δ2 ,⋯,δr

Corollary 3.9. Id (X)((r−1)(d−1)) ⊂ Ir(d−1)+1 (σr (X)). Proof. By Proposition 3.8, Ir(d−1)+1 (σr (X)) =

⋂ δ1 +δ2 +⋯+δr =r(d−1)+1, δ1 ≥δ2 ≥⋯≥δr

⋂

⊃

δ1 +δ2 +⋯+δr =r(d−1)+1, δ1 ≥δ2 ≥⋯≥δr

(Aδ,1 Pδ−1 ⃗ + ⋯ + Aδ,r ⃗ ) 1 ,δ2 ,⋯,δr Pδ−1 (Aδ,1 ⃗ ). 1 ,δ2 ,⋯,δr

(r(d−1)+1−δ1 ) (Aδ,1 By similar arguments as Proposition 3.2, Pδ−1 . ⃗ ) = Iδ1 (X) 1 ,δ2 ,⋯,δr

Since δ1 ≥ d, by Proposition 3.5, Id (X)((r−1)(d−1)) ⊂ Iδ1 (X)(r(d−1)+1−δ1 ) , therefore Id (X)((r−1)(d−1)) ⊂ Ir(d−1)+1 (σr (X)).  Remark 3.10. Proposition 3.8 and Corollary 3.9 implies Theorem 3.3. Proof. First, by Proposition 3.8, Ir(d−1) (σr (X)) =

⋂ δ1 +δ2 +⋯+δr =r(d−1)

Pδ−1 (Aδ,1 ⃗ + ⋯ + Aδ,r ⃗ ). 1 ,δ2 ,⋯,δr

−1 In particular, when δ1 = δ2 = ⋯ = δr = (d−1), Aδ,i ⃗ = 0 for i = 1, ⋯, r, so Pδ1 ,δ2 ,⋯,δr (Aδ,1 ⃗ +⋯+Aδ,r ⃗ )= 0. Therefore Ir(d−1) (σr (X)) = 0. Second, by Proposition 3.8,

Ir(d−1)+1 (σr (X)) =

⋂

δ1 +δ2 +⋯+δr =r(d−1)+1

Pδ−1 (Aδ,1 ⃗ + ⋯ + Aδ,r ⃗ ). 1 ,δ2 ,⋯,δr

In particular, when δ1 = d, δ2 = ⋯ = δr = d − 1, Aδ,i ⃗ = 0 for i = 2, ⋯, r. so −1 ((r−1)(d−1)) (Aδ,1 Pδ−1 . ⃗ + ⋯ + Aδ,r ⃗ ) = Pδ1 ,δ2 ,⋯,δr (Aδ,1 ⃗ ) = Id (X) 1 ,δ2 ,⋯,δr

EQUATIONS FOR SECANT VARIETIES OF CHOW VARIETIES

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Therefore Ir(d−1)+1 (σr (X)) ⊂ Id (X)((r−1)(d−1)) . On the other hand, by Corollary 3.9, Id (X)((r−1)(d−1)) ⊂ Ir(d−1)+1 (σr (X)), so equality holds.  Conjecture 3.11. Let X ∈ P W ∗ be an algebraic variety, and δ = kr + l with 0 ≤ l < r, take δ⃗ such that δ1 = ⋯ = δl = k + 1 and δl+1 = ⋯ = δr = k, then (Aδ,1 Iδ (σr (X)) = Pδ−1 ⃗ + ⋯ + Aδ,r ⃗ ). 1 ,δ2 ,⋯,δr Example 3.12. Consider Ch3 (V ∗ ), by Proposition 2.9, Proposition 2.10 and (4), I3 (Ch3 (V ∗ )) = 0 and I4 (Ch3 (V ∗ )) = S(7,3,2) V + S(6,2,2,2) V + S(5,4,2,1) V . Consider the polarization maps Fδ,8−δ ∶ S 8 (S 3 V ) → S δ (S 3 V ) ⊗ S 8−δ (S 3 V ). By Propositions 3.8 and 3.5, I8 (σ2 (Ch3 (V ∗ ))) =

8

δ 3 ∗ ∗ 8−δ 3 −1 ⋂ (Fδ,8−δ (S (S V ) ⊗ I8−δ (Ch3 (V )) + Iδ (Ch3 (V )) ⊗ S (S V ))

δ=4 8

=

∗ 4 3 4 3 ∗ −1 ∗ (8−δ) ⋂ Iδ (Ch3 (V )) ⋂ F4,4 (I4 (Ch3 (V )) ⊗ S (S V ) + S (S V ) ⊗ I4 (Ch3 (V )))

δ=5

(5)

−1 (I4 (Ch3 (V ∗ )) ⊗ S 4 (S 3 V ) + S 4 (S 3 V ) ⊗ I4 (Ch3 (V ∗ ))). = I5 (Ch3 (V ∗ ))(3) ⋂ F4,4

3.2. Partial derivatives and prolongations. Definition 3.13. Let V = span{e1 , ⋯, en }, S d V has a natural basis {eα1 1 ⋯eαnn = eα }α1 +⋯+αn =d . Assume e1 > e2 > ⋯ > en . Define a partial order on the natural basis of S d V such that eα > eβ ⇔ one can get eα from eβ via raising operators. Definition 3.14. Let f ∈ W(a1 ,⋯,an ) ⊂ S k (S d V ), let α be the index of the last d elements in (a1 , ⋯, an ), then ∂e∂α is the lowest possible partial derivative of f with respect to the partial order in Definition 3.13. Example 3.15. Let f ∈ W(5,4,4,2) ⊂ S 5 (S 3 V ), then α = (0, 0, 1, 2) and the lowest possible partial derivative of f is ∂e∂fe2 . 3 4

α

α

j+1 Definition 3.16. Let eα = eα1 1 ⋯ej j ej+1 ⋯eαnn , define the normalized lowering operators

˜ j+1 eα = eα1 ⋯eαj −1 eαj+1 +1 ⋯eαnn . E 1 j j j+1 Where j = 1, ⋯, n − 1. The following proposition tells the relation between raising operators and partial derivatives of polynomials in S k (S d V ). Proposition 3.17. Let f ∈ S k (S d V ) and eα be a basis vector of S d V , then [

∂f ∂ j , Ej+1 ]f = (1 + αj+1 ) . α ∂e ∂(E˜jj+1 eα )

˜ j+1 (j = 1, ⋯, n − 1) are the normalized lowering operators. Where E j

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YONGHUI GUAN

Proof. Since all the operators are linear here, we only to prove the case when f is a monomial. α αj+1 ˜ j+1 eα = eα1 ⋯eαj −1 eαj+1 +1 ⋯eαnn = eβ . Write f = g(eα )m (eβ )n , Let eα = eα1 1 ⋯ej j ej+1 ⋯eαnn , so E 1 j j j+1 where g can not be divided by eα or eβ . Then Ejj+1 f

= (Ejj+1 g)(eα )m (eβ )n + gEjj+1 ((eα )m )(eβ )n + g(eα )m Ejj+1 ((eβ )n ) = (Ejj+1 g)(eα )m (eβ )n + mg(eα )m−1 Ejj+1 (eα )(eβ )n + n(1 + αj+1 )g(eα )m+1 (eβ )n−1

So ∂(Ejj+1 f ) ∂eα

= m(Ejj+1 g)(eα )m−1 (eβ )n + m(m − 1)g(eα )m−2 Ejj+1 (eα )(eβ )n + n(m + 1)(1 + αj+1 )g(eα )m (eβ )n−1 .

On the other hand

Ejj+1 (

∂f = mg(eα )m−1 (eβ )n . ∂eα

∂f ) = m(Ejj+1 g)(eα )m−1 (eβ )n + m(m − 1)g(eα )m−2 Ejj+1 (eα )(eβ )n + ∂eα nm(1 + αj+1 )g(eα )m (eβ )n−1

So ∂(Ejj+1 f )

(6)

∂eα

j − Ej+1 (

∂f ∂f . ) = n(1 + αj+1 )g(eα )m (eβ )n−1 = (1 + αj+1 ) α ˜ ∂e ∂(Ejj+1 eα ) 

In particular if f ∈ S k (S d V ) is a highest weight vector of some module, then j Ej+1 (

(7)

∂f ∂f . ) = −(1 + αj+1 ) α ∂e ∂(E˜jj+1 eα )

Therefore Lemma 3.18. If f ∈ S k+1 (S d V ) is a highest weight vector for some module S(a1 ,⋯,an ) V = Sa V , ∂f then the lowest possible partial derivative ∂e α (see Definition 3.14) is killed by all the raising ∂f operators, i.e. either ∂eα is 0 or a highest weight vector of Sa−α V ⊂ S k (S d V ). By induction, we get Proposition 3.19. If f ∈ S k+1 (S d V ) is a highest weight vector for some module S(a1 ,⋯,an ) V = ∂f Sa V , then there exists a basis vector eβ of S d V such that ∂e β is a highest vector of Sa−β V ⊂ k d S (S V ). Remark 3.20. Let f ∈ S k+1 (S d V ) be a highest weight vector for some module S(a1 ,⋯,an ) V = Sa V , ∂f Proposition 3.19 tells us if we can find all the eβ such that ∂e β is a highest vector of Sa−β V ⊂ k d S (S V ), the sum of all these modules is the smallest possible module such that Sa V lies in its first prolongation. For simplicity, write

∂f ∂eβ

= feβ from now on.

Example 3.21. Let f be the highest weight vector of S(7,3,2) V ⊂ S 4 (S 3 V ) in Example 2.3, then fe2 e2 = (e31 )2 (e1 e22 ) − e31 (e21 e2 )2 , which is a highest weight vector of S(7,2) V ⊂ S 3 (S 3 V ). 3

EQUATIONS FOR SECANT VARIETIES OF CHOW VARIETIES

9

The following proposition, on the other hand, tells us which prolongation a given module lies in. Proposition 3.22. If (Sa V )⊕ma ⊂ S k+1 (S d V ), let Ma = {b∣(Sb V )⊕mb ⊂ S k (S d V ) with mb > 0 and Sa V ⊂ Sb V ⊗ S d V as abstract modules}. then (Sa V )⊕ma ⊂ ( ⊕ (Sb V )⊕mb )(1) . b∈Ma

In particular, ma ≤ ∑ mb b∈Ma

Proof. Consider the polarization map Pk,1 ∶ S k+1 (S d V ) → S k (S d V ) ⊗ S d V. By Schur’s lemma Pk,1 ((Sa V )⊕ma ) ⊂ ( ⊕ (Sb V )⊕mb ) ⊗ S d V. b∈Ma

By Proposition 3.2 (Sa V )⊕ma ⊂ ( ⊕ (Sb V )⊕mb )(1) . b∈Ma

Since Pk,1 is injective, ma ≤ ∑ mb b∈Ma

 Proposition 3.23. S(5,4,4,2) V ⊂ (S(5,4,2,1) V ⊕ S(4,4,4) V )(1) . Let f ∈ S(5,4,4,2) V ⊂ S 5 (S 3 V ) be a highest weight vector, then fe1 e2 is a highest weight vector of S(4,4,4) V ⊂ S 4 (S 3 V ) and fe2 e4 is 4 3 a highest weight vector of S(5,4,2,1) V ⊂ S 4 (S 3 V ). Therefore S(5,4,4,2) V is not contained in the prolongation of S(4,4,4) V or S(5,4,2,1) V . Proof. Since S 4 (S 3 V ) =

S(12) V + S(10,2) V + S(9,3) V + S(8,4) V + S(8,2,2) V + S(7,4,1) V + S(7,3,2) V + S(6,6) V + S(6,4,2) V + S(6,2,2,2) V + S(5,4,2,1) V + S(4,4,4) V.

By Proposition 3.22, S(5,4,4,2) ⊂ (S(5,4,2,1) V ⊕ S(4,4,4) V )(1) . By induction fe1 e2 and fe2 e4 are 4 3 killed by all raising operators. Let h1 be a highest weight vector of S(4,4,4) V ⊂ S 4 (S 3 V ) and h2 be a highest weight vector of S(5,4,2,1) V ⊂ S 4 (S 3 V ). Set fe1 e2 = c1 h1 and fe2 e4 = c2 h2 , where c1 4 3 and c2 are constants, then c1 , c2 can not be both 0 by Proposition 3.19. Since fe3 ∈ S(5,4,2,1) V ⊕ S(4,4,4) V with weight (5,4,1,2), I claim that fe3 = c3 E34 fe2 e4 , where c3 3 3 3 is a constant. By (7), E43 fe3 = −fe2e4 , so c3 E43 E34 fe2e4 = −fe2 e4 , which implies c3 (E33 − E44 )fe2 e4 = 3 3 3 3 3 −fe2e4 , so c3 = −1. Since (fe1 e2 )e3 = (fe3 )e1 e2 , 3

4

3

3

c1 (h1 )e3 3

4

= (−E34 fe2 e4 )e1 e2 3

4

= −c2 (E34 h2 )e1 e2 4

= −c2 (E34 (h2 )e1 e2 − (h2 )e1 e3 e4 ) 4

= c2 (h2 )e1 e3 e4

10

YONGHUI GUAN

By Proposition 2.5, (h1 )e3 and (h2 )e1 e3 e4 are both highest weight vectors of S(4,4,1) V ⊂ S 4 (S 3 V ), 3 by rescaling, we may assume they are equal, so c1 = c2 , so c1 and c2 are both nonzero, therefore fe1 e2 is a highest weight vector of S(4,4,4) V ⊂ S 4 (S 3 V ) and fe2 e4 is a highest weight vector of 4 3 S(5,4,2,1) V ⊂ S 4 (S 3 V ).  4. The case when the degree is 3 Consider dim V = 6 and σ2 (Ch3 (V ∗ )) , Proposition 4.1. When dim V = 6, we have I4 (Ch3 (V ∗ ))(1) = S(7,2,2,2,2) V ⊕ S(6,4,2,2,1) V ⊕ S(5,5,3,1,1) V,

I4 (Ch3 (V ∗ ))(2) = S(8,2,2,2,2,2) V ⊕ S(7,4,2,2,2,1) V ⊕ S(6,5,3,2,2,1) V ⊕ S(5,5,5,1,1,1) V,

I4 (Ch3 (V ∗ ))(3) = 0. Proof. First want to prove

I4 (Ch3 (V ∗ ))(1) = S(7,2,2,2,2) V ⊕ S(6,4,2,2,1) V ⊕ S(5,5,3,1,1) V.

(8) By (4), Recall

I4 (Ch3 (V ∗ )) = S(7,3,2) V + S(6,2,2,2) V + S(5,4,2,1) V. S 4 (S 3 V ) =

S(12) V + S(10,2) V + S(9,3) V + S(8,4) V + S(8,2,2) V + S(7,4,1) V + S(7,3,2) V + S(6,6) V + S(6,4,2) V + S(6,2,2,2) V + S(5,4,2,1) V + S(4,4,4) V.

and S 5 (S 3 V ) = S(15) V + S(13,2) V + S(12,3) V + S(11,4) V + S(11,2,2) V + S(10,5) V + S(10,4,1) V + S(10,3,2) V + S(9,6) V + 2S(9,4,2)V + S(9,2,2,2)V + S(8,6,1) V + S(8,5,2)V + S(8,4,3) V + S(8,4,2,1)V + S(8,3,3,2) V + S(7,6,2) V + S(7,5,2,1) V + S(7,4,4) V + S(7,4,3,1)V + S(7,4,2,2) V + S(7,2,2,2,2) V + S(6,6,3) V + S(6,5,2,2)V + S(6,4,4,1) V + S(6,4,2,2,1) V + S(5,5,3,1,1) V + S(5,4,4,2)V.

Since I4 (Ch3 (V ∗ )) contains all the modules with length 4 in S 4 (S 3 V ), by Proposition 3.22 any module with length 5 in S 5 (S 3 V ) is in I4 (Ch3 (V ∗ )(1) . On the other hand, we can prove the other modules with length no more than 4 in S 5 (S 3 V ) are not in I4 (Ch3 (V ∗ )(1) one by one. By Proposition 3.19, for any module with length no more than 4 in S 5 (S 3 V ), one can find a partial derivative of a highest weight vector of this module such that it is a highest weight vector of a module in S 4 (S 3 V ) but not in I4 (Ch3 (V ∗ ). For most modules, we can check directly, but for some modules, we need to verify carefully, for instance, By Proposition 3.23, S(5,4,4,2) ⊂ (S(5,4,2,1) ⊕ S(4,4,4) V )(1) , but S(5,4,4,2) is not in the in the prolongation of any of the two modules, so S(5,4,4,2) is not not in I4 (Ch3 (V ∗ )(1) , the reason is fe1 e2 is a highest weight vector of S(4,4,4) V ⊊ I4 (Ch3 (V ∗ )). We conclude 4

I4 (Ch3 (V ∗ )) = S(7,3,2) V + S(6,2,2,2) V + S(5,4,2,1) V. Similarly, by studying the modules in S 6 (S 3 V ) and S 7 (S 3 V ) one by one, we conclude I4 (Ch3 (V ∗ ))(2) = S(8,2,2,2,2,2) V ⊕ S(7,4,2,2,2,1) V ⊕ S(6,5,3,2,2,1) V ⊕ S(5,5,5,1,1,1) V, I4 (Ch3 (V ∗ )(3) = 0.



EQUATIONS FOR SECANT VARIETIES OF CHOW VARIETIES

11

Therefore by Proposition 4.1 and Theorem 3.3, Theorem 4.2. (restatement of Theorem 1.2) I7 (σ2 (Ch3 (V ∗ ))) = I4 (Ch3 (V ∗ ))(3) = 0. Also Theorem 4.3. (restatement of Theorem 1.3) I8 (σ2 (Ch3 (V ∗ ))) ⊃ S(5,5,5,5,3,1) V. Proof. By Example 3.8, −1 (I4 (Ch3 (V ∗ )) ⊗ S 4 (S 3 V ) + S 4 (S 3 V ) ⊗ I4 (Ch3 (V ∗ ))). I8 (σ2 (Ch3 (V ∗ ))) = I5 (Ch3 (V ∗ ))(3) ⋂ F4,4

Since all the modules in S 5 (S 3 V ) with the first row 5 is contained in I5 (Ch3 (V ∗ )), by Proposition 3.2 and Schur’s lemma, S(5,5,5,5,3,1) V ⊂ I5 (Ch3 (V ∗ )(3) .

(9) Consider the map

F4,4 ∶ S 8 (S 3 V ) → S 4 (S 3 V ) ⊗ S 4 (S 3 V ). Since I4 (Ch3 (V ∗ ))

c

= S(12) V + S(10,2) V + S(9,3) V + S(8,4) V + S(8,2,2) V + S(7,4,1) V + S(6,6) V + S(6,4,2) V + S(4,4,4) V,

and S(5,5,5,5,3,1) V ⊈ S(4,4,4) V ⊗ S(4,4,4) V by the Littlewood-Richardson rule, c

c

S(5,5,5,5,3,1) V ⊈ I4 (Ch3 (V ∗ )) ⊗ I4 (Ch3 (V ∗ )) . Therefore by Schur’s lemma −1 S(5,5,5,5,3,1) V ⊂ F4,4 (I4 (Ch3 (V ∗ )) ⊗ S 4 (S 3 V ) + S 4 (S 3 V ) ⊗ I4 (Ch3 (V ∗ ))).

The result follows.



Remark 4.4. Since σ2 (Ch3 (C5∗ )) ⊆ S 3 (C5∗ ), by inheritance (see [16]), the ideal of σ2 (Ch3 (V ∗ )) should contain modules with length 5. So S(5,5,5,5,3,1) V is not enough to cut out σ2 (Ch3 (V ∗ )) set-theoretically. One can get length 5 modules with high degree in the ideal of σ2 (Ch3 (V ∗ )) by Koszul Young flattenings [10], but I still do not know whether they are enough to define σ2 (Ch3 (V ∗ )) set-theoretically. dim S(5,5,5,5,3,1) V = 1134 and codim σ2 (Ch3 (V ∗ )) = 24, therefore σ2 (Ch3 (V ∗ )) is very far from being a complete intersection. The next question is: what is the difference between the dimension of σ2 (Ch3 (V ∗ )) and the zero set of S(5,5,5,5,3,1) V ? 5. The case when the degree is 4 Consider σr (Ch4 (V ∗ )) ⊂ S 4 (V ∗ ), where dim V ≥ 4r, one can directly get nontrivial modules in the ideal of σr (Ch4 (V ∗ )) via prolongations. Theorem 5.1. (restatement of Theorem 1.4) When dim V ≥ 4r, I4r+1 (σr (Ch4 (V ∗ ))) = I5 (Ch4 (V ∗ ))(4r−4) and S(6,6,44r−2 ) V ⊂ I4r+1 (σr (Ch4 (V ∗ ))). Proof. By Proposition 2.7, Proposition 2.9 and Proposition 2.10, I4 (Ch4 (V ∗ )) = 0 and I5 (Ch4 (V ∗ )) = S 5 (S 4 V ) − S 4 (S 5 V ) as abstract modules. By Theorem 3.3, I4r+1 (σr (Ch4 V ∗ ) = I5 (Ch4 V ∗ )(4r−4) .

12

YONGHUI GUAN

Since S(6,6,6,2) ⊂ S 4 (S 5 V ) has the lowest highest weight vector with respect to the lexicographic order among all the modules in S 4 (S 5 V ), consider the polarization map F4r−4,5 ∶ S 4r+1 (S 4 V ) → S 4r−4 (S 4 V ) ⊗ S 5 (S 4 V ). By Proposition 3.2, −1 (S 4r−4 (S 4 V ) ⊗ I5 (Ch4 (V ∗ ))). I5 (Ch4 V ∗ )(4r−4) = F4r+1,4

By Schur’s lemma and the Littlewood-Richardson rule, S(6,6,44r−2 ) V ⊂ I5 (Ch4 V ∗ )(4r−4) = I4r+1 (σr (Ch4 (V ∗ )).  Remark 5.2. Consider r = 2 and dim V = 8. Since σ2 (Ch4 C4∗ )) ⊆ S 3 (C4∗ ), by inheritance (see [16]), the ideal of σ2 (Ch4 (V ∗ )) should contain modules with length 4. So S(6,6,4,4,4,4,4,4) V is not enough to cut out σ2 (Ch4 (V ∗ )) set-theoretically. One can get a length 4 module with high degree in the ideal of σ2 (Ch4 (V ∗ )) by Koszul Young flattenings [10], but I still do not know whether they are enough to define σ2 (Ch4 (V ∗ )) set-theoretically. We know that dim S(6,6,4,4,4,4,4,4) V = 336 and codim σ2 (Ch3 (V ∗ )) = 272, therefore σ2 (Ch4 (V ∗ )) is far from being a complete intersection. The next question is: what is the difference between the dimension of σ2 (Ch4 (V ∗ )) and the zero set of S(6,6,4,4,4,4,4,4) V ? 6. General case for even degrees Let λ be a partition of order kd, a semi-standard tableau of shape λ and content k × d is a semi-standard filling of a Young diagram associated to λ with #1 = ⋯ = #k = d. Proposition 6.1. [3] Let λ be a partition with order kd with d odd , then the multiplicity of λ in S k (S d V ) is less than or equal to the number of semi-standard tableaux of shape λ and content k × d with the additional property : for each pair (i, j), 1 ≤ i ≠ j ≤ k, the set of columns of i is not exactly the columns of j. Proposition 6.2. [20] Let λ be a partition with order kd and u be even, mult(Sλ V, S k (S d V )) = mult(Sλ+(uk ) V, S k (S d+u V )).

Theorem 6.3. S((2m+2)2m−1 ,2) V ⊂ S 2m (S 2m+1 V ), with multiplicity 1, and S((2m+2)2m−1 ,2) V is the smallest module with respect to the lexicographic order among all the modules in the decomposition of S 2m (S 2m+1 V ). Proof. First, let λ = (λ1 , ⋯, λ2m ) be a partition with order 4m2 + 2m and smaller than ((2m + 2)2m−1 , 2) , then λ1 ≤ 2m + 2 and λ2m ≥ 3. Consider the semi-standard tableaux with content λ1 −3 ) ≤ 2m − 1 2m × (2m + 1), the first 3 columns should be 1 to 2m. Therefore there are (2m−2 possible sets of columns, but there are 2m numbers to be filled in the semi-standard tableaux, by Proposition 6.1, mult(Sλ V, S 2m (S 2m+1 V )) = 0. Second, consider the partition λ = ((2m+2)2m−1 , 2), by Proposition 6.2, mult(Sλ V, S 2m (S 2m+1 V )) = mult(S(2m2m−1 ) V, S 2m (S 2m−1 V )). By [15] formula number (80), mult(S(2m2m−1 ) V, S 2m (S 2m−1 V )) = 1, the only filling is the following (I take m=3 as an example). 1 2 3 4 5

1 2 3 4 6

1 2 3 5 6

1 2 4 5 6

1 3 4 5 6

2 3 4 5 6

EQUATIONS FOR SECANT VARIETIES OF CHOW VARIETIES

13

 Let d = 2m and dim V ≥ 2mr, consider the variety σr (Ch2m (V ∗ )) ⊂ S 2m V ∗ . Theorem 6.4. (restatement of Theorem 1.5) The isotypic component of S((2m+2)m ,(2m)2mr−m ) V

is contained in I2m+1 (Ch2m (V ∗ ))(2m(r−1)) ⊂ I2mr+1 (σr (Ch2m (V ∗ ))). Moreover any module with even partition and smaller than ((2m + 2)m , 2) (with respect to the lexicographic order) is in I2mr+1 (σr (Ch2m (V ∗ ))). Proof. By Theorem 6.3, S(2m+2m ,2) V is the smallest module (with respect to the lexicographic order) in the decomposition of S 2m (S 2m+1 V ). Therefore by Corollary 2.8, any module smaller than S(2m+2)m ,2) V (with respect to the lexicographic order) is not in I2m+1 (Ch2m V ∗ ))c ⊂ S 2m+1 (S 2m V ). Consider the polarization map F2mr−2m,2m+1 ∶ S 2mr+1 (S 2m V ) → S 2mr−2m (S 2m V ) ⊗ S 2m+1 (S 2m V ). By Proposition 3.2, −1 I2mr+1 (Ch2m V ∗ )(2m(r−1)) = F2mr−2m,2m+1 (S 2mr−2m (S 2m V ) ⊗ I2m+1 (Ch2m V ∗ )).

By Schur’s lemma and the Littlewood-Richardson rule, the whole isotypic component of −1 S((2m+2)m ,(2m)2mr−m ) V ⊂ F2mr−2m,2m+1 (S 2mr−2m (S 2m V )⊗I2m+1 (Ch2m V ∗ ) = I2mr+1 (Ch2m V ∗ )(2m(r−1)) . By Corollary 3.9, I2mr+1 (Ch2m V ∗ )(2m(r−1)) ⊂ I2mr+1 (σr (Ch2m V ∗ )). In fact any module in S 2mr+1 (S 2m V ) with even partition and smaller than ((2m + 2)m , 2) (with respect to the lexicographic order) is in I2mr+1 (σr (Ch2m (V ∗ ))).  7. A property about Plethysm Lemma 7.1. [19, 5, 20] mult(Sλ V, S k (S 2l V )) = mult(SλT V, S k (Λ2l V )), and mult(Sλ V, S k (S 2l+1 V )) = mult(SλT V, Λk (Λ2l V )). Theorem 7.2. Let d be even, if S(a1 ,⋯,ap ) ⊂ S k (S d V ) and S(b1 ,⋯,bq ) ⊂ S l (S d V ) with ap ≥ b1 , then S(a1 ,⋯,ap ,b1 ,⋯,bq ) ⊂ S k+l (S d V ) as long as dim V ≥ k + l. Proof. Let λ = (a1 , ⋯, ap ) and µ = (b1 , ⋯, bq ). By Lemma 7.1, mult(SλT V, S k (Λd V )) > 0 and mult(SµT V, S l (Λd V )) > 0, so mult(SλT +µT V, S k+l (Λd V )) > 0. By Lemma 7.1 again, mult(S(λ,µ) V, S k+l (S d V )) > 0.  Remark 7.3. This is false when d is odd: C.Ikenmeyer gave a counterexample for d = 3. There exists k0 such that S6k0 V ⊂ S 2k0 (S 3 V ) but S6k0 +1 V ⊊ S 2k0 +2 (S 3 V ).

14

YONGHUI GUAN

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