www.theoryofcomputing.org
Explicit Polynomial Sequences with
Maximal Spaces of Partial Derivatives and a
Question of K. Mulmuley
Fulvio Gesmundo
∗Joseph M. Landsberg
†Received June 18, 2017; Revised December 31, 2018; Published September 12, 2019
Abstract: We answer a question of K. Mulmuley. Efremenko et al. (Math. Comp., 2018) have 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.
ACM Classification:F.1.3
AMS Classification:68Q15, 15A69
Key words and phrases: computational complexity, Waring border rank, shifted partial derivatives, Koszul flattening
1
Introduction
LetSdCN denote the space of homogeneous polynomials of degreed inNvariables and let p∈SdCN. LetSeCN∗denote the space of homogeneous differential operators of orderewith constant coefficients,
∗Supported by the European Research Council (ERC Grant Agreement no. 337603) and VILLUM FONDEN via the QMATH Centre of Excellence (Grant no. 10059).
which acts onSdCN whene≤d. Thee-th partial derivative mapofp(ore-th flattening of p) is
pe,d−e:SeCN∗→Sd−eCN (1.1)
D7→D(p).
We call the image ofpe,d−ethee-thspace of partial derivativesofp; it is straightforward to verify that rank(pe,d−e) =rank(pd−e,e)and that givene0≤e≤d/2, ifpe,d−ehas full rank thenpe0,d−e0 has full rank.
LetM≥N. Choose a linear inclusionCN⊆CM, so that polynomials inNvariables can be regarded as polynomials inM variables which happen to use only N of them. A polynomial p∈SdCN is a degenerationofq∈SdCM, if p∈GLM·q⊆SdCM, where the overline denotes closure, equivalently in the usual (Euclidean) topology or in the Zariski topology. Similarly, p is aspecializationof qif p∈EndM·q⊆SdCM. Notice that ifpis a specialization ofqthen it is a degeneration ofq. In complexity theory, one is interested in finding obstructions to specialization of a polynomialpto a polynomialq.
A common strategy to determine obstructions to degeneration (hence, to specialization) of a polyno-mialqto a polynomialpis to find closed conditions thatqand every element ofGLM·qsatisfy, butp does not satisfy. Typical examples of this method are flattening techniques (see, e. g., [4]), which exploit the semi-continuity of matrix rank, namely that{X∈Mat`×`: rank(X)≤r}is a closed subset of Mat`×`
(both in the usual and in the Zariski topology).
The method of partial derivatives is one such example: since the entries ofqe,d−e (as a matrix expressing the mapSeCN∗→Sd−eCN) are continuous in the coefficients ofq, wheneverpis a degeneration ofq, semi-continuity of matrix rank guarantees that rank(pe,d−e)≤rank(qe,d−e)for alle. Therefore, comparing the ranks of the partial derivatives maps ofpandqfor anye, one can prove that pis not a degeneration ofq(and thus nor is pa specialization ofq).
The method of partial derivatives dates back to Sylvester in 1852 [32], who called the maps (1.1) 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., [2, 16]). Thesymmetricor Waring rankof a polynomial p∈SdCN is the smallest r such that p=∑rj=1`dj where`j∈CN are linear forms. One writesRS(p) =r. ThesymmetricorWaring border rankofpis the smallestrsuch thatpis a limit of polynomials of Waring rankr, and one writesRS(p) =r. The ranks of the partial derivatives maps give lower bounds for the symmetric border rank ofp: RS(p)≥maxe{rank(pe,d−e)}. From a complexity theory perspective, Waring rank captures the complexity of a polynomial in the model of depth-three powering circuits, orΣΛΣcircuits, namely depth-three arithmetic circuit whose first and third layers
Let pn,d =xd1+· · ·+xnd denote the power sum polynomial of degree d inn variables and hn,d =
∑|α|=dxα11 · · ·xαnn the complete symmetric polynomial of degreedinnvariables. For the following polynomial sequences, all partial derivatives maps have full rank:
• PBier,n,d:=∑|α|=d(α1x1+· · ·+αnxn)d whereαranges over all multi-indices(α1, . . . ,αn)of non-negative integers such that α1+· · ·+αn=d, i. e., exponents of monomials of degree d in n variables (Bierman [1], Reznick [28]);
• fn,k:= (pn,2)k∈S2kCn (Reznick [28], a proof is given in §2); • efn,k:=pn,1fn,k∈S2k+1Cn (Theorem 2.5);
• hn,d∈SdCn (Theorem 5.6).
Whenn,dare polynomially related,hn,d∈VPs, 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. On the other hand, the polynomials fn,kand efn,kbelong to the complexity class
VPeof polynomials admitting a polynomial size formula. The fact that there are elements inVPehaving partial derivatives map of full rank suggests that theΣΛΣmodel is quite weak.
Themethod of shifted partial derivativesis a variant of the method of partial derivatives. Kayal introduced it in [18] and Gupta et al. [14] exploited it to prove super-polynomial complexity lower bounds for depth-four circuits for the permanent (and determinant). In the same paper the authors ask if the method could be used to approach Valiant’s conjecture, that is to separate the classVPfrom the class
VNP.
For p ∈SdCN the method of shifted partials is based on the study of the following maps (for judiciously choseneandτ):
p(e,d−e)[τ]:SeCN
∗
⊗Sτ
CN →Sd−e+τCN
D⊗q7→qD(p).
Notice that ifτ=0, thenp(e,d−e)[τ]is the partial derivative map defined in (1.1). Let
h∂=epi=τ:=p(e,d−e)[τ](SeCN
∗
⊗Sτ
CN).
Again, semi-continuity of matrix rank guarantees that ifpis a degeneration ofq, then dimh∂=eqi=τ≥
dimh∂=epi=τ for allτand the method of shifted partials can be used to prove thatpis not a degeneration
ofqby showing that dimh∂=epi=τ >dimh∂=eqi=τ for someτ.
The results of [6] show that the method of shifted partial cannot be used to separateVPsfromVNP in the classical formulation of Valiant’s conjecture, where one seeks for a super-polynomial lower bound on the so-called determinantal complexity of thepaddedpermanent polynomial, namely on the smallest possiblen(m)such thatzn(m)−mpermmis a specialization of detn(m). Informally, the proof of [6] exploits
the paddingzn−mto prove that the shifted partial spaces ofzn−mpermmdo not grow fast enough to give a super-polynomial separation from the shifted partial spaces of detn.
This motivates the question on whether the method of shifted partials can be used to achieve a super-polynomial lower bound in a model which does not require padding.
Definition 1.1. Givenn,d, letXα= ((ξα)
i
j)i,j=1,...,nben×nmatrices of indeterminates forα=1, . . . ,d. The(n,d)-iterated matrix multiplicationpolynomial IMMdn∈Sd(Cdn
2 )is
IMMdn:(X1, . . . ,Xd)7→trace(X1· · ·Xd).
Let (ξij)i,j=1,...,n be an n×n matrix of indeterminates. The (n,d)-matrix powering polynomial Powdn∈SdCn
2
is
Powdn:X 7→trace(X d).
By Nisan [24], the polynomials IMMdn and Powdn can be used to defineVPs-complete sequences without the use of padding. More precisely, a sequence of homogeneous polynomials{fm}m∈Nwith
fm∈Sdm
CMm (withdm,Mmgrowing polynomially inm) is inVPsif and only if either of the following two equivalent conditions holds:
• there exists a functionn(m)growing polynomially inm, such thatfmis a specialization of IMMdnm(m);
• there exists a function n0(m) growing polynomially in m, such that fm is a specialization of Powdm
n0(m)).
Remarkably, in this case, the model does not require padding and allows one to compare IMMdnand Powdn directly with the sequence of polynomials.
In particular, Valiant’sVPs6=VNPconjecture can be rephrased by stating that there is no polynomially bounded functionn(m)such that the permanent polynomial permmis a specialization of IMM
m
n(m)(or of
Powmn(m)). Note that Powmn is a specialization of IMMmn.
We prove that the method of shifted partials cannot be used to achieve a super-polynomial separation between permmand IMMmn:
Theorem 1.2. If n>m5, thenpermmcannot be separated fromIMMmn by the method of shifted partial derivatives. More precisely, given any linear inclusionCm
2
⊆Cmn2 consideringperm
m∈SmCmn
2
as a polynomial that just involves m2of the mn2variables, then for all choices of e,τ,
dimh∂=e(permm∈S m
Cmn
2
)i=τ≤dimh∂=
e
Additional results
We givea prioriupper bounds for the utility of Koszul flattenings, another variant of the partial derivatives map, in comparing the complexity of polynomials (Proposition 6.1). We show that these bounds are sharp for the first Koszul flattenings in low dimensions and degree (Remark 6.6). We obtain explicit (but not sharp) lower bounds for the Koszul flattenings of efn,k, showing that one obtains better Waring border rank lower bounds with this method than by the method of partial derivatives (Proposition 6.4). Ironically, now the simple polynomial efn,khas the highest Waring border rank lower bound of all explicit polynomials of odd degree.
Related work
Leten,d=∑1≤i1<i2<···<id≤nxi1· · ·xid ∈S
d
Cndenote the elementary symmetric polynomial of degreedin nvariables. The complexity ofen,dhas been well studied: its symmetric border rank is bounded below by
n bd/2c
because this is the rank of(en,d)bd/2c,dd/2e (see, e. g., [25]); Lee [22] proved that its symmetric rank is∑bdi=/02c
n i
whendis odd and there is a similar formula providing a lower bound whendis even. Its padded version can be computed by a homogeneous depth-three (ΣΠΣ) circuit of sizen2 (due to
Ben-Or), and when log(n)≤d≤2n/3 , one has the lower bound of max{Ω(n2/d),Ω(nd)}by Shpilka and Wigderson [29] 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∏ni=1(1−xit)−1. The gap between the padded and unpadded depth-three circuit
complexity may have led researchers to think the results of [6] might fail in a model without padding. For this reason, K. Mulmuley suggested that we investigate the barriers of the method of shifted partials in the unpadded setting (although Mulmuley himself anticipated our answer). Similar concerns were shown after the results by Ikenmeyer and Panova [17] and Bürgisser, Ikenmeyer and Panova [3] on the Geometric Complexity Theory program, and were partially addressed in [10], exploiting the same homogenization result of Nisan [24] that we use in this work.
The shifted partial derivative complexity of elementary symmetric polynomials is studied by Fournier et al. [7], where strong lower bounds are proved, which in turn give complexity lower bounds for depth-four circuits.
Acknowledgements
We thank Catherine Yan for discussions on the Gessel-Viennot method, Bruce Reznick and Zach Teitler for historical information, and Noah Stein and Mathoverflow for the proof ofLemma 5.3. We thank the Santa Fe Institute and the organizers of the working group inGeometric Complexity Theoryin December 2016 that inspired this work. We also thank the anonymous referees for numerous suggestions to improve the exposition and on the estimate in Case 2 of the proof ofTheorem 1.2.
2
Proofs that
f
n,k,
f
e
n,khave maximal partial derivatives
In this section, we prove that fn,k = (pn,2)k and fen,k = pn,1fn,k have partial derivatives maps of maximal rank, or equivalently that their spaces of partial derivatives have the maximal possible dimension. Although the result for fn,k already appeared in [28, Thm. 8.15], we include our proof in this section because it seems less involved and more accessible than the original one; moreover the proof of the result for efn,k relies on the method that we use to prove the result for fn,k(seeRemark 2.3).
Proposition 2.1. The space of first derivatives of homogeneous degree d+2polynomials of the form hqn, where h runs over the space of homogeneous degree d polynomials, equals the space of all homogeneous polynomials of degree d+1. In symbols, setting L={∂(hqn):∂ ∈(Cn)∗,h∈SdCn}, we have L=Sd+1Cn.
Proof. Ifd=0, the statement holds as
∂
∂xiqn=2xi.
Letd≥1. Consider a monomialxα of degreed−1, where
α is a multi-index,|α|=d−1. For
i∈ {1, . . . ,n}, define
gi:=
∂ ∂xi
(qnxixα) =2xi2xα+ (αi+1)qnxα=
2x2i + (αi+1)qn
xα.
For everyi,gi∈L. Letg= (g1, . . . ,gn)T andp= (x21, . . . ,x2n)T be column vectors. We have
g= [(2Id+A)p]xα
whereAis the(n+1)×(n+1)rank 1 matrix whose entries in thei-th row areαi+1 fori=1, . . . ,n. Let1= (1, . . . ,1)T and a= (α1+1, . . . ,αn+1)T and noticeA=1aT. In particular by the Sylvester determinant identity (see, e. g., [27, Chap. 1, Prob. 3.1]) det(2Id+A)6=0, so 2Id+Ais invertible.
Therefore,x2ixα∈(
Cn)∗· qnSdCnfor every monomialxα∈Sd−1Cn. This shows that every non-square-free monomial belongs to(Cn)∗·(qnSdCn).
Now letxβ∈Sd+1
Cnbe a square-free monomial; supposeβ1=1 and letγbe the multi-index with γ1=0 andγj=βjfor j≥2, so thatxγ =xβ/x1. Then
2xβ = ∂
∂x1
(qnxγ).
This concludes the proof.
Theorem 2.2(Reznick, [28]). The space of e-th partial derivatives of fn.k consists of all multiples of qk−en of degree2k−e. In symbols, for all n,k and e≤k,(fn,k)e,2k−e(Se(Cn)∗) =qk−en SeCn. In particular, for all n,k,e, the flattening(fn,k)e,2k−ehas full rank.
Proof. We proceed by induction one. Fore=0 there is nothing to prove. Lete≥1. By the induction hypothesis, the(e−1)-st flattening surjects ontoqnk−e+1Se−1Cn. It suffices to show that
(Cn)∗·(qk−en +1S e−1
Cn) =qk−en S e
Notice that, for a monomialxα∈Se−1
Cn,
∂ ∂xi
(qk−en +1xα) =2(k−e+1)xiqk−e
n xα+q k−e+1
n
∂xα ∂xi
=qk−en
2(k−e+1)xixα+qn∂x α
∂xi
.
Up to rescalingqnand the differential operators, the term in parenthesis is ∂ ∂xi(x
αqn). These terms span
SeCnbyProposition 2.1.
Remark 2.3. The results ofProposition 2.1andTheorem 2.2hold for every non-degenerate quadratic formg∈S2Cn. Indeed every quadric can be diagonalized by the action ofGLnand the proof only uses the fact thatqnis diagonal. In particular, for everyg∈S2Cn, and everye≤k, we have
(SeCn)∗·(gk) =gk−eSeCn.
In order to prove the analog ofProposition 2.1for efn,k, we will exploit the decomposition ofSdCn into Specht modules under the action of symmetric groupSnwhich permutes the variables. In particular, ifCn= [n]⊕[n−1,1]where[n] =h`niis an invariant subspace and[n−1,1] =hxi−xj:i,j=1, . . . ,ni is isomorphic to the standard representation ofSn. In particular, a homogeneous polynomial f ∈SdCn can be written as
f =
d
∑
e=0
`egd−e (2.1)
wheregd−e∈Sd−e[n−1,1]; this is the decomposition of f as sum of bi-homogeneous terms according to the decompositionCn= [n]⊕[n−1,1].
We redefine our basis of Cn and of Cn∗ in accordance to the splitting [n]⊕[n−1,1]: consider
[n] =h`niand[n−1,1] =hxi−x1:i=2, . . . ,niand in the dual space
`∗n= 1 n
∂ ∂x1
+· · ·+ ∂
∂xn
and ∆i=
1 2
∂ ∂xi
− ∂
∂x1
.
In particular,h∆i:i=2, . . . ,ni '[n−1,1]andh`∗ni '[n]. Since∆i(`n) =0 and`∗n·[n−1,1] =0, this gives the splittingCn∗=h`ni∗⊕[n−1,1]∗, where we identifyh`ni∗and[n−1,1]∗as subspace ofCn.
Sinceqnis anSn-invariant, we deduce thatqnis a linear combination ofgnand`2n: indeed,qn=
(1/n)(`2n+gn).
We can now prove the following result.
Theorem 2.5. For every n,k,e, the flattening(efn,k)e,2k+1−ehas full rank.
Proof. As before, writeCn=h`ni ⊕[n−1,1]. Up to rescalingqn, writeqn=`2n+gnas inRemark 2.4. We have
e
fn,k=`nqkn=`n(`2n+gn) k=`
n∑k0
k j
`2n(k−j)gk−n j=∑ k
0` 2(k−j)+1
n gnj
Lete≤k. In order to calculate rk(efe,2k+1−e), we study the image of efe,2k+1−eas anSn-module. We haveSeCn∗=Se(h`∗ni ⊕[n−1,1]∗) =
Le
j=0h(`∗n)ji ⊗Se−j[n−1,1]∗, so Se(h`∗ni ⊕[n−1,1]∗)·(`nqkn) =
e M
j=0
h(`∗n)ji ⊗Se−j[n−1,1]∗
·(`nqkn)
=
e
∑
j=0
h(`∗n)ji(Se−j[n−1,1]∗·(`nqkn)).
(2.2)
To conclude, we show that the last summation is in fact a direct sum. In the j-th summand, we have
h(`∗n)ji ·(Se−j[n−1,1]∗·(`nqkn)) =Se−j[n−1,1]∗·h`∗jn ·∑k i=0
k i
`2ni+1gk−in i
=
=Se−j[n−1,1]∗· k
∑
i=dj−21e k i
`2ni+1−jgk−in
. (2.3)
If∆∈Se−j[n−1,1]∗, we have
∆
k
∑
i=dj−1 2 e
k i
`2ni+1−jg k−i n = k
∑
i=dj−1 2 e
k i
`2ni+1−j∆(g k−i n ).
In particular the summands withi>k−e+jare 0; moreover, thei-th term of the summation, varying
∆, ranges on the entire`2ni+1−jg
k−i−(e−j)
n Se−j[n−1,1]by Theorem 2.2andRemark 2.3. We deduce
that the leading term (in`n) of the last line of (2.3) ranges on the whole`n2(k−e+j)+1−jSe−j[n−1,1] =
`2nk−2e+j−1Se−j[n−1,1]. This shows that the summands in (2.2) are linearly independent because their leading terms have different degrees, the j-th one having degree in`nequal to 2k−2e+j−1.
From (2.2), we deduce that rank of efe,2k+1−eis dim(SeCn∗·fee,2k+1−e) =
e
∑
j=0
dimSe−j[n−1,1] =
=
e
∑
j=0
e−j+n−2 e−j
=
=
e
∑
j=0
n−2+j j
=
e+n−1 e
hence, efe,2k+1−e is injective and this conclude the proof whene≤k. Since fe,2k+1−eis the transpose of f2k+1−e,e, the proof is complete.
3
Two auxiliary results
Proposition 3.1. Let n(m),k(m)be polynomially bounded functions of m. Then the sequences{fn,k}m and{efn,k}mare in the algebraic complexity classVPeof sequences admitting polynomial size formulas.
Proof. The standard expression
fn,k=qn· · ·qn (ktimes), with
qn=x1·x1+· · ·+xn·xn
gives a formula of sizek(3n−1)for fn,k. The additional`n=x1+· · ·+xnand efn,k=`n·fn,kprovides a formula of sizek(3n−1) +nfor efn,k.
Proposition 3.2. If n=2m then the polynomial fn,kis a specialization of the matrix powering polynomial Pow2mk+1 and the polynomial efn,k is a specialization of the iterated matrix multiplication polynomial IMM2mk++21. If n=2m+1then the polynomial fn,k is a specialization of the matrix powering polynomial Pow2mk+2 and the polynomial efn,k is a specialization of the iterated matrix multiplication polynomial IMM2mk++31.
Proof. Let n=2m+1 and sety±j =x2j−1±
√
−1x2j for j=1, . . . ,m. Consider the specialization of Pow2mk+2to the matrix
Qm=
0 y+1 · · · y+m xn y−1
..
. 0
y−m xn
,
of sizem+2.
We show that Pow2mk+2(Qm) = fn,k, up to scale. The characteristic polynomial ofQmis det(Qm−tIdm+2) = (−1)m+2tm
t2−∑jy+jy −
j −x2n
= (−1)m+2tm t2−qn.
Thus, the nonzero eigenvalues ofQm(as functions ofx1, . . . ,xn), are±
√
qn. In particular, Powdm+2(Qm) =
(√qn)d+ (−√qn)dis 0 ifd is odd and it is 2qkn=2fn,kifd=2kis even.
Ifn=2mis even, apply the same argument to the matrix obtained fromQmby removing the last row and the last column.
Similarly, efn,k is a specialization of IMM2mk++21or IMM2
k+1
m+3 (depending on the parity ofn) by making
4
Proof of
Theorem 1.2
Sincen≥m5, we may choose a linear inclusionCm
2
⊆Cmn2
and regard permm∈SmCmn2. Our goal is to show that for everys,τ
dimh∂=sIMMmni=τ ≥dimh∂=
s
permmi=τ.
We split the proof into three cases. In the first and in the second case, we degenerate IMMmn to fn,k if m=2kis even and to efn,kifm=2k+1 is odd. This is possible byProposition 3.2. WriteFm,nfor either fn,k or efn,kin what follows. Since IMMmn degenerates toFm,n, we have dimh∂=sIMMmni=τ≥dimh∂=sFm,ni=τ.
In the third case, we specialize IMMmn to the power sum polynomial of degree minm2 variables ym1+· · ·+ymm2by specializing every argument of IMM
m
n to the diagonal matrix of sizen×nwithy1, . . . ,ym2
in the firstm2diagonal entries and 0 elsewhere.
Case 1:s≥ dm/2e. We show that dimh∂=sFm,ni=τ≥dimh∂=spermmi=τ whens≥ dm/2e. Up to the
action ofGLmn2, assumeCm 2
⊆Cn⊆ Cmn
2
, whereCm
2
is the space spanned by the variables of permm andCnis the space spanned by the variables ofFm,n. It will suffice to prove
dimh∂=s(permm∈S m
Cn)i=τ ≤dimh∂
=s
Fm,n∈SmCni=τ
because the remainingmn2−nvariables will contribute the same growth to the idealsJs(permm)and
Js(Fm,n). Since s≥ dm/2e, (Fm,n)s,m−ssurjects onto Sm−sCn by Theorem 2.2and Theorem 2.5, and thus, for every shiftτ, the shifted partial derivative map surjects ontoSm−s+τCnfor allτ. This shows
h∂=sFm,n∈SmCni=τ=Sm−s+τCnand proves this case.
Case 2:s<dm/2eandτ<2m3. Again, it suffices to prove
dimh∂=spermm∈SmCni=τ≤dimh∂
=s
Fm,n∈SmCni=τ.
Sinces≤ dm/2e,(Fm,n)s,m−sis injective, and its partials have image of dimension n+s−s 1. Corollary 2.4 of [6] states that any subspace ofSm−sCnof dimension n+s−s 1
generates an ideal that in degreem−s+τ
has dimension at least n+s+τ−1
s+τ
; this is a consequence of a general result on the growth of ideals known as Macaulay’s Theorem (see, e. g., [30]). Thus
dimh∂=sFm,n∈SmCni=τ≥
n+s+τ−1
s+τ
,
(and equality holds in the casemis even).
We compare this with the crude estimate for permmthat ignores syzygies of itss-th Jacobian ideal. The spaceh∂=spermm⊆SmCni=0has dimension ms
2
, because s-th partial derivatives of permmare subpermanents of sizem−s, so there is one for every choice ofsrows ands columns of the matrix. Ignoring syzygies ofJs(permm),
m
s 2
n+τ−1
τ
≥dimh∂=s(permm⊆S m
Cn)i=τ.
We will conclude that
n+s+τ−1
s+τ
> m s 2
n+τ−1 τ
in the range we consider. This is equivalent to
(n+s+τ−1)(n+s+τ−2)· · ·(n+τ)
(τ+s)(τ+s−1)· · ·(τ+1) >
m
s 2
. (4.1)
The left hand side is bounded from below by(n+τ)s/(τ+s)sand the right hand side is bounded from above bym2s, so that a sufficient condition for (4.1) is
n+τ τ+s >m
2.
which holds whenn≥m5andτ≤2m3ass≤m.
Case 3:s<m/2 andτ>m3. Here set all matrices(X1, . . . ,Xm)equal to a matrix that is zero except
for the firstm2entries on the diagonal, call them y1, . . . ,ym2. The resulting degeneration of IMMmn is
ym1+· · ·+ymm2. As in Case 1, it will suffice to prove the result for the shifted partials of both polynomials
inm2variables because the remainingmn2−m2variables will contribute the same growth to both ideals. The space of partial derivatives of ordersishym−s1 , . . . ,ym−s
m2 i. The image of theτ-th shifted partial map
consists of all polynomials inm2variables of degreem−s+τas soon asm−s+τ >m2(m−s−1) +1,
so that every monomial of degreem−s+τ is divisible by at least one power of orderm−s. In particular, the shifted partials derivative map is surjective whenever whenτ >m3.
5
Complete symmetric functions
Recall thathn,dis the complete symmetric function of degreedinnvariables:
hn,d=
∑
|α|=dxα,
where the summation is over all multi-indicesα= (α1, . . . ,αn)withα1+· · ·+αn=d.
Proposition 5.1. For every monomialxβ with|
β|=e≤d, we have
∂e ∂xβ
hn,d=β!·hn+e,d−e(x1, . . . ,xn,x(β)), where
x(β)= (x
1, . . . ,x1
| {z }
β1
, . . . ,xn, . . . ,xn | {z }
βn )
andβ!=β1!· · ·βn!. In particular the image of the flattening(hn,d)e,d−eis the space D
hn+e,d−e(x1, . . . ,xn,x(β)):|β|=e
E
Proof. We proceed by induction one. Ife=1, supposexβ=x
nand write
hn,d= d
∑
j=0
xnjhn−1,d−j(x1, . . . ,xn−1)
so that
∂
∂xnhn,d= d
∑
j=0
j·xnj−1hn−1,d−j(x1, . . . ,xn−1) =
=
d−1
∑
`=0
(`+1)·x`
nhn−1,d−`−1(x1, . . . ,xn−1) =
= "
d−1
∑
`=0
x`nhn−1,d−`−1(x1, . . . ,xn−1)
#
+xn "
d−2
∑
`=0
(`+1)·x`nhn−1,d−`−2(x1, . . . ,xn−1)
#
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 tohn,d−1(x1, . . . ,xn); by repeating this on the second summation we obtain
hn,d−1(x1, . . . ,xn) +xnhn,d−2(x1, . . . ,xn) +x2n "d−
3
∑
`=0
(`+1)·x`nhn−1,d−`−3(x1, . . . ,xn−1)
#
and iterating this process we obtain∑dj=0x
j
nhn,d−1−j(x1, . . . ,xn) =hn+1,d−1(x1, . . . ,xn,xn) proving the base case.
Lete≥1 and supposeβ1≥1. Letγ= (β1−1,β2, . . . ,βn). We have
∂e ∂xβ
hn,d=
∂ ∂x1
∂e−1 ∂xγ hn,d=
=γ!· ∂ ∂x1
hn+e−1,d−e+1(x1, . . . ,xn,x(γ)). By chain rule and by symmetry
∂ ∂x1
hn+e−1,d−e+1(x1, . . . ,xn,x(γ)) =
= (γ1+1)
∂ ∂y1
(
x1,...,xn,x(γ),x
1)hn+e−1,d−e+1(y1, . . . ,yn+e−1) = =β1hn+e,d−e((x1, . . . ,xn,x(γ),x1)),
where we used the casee=1 again. Sinceγ!·β1=β! , we conclude.
Proposition 5.2. For any choice of multi-indicesβ,γwith|β|=p and|γ|=e, the coefficient ofxγ in
hn+p,e(x1, . . . ,xn,x(β))is
n
∏
i=1
βi+γi
γi
Proof. Write[f]γ for the coefficient ofxγ in the polynomial f.
We use induction onp. Ifp=0, thenhn+p,e(x1, . . . ,xn,x(β)) =hn,e and for everyγwe have
[hn,e]γ=1=∏
n i=1
γi
γi
.
Let p ≥1 and suppose β1 ≥1. Write hn+p,e(y) = ∑ej=0y
j
1hn+p−1,e−j(y2, . . . ,yn+p). Let η1 =
(1,0, . . . ,0)∈Zn≥0. We have
[hn+p,e(x1, . . . ,xn,x(β))]γ = γ1
∑
j=0
[hn+p−1,e−j(x1, . . . ,xn,x(β−η1))]γ−jη1.
Apply the inductive hypothesis to the summands of the right hand side to get
[hn+p,e(x1, . . . ,xn,x(β))]γ= γ1
∑
j=0
"
γ1−j+β1−1
γ1−j
·
n
∏
i=2
βi+γi
γi # = = γ1
∑
j=0
β1−1+j
j
! n
∏
i=2
βi+γi
γi = = n
∏
i=1
βi+γi
γi
.
Proposition 5.2shows that the entries of the matrix representing the partial derivatives map ofhn,d in the monomial basis are products of binomial coefficients with a special combinatorial structure. Matrices with this structure are the object of study of the Lindström-Gessel-Viennot theory on totally nonnegative matrices (see, e. g., [11]). We will provide some results on matrices with this structure which will be used to prove that the matrices described inProposition 5.2have full rank.
Leta1, . . . ,aN∈Z≥0be nonnegative integers and letG(a1, . . . ,aN)be theN×Nsymmetric matrix
whose(i,j)-th entry is ai+aj
ai
. The Lindström-Gessel-Viennot Lemma (see [11, §2]) guarantees that G(a1, . . . ,aN)is a totally nonnegative matrix (in the sense that every minor is nonnegative), and its rank is equal to the number of distinctai’s. In particular,G(a1, . . . ,aN)is always positive semidefinite and it is
positive definite if and only if theai’s are distinct. Moreover ifai1=ai2 for somei1,i2, then thei1-th and
i2-th rows are equal.
Given two matricesA,Bof the same size, defineABto be theHadamard productofAandB. For vectorsa1, . . . ,aN∈Zm≥0, withai= (ai j)j=1,...,mdefine
G(a1, . . . ,aN):= m K
i=1
G(a1,i, . . . ,aN,i).
Our goal is to prove thatG(a1, . . . ,aN)is positive definite if theaiare distinct.
We will need the following two technical results. Given a matrixAwe denote byAi•thei-th row and byA•i thei-th column ofA, respectively, and byAIJ the submatrix consisting of rows in the setI of indices and columns in the setJof indices.
Lemma 5.3(Stein). Let A be symmetric, positive semidefinite. Let I={i1, . . . ,ir}be a set of indices such that the r vectors{A•
i}i∈Iare linearly independent. Then the principal submatrix AIIof A has full rank.
Proof. Without loss of generality, supposeI={1, . . . ,r}and letR=AII. We want to prove thatRis full rank, namely thatRu=0 for someu∈Rrimpliesu=0. Letv∈
Rnsuch thatvi=uiifi≤randvi=0 ifi>r. SinceAis positive semidefinite, writeA=BTB. We have
0=uTRu=vTAv=vTBTBv=kBvk.
In particularBv=0, thereforeAv=0; since the firstrcolumns ofAare linearly independent, we deduce v=0, so thatu=0 andRis nonsingular.
Lemma 5.4. Let A,B be symmetric N×N matrices such that A is positive definite and B is positive semidefinite with strictly positive diagonal entries. Then AB is positive definite.
Proof. Given a symmetric matrixC, denote byC(k)thek-th leading principal submatrix ofC, namely the
k×ksubmatrix consisting of the firstkrows andkcolumns ofC. ThenCis positive definite if and only if det(C(k))is positive for everyk. Therefore it suffices to show that the leading principal minors ofAB
are positive.
Let k0 be the smallestk such that det(B(k)) =0. From Schur’s Product Theorem (see, e. g., [27, Ex. 36.2.1]), the Hadamard product of positive definite matrices is positive definite. For everyk<k0, we
have thatB(k)is positive definite, so(AB)(k)=A(k)B(k)is positive definite as well and in particular
it has positive determinant.
Ifk≥k0, from [23, Eqn. 1.11], we have
det(A(k)B(k)) +det(A(k)B(k))≥det(A(k))∏ki=1bii+det(B(k))∏ki=1aii.
Now, fork≥k0,B(k)is a positive semidefinite matrix andB(k0)is a principal submatrix with det(B(k0)) =0,
so det(B(k)) =0 as well. Therefore
det((AB)(k))≥det(A(k))∏ki=1bii>0
proving that(AB)(k)has positive determinant.
Proposition 5.5. Leta1, . . . ,aN ∈Z≥m0. Then the rank of G(a1, . . . ,aN) equals the number of distinct m-tuplesa1, . . . ,aN.
Proof. We proceed by induction onm. Form=1, the statement follows from the Lindström-Gessel-Viennot Lemma.
Ifm≥2, for everyi=1, . . . ,N, writeai= (a0i,ai,m). LetA=G(a01, . . . ,a0N)andB=G(am,1, . . . ,am,N). By the induction hypothesisAis positive semidefinite and its rank is equal to the number of distincta0i’s. Similarly forB.
LetC=G(a1, . . . ,aN) =AB. If two pairsai=aj, thenA•i =A•j andB•i =B•j so that the corre-sponding two columns ofCare equal. Conversely, if two columnsCi•,C•j are equal, we show that the correspondingm-tuplesaiandajare equal. Consider the principal 2×2 submatrix obtained from these two columns:
IfA•i 6=A•j, then they are linearly independent by the induction hypothesis, and by Lemma 5.3 the submatrixAi ji j is positive definite. The submatrixBi ji j is positive semidefinite and has strictly positive
diagonal entries, therefore byLemma 5.4Ci ji jis positive definite, in contradiction with the assumption. This shows that ifCi•=C•j , thenA•i =A•j and thereforeB•i =B•j. In particularai=aj.
Therefore, we may assume thatChas distinct columns and our goal is to show thatChas full rank. Suppose by contradiction thatCdoes not have full rank and let
0=α1C1•+· · ·+αNC
•
N (5.1)
be a vanishing linear combination of the columns ofC.
Up to conjugation by a permutation matrix, suppose there exist 0=k0<k1<· · ·<kr=N such
thata0i=a0j ifks<i,j≤ks+1anda0i6=a0jotherwise. Notice that if theai’s are distinct, thenAhas full rank and so doesC fromLemma 5.4becauseBis positive semidefinite with strictly positive entries. Therefore, supposek1≥2 and up to reducing to a principal submatrix suppose thatαk1 6=0. Since the
firstk1columns (and rows) ofAare equal, the firstk1columns (and rows) ofBare linearly independent,
otherwise two of them would be equal, providing that twom-tuplesaiandajfori,j≤k1would be equal.
ByLemma 5.3, the principal submatrixB1,...,k1
1,...,k1 is positive definite.
The linear combination (5.1) can be written as
0=A•1(α1B•1+· · ·+αk1B
•
k1) +∑i>k1αi(A
• i B
• i).
DefineAeto be the matrix obtained fromAby removing the firstk1−1 rows and columns, that is e
A=G(a0k 1, . . . ,a
0
N).Aehas the same rank asA. LetB0=PTBP, for
P=
1 α1
. .. ... 1 ...
αk 1 . .. 1 ;
notice thatB0is obtained fromBby performing row and column operations. In particularB0has the same signature asB; moreover, from the block structure ofP, we deduce that the submatrixB01,...,k1
1,...,k1 has the
same signature asB1,...,k1
1,...,k1, namely it is positive definite. This shows that thek1-th diagonal entry ofB
0 is
strictly positive. DefineBeto be the submatrix obtained fromB0by removing the firstk1−1 rows and columns. DefineCe=AeB. The linear combination of (e 5.1) induces a vanishing linear combination among the columns ofC.e
By repeating this procedure at mostrtimes, we find a singularr×rmatrixe e C=e
e
AeBewithe e
Apositive
semidefinite and of full rank (so positive definite) ande e
Using these results, we can finally prove:
Theorem 5.6. For every n,d,e, the flattening(hn,d)e,d−e:Se(Cn)∗→Sd−eCnof the complete symmetric function hn,dhas full rank.
Proof. First, we consider the cased=2keven. It suffices to prove the result fore=k.
(hn,d)k,k(Sk(Cn)∗) = *
∂|β| ∂xβ
hn,d:|β|=k +
=
=Dhn+k,k(x1, . . . ,xn,x(β)):|β|=k
E
.
Define two column vectors
h= hn+k,k(x1, . . . ,xn,x(β)):|β|=k
T
,
b= xβ:|
β|=kT.
FromProposition 5.1andProposition 5.2, we haveh=Abwhere the(β,γ)-th entry ofAis
Aβ,γ =∏ni=1
βi+γi
γi
namelyA=G(β :|β|=k). Since the multi-indicesβ are all distinct, by applyingProposition 5.5we
deduce thatAis nonsingular and therefore the entries ofbare linear combinations of the entries ofh. This shows that(hn,d)k,kis full rank.
Now considerd=2k+1 odd. It suffices to prove the result fore=k+1. Let
g= ∂
∂x1
hn,d=hn+1,d−1(x1,x1,x2, . . . ,xn)∈S2kCn.
The image of the flatteninggk,k:Sk(Cn)∗→SkCnis contained in the image of(hn,d)k+1,k. To conclude, we will show thatgk,kis full rank.
Leth=hn+1,d−1(y1, . . . ,yn+1). By the result in the case of even degree, we know that(h)k,k has full rank, namely
D
hn+1+k,k(y1, . . . ,yn+1,y(β)):|β|=k
E
=SkCn+1.
In particular, the image of this space under the specialization(y1, . . . ,yn+1) = (x1, , . . . ,xn,x1)isSkCn. On the other hand, notice that,
hn+1+k,k(y1, . . . ,yn+1,y(β))|(y1,...,yn+1)=(x1,...,xn,x1)=hn+k+1,k(x1, . . . ,xn,x
(γ))
whereγ1=β1+βn+1+1 andγi=βi fori=2, . . . ,n(indeed|γ|=k+1). This shows that the image of
Corollary 5.7. For every n,d,
RS(hn,d)≥RS(hn,d)≥
n+bd/2c −1 bd/2c
.
For readers familiar with cactus rank and cactus border rank, by [16, Thm. 5.3D], we obtain the same lower bounds for cactus rank and cactus border rank.
6
Koszul flattenings
We recall the definition of Koszul flattening introduced in [21]. A Koszul flattening map ofp∈SdCN depends on two parameterss,q: it is obtained by tensoring thes-th partial derivative map (1.1) with the identity map on the spaceΛqCN(for someq) and then applying the exterior derivative map. In symbols:
ΛqCN⊗Ss(CN)∗
id⊗ps,d−s
−−−−−→ ΛqCN⊗Sd−sCN
δq,d−s
−−−→ Λq+1CN⊗Sd−s−1CN X⊗D 7−−−−−→ X⊗D(p)
X⊗g 7−−−→ ∑N1
(xi∧X)⊗∂∂xg
i
Letp∧qs,d−s:ΛqCN⊗SsCN∗→Λq+1CN⊗Sd−s−1CN denote the composition of the two maps. Explic-itly, writingxI=xi1∧ · · · ∧xiq for aq-tupleI= (i1, . . . ,i|I|)and
∂s ∂xJ =
∂ ∂xj1
· · · ∂
∂xjs
for ans-tupleJ= (j1, . . . ,js), the map is p∧qs,d−s:xI⊗ ∂
s
∂xJ
7→xk∧xI⊗ ∂ s+1p
∂xk∂xJ
.
Then, by [21, Prop 4.1.1], one obtains the border rank lower bound
RS(p)≥
rank(p∧qs,d−s)
rank((`d)∧q s,d−s)
= rank(p
∧q s,d−s) N−1
q
. (6.1)
When ps,d−sis of maximal rank for alls, Koszul flattenings can only give a better Waring border rank lower bound than the partial derivative maps whend=2k+1 is odd ands=k. For example, if d=2kis even, then the rank of the Koszul flattening is bounded above by dim ΛqCN⊗Sk−1CN so the Waring border rank lower bound from Koszul flattenings is bounded above by N+k−k−12 N
N−q whereas from flattenings alone, one already gets the larger lower bound of N+kk−1
. Similarly, ifd is odd but N6=2q+1, the bound from Koszul flattenings is lower than the one from standard flattenings.
Supposed=2k+1 andpk,k+1is of maximal rank. Viewed naïvely, one might think Koszul flattenings
could prove Waring border rank lower bounds of up to
dim(Λq
CN⊗SkCN∗) N−1
q
=
N+k−1 k
but this is not possible because the exterior derivative map is aGLN-module map that is not surjective unlessq=N,N−1. Indeed, we have the decompositions
ΛqCN⊗Sk+1CN=Sk+1,1qCN⊕Sk+2,1q−1CN, Λq+1CN⊗SkCN=Sk,1q+1CN⊕Sk+1,1qCN
so, by Schur’s Lemma,Sk+2,1q−1CN=ker(δ)and image(δ) =Sk+1,1qCN.
The following result gives ana prioriupper bound for the rank of the Koszul flattening, by determining a lower bound for the dimension of ker(p∧qk,k+1).
Proposition 6.1. Let p∈S2k+1CN. Then for every q
rank(p∧qk,k+1)≤ k
∑
j=0
(−1)jdim(Λq−jCN⊗Sk−jCN∗). (6.2)
Proof. We will prove that
dim image(p∧e−(q−1,d−e1) +1)≤dim ker(p∧qe,d−e)
and conclude that the estimate holds for everyqvia an induction argument. Indeed,
image
p∧e−(q−1,d−e1) +1
⊆image(IdΛqCN⊗pe,d−e)
because the exterior derivative map takes derivatives on the factorSd−e+1CN. Moreover, sinceδ2=0,
we have image(p∧q−e−1,1d−e+1)⊆ker(δq,d−e). Therefore, passing to dimensions
dim(image(p∧q−e−1,1d−e+1))≤dim(ker(δq,d−e|image(IdΛqCN⊗pe,d−e))≤dim ker(p
∧q e,d−e).
Now,
rank(p∧qk,k+1) =dim(ΛqCN⊗SkCN∗)−dim(ker(p∧qk,k+1))
≤dim(ΛqCN⊗SkCN∗)−rank(pq−k−11,k+2)
and we conclude by induction.
Remark 6.2. This is still not the end of the story: whenN=2q+1 withqodd, then the linear map p∧qk,k+1:ΛqCN⊗SkCN∗→Λq+1CN⊗SkCN'ΛqCN∗⊗SkCN was observed in [21], at least in certain cases, to be skew-symmetric. In particular, in this case, if the bound in (6.2) is odd, it cannot be attained.
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 6.4. For all n>2and all q<n/2, the Koszul flattening(fne,k)∧qk,k+1has rank at least
n−1 q
n+k−1
k
+q−1
.
In particular,RS(efn,k)≥ n+k−k 1
+q−1, which is greater than the lower bound obtainable by flattenings.
Proof. For fixedn, recall the uniqueSn-invariantg:=gn∈S2[n−1,1]from the proof ofTheorem 2.5. Let[n]be the span of`:=`n. For every s, write ps:= (`∗)s−1·(`qs−1), which from Eqn. (2.3) is a polynomial of degrees with non-zero projection onto[n]s. From Eqn. (2.2), the image of the k-th flattening map of efn,kis
image(fen,k)k,k+1= k M
s=0
ps+1Sk−s[n−1,1]⊆Sk+1Cn.
We will give a lower bound for the dimension of the image ofΛqCn⊗image(fne,k)k,k+1under the
exterior derivative map. We haveΛqCn=Λq[n−1,1]⊕ [n]∧Λq−1[n−1,1]as aSn-module.
Consider the image of([n]∧Λq−1[n−1,1])⊗(ps+1Sk−s[n−1,1])under the exterior derivative. The
Sn-equivariant projection of this space onto([n]∧Λq−1[n−1,1])⊗([n]s+1⊗Sk−s[n−1,1])commutes
with the exterior derivative, which isGLn-equivariant and thereforeSn-equivariant. The image of([n]∧
Λq−1[n−1,1])⊗([n]s+1⊗Sk−s[n−1,1])under the exterior derivative is (whens≤k−1 and after reorder-ing the factors) the subspaceS(k−s,1q−1)[n−1,1]⊗[n]⊗[n]s+1⊆[n]∧Λq[n−1,1]⊗[n]s+1Sk−s−1[n−1,1].
Now consider the image of Λq[n−1,1]⊗ps+1Sk−s[n−1,1]. Again, consider its projection onto
Λq[n−1,1]⊗([n]s+1⊗Sk−s[n−1,1]). By applying the exterior derivative map, we obtain a subspace
of ([n]∧Λq[n−1,1])⊗([n]s⊗Sk−s[n−1,1])
⊕ Λq+1[n−1,1]⊗([n]s+1⊗Sk−s−1[n−1,1])
; consider its projection to the second summandΛq+1[n−1,1]⊗[n]s+1Sk−s−1[n−1,1]whens≤k−1. For the same
reason as above, the image of this projection isS(k−s,1q)[n−1,1]⊗[n]s+1up to reordering the factors.
Note thatS(k−t+1,1q−1)[n−1,1]⊕S(k−t,1q)[n−1,1] =Sk−t[n−1,1]⊗Λq[n−1,1]as aGL([n−1,1]) -module. Consider the summands forsranging from 0 tok−2 in the first case and 1 tok−1 in the second, we obtain componentsSk−t[n−1,1]⊗Λq[n−1,1]fort from 1 tok−1. We obtain a subspace in the
image of the Koszul flattening that is isomorphic as aGL([n−1,1])-module to
Mk−1 t=1
Sk−t[n−1,1]⊗Λq[n−1,1].
The first factor of the space above has the same dimension asSk−1Cnminus dim(S0[n−1,1]) =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 factorΛq[n−1,1]. The exterior derivative map is
Λq[n−1,1]⊗([n]⊗Sk[n−1,1])→Λq[n−1,1]∧[n]⊗Sk[n−1,1]⊕Λq[n−1,1]⊗([n]⊗Sk−1[n−1,1]).
This projects isomorphically onto the first term in the target and(Λq[n−1,1]∧[n])⊗Sk[n−1,1]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+kk−1 n−q1. We now have a contribution of
n−1
q
n+k−2
k−1
−1+
n+k−2 k
=
n−1 q
n+k−1
k
−1
to the rank.
Finally consider the s=k and s=k−1 terms in the term with a factor[n]∧Λq−1[n−1,1]: the
sources are respectively[n]∧Λq−1[n−1,1]⊗ hpk+1iand([n]∧Λq−1[n−1,1])⊗pk[n−1,1]. Consider the projections respectively to([n]∧Λq−1[n−1,1])⊗`k−1S2[n−1,1](the second factor is`k−1g) and ([n]∧Λq−1[n−1,1])⊗`k−2S3[n−1,1](the second factor is`k−2g[n−1,1]). Now, applying the exterior
derivative, these spaces map injectively to([n]∧Λq[n−1,1])⊗[n]k−1[n−1,1]and([n]∧Λq[n−1,1])⊗
`k−2S2[n−1,1]respectively. These targets do not appear in other terms analyzed above, so we pick up
n−1 q−1
=q
n
n−1 q
and (n−1)
n−1 q−1
=q(n−1)
n
n−1 q
additional contributions to the rank.
Collecting all the contributions together, we conclude.
Remark 6.5. A more careful analysis of theq=1 case shows it also improves the flattening lower bound.
Remark 6.6. Computer experiments performed in Macaulay2 [12], using a variant of the code of [26], indicate that the situation may be significantly better.
Forn=3, . . . ,6,k=1, . . . ,6, letp∈S2k+1Cnbe generic. The Koszul flattening p∧k,1k+1has rank equal to the bound in (6.1) except ifn=3 andkis even (in accordance withRemark 6.2).
Forn=3, . . . ,6 andk=1, . . . ,6, let p=hn,2k+1 or p= efn,k. The Koszul flattenings of p∧k,1k+1has rank equal to the bound in (6.1) ifkis odd and one less than the bound in (6.1) ifkis even.
Notice that in the casesn=4,5,6 withkeven, the Koszul flattenings give a border rank lower bound forhn,2k+1and efn,k that is one less than the bound for a generic polynomial.
Question 6.7. What are the ranks of the Koszul flattenings for efn,k andhn,2k+1in general?
References
[2] WERONIKABUCZY ´NSKA ANDJAROSŁAVBUCZY ´NSKI: Secant varieties to high degree Veronese reembeddings, catalecticant matrices and smoothable Gorenstein schemes. J. Algebraic Geom., 23(1):63–90, 2014. [doi:10.1090/S1056-3911-2013-00595-0,arXiv:1012.3563] 2
[3] PETERBÜRGISSER, CHRISTIANIKENMEYER,ANDGRETAPANOVA: No occurrence obstructions in geometric complexity theory. J. Amer. Math. Soc., 32(1):163–193, 2019. [doi:10.1090/jams/908,
arXiv:1604.06431] 5
[4] XICHEN, NEERAJKAYAL,ANDAVIWIGDERSON: Partial derivatives in arithmetic complexity and beyond. Found. Trends Theor. Comput. Sci., 6(1-2):1–138, 2010. [doi:10.1561/0400000043] 2 [5] KLIM EFREMENKO, ANKIT GARG, RAFAEL OLIVEIRA, ANDAVIWIGDERSON: Barriers for rank methods in arithmetic complexity. In Proc. 9th Innovations in Theoret. Comp. Sci. conf. (ITCS’18), pp. 1:1–1:19. Leibniz Zentrum für Informatik, 2018. [doi:10.4230/LIPIcs.ITCS.2018.1,
arXiv:1710.09502] 18
[6] KLIM EFREMENKO, JOSEPH M. LANDSBERG, HAL SCHENCK, ANDJERZY WEYMAN: The
method of shifted partial derivatives cannot separate the permanent from the determinant. Mathe-matics of Computation, 87(312):2037–2045, 2018. [doi:10.1090/mcom/3284,arXiv:1609.02103] 4, 5,10
[7] HERVÉFOURNIER, NUTANLIMAYE, MEENAMAHAJAN,ANDSRIKANTHSRINIVASAN: The shifted partial derivative complexity of elementary symmetric polynomials. Theory of Computing, 13(9):1–34, 2017. Preliminary version inMFCS’15. [doi:10.4086/toc.2017.v013a009] 5
[8] WILLIAMFULTON ANDJOEHARRIS: Representation Theory: A First Course. Springer, 1991. [doi:10.1007/978-1-4612-0979-9] 7
[9] MACIEJGAŁ ˛AZKA: Vector bundles give equations of cactus varieties.Lin. Alg. Appl., 521:254–262,
2017. [doi:10.1016/j.laa.2016.12.005,arXiv:1605.08005] 18
[10] FULVIOGESMUNDO, CHRISTIANIKENMEYER,ANDGRETAPANOVA: Geometric complexity the-ory and matrix powering. Diff. Geom. Appl., 55:106–127, 2017. [doi:10.1016/j.difgeo.2017.07.001,
arXiv:1611.00827] 5
[11] IRAGESSEL ANDGÉRARDVIENNOT: Binomial determinants, paths, and hook length formulae. Adv. in Math., 58(3):300–321, 1985. [doi:10.1016/0001-8708(85)90121-5] 13
[12] DANIELR. GRAYSON ANDMIKEE. STILLMAN: Macaulay2, a software system for research in algebraic geometry. Available here. 20
[13] EDWARDL. GREEN: Complete intersections and Gorenstein ideals. J. Algebra, 52(1):264–273, 1978. [doi:10.1016/0021-8693(78)90274-0] 2
[14] ANKITGUPTA, PRITISHKAMATH, NEERAJKAYAL,ANDRAMPRASADSAPTHARISHI:
[15] ANTONYIARROBINO ANDJACQUESEMSALEM: Some zero-dimensional generic singularities; finite algebras having small tangent space. Compositio Math., 36(2):145–188, 1978. NUMDAM.2 [16] ANTONYIARROBINO ANDVASSILKANEV:Power Sums, Gorenstein Algebras, and Determinantal
Loci. Springer, 1999. [doi:10.1007/BFb0093426] 2,17
[17] CHRISTIANIKENMEYER ANDGRETAPANOVA: Rectangular Kronecker coefficients and plethysms in geometric complexity theory.Adv. Math., 319:40–66, 2017. Preliminary version inFOCS’16. [doi:10.1016/j.aim.2017.08.024,arXiv:1512.03798] 5
[18] NEERAJKAYAL:An exponential lower bound for the sum of powers of bounded degree polynomials.
Electr. Coll. Comp. Compl. (ECCC), 81, 2012.3
[19] JOSEPH M. LANDSBERG: Tensors: Geometry and Applications. Amer. Math. Soc., 2012. [doi:10.1090/gsm/128] 7
[20] JOSEPHM. LANDSBERG:Geometry and Complexity Theory. Cambridge University Press, 2017. [doi:10.1017/9781108183192] 2
[21] JOSEPH M. LANDSBERG AND GIORGIO OTTAVIANI: Equations for secant varieties of Veronese and other varieties. Annali di Matematica Pura ed Applicata, 192(4):569–606, 2013. [doi:10.1007/s10231-011-0238-6,arXiv:1111.4567] 17,18
[22] HWANGRAELEE: Power sum decompositions of elementary symmetric polynomials. Lin. Alg. Appl., 492:89–97, 2016. [doi:10.1016/j.laa.2015.11.018,arXiv:1508.05235] 5
[23] M. STUART LYNN: On the Schur product of H-matrices and non-negative matri-ces, and related inequalities. Math. Proc. Cambridge Phil. Soc., 60(3):425–431, 1964. [doi:10.1017/S0305004100037932] 14
[24] NOAMNISAN: Lower bounds for non-commutative computation. InProc. 23rd STOC, pp. 410–418. ACM Press, 1991. [doi:10.1145/103418.103462] 4,5
[25] NOAM NISAN AND AVI WIGDERSON: Lower bounds on arithmetic circuits via partial derivatives. Comput. Complexity, 6(3):217–234, 1996. Preliminary version in FOCS’95. [doi:10.1007/BF01294256] 5
[26] LUKE OEDING ANDGIORGIOOTTAVIANI: Eigenvectors of tensors and algorithms for Waring decomposition.J. Symb. Comput., 54:9–35, 2013. [doi:10.1016/j.jsc.2012.11.005,arXiv:1103.0203] 20
[27] VIKTOR V. PRASOLOV: Problems and Theorems in Linear Algebra. Amer. Math. Soc., 1994. [doi:10.1090/mmono/134] 6,14
[29] AMIR SHPILKA AND AVI WIGDERSON: Depth-3 arithmetic circuits over fields of char-acteristic zero. Comput. Complexity, 10(1):1–27, 2001. Preliminary version in CCC’99. [doi:10.1007/PL00001609] 5
[30] RICHARD P. STANLEY: Hilbert functions of graded algebras. Adv. in Math., 28(1):57–83, 1978.
[doi:10.1016/0001-8708(78)90045-2] 10
[31] NOAHSTEIN: Full rank submatrices of positive semidefinite matrix, 1917. MathOverflow. 13 [32] JAMESJ. SYLVESTER: On the principles of the calculus of forms.Cambridge and Dublin Math. J.,
7:52–97, 1852.2
AUTHORS
Fulvio Gesmundo Postdoc at QMath
University of Copenhagen Denmark
fulges math ku dk
Joseph M. Landsberg Professor
Dept. of Mathematics
Texas A&M University, College Station, TX jml math tamu edu
http://www.math.tamu.edu/~jml
ABOUT THE AUTHORS
