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In [12] P. Comon and G. Ottaviani studied the real case for bivariate symmetric tensors. Even in this case there are many open conjectures and, to now, few cases are completely settled ([12], [9], [2], [7]). In this paper we want to study the relation among rC(P ) and rR(P ) in the special circumstance in which rC(P ) < rR(P ). In particular we will show that, in a certain range (say, rC(P )+ rR(P ) ≤ 3 deg(P ) − 1), all homogeneous polynomials P of that degree with rR(P ) 6= rC(P ) are characterized by the existence of a curve with the property that the sets evincing the real and the complex ranks coincide out of it (see Theorem 1 for the precise statement). More precisely: let P ∈ SdRm+1 be a real homogeneous polynomial of degree d in m + 1 variables such that rC(P ) < rR(P ) and rC(P )+ rR(P ) ≤ 3 deg(P ) − 1; therefore its real and complex decomposition are d d d d P = a1L1 + ··· + akLk + c1M1 + ··· + csMs , d d d d P = N1 + ··· + Nh + c1M1 + ··· + csMs 1 m+1 respectively, with k>h, ai ∈ {−1, 1}, cj ∈ {−1, 1}, M1,...,Ms ∈ S R , h+k> ′ m d+2. Moreover there exists a curve C ⊂ P such that [L1],..., [Lk], [N1],..., [Nh] “ depends only from the variables of C′ ” and C′ is either a line or a reduced conic or a disjoint union of two lines. If C′ is a line (item (a) in Theorem 1) then both the ′ Li’s and the Ni’s are linear forms in the same two “variables”. If C is a conic, then ′ Li’s and Ni’s depend on 3 “variables” and their projectivizations lie on C . See item (c) of Theorem 1 for the geometric interpretation of the reduction of L1,...,Lk and ′ ′ N1,...,Nh to bivariate forms involved with C when C = l ⊔ r is a disjoint union of two lines l and r (we have two sets of bivariant forms, one for the variables of l and one for the variables of r).

1. Notation and statements Before giving the precise statement of Theorem 1 we need to introduce the main algebraic geometric tools that we will use all along the paper. m N m+d Let νd : P → P , N := d  − 1, denote the degree d Veronese embedding of m m N P (say, defined over C). Set Xm,d := νd(P ). For any P ∈ P , the symmetric rank or symmetric tensor rank or, just, the rank rC(P ) of P is the minimal cardinality m of a finite set S ⊂ P (C) such that P ∈ hνd(S)i, where h i denote the linear span (here the linear span is with respect to complex coefficients), and we will say that S evinces rC(P ). Notice that the Veronese embedding νd is defined over R, i.e. m N N νd(P (R)) ⊂ P (R). For each P ∈ P (R) the real symmetric rank rR(P ) of P is m the minimal cardinality of a finite set S ⊂ P (R) such that P ∈ hνd(S)iR, where h iR means the linear span with real coefficients, and we will say that S evinces m N rR(P ). The integer rR(P ) is well-defined, because νd(P (R)) spans P (R). Let us fix some notation: If C ⊂ Pm is either a curve or a subspace and S ⊂ Pm is a finite set, we will use the following abbreviations:

SC := S ∩ C,

SCˆ := S \ (S ∩ C). N Theorem 1. Let P ∈ P (R) be such that rC(P )+ rR(P ) ≤ 3d − 1 and rC(P ) 6= m m rR(P ). Fix any set SC ⊂ P (C) and SR ⊂ P (R) evincing rC(P ) and rR(P ) respectively. Then one of the following cases (a), (b), (c) occurs: (a) There is a line l ⊂ Pm defined over R and with the following properties: SYMMETRIC TENSOR RANK 3

(i) SC and SR coincide out of the line l in a set Sˆl:

SC \ SC ∩ l = SR \ SR ∩ l =: Sˆl;

(ii) there is a point Pl ∈ hνd(SC,l)i ∩ hνd(SR,l)i such that SC,l evinces rC(Pl) and SR,l evinces rR(Pl); (iii) ♯(SC,l ∪ SR,l) ≥ d +2 and ♯(SC,l) < ♯(SR,l). (b) There is a conic C ⊂ Pm defined over R and with the following properties:

(i) SC and SR coincide out of the conic C in a set SCˆ :

SC \ SC,C = SR \ SR ∩ C =: SCˆ ;

(ii) there is a point PC ∈ hνd(SC,C )i ∩ hνd(SR,C )i such that SC,C evinces rC(PC ) and SR,C evinces rR(PC ); (iii) ♯(SC,C ∪ SR,C) ≥ 2d +2 and ♯(SC,C ) < ♯(SR,C ); (iv) if C is reducible, say C = l1 ∪ l2 with Q = l1 ∩ l2, then ♯((SC ∪ SR) ∩ (li \ Q)) ≥ d +1 for i ∈{1, 2}. (c) m ≥ 3 and there are 2 disjoint lines l, r ⊂ Pm defined over R with the following properties:

(i) SC and SR coincide out of the union Γ := l ∪ r in a set SΓˆ :

SC \ SC ∩ (l ∪ r)= SR \ SR ∩ (l ∪ r) := SΓˆ ;

(ii) ♯(SC,l ∪ SR,l) ≥ d +2 and ♯(SC,r ∪ SR,r) ≥ d +2; N (iii) the set hνd(SΓˆ )i ∩ hνd(Γ)i is a single point, OΓ ∈ P (R); SC,Γ evinces rC(OΓ) and SR,Γ evinces OΓ; (iv) the set h{OΓ} ∪ νd(l)i ∩ hνd(l)i (resp. h{OΓ} ∪ νd(r)i ∩ hνd(r)i) is formed N N by a unique point Ol ∈ P (R) (resp. Or ∈ P (R)); SC,l (resp. SC,r) evinces rC(Ol) (resp. rC(Or)); SR,l (resp. SR,r) evinces rR(Ol) (resp rR(Or)).

2. The proof Remark 1. Let S ⊂ PN (R). It will be noteworthy in the sequel that S can be N N used to span both a real space hSiR ⊂ P (R) and a complex space hSiC ⊂ P (C) N of the same dimension and hSiC ∩ P (R) = hSiR. In the following we will always use h i to denote h iC. N N Remark 2. Fix P ∈ P and a finite set S ⊂ P such that S evinces rC(P ). Fix any E ( S. Then the set h{P } ∪ Ei ∩ hS \ Ei is a single point (call it P1) and S \ E N N N evinces rC(P1). Now assume P ∈ P (R) and S ⊂ P (R). Then P1 ∈ P (R). If S evinces rR(P ), then S \ E evinces rR(P1). Lemma 1. Let C ⊂ Pm be a reduced curve of degree t with t =1, 2. Fix finite sets A, B ⊂ Pm. Fix an integer d>t such that: h1(I (d − t))=0. (A∪B)Cˆ ′ ′ Assume the existence of P ∈ hνd(A)i ∩ hνd(B)i and P/∈ hνd(S )i for any S ( A and any S′ ( B. Then

ACˆ = BCˆ . Proof. The case t = 1 is [4, Lemma 8]. If t = 2, then either C is a conic or m ≥ 3 0 and C is a disjoint union of 2 lines. In both cases we have h (IC (t)) > 0 and the linear system |IC (t)| has no base points outside C. Since A ∪ B is a finite set, there 4 EDOARDOBALLICO,ALESSANDRABERNARDI is M ∈ |IC (t)| such that M ∩ (A ∪ B) = C ∩ (A ∪ B). Look at the residual exact sequence (also called the Castelnuovo’s exact sequence): (2) 0 → I (d − t) → I (d) → I (d) → 0 (A∪B)Cˆ A∪B (A∪B)∩M,M We can now repeat the same proof of [4, Lemma 8] but starting with (2) instead of the exact sequence used there (cfr. first displayed formula in the proof of [4, Lemma 8]).

We will therefore get AMˆ = BMˆ . Now, since M ∩ (A ∪ B) = C ∩ (A ∪ B), we are done.  We are now going to prove Theorem 1.

Proof of Theorem 1 N Fix P ∈ P (R) such that rC(P )+ rR(P ) ≤ 3d − 1 and rC(P ) 6= rR(P ). m m Fix any set SC ⊂ P (C) evincing rC(P ) and any SR ⊂ P (R) evincing rR(P ). By applying [3], Lemma 1, we immediately get that 1 h (ISC∪SR (d)) > 0. m Since ♯(SC)+ ♯(SR) ≤ 3d − 1, either there is a line l ⊂ P such that ♯(SC,l ∪ SR,l) ≥ d + 2 or there is a conic C such that ♯(SC,C ∪ SR,C ) ≥ 2d + 2 ([14], Theorem 3.8). We are going to study separately these two cases in items (1) and (2) below. (1) In this step we assume the existence of a line l ⊂ Pm such that

♯(SC,l ∪ SR,l) ≥ d +2.

This hypothesis, together with rR(P ) 6= rC(P ), immediately implies property (aiii) of the statement of the theorem. 1 We are now going to distinguish the case h (ISC∪SR (d − 1)) = 0 (item (1.1) 1 below) from the case h (ISC∪SR (d − 1)) > 0 (item (1.2) below). 1 (1.1) Assume h (ISC∪SR (d − 1)) = 0. First of all, observe that the line l ⊂ Pm is well defined over R since it contains at least 2 points of SR (Remark 1). Then, by Lemma 1, we have that SC and SR have to coincide out of the line l:

SC \ SC,l = SR \ SR,l := Sˆl, and this proves (ai) of the statement of the theorem in this case (1.1).

The fact that ♯(SC) < ♯(SR) implies that ♯(SR,l) > (d+2)/2 and ♯(Sˆl) ≤ d, hence 1 h (ISlˆ∪l(d))=0. Therefore we have that dim(hνd(Sˆl ∪l)i)= ♯(Sˆl)+d+1, dim(hνd(Sˆl)i)= ♯(Sˆl)−1 and dim(hνd(l)i)= d + 1, and Grassmann’s formula gives hνd(Sˆl)i ∩ hνd(l)i = ∅.

Since P ∈ hνd(Sˆl ∪ SC)i and SC,l ⊂ l, the set hνd(Sˆl) ∪{P }i∩hνd(l)i is a single N point, Pl ∈ P (R).

′ ′ Since P ∈ hνd(SC)i and P/∈ hνd(SC)i for any S ( SC, the set hνd(Sˆl) ∪{P }i ∩ hνd(SC,l)i is a single point, PC (Remark 2). Then obviously: N PC = Pl ∈ P (R).

Since SC evinces rC(P ), then SC,l evinces Pl (Remark 2). In the same way we see that hνd(Sˆl) ∪{P }i∩hνd(SR,l)i = {Pl} and that SR,l evinces rR(Pl). This proves (aii) of Theorem 1 in this case (1.1). SYMMETRIC TENSOR RANK 5

1 (1.2) Assume h (ISC∪SR (d − 1)) > 0. m First of all, observe that there exists a line r ⊂ P such that ♯(r ∩ (SC ∪ SR)ˆl) ≥ d + 1, because ♯(SC ∪ SR)ˆl ≤ 3d − 1 − d − 2 ≤ 2(d − 1) + 1. By the same reason, if we write: C := l ∪ r, we get that ♯(SC ∪ SR)Cˆ ≤ 3d − 1 − d − 2 − d − 1 ≤ d − 2 and hence h1(I (d − 2)) = 0 (e.g. by [6], Lemma 34, (SC∪SR)Cˆ or by [14], Theorem 3.8). Lemma 1 gives:

(3) SC,Cˆ = SR,Cˆ . – Assume for the moment l ∩r 6= ∅. In this case, Remark 2 indicates that we can consider case (b) of the statement of the theorem. Therefore (3) proves (bi) in the case that the conic C in (b) in the statement of the theorem is reduced. Moreover, condition (biv) is satisfied, because ♯(r ∩ (SC ∪ SR)ˆl) ≥ d + 1. – Now assume l ∩ r = ∅. We will check that we are in case (a) with respect to the line l if ♯(SC,r ∪ SR,r)= d + 1, while we are in case (c) with respect to the lines l and r if ♯(SC,r ∪ SR,r) ≥ d +2, and the case ♯(SC,l ∪ SR,l)= ♯(SC,r ∪ SR,r)= d +1 cannot occur. Set Γ := l ∪ r. Since r ∩ l = ∅, we have dimhΓi = 3 and hence m ≥ 3. 1 — Assume for the moment m ≥ 4. Hence ♯(SC∪SR) c ≤ d and h (I SC SR d (d− hΓi ( ∪ )hΓi 1)) = 0. Therefore:

c c (4) SC,hΓi = SR,hΓi

c and the set h{P } ∪ νd(SC,hΓi)i ∩ hνd(hΓi)i is a single real point:

N c (5) O := h{P } ∪ νd(SC,hΓi)i ∩ hνd(hΓi)i ∈ P (R),

SC,Γ evinces rC(O) and SR,Γ evinces rR(O). Now, (4) implies that if we are either in case (a) or in case (c) of the theorem, we can simply study what happens at SC,hΓi and at SR,hΓi, which means that we can reduce our study to the case m = 3, since hΓi = P3. — Until step (2) below, we will assume m = 3. The linear system |IΓ(2)| on hΓi has no base points outside Γ itself. Since SC ∪SR is finite, there is a smooth quadric surface W containing Γ such that

SC,W ∪ SR,W = SC,Γ ∪ SR,Γ. Moreover, such a W can be found among the real smooth quadrics, since l and r are real lines.

1 Since ♯(SC ∪ SR ) ≤ d − 1, we have h (I (d − 2)) = 0. Hence ,hΓi ,hΓi Wˆ (SC∪SR)Wˆ Lemma 1 applied to the point O defined in (5), gives:

(SC,hΓi)Wˆ = (SR,hΓi)Wˆ , h{O} ∪ νd(SC,hΓ,i)Wˆ i ∩ hνd(W )i is a single real point ′ N (6) O = h{O} ∪ νd(SC,hΓ,i)Wˆ i ∩ hνd(W )i ∈ P (R), ′ ′ and SC,W evinces rC(O ). If (O ,SC,W ,SR,W ) is either as in case (a) or in case (c) of the statement of the theorem, then (O,SC,hΓi,SR,hΓi) is in the same case. Consider the system |(1, 0)| of lines on the smooth quadric surface W containing Γ. We have 6 EDOARDOBALLICO,ALESSANDRABERNARDI

1 0 that h (W, OW (d − 2, d)) = 0, and hence the restriction map H (W, OW (d)) → 0 H (Γ, OΓ(d)) is surjective. Therefore 1 1 1 h (W, IW ∩(SC∪SR)(d)) = h (l, Il∩(SC∪SR),l(d)) + h (r, Ir∩(SC∪SR),r(d)).

Now, this last equality, together with the facts that νd(SC,W ) and νd(SR,W ) are linearly independent and (SC ∪ SR)W ⊂ Γ, gives:

dim(hνd(SC,W )i ∩ hνd(SR,W )i)= ♯(SC ∩ SR)l + ♯(SC ∩ SR)r+ 1 1 (7) h (l, Il∩(SC∪SR),l(d)) + h (r, Ir∩(SC∪SR),r(d)).

(1.2.1) Observe that (7) implies that the case ♯(SC ∪ SR)r = ♯(SC ∪ SR)l = 1 d + 1 cannot happen because there is no contribution from h (l, Il∩(SC∪SR),l(d)) + 1 h (r, Ir∩(SC∪SR),r(d)) since both terms, in this case, are equal to 0. So, we can assume that at least ♯(SC ∪ SR)l > d + 1.

(1.2.2) Assume ♯(SC ∪ SR)r = d + 1 and ♯(SC ∪ SR)l > d + 1. To prove that we are in case (a) with respect to l it is sufficient to prove SC,r = SR,r. 1 We have dimhνd(SC∪SR)i = ♯(SC ∪SR)−1−h (ISC∪SR (d)). Since νd(SC) and νd(SR) are linearly independent, ♯(SC ∪ SR)= ♯(SC)+ ♯(SR) − ♯(SC ∩ SR), ♯(SC,l ∪ SR,l)= 1 1 ♯(SC,l)+ ♯(SR,l) − ♯(SC,l ∩ SR,l) and h (ISC∪SR (d)) = h (l, ISC,l∪SR,l ), Grassmann’s formula gives that hνd(SC)i ∩ hνd(SR)i is generated by hνd(SC,l)i ∩ hνd(SR,l)i. Since ′ ′ P/∈ hνd(S )i for any S ( SC, we get SC = SC,l ∪ (SC ∩ SR)ˆl, i.e. Sˆl = SR \ SC,l. Hence we can consider case (a), and (3) proves property (ai) also for the case (1.2) that we are treating. The point Pl that we need to get (aii) can be identified with the point O′ defined in (6) while (aiii) comes from our hypotheses. This gives all cases (a) of Theorem 1.

(1.2.3) Assume that both ♯(SC ∪ SR)r ≥ d + 2 and ♯(SC ∪ SR)l ≥ d + 2.

We need to prove that we are in case (c). Recall that SC,Γˆ = SR,Γˆ and that 1 h (ISC,Γˆ ∪Γ(d)) = 0. The latter equality implies, as in Remark 2, that h{P } ∪ νd(SC,Γ)i∩hνd(Γ)i is a single real point O1, that SC,Γ evinces rC(O1) and that SR,Γ evinces O1. Now O1 plays the role of OΓ of case (ciii) in Theorem 1. Since hνd(l)i∩hνd(r)i = ∅ and O1 ∈ hνd(Γ)i, the sets h{O1}∪νd(l)i∩hνd(l)i (resp. h{O1}∪νd(r)i∩hνd(r)i) are formed by a unique point O2 (resp. O3). Remark 2 gives N that Oi ∈ P (R), i = 1, 2, SC,l evinces rC(O2), SR,l evinces rR(O2), SC,r evinces rC(O3) and SR,r evinces rR(O3). The hypotheses of the case (1.2.3) coincide with (cii) of the statement of the theorem, while (3) gives also property (ci). Moreover O2 and O3 defined above coincide with Ol and Or in (civ) of Theorem 1, therefore we have also proved case (c) of Theorem 1. (2) Now assume the existence of a conic C ⊂ Pm such that

(8) deg(SC ∪ SR)C ≥ 2d +2.

Since ♯(SC ∪ SR) ≤ 3d − 1 − 2d − 2 ≤ d − 1, we have h1(I (d − 2)) = 0. C (SC∪SR)Cˆ By Lemma 1 we have

(9) SC,Cˆ = SR,Cˆ , the set h{P } ∪ νd(SC,Cˆ )i ∩ hνd(hCi)i is a single point: ′ (10) P := h{P } ∪ νd(SC,Cˆ )i ∩ hνd(hCi)i ′ ′ N and SC,C evinces rC(P ). Moreover, if C is defined over R, then P ∈ P (R) and ′ SR,C evinces rR(P ). Hence ♯(SC,C ) < ♯(SC ∩ SR). SYMMETRIC TENSOR RANK 7

(2.1) Assume that C is smooth. Therefore (9) proves (bi) of the statement of the theorem in the case where C is smooth. Since the reduced case is proved above (immediately after the displayed formula (3)), we have concluded the proof of (bi). Moreover the hypothesis (8) coincides with (biii) of the statement of the theorem since ♯(SC,C ) is obviously strictly smaller than ♯(SR,C ). This concludes (biii). The fact that ♯(SC,C ) < ♯(SR,C ) also implies that ♯(SR,C ) ≥ 5. Since each ′ point of SR is real, C is real. Remark 2 gives that νd(SR,C ) evinces rR(P ). Since ′ SR,C ⊂ C, SR,C also evinces the real symmetric tensor rank of P with respect to ′ the degree 2d rational normal curve νd(C). The point P defined in (10) plays the role of the point PC appearing in (bii) of the statement of the theorem. Therefore, we have just proved (bii) of Theorem 1. We treat the case (2.2) below for sake of completeness, but we can observe that this concludes the proof of Theorem 1.

(2.2) Assume that C is reducible, say C = L1 ∪ L2 with L1 and L2 lines and

♯((SC ∪ SR)L1 ) ≥ ♯((SC ∩ SR)L2 ). If ♯((SC ∪ SR) ∩ (L2 \ L2 ∩ L1)) ≤ d, then we proved in step (1) that we are in case (a) with respect to the line L1. Hence we may assume ♯((SC ∪ SR) ∩ (L2 \ L2 ∩ L1)) ≥ d + 1. Thus even condition (biv) is satisfied as already remarked above after the displayed formula (3).

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University of Trento, Dept. of Mathematics, 38123 Povo (TN), Italy

University of Torino, Dept. of Mathematics “Giuseppe Peano”, 10123 Torino, Italy E-mail address: [email protected], [email protected]