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HAL Id: hal-03193911

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On tensors that are determined by their singular tuples

Ettore Turatti

To cite this version:

Ettore Turatti. On tensors that are determined by their singular tuples. 2021. �hal-03193911�

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On tensors that are determined by their singular tuples

Ettore Turatti

Abstract

In this paper we study the locus of singular tuples of a complex valued multisymmetric tensor. The main problem that we focus on is: given the set of singular tuples of some general tensor, which are all the tensors that admit those same singular tuples. Assume that the triangular inequality holds, that is exactly the condition such that the dual variety to the Segre-Veronese variety is an hypersurface, or equivalently, the hyperdeterminant exists. We show in such case that, when at least one component has degree odd, this tensor is projectively unique. On the other hand, if all the degrees are even, the ber is an1-dimensional space.

1 Introduction

Let V1, . . . , Vk be vector spaces over C of dimension m1+ 1, . . . , mk+ 1, and let ql be a quadratic form onVl that, after a choice of basis for Vl, is given in coordinates by ql=x20+

· · ·+x2m

lthat denes the distance function inVl. Given a tensort∈Symd1V1⊗· · ·⊗SymdkVk, we say that a rank-one tensorv=v1⊗ · · · ⊗vkis a singular tuple oftif we have that for each atteningt:Symd1V1⊗· · ·⊗Symdl−1Vl⊗· · ·⊗SymdkVk→Vl,t(v1d1⊗· · ·⊗vldl−1⊗· · ·⊗vdkk) = λvl. In the particular case thatk= 1we callvan eigentensor oft, and it satisest(vd−1) =λv. The notion of eigentensor has been introduced in 2005by Qi for symmetric tensors [6], [17]

and it was generalized by Lim [13] for tensors as singular tuples in the same year. Moreover, singular tuples extend the notion of eigenvector of a matrix to a tensor of any order. The rst interesting result on singular tuples is proven in [13], the singular tuples of a given tensor t are the critical points of the distance function between tand the Segre-Veronese variety of rank-1tensors. This opens a new perspective to tensor optmization, and a possible direction to obtain a general notion of the Eckart-Young Theorem. Indeed, this theme has been studied in many works, we suggest [1], [4], [5], [7], [8], [14], [15], [16], [19], [20] for a clearer picture of the topic.

Università di Firenze - DIMAI, Viale Morgagni, 67/A, 50134 Firenze, Italy - ettore.teixeiraturatti@uni.it 2020Mathematics Subject Classication. 14N07;14M17;15A18;15A69;58K05.

Key words: Tensor; Eigentensor; Singular tuple; Critical points.

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A natural question to be posed is, given the singular tuples of a tensor t, which are all the tensors that admit such conguration of singular tuples? The answer for the matrix case is described by the singular value decomposition, where changing the singular values in the decomposition gives all the matrices with same singular tuples, this is described in Example 2.3. The problem for symmetric tensors was rst studied on [2] for the particular case of SymdC3 and d odd, later in [3] the result obtained has been revisited for general d, and Beorchia, Galuppi and Venturello stated the theorem that we show next. We denote by edX

theED-degree of the Veronese varietyX. LetΣnbe then-symmetric group of permutations, we denote by PV(n) = (PV)nn the symmetric cartesian product such that the points are unordered, let Eig(f) ⊂ PV(edX) be the set of eigentensors of a general polynomial f and q=x20+· · ·+x2m the distance function in a basis of the vector space V.

Theorem (Beorchia-Galuppi-Venturello [3]) LetV be a 3-dimensional vector space, let τ :P SymdV

99KPV(edX), f 7→Eig(f)

be the map that associates a general polynomial f to the set of its eigentensors. Then, if f ∈P(SymdV) is a general polynomial, we obtain that

τ−1(τ(f)) =

[f], if d is odd;

{[f +cqd2]|c∈C}, if dis even.

Moreover, the image of τ has dimension

dim(Im(τ)) =

d+2 2

−1, if dis odd;

d+2 2

−2, if dis even.

The approach utilised in the proof of this result was dicult to generalize to polynomials in more than three variables, so we opted for a dierent technique. We notice that we can decompose the mapτ =ψ◦ϕinto two parts, the rst map that we use is the projectivization of the linear map ϕ:SymdV → H0(Q(d−1)), whereQ is the quotient bundle and a global sectionsf associated to a polynomialf is dened byϕ(f) =sf =

"

∇f(x) x

#

; we allow an abuse of notation for simplicity and we denote the projectivizationP(ϕ) also byϕ. The second map ψ:P(H0(Q(d−1)))99KPV(edX)takes the zero locus of the sectionsf that is exactly the zero dimensional eigenscheme Eig(f). With this approach we were able to generalize this result to polynomials in any number of variables.

Theorem 1.1 Let V be a vector space of dimension m+ 1. Let d ≥ 3 be an integer, and

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f ∈P(SymdV) be a general polynomial. Let τ :P SymdV

99KPV(edX), f 7→Eig(f) be the map that associates f to its eigentensors locus Eig(f). Then

τ−1(τ(f)) =

[f], ifd is odd;

{[f +cqd2]|c∈C}, ifd is even. Moreover, the image of the map τ has dimension

dim(Im(τ)) =

d+m d

−1, ifd is odd;

d+m d

−2, ifd is even.

The next natural step is to understand what happens in the case of multisymmetric tensors. Altough several new technical lemmas are required, the approach to generalize this result is similar to the symmetric tensor case. Let X ⊂ P Symd1V1⊗ · · · ⊗SymdkVk

be the Segre-Veronese variety, πi : X → PVi be the projection on the i-th coordinate and τ : P Symd1V1⊗ · · · ⊗SymdkVk

99K (PV1 × · · · ×PVk)(edX) be the map that associates a multisymmetric tensor T to the its eigenschemeEig(T), i.e. the locus of its singular tuples.

We construct the bundleE =L

πiQi(d1, . . . , di−1, di−1, di+1, . . . , dk) and use the fact that the zero locus of a global sectionsT ∈H0(E) associated to a multisymmetric tensor T is the singular tuple locus of T, as described in more details in both [8] and [9], to split the map τ = ψ◦ϕ as in the symmetric tensor case, where here we consider the projectivization of ϕ:Symd1V1⊗ · · · ⊗SymdkVk→H0(E) andψ:P(H0(E))→(PV1× · · · ×PVk)(edX).

Theorem 1.2 LetV1, . . . , Vk be vector spaces of dimensionm1+ 1, . . . , mk+ 1. Let d1, . . . , dk be positive integers, andT ∈P Symd1V1⊗ · · · ⊗SymdkVk

be a general tensor. Let τ :P Symd1V1⊗ · · · ⊗SymdkVk

99K(PV1× · · · ×PVk)(edX), T 7→Eig(T),

be the map that associates a tensor T to its singular tuples locus Eig(T). If k ≥ 3 and suppose that ml ≤ P

j6=lmj whenever dl = 1, and for k = 2 we include the hypothesis that (d1, d2)6= (1,1), then

τ−1(τ(T)) =

[T], if di is odd for some i;

{[T+cq

d1 2

1 ⊗ · · · ⊗q

dk 2

k ]|c∈C}, if dl is even for all l.

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Moreover, the image of the map τ has dimension

dim(Im(τ)) =

 Qk

l=1 dl+ml

d

−1, if di is odd for some i;

Qk l=1

dl+ml d

−2, if dl is even for all l.

The hypothesis of the triangular inequality ml ≤ P

i6=lmi, that also appears in the de- scription of the codimension of the critical space in [8], could seem unnatural at rst glance, but this condition can be understood in terms of the dual variety of the Segre-Veronese variety, as described in the next theorem.

Theorem (Gelfand-Kapranov-Zelevinsky, Corollary 5.11, [11]) SupposeXl for l= 1, . . . , kis the projective space Pml in the Veronese embedding into P(SymdlVl). Then the dual variety (X1× · · · ×Xk) is an hypersurface if and only if ml≤P

i6=lmi hold for alllsuch that dl = 1. The case when we have that dl = 1 and the equality on the triangular inequality ml ≤ P

i6=lmi holds is called boundary format case.

The article is divided into three parts. Section two is a preliminaries section, where we introduce with more details the singular tuples and we give the cohomological tools that are necessary for our results. In the third section we work on the symmetric tensor case and prove Theorem 1.1. In the nal section we analyse the multisymmetric tensor case and prove Theorem 1.2.

Acknowledgement

The author would like to thank Giorgio Ottaviani for proposing this work and for the valuables discussions, suggestions and encouragement.

This work has been supported by European Union's Horizon 2020 research and innovation programme under the Marie Skªodowska-Curie Actions, grant agreement 813211 (POEMA).

2 Preliminaries

2.1 Eigentensors and singular tuples

We suggest both [12] and [18] as references for a deeper understanding of the notions presented in this section.

Denition 2.1 Supposef ∈SymdV is a symmetric tensor, i.e., an homogenous polynomial and q=x20+· · ·+x2m is a quadratic form onV in coordinates for a choice of basis ofV, the eigenvectors off are dened in [17] as the vectorsx∈V such that

f(xd−1) =λx,

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for λ∈ C. In other words, we can dene the eigentensors of f as the xed points of ∇f =

∂f

∂x0(x), . . . ,∂x∂f

m(x)

, the gradient vector, i.e. the solutions of the equation

∇f(x) =λx.

In conclusion, the equations of the locus of eigentensors Eig(f) is determined by the 2×2- minors of the matrix "∂f

∂x0 . . . ∂x∂f

m

x0 . . . xm

# .

Denition 2.2 Let T ∈ Symd1V1 ⊗ · · · ⊗SymdkVk be a tensor and qi = x20+· · ·+x2mi a quadratic form on Vi for a choice of basis for eachVi. We can dene the singular tuples ofT as the tuple (v1, . . . , vn), such that each attening

Ti:Symd1V1⊗ · · · ⊗Symdi−1Vi⊗ · · · ⊗SymdkVk→Vi

satises

Ti(v1d1⊗ · · · ⊗vidi−1⊗ · · · ⊗vkdk) =λvi.

We dene the zero dimensional scheme Eig(T) to be the locus of singular tuples of T. Example 2.3 The relation between a given set of singular tuples and the matrices that have such singular tuple locus conguration is described by the Singular Value Decomposition, since the singular tuples of a matrix are given by the rst columns of the orthogonal matrices on the decomposition. We briey describe it next.

LetA∈Hom(Y, W), whereY, W are vector spaces of dimensionsdimY =n, dimW =m, we recall that the singular value decomposition tells us that A=Udiag(σ1, . . . , σmin{m,n})Vt. If we letui andvi be the columns ofU andV, as described by Ottaviani and Paoletti in [15], we have that for 1≤i≤m= min{m, n}, Avi =ui and Atui =vi, in other words, the pairs (ui, vi) are the singular pairs of A. Let τ : Hom(Y, W)→ (Y ×W)(m), A7→Eig(A), where Eig(A)is the set consisting of the singular tuples ofA. Therefore, given a singular tuple locus Z ={(ui, vi)}mi=1, and orthogonal matrices U, V such that the rst m columns are ui and vi

we have that

τ−1(Z) ={B ∈Hom(Y, W)|B =Udiag(σ1, . . . , σm)Vt, σi∈C}.

Let V1, . . . , Vk be vector spaces of respective dimension m1+ 1, . . . , mk+ 1, and let T ∈ Symd1V1⊗ · · · ⊗SymdkVk.

Denition 2.4 Let X = PV1× · · · ×PVk be the Segre-Veronese variety of rank 1 tensors embedded with O(d1, . . . , dk) in P(Symd1V1⊗ · · · ⊗SymdkVk). Let πl : PX → PVl be the

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projection on thel-th component, and letQlbe the quotient bundle, whose bers over a point vl ∈ Vl are Vl/hvli. Let El = πlQl⊗ O(d1, . . . , dl −1, . . . , dl), we can construct the vector bundle

E =

k

M

l=1

El.

A tensorT ∈P(Symd1V1⊗ · · · ⊗SymdkVk) leads to a global section of El which over a point v= (v1, . . . , vk)is the map sendingv to the natural pairing off with(v1d1)· · ·(vldl−1)· · ·(vkdk) modulohvli, that is a vector inVl/hvli.

The reason to consider this particular bundle is that in [8] and [9] it is proven that if we consider the section sT associated to a multisymmetric tensor T, then the zero locusZ(sT) is equal to the locus of singular tuples ofT, that is Z(sT) =Eig(T). In particular Friedland and Ottaviani [9] used this fact to compute the number of singular tuples of a general tensor as the top Chern class of the bundle E in Theorem 2.6.

Denition 2.5 TheED-degree of a subvarietyX ⊂P(Symd1V1⊗ · · · ⊗SymdkVk)is dened as the number of critical points of the functiondT :X →R, wheredT is the distance function between a general tensorT ∈P(Symd1V1⊗ · · · ⊗SymdkVk) and X. We consider the natural extension of this function to the complex numbers for our porposes.

The ED-degree has been studied in [7], and we suggest it as a reference for a better comprehension. In particular, if we consider the variety X to be the Segre-Veronese variety, we have that the ED-degree counts the number of singular tuples of a general tensor. We are going to denote the ED-degree of the Segre-Veronese variety by edX. This particular ED-degree has been studied before in [9], where the next theorem is presented.

Theorem 2.6 ([9], Theorem15) LetV1, . . . , Vkbe vector spaces of dimensionm1+1, . . . , mk+ 1. The number of singular tuples of a general tensor T ∈ P(Symd1V1 ⊗ · · · ⊗SymdkVk), is equal to the coecient of tm11· · ·tmkk in the polynomial

k

Y

l=1

l ml+1

−tml l+1l−tl where ˆtl= (Pk

i diti)−tl.

Example 2.7 Assume that di =mi = 1, for all i= 1, . . . , k, in this setting we can compute using the previous formula that the ED-degree of the Segre variety X is k!.

Example 2.8 Furthermore, the number of singular tuples stabilises at the boundary format case, that is, suppose thatdk= 1andN =Pk−1

i=1 mi ≤mk, then the number of singular tuples of a general tensor is constant asmk increases.

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Consider the map

τ :P Symd1V1⊗ · · · ⊗SymdkVk 99K

P(V1)× · · · ×P(Vk)(edX)

that for a general tensorT ∈Symd1V1⊗ · · · ⊗SymdkVk it associates its singular tuples locus Eig(T). Studying this map is dicult in general, but decomposing it through the bundle E is advantageous. We decomposeτ in the following manner

P Symd1V1⊗ · · · ⊗SymdkVk

P(V1)× · · · ×P(Vk)(edX)

P(H0(E))

ϕ

τ

ψ

In this diagram the map ϕ associates a tensor T to the global section sT described before in the denition 2.4. The map ψ sends a global section s∈H0(E) to its zero locusZ(s), in particular the codomain is well dened for a sectionsT when the singular tuples ofT consists of exactly edX points.

Example 2.9 Notice that for k = 1 we obtain the symmetric tensor case, in such case E is simply equal to Q(d−1) and the map ϕcan be described as

ϕ:SymdV →H0(Q(d−1)), f 7→sf =

"

∇f(x) x

#

The other interesting case is when dl = 1for all l= 1, . . . , k. In such case, X is the Segre variety and the mapϕ can be described by means of the attenings of the tensor T, that is

ϕ:V1⊗ · · · ⊗Vk

k

M

l=1

Hom(V1⊗ · · · ⊗Vcl⊗ · · · ⊗Vk, Vl), T 7→(T1, . . . , Tk);

In our notation Tl represents thel-attening of T, namely Tl:V1⊗ · · · ⊗Vˆl⊗ · · · ⊗Vk→Vl.

In the general case X is the Segre-Veronese variety; this is the case of multisymmetric tensors. The map ϕ in this case is a combination of the previous two, that is, in each l-th component the maps acts as the contraction in the l-th coordinate and the evaluation in the others.

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2.2 Cohomological ingredients

We recall the next classical concepts and results that will be utilised in the course of this article, we suggest [21] for more details.

Theorem (Künneth's formula) Let Bi be vector bundles on PVi, i= 1, . . . , k and q a non- negative integer, then

Hq k

O

i=1

πiBi

∼= M

q1+···+qk=q

O

i

Hqi(Bi).

where the sum goes over all tuples of non-negative integers summingq.

Let G be a semisimple simply connected group, let P ⊂ G be a parabolic subgroup.

Let Φ+ be the set of positive roots of G. Let δ = P

λi be the sum of all the fundamental weights and letλbe a weight. LetEλ be the homogeneous bundle arising from the irreducible representation of P with highest weight λand( , ) be the Killing form.

Denition 2.10 The weight λ is called singular if there exists a root α ∈ Φ+ such that (λ, α) = 0. Otherwise, if(λ, α)6= 0 for all the rootsα∈Φ+, we say thatλis regular of index p if there exists exactlyp rootsα1, . . . , αp ∈Φ+ such that(λ, α)<0.

Theorem (Bott) If δ+λis singular, then Hi(G/P, Eλ) = 0 for all i. If δ+λ is regular of index p, then Hi(G/P, Eλ) = 0 for i6=p.

3 Symmetric Tensors

Lemma 3.1 LetV be a vector space of dimensionm+ 1,q =x20+· · ·+x2m a quadratic form on V, and da positive integer. If dis odd, the mapϕ:SymdV →H0(Q(d−1))is injective.

If dis even, ϕ has a 1-dimensional kernel, namely, kerϕ=hqd/2i. Proof. We recall that SymdV splits as SO(V)-modules as

SymdV =Hd⊕Hd−2⊕ · · · ⊕

H1 if dis odd H0 if dis even,

whereHd−2j ={f qj|f is an harmonic polynomial of degreed−2j} is an irreducibleSO(V)- module.

Therefore we can restrict ϕto each Hj, in such way we have ϕ:Hj →Wj ⊂H0(Q(d−1)),

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whereWj = Im(ϕ)|Hj. This map is either an isomorphism or zero by Schur's lemma. Let j be such that d−2j≥1, then we have that for g= (x0+ix1)d−2jqj ∈Hd−2j it is mapped by ϕto

sg =

"∂g

∂x0 . . . ∂x∂g

m

x0 . . . xm

#

that has not rank 1everywhere. Indeed

∂g

∂x0x1− ∂g

∂x1x0 = (d−2j)(x0+ix1)d−2j−1

x1−ix0 6≡0.

On the other hand, H0 ={λqd2|λ∈C}. In such case we have for an element of H0 that

∂λqd2

∂xi

xj−∂λqd2

∂xj

xi=λ(2xixjqd2−1−2xixjqd2−1) = 0, ∀ i, j∈ {0, . . . , m}.

We conclude that ifdis odd, the mapϕis an isomorphism in each irreducible representation;

ifdis even, it is an isomorphism in each of them, with the exception ofH0, as we wished.

Lemma 3.2 Let Z be the zero locus of a section in Q(d−1), and assume that d≥3. Then the natural map from the Koszul complexH0(EndQ)→H0(IZ⊗Q(d−1))is an isomorphism of 1-dimensional spaces.

Proof. Indeed, consider the Koszul complex

0−−→ϕm

m

^Q(m(1−d))−−−→ϕm−1 . . .−→ϕ2

2

^Q(2(1−d))→Q(1−d)→ IZ→0, tensoring it byQ(d−1)we obtain the exact sequence

0→

m

^Q⊗Q((m−1)(1−d))→ · · · →

2

^Q⊗Q(1−d)→End(Q)→ IZ⊗Q(d−1)→0.

Let Fr to be dened as the quotient Fr =Vr

Q(r(1−d))/Imϕr. Thus we obtain short exact sequences

0→ F2 → Q(1−d)→ IZ→0 0→ Fr+1→Vr

Q(r−rd)→ Fr→0, for r= 2, . . . , m.

Tensoring the second short exact sequence byQ(d−1)we obtain the long exact sequence of cohomologies

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· · · →Hr(

r

^Q⊗Q((r−1)(1−d)))→Hr−2(Fr⊗Q(d−1))→Hr−1(Fr+1⊗Q(d+ 1))→

→Hr−1(^

Q⊗Q((r−1)(1−d)))→Hr−1(Fr⊗Q(d−1))→Hr(Fr+1⊗Q(d+ 1))→. . . We have that Vr

Q ⊗Q((r −1)(1−d)) = Vm−r

Q⊗Q((r −1)(1−d)−1), so if we have thatr ≥2, we obtain that Hr−2(Vm−r

Q⊗Q((r−1)(1−d)−1)) = 0. Also, ifd≥3, Hr−1(Vm−r

Q⊗Q((r−1)(1−d)−1)) = 0. This means that

H0(F2⊗Q(d−1))∼=H1(F3⊗Q(d−1))∼=. . .∼=Hm−1(Fm+1⊗Q(d−1)) = 0 H1(F2⊗Q(d−1))⊂H2(F3⊗Q(d−1))⊂ · · · ⊂Hm(Fm+1⊗Q(d−1)) = 0 Applying the long exact sequence of cohomologies to

0→ F2⊗Q(d−1)→End(Q)→ IZ⊗Q(d−1)→0 gives the desired result.

We would like to add a remark that, altough already utilised, the vanishing of the coho- mology Hq(VrQ ⊗Q(t)) is carefully done in the next section on Lemma 4.3. We decide in favour of postponing those computations because the full usefulness of such cohomologies appears in the multisymmetric case.

Corollary 3.3 Letf, g∈SymdV be two general polynomials such thatZ(sf) =Z(sg), d≥3.

Thensf =αsg for some α∈C.

Proof. The hypothesis that Z(sf) = Z(sg) implies that sf ∈ H0(IZ(sg)⊗Q(d−1)). Since this space is one-dimensional we have that sf =αsg.

We conclude this section observing that sinceτ =ψ◦ϕ, then the Theorem 1.1 is obtained just as the combinination of the Lemma 3.1 with the Corollary 3.3.

4 Multisymmetric Tensors

Now that the pre-image of the mapτ is completely analysed for symmetric tensors, we can go through to the next step, that is, we consider the Segre-Veronese variety Symd1V1⊗ · · · ⊗ SymdkVkand we anylise the mapτ :P Symd1V1⊗· · ·⊗SymdkVk

P(V1)×· · ·×P(Vk)(edX)

that associates a tensor T to its singular tuples Eig(T). We begin the multisymmetric case with the generalization of the Lemma 3.1 to Segre-Veronese varieties.

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Theorem 4.1 LetV1, . . . , Vk be vector spaces of dimension m1+ 1, . . . , mk+ 1, and we recall that qi = x20+· · ·+x2mi is the quadratic form on Vi that denes the distance function for i= 1 . . . , k. We consider the map

ϕ:Symd1V1⊗ · · · ⊗SymdkVk→H0(E),

where E is dened in the Denition 2.4. Then ϕis injective if at least one di is odd. In the case that all the di are even, we have that the kernel of ϕ is one dimensional and it is given by

kerϕ=hq

d1 2

1 i ⊗ · · · ⊗ hqkdk2 i Proof. Since we have that

SymdlVl∼=Hdl⊕Hdl−2⊕ · · · ⊕

H1 if dl is odd H0 if dl is even,

and that eachHdj−2tjis an irreducibleSO(Vl)-representation, then alsoHd1−2t1⊗· · ·⊗Hdk−2tk is an irreducible SO(V1)× · · · ×SO(Vk)-representation, we need to show that ϕis non zero whendj−2tj >0 for at least onej, and that it is zero when we have dj −2tj = 0 for allj.

Indeed, in the rst case we consider the element

g=g1⊗ · · · ⊗gk, gj = (x0+ix1)k−2jqtj,

thenϕ(g) =sg = (sg1 ⊗1)⊕ · · · ⊕(1⊗sgk), where sgj⊗1∈ Ej is non zero as seen before in the symmetric tensor case. Therefore by Schur's lemma we have that in this restriction the map is an isomorphism, thus ifdj−2tj >0for somej,sg does not belong to the kernel ofϕ. On the other hand, if alldj−2tj = 0, thengj =cqdj2 , wherec∈C, thensgj = 0, therefore summing all together we obtain thatsg = 0, so by Schur's Lemma the restriction ofϕon this subrepresentations is the zero map, as wished.

With this result we understand the rst map ϕin the decomposition τ =ψ◦ϕ. Now we can aim to understand better the mapψ, we will show that, under the hypothesis of Theorem 1.2, when two section s, t have the same image under the map ψ, where s, t are sections coming from tensors S, T ∈Symd1V1⊗ · · · ⊗SymdkVk, thens=λt.

The rst step to achieve this goal is to prove a similar result to Lemma 3.2 for the case of multisymmetric tensors, in order to do that we prove a series of technical lemmas.

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Lemma 4.2 Let E=Lk

l=1Ql(−d1, . . . ,−dl+ 1, . . . ,−dk), then, for j= 1, . . . , k, then

r

^E⊗Qj(d1, . . . , dj−1, . . . , dk) =

= M

r1+···+rk=r k

O

l=1,l6=j

rl

Pml(2rl−dl(r−1))⊗

mj−rj

^ Qj⊗Qj(−dj(r−1) +rj−2).

Proof. We have that

E=M

Ql(−d1, . . . ,−dl+ 1, . . . ,−dk), thus,

r

^E= M

r1+···+rk=r k

O

l=1 rl

^Ql

−rld1, . . . ,−rl(dl−1), . . . ,−rldk) ,

by separating the terms we obtain that

r

^E = M

r1+···+rk=r k

O

l=1 rl

^Ql(−rdl+rl).

We now tensor it byQj(d1, . . . , dj−1, . . . , dk), so we have thatVrE⊗Qj(d1, . . . , dj−1, . . . , dk) is equal to

M

r1+···+rk=r k

O

l=1,l6=j rl

^Ql(−rdl+rl+dl)⊗

rj

^Qj ⊗Qj(−rdj +rj+dj −1).

We now use the facts that Ωrl(rl) =Vrl

(Ω1(1)),Ω1(1) =Q, and VrjQj =Vmj−rjQj(−1), to obtain that VrE⊗Qj(d1, . . . , dj−1, . . . , dk) is equal to

M

r1+···+rk=r k

O

l=1,l6=j

rl

Pml(2rl−dl(r−1))⊗

mj−rj

^ Qj⊗Qj(−dj(r−1) +rj−2).

Lemma 4.3 (Bott's Theorem) The cohomology Hq(Vmj−rjQj⊗Qj(t))is non vanishing for

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the following cases

Hq

mj−rj

^ Qj⊗Qj(t)

6= 0, if





















q = 0, t≥0,

q =rj−1, t=−rj, 1≤rj ≤mj, q =rj, t=−rj−1, 0≤rj ≤mj−1, q =mj−1, t=−mj−1, 0≤rj ≤mj−1, q =mj, t≤ −mj−2.

(1)

Proof. The associated weight will be calculated in three cases depending on therj; the cases arerj = 0,1≤rj ≤mj−1and rj =mj.

For the case1≤rj ≤mj−1, we have thatVmj−rj

Qj⊗Qj(t) is not irreducible, therefore we have that the associated weight λis given by two parts

λ=λ(1)⊕λ(2). whereλ(1)rj+1mj+tλ1 andλ(2)rj+tλ1.

For λ(1) we have that

(1)+δ, α1+· · ·+αs) =









s+tif s≤rj,

s+t+ 1if rj + 1≤s≤mj−1, s+t+ 2if s=mj.

This implies the following cases:

1. t≥0, then index 0.

2. −1≥t≥ −rj, then it is singular (s=−tgives the vanishing).

3. Ift=−rj−1, then index rj.

4. If−rj −2≥t≥ −mj, then it is singular (s=−t−1 gives the vanishing).

5. if t=−mj−1, then index mj−1. 6. if t=−mj−2, it is singular (s=mj).

7. if t≤ −mj−3, then index mj. For λ(2) we have that

(2)+δ, α1+· · ·+αs) =

s+tif s≤rj−1, s+t+ 1if s≥rj.

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That implies the following cases:

1. Ift≥0, index 0.

2. If−1≥t≥ −(rj−1), singular for s=−t. 3. Ift=−rj, indexrj−1.

4. If−rj −1≥t≥ −mj−1, singular for s=−t−1. 5. Ift≤ −mj−2, index mj.

Forrj =mj we haveQj(t), therefore the associated weightλis λ=λmj+tλ1,thus we have

(λ+δ, α1+· · ·+αs) =

s+tif s≤mj−1, s+t+ 1if s=mj. This implies the following cases

1. Ift≥0, we have index0.

2. If−1≥t≥ −mj+ 1, then it is singular for s=−t. 3. Ift=−mj, then indexmj−1.

4. Ift=−mj−1, then it is singular fors=mj. 5. Ift≤ −mj−2, then index mj.

The nal case is when rj = 0, then we have Qj(t+ 1) and the associated weight λ is λmj+ (t+ 1)λ1, therefore

(λ+δ, α1+· · ·+αs) =

s+t+ 1ifs≤mj −1, s+t+ 2ifs=mj. This implies the following cases

1. Ift≥ −1, we have index0.

2. If−2≥t≥ −mj, then it is singular fors=−t−1. 3. Ift=−mj−1, then index mj−1.

4. Ift=−mj−2, then it is singular fors=mj. 5. Ift≤ −mj−3, then index mj.

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Lemma 4.4 (Bott's Theorem)

Hqrl(t) 6= 0 if









q= 0, t > rl q=rl, t= 0 q=n, t < rl−n

(2)

Lemma 4.5 Let ml= dimPVl andk≥3. Suppose that ml≤P

i6=lmi holds for everyl such thatdl= 1. Letr≥2be an integer,q1, . . . , qk be non negative integers such thatP

ql ≤r−1, and letr1, . . . , rk be non negative integers such that P

rl=r, then

k

O

l=1,l6=j

Hql(Ωrl

Pml(2rl−dl(r−1)))O Hqj

mj−rj

^ Qj⊗Qj(−dj(r−1) +rj−2)

= 0, for every j∈ {1, . . . , k}.

Furthermore, if k= 2 and we add the hypothesis that(d1, d2)6= (1,1)the result still holds.

Proof. Suppose that the cohomology of the tensor product is non vanishing. We x that the index j will associated to the unique case coming from the cohomology table (1), if not said otherwise.

Not all the cases can come from the third, fourth or fth line of (1) and from the second and third lines of (2). Suppose that one case comes from either the third, fourth or fth lines of (1), and all the remaining cases come from the second and third line (2), this means that ql≥rl, and we have thatr > q=P

ql≥P rl =r.

So at least one cohomology case must come from the other lines in (1) or(2).

No case can come from the rst line of(1). Suppose that we have that the only case of (1)comes from the rst line, this means that−dj(r−1) +rj−2≥0, so we obtain that

rj ≥(r−1)dj+ 2> dj(r−1) + 1≥(r−1) + 1 =r, that is rj > r, a contradiction.

No case can come from the st line of (2). Suppose that we have one case coming from the rst line of (2)for a xed l, we have thatrl> dl(r−1), then the only possiblity is thatrl=r and all other ri = 0, for i6=land dl = 1. In such case, for i6=l we have that the other cohomologies can not be on the rst line, otherwise it would be 0. Let j be the only case coming from (1), then for i6=l, j we have that it can not be on the second line of (2), because 0 =ri = qi =−di(r−1) and r−1, di >0. For j we have that the second line of (1) does not apply since qj =rj−1 = −1 and the third line of (1) implies 0 = qj =rj and

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−dj(r−1)−2 =−1, thendj(r−1) =−1, that is a contradiction since both terms on the left side are non negative. So in those cases we have the vanishing of the cohomology, therefore we have that one case is either on the fourth or fth line of (1) and all the remaining cases are on the third line of (2). If one case is on the fth line of (1) and all the others on the third line of(2), we have that qi=mi andqj =mj for i6=l. This means

ml≥rl=r >X

i6=l

qi =X

i6=l

mi,

this implies that ml >P

i6=lmi, that is a contradiction since dl = 1. The case coming from (1) can not be on the fourth line of (1), and all the others coming from the third line of(2) either, because in such case we have that−dj(r−1)−2 =−mj−1, that is,mj−1 =dj(r−1), but since r >P

i6=lqi =P

i6=l,j(mi) +mj−1 ≥mj, we have that r > mj, and the equality can not be satised sincedj ≥1. In the casek= 2, notice thatr ≥mj and since dl = 1, we must havedj ≥2. Again the wished equality mj −1 =dj(r−1) can not hold. This implies that no cohomology can come from the rst line of (2).

No case can come from the second line of (1). The last remaining possibility is to have the only case of (1) coming from the second line. In such case we notice that we have qj = rj −1 and no case on (2) comes from the rst line, thus ql ≥ rl for l 6=j. This, together with the fact that Pk

i=1ql < r, implies that ql = rl for l 6= j. We have that

−2(rj−1) =−dj(r−1), thereforerj =r and dj = 2, or rj < r and dj = 1.

In the rst case we have that rj =r implies that ri = 0for every i6=j. This means that we have Ωri(2ri−di(r−1)) = OPmi(−di(r−1)). Since −di(r−1) < 0, we have that the cohomology Hqi(OPmi(−di(r−1))) does not vanish just for qi = mi, but since mi >0, we have thatq=P

i6=jqi+qj =P

i6=jqi+r−1≥r, therefore our cases of interest have vanishing cohomology.

The second possibility for this cohomology to be non vanishing is that we haverj < rand dj = 1. Suppose that one of these non vanishing cohomologies comes from the second line of (2)for somel. From the conditions on(1)and(2)respectively, we have that2(rj−1) =r−1 and 2rl−dl(r−1) = 0, since dl≥1this implies that

rj = r 2 +1

2, rl≥ r 2 −1

2,

thereforerj+rl≥r. Sincekis at least3, we have another casei, that comes either from the second or third line of(2), and we must haveri = 0. If it is on the second line we have that 2ri−di(r−1) = 0, thus di(r−1) = 0, that is a contradiction. It can not be on the third line either, since ri =mi = 0 is also a contradiction. Otherwise, in case k = 2, we assume, without loss of generality, that j= 1 and l= 2, thend1 = 1and d2 ≥2, thusr2 ≥r−1, so

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we obtain

r1+r2 ≥ r 2+1

2 +r−1≥r+1 2 > r,

that is a contradiction. Therefore, no case can come from the second line on(2).

This means that we have that all the other cases must come from the third line of (2), that is ql=ml=rl for l6=j. We notice thatrj > r2 implies that rj >P

l6=jrl, thus mj ≥rj >X

l6=j

rl =X

l6=j

ml,

this is a contradiction since dj = 1.

Corollary 4.6 On the hypothesis of Lemma 4.5 we have that Hq((

r

^E)⊗ E) = 0.

Theorem 4.7 On the hypotesis of Lemma 4.5, the induced homomorphism E⊗ E → IZ⊗ E

induces an isomorphism at the level of global sections, where Z is the zero locus of a section s∈ E.

Proof. We have the following Koszul complex

0 =

N+1

^ E −−→ϕN

N

^E −−−→ϕN−1 . . .−→ϕ2

2

^E→ E → IZ→0.

Let Fr to be dened as the quotient Fr = VrE/Imϕr. Thus we obtain short exact sequences

0→ F2 → E → IZ →0 0→ Fr+1→Vr

E → Fr→0, for r= 2, . . . , N.

Tensoring the second short exact sequence by E, we obtain the long exact sequence of cohomologies

· · · →Hr−2(

r

^E⊗ E)→Hr−2(Fr⊗ E)→Hr−1(Fr+1⊗ E)→

→Hr−1(

r

^E⊗ E)→Hr−1(Fr⊗ E)→Hr(Fr+1⊗)→. . .

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By the previous lemma we have that both terms on the left are zero, therefore we have that Hr−2(Fr⊗ E)∼=Hr−1(Fr+1⊗ E), Hr−1(Fr⊗ E)⊂Hr(Fr+1⊗ E).

This implies that

H0(F2⊗ E)∼=. . .∼=HN−1(FN+1⊗ E) = 0 H1(F2⊗ E)⊂ · · · ⊂HN(FN+1⊗ E) = 0.

If we consider now the long exact sequence of cohomologies from 0 → F2⊗ E → E⊗ E → IZ⊗ E →0, we obtain

H0(F2⊗ E)→H0(E⊗ E)→H0(IZ⊗ E)→H1(F2⊗ E), since the end terms are zero, we obtain the desired isomorphism.

To make a comparison with the symmetric case, our next objective is to prove the extension of Corollary 3.3 to the multisymmetric case. That is, we will show if two sections s, t, that arise from the respective tensors S, T ∈ Symd1V1 ⊗ · · · ⊗SymdkVk, have the same image underψ, that is, we have the equality of the zero locus Z(s) =Z(t), thens=λt.

Lemma 4.8 Let EiiQi(d1, . . . , di−1, . . . , dk). If dimPVj ≥2 for all j, then H0(Hom(E,Ej)) =H0(Hom(Ej,Ej)) =C.

Moreover, if we assume that i6=j, then

H0(Hom(Ei,Ej)) = 0.

Proof. For the second equality we have that

Hom(Ei,Ej) =πiQi ⊗πjQj(0, . . . ,0,1,0, . . . ,0,−1,0, . . . ,0) = OPV1⊗ · · · ⊗πiQi(1)⊗ · · · ⊗πjQj(−1)⊗ · · · ⊗ OPVk.

We notice that, for alli, we have thatH0(Qi(1)) =H0(Ω1(2))6= 0. Meanwhile, if dimPVj ≥ 2, thenH0(Qj(−1)) = 0, thus by the Künneth's formula we have that

H0(Hom(Ei,Ej)) = 0.

On the other hand,

Hom(Ej,Ej) =OPV1 ⊗ · · · ⊗ πj(Qj ⊗Qj)

⊗ · · · ⊗ OPVk = Hom(Qj, Qj),

(20)

since the bundleQj is simple we obtain the desired result.

Lemma 4.9 Let ρ ∈ End(H0(E)) be a endomorphism of H0(E), suppose that f, g ∈ Symd1V1⊗ · · · ⊗SymdkVk are tensors such that ρ(sf) =ρ(sg), then sf =λsg for λ∈C.

Proof. LetI1 be the set of indices such thatdimPVi = 1andI2be the set of indices such that dimPVi ≥2. By the previous lemma, we have that H0(Hom(Ei,Ej)) = 0, whenever j ∈I2, that is, no map can act there besides its own endomorphism.

Now we consider a section sf coming from a tensor f ∈Symd1V1⊗ · · · ⊗SymdkVk. We recall that the map ϕ associates f to the diagonal map of its attenings in each coordinate l, that is, f :Symd1V1⊗Symdl−1Vl⊗ · · · ⊗SymdkVk →Vl. This means thatϕ(f) =sf can be interpreted as the diagonal element sf = (f, . . . , f), where f in the l entry of this vector means the section of El corresponding to f.

Suppose that the rst l indices are in I1 and the others in I2. Applying ρ to ϕ(f) we obtain that

ρ(ϕ(f)) = (M1(f), . . . , Ml(f), λl+1f, . . . , λkf) = (g, . . . , g) =sg,

whereg∈Symd1V1⊗ · · · ⊗SymdkVk is a tensor, thus from the previous lemma we have that all the mapsλi must be multiplication by scalars, this means that λf =g, for some λ∈C.

It remains the case when I2=∅. In such case we notice that

Hom(Ei,Ej) = (0, . . . ,0, Qi(1),0, . . . ,0, Qj(−1),0, . . . ,0);

since the dimension of eachPViis1, we haveQj(−1) =OP1, moreoverQi(1) = Ω1

P1(2) =OP1. We recall that both of those bundles are1-dimensional at the level of global sections, that is, dimH0(O

P1) = 1, therefore dimH0(Hom(Ei,Ej)) = 1. This implies that if ρ(sg) =sf, then sf =λsg is the only possible image.

Combining the previous results together, we obtain the next theorem.

Theorem 4.10 Let S, T ∈Symd1V1⊗ · · · ⊗SymdkVk be two general tensors. Assume that ml ≤P

i6=lmi holds for everyl such thatdl= 1,k≥3, and thatmj ≥1for allj. Lets, t∈ E be the sections coming from the tensors,S andT, and assume thatZ(s) =Z(t), thens=λt, for λ∈C.

Additionally, if k= 2 and we also consider the hypothesis that (d1, d2) 6= (1,1), then the result still holds.

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Proof. The Theorem 4.7 says that the mapH0(End(E))→H0(IZ⊗E)dened byρ7→ρ(sg) is an isomorphism. This means that ifs, tare two tensors such thatZ(s) =Z(t), then there exists a morphism ρ∈EndE such that

ρ(t) =s.

Furthermore, from the Lemma 4.9 we obtain thats=λt.

With all those results in mind, we can conclude with a note that combining together Theorem 4.1 with the Theorem 4.10 we obtain the Theorem 1.2. This nishes the proof of the main results.

4.1 A remark on sections coming from tensors

We make a brief remark that the Lemma 4.9 does not mean that all the morphisms in End(E)are multiplication by scalars, since the mapϕis not surjective in general. Indeed, for each spaceVl, we can compute what is the image of ϕl :SymdlVl→H0(Ql(dl−1)).

We have the Euler exact sequence

0→H0(O(dl−2))→H0(O(dl−1)⊗Vl)→H0(Q(dl−1))→0.

Moreover we have an isomorphism

H0(O(dl−1)⊗Vl)∼=Symdl−1Vl⊗Vl andH0(O(dl−2))∼=Symdl−2Vl In terms of Young diagrams,Symdl−2Vl has the representation

and Symdl−1Vl⊗Vl is represented by

O = M

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