A LOW GENUS PROJECTIVE CURVE
EDOARDO BALLICO
LetX⊂P2k+1 be a smooth, non-degenerate and real projective curve. For each P ∈ P2k+1(R) we study the set of all integers ](S∩X(R)) for all possible sets S⊂X(C) such thatP ∈ hSiand](S) =k+ 1 (mainly whenpa(X)≤1).
AMS 2010 Subject Classification: 14N05; 14H99; 14P99.
Key words: ranks, border ranks, real curve, real elliptic curve.
1. INTRODUCTION
Let X ⊆ Pn be an integral and non-degenerate variety defined over C.
For anyP ∈Pn(C) theX-rankrX(P) ofPis the minimal cardinality of a finite set S ⊂ X(C) such that P ∈ hSi, where h i denote the linear span. In the applications whenX is a Veronese embedding ofPm theX-rank is also called the “structured rank” or “symmetric tensor rank” (this is related to the virtual array concept considered in sensor array processing ([5], [8], [9])). On this topic there was the 2008 AIM workshop Geometry and representation theory of tensors for computer science, statistics and other areas and a summer school:
“Geometry of tensors and applications”, 14–18, June 2010, at the Sophus Lie Conference Center, Nordfjordeid, Norway. In the applications quite often the base field is the real field ([9], [6] and references therein). In this case there are several possible related questions and here we introduce another one, different from the notion of real rank considered in [6].
For any algebraic scheme Y defined over C let Y(C) denote the set of all C-points with the euclidean topology. We will explicitly say when we will use the Zariski topology onY(C). ThusY(C) is Hausdorff. If Y is projective, then Y(C) is compact. Now assume thatY is defined over R. Any subset of Y(C) (and in particular the real locusY(R)⊂Y(C)) inherits a topology. The complex conjugation σ is a homeomorphism σ : Y(C) → Y(C) and Y(R) = {P ∈ Y(C) :σ(P) = P}. With this topology if Y is smooth and projective, thenY(R) is a compact real manifold with pure dimension dim(Y) (sometimes empty). Thus if Y is a smooth and projective curve, then Y(R) is a disjoint union of finitely many circles.
MATH. REPORTS13(63),4 (2011), 329–335
For anyP ∈Pn(C) letS(X, P) denote the set of allS ⊂X(C) computing rX(P), i.e., such that](S) =rX(P) and P ∈ hSi. Set EX(k) :={P ∈Pn(C) : rX(P) = k}. For any integert≥1 let σt(X) or σt(X)(C) denote the closure in Pn(C) of the union of all (t−1)-dimensional linear subspaces spanned by t points of X(C). For any P ∈ Pn(C) let bX(P) denote the border X-rank of P, i.e., the first integer s > 0 such that P ∈ σs(X). Let zX(P) be the minimal integer deg(Z) for some zero-dimensional scheme Z ⊂ X(C) such that P ∈ hZi. Notice that if Z computes zX(P), then P /∈ hZ0i for any Z0 ⊂ Z, Z0 6= Z. In this paper we only use the function zX when X is a smooth curve of genus zero or one. In these caseszX =bX ([3], Proposition 1, [4], Lemma 2.1.5). Set m := dim(X). We always have rX(P) ≤ n−m+ 1 ([8], Proposition 5.1). The generic X-rank ρX is the X-rank of a non-empty open subset of Pn(C), i.e., the first integer t ≥ 1 such that σt(X) = Pn (as irreducible varieties), i.e., the first integert such thatσt(X(C)) =Pn(C). Set UX :=EX(ρX).
FixX ⊂Pndefined overRandP ∈Pn(R). Recall thatrX(P) will always mean the complex X-rank of X, i.e., the rank with respect to X(C). Since X and P are real, σ(S(X, P)) = S(X, P). Set S(X, P)σ := {S ∈ S(X, P) : σ(S) = S}. Each element of S(X, P)σ is a disjoint union of some points of X(R) and some complex conjugate pairs of points of X(C)\X(R). For any integer t such that t ≡ rX(P) (mod 2) let S(X, P;R)σt denote the set of all S ∈ S(X, P)σ which are unions oftpoints ofX(R) and (rX(P)−t)/2 complex conjugate pairs of points of X(C).
The complex conjugationσ acts on S(X, P). If σ :S(X, P) → S(X, P) has no fixed point, then we say that the signature ofP with respect to the real curve X is ∅ and that the extended signature is {∅}. If there is at least one such fixed point, then the signature τ(X,R, P) ofP with respect to the real curve X is the pair (rX(P), U), where U := {t ∈N :S(X, P;R)σt 6= ∅}. We are interested in the possible signatures of the points ofUX(R) :=UX∩Pn(R).
Notice that points ofUX(R) with different signatures lie in different connected components ofUX(R) for the euclidean topology. If there isS ∈ S(X, P) such that σ(S) 6= S, then we call τ(X,R, P)0 := τ(X,R, P)∪ {∅} the extended signature of P. If there is no such S then we call τ(X,R, P)0 := τ(X,R, P) the extended signature of P. If X(R) has several connected components, say T1, . . . , Ts, then we may refine the signature of τ(X,R, P) of P with respect to the real variety Xby prescribing how many of the real points are contained in each component T1, . . . , Ts ofX(R).
In this note we prove the following results.
Theorem 1.1. Let X ⊂ Pn be a real and non-degenerate embedding of the real integral curve X. Assume Xreg(R) 6=∅. We have ρX =b(n+ 2)/2c.
Each non-empty admissible pair (b(n+ 2)/2c, t), 0 ≤ t ≤ b(n+ 2)/2c, t ≡
b(n+ 2)/2c (mod 2) arises as the signature of a non-empty open subset (for the euclidean topology) of UX(R).
Theorem 1.2. LetX ⊂P2k+1,k≥1, be a rational normal curve defined over R. We have ρX =k+ 1, every P ∈ UX(R) has a unique signature and each non-empty admissible pair (k+ 1, t), 0≤ t≤ k+ 1, t ≡k+ 1 (mod 2) arises as the signature of a non-empty open subset (for the euclidean topology) of UX(R). {∅}is not in the extended signature of any P ∈UX(R).
The last assertion of Theorem 1.2 shows that sometimes in the statement of Theorem 1.1 we need to exclude the ∅signature.
Theorem 1.3. Let X ⊂P2k+1, k≥1, be a real linearly normal embed- ding of a smooth curve of genus 1. Assume X(R)6=∅. Set U0X :={P ∈UX : OX(1) OX(2Z) for all Z ∈ S(X, P)}. U0X contains a non-empty Zariski open subset of UX.
(a)Every P ∈U0X(R) has at most two signatures.
(b) Fix integers t1, t2 such that 0 ≤t1 ≤ t2 ≤ k+ 1 and t1 ≡ t2 ≡k+
1 (mod 2). Then there exists a non-empty open subsetA(t1, t2) ofU0X(R) (for the euclidean topology) such that every P ∈A(t1, t2) has (k+ 1,{t1, t2}) as its only extended signatures.
Notice that in part (b) of Theorem 1.3 we allow the caset1 =t2: every P ∈A(t1, t1) has (k+ 1,{t1}) as its only extended signature.
2. THE PROOFS
We lift from [2] the following lemma and its proof.
Lemma 2.1. Fix any P ∈ Pn(C) and two zero-dimensional subschemes A,B of X(C) such thatA6=B, P ∈ hAi,P ∈ hBi, P /∈ hA0i for any A0 ⊂A, A06=A, and P /∈ hB0i for any B0 ⊂B, B06=B. Then h1(Pn,IA∪B(1))>0.
Proof. SinceA and B are zero-dimensional, we have h1(Pn,IA∪B(1))≥ max{h1(Pn,IA(1)), h1(Pn,IB(1))}. Thus we may assume h1(Pn,IA(1)) = h1(Pn,IB(1)) = 0, i.e., dim(hAi) = deg(A)−1 and dim(hBi) = deg(B)−1.
SetD:=A∩B (scheme-theoretic intersection). Thus deg(A∪B) = deg(A) + deg(B) − deg(D). Since D ⊆ A and A is linearly independent, we have dim(hDi) = deg(D)−1. Since h1(Pn,IA∪B(1)) > 0 if and only if dim(hA∪ Bi)≤deg(A∪B)−2, we geth1(Pn,IA∪B(1))>0 if and only ifhDi ⊂ hAi∩hBi and hDi 6=hAi ∩ hBi. SinceA 6=B, D⊂A,D 6=A. Hence P /∈ hDi. Since P ∈ hAi ∩ hBi, we are done.
Lemma 2.2. Let C ⊂ P2k+1, k ≥ 1, be a rational normal curve. Fix P ∈P2k+1 such that there is a zero-dimensional subscheme A ⊂C such that
P ∈ hAi, P /∈ hA0i for any A0 ⊂ A, A0 6= A. and deg(A) ≤ k+ 1. Let B ⊂Cbe a zero-dimensional scheme such thatP ∈ hBianddeg(B)≤deg(A).
Then B =A.
Proof. Use Lemma 2.1, the inequality deg(A∪B)≤deg(C) + 1 and that any zero-dimensional scheme with degree ≤ c+ 1 of a rational normal curve of Pc is linearly independent.
Remark2.3. Let X ⊂Pn be an integral and non-degenerate curve. We have dim(σt(X)) = min{n,2t−1}) for all integer t ≥ 1 ([1], Remark 1.6).
Hence ρX =b(n+ 2)/2c andUX =EX(b(n+ 2)/2c).
For any schemeX⊆Pn letXb ⊆An+1 denote its affine cone.
Lemma 2.4. Let X, Y ⊂Pn be complex integral subvarieties. Fix P ∈ Xreg(C) andQ∈Yreg(C)such thatTPX∩TQY =∅. Then the joinJ(X, Y)of X and Y has a smooth branch of dimensiondim(X) + dim(Y) + 1at a general point E of the line h{P, Q}i and the natural addition mapXb+Yb → J(X, Y\) considered in [1], §1, is a submersion at E.
Proof. Let α : An+1 ×An+1 → An+1 denote the addition. Set u :=
α|(Xbreg)×(Ybreg). Since dim(Xb×Yb) = dim(X) + dim(Y) + 2, the differential of the map u : (Xbreg)×(Ybreg) → An+1 has always rank at most dim(X) + dim(Y)+2. Our assumption means that it achieves its maximal value at (P ,b Q)b ([1], Corollary 1.11). Hence Im(u) has a smooth branch of dimension dim(X)+
dim(Y) + 2 in a neighborhood of u(P ,b Q), proving the last assertion. Takingb the projectionAn+1\ {O} →Pn we get the other statements of the lemma.
Lemma 2.5. Let X ⊂ Pn be an integral and non-degenerate complex projective curve. Fix an integer t ≥ 2 and P1, . . . , Pt ∈ Xreg(C) such that dim(hTP1X∪ · · · ∪TPtXi) = 2t−1. Then dim(h{P1, . . . , Pt}i) = t−1 and a general point of h{P1, . . . , Pt}iis contained in a smooth branch of σt(X).
Proof. Since hTP1X∪ · · · ∪TPtXi is generated by h{P1, . . . , Pt}i and t points (one for each tangent line), we obviously have dim(h{P1, . . . , Pt}i) = t−1. Ift= 2, then the second assertion is Lemma 2.4 applied to (X, Y, P, Q) :=
(X, X, P1, P2). Now assume t ≥ 3. We use induction on t. We apply Lemma 2.4 to (X, Y, P, Q) := (X, σt−1(X), P1, Q) where Q is a general ele- ment ofh{P2, . . . , Pt}i.
Lemma 2.6. Let X ⊂ Pn be an integral and non-degenerate projective curve defined over R. Call T1, . . . , Ts the connected components of Xreg(R).
Fix (α, β) ∈N2\ {(0,0)} such that 2α+ 4β ≤ n+ 1. If s= 0, then assume α = 0. If s >0, then for eachi∈ {1, . . . , α} fix an integer γ(i)∈ {1, . . . , s}.
There are α real points P1, . . . , Pα of X with Pi ∈ Tγ(i) for all i and β pairs
Q0i, Q00i, 1 ≤ i ≤ β, of complex conjugate points of X(C) \X(R) such that dim(hSα
i=1TPiX∪Sβ i=1(TQ0
iX∪TQ00
iX)i= 2α+ 4β−1.
Proof. By Terracini’s lemma ([1], Corollary 1.11) and by [1], Remark 1.6, there is a Zariski open dense subset U of Xα+2β(C) such that
dim(hSα+2β
i=1 TQiXii) = 2α+ 4β−1
for every (Q1, . . . , Qα+2β) ∈ U. For any x < α+ 2β and fixed Q1, . . . , Qx
such that dim(hSx
i=1TQiXii) = 2x−1 the set of all (Qx+1, . . . , Qα+2β) that we may add to (Q1, . . . , Qx) and have dim(hSα+2β
i=1 TQiXii) = 2α+ 4β−1 is a non-empty open subset of Xα+2β−x. Since eachTi is Zariski dense inX, the previous observation reduces the proof of the lemma for the pair (α, β) to the proof of the lemma for the pair (0, β). To prove the case (0, β) it is sufficient to use induction on β and the observation that the set of all pairs (Q, σ(Q)) are Zariski dense in (Xreg(C)\Xreg(R))×(Xreg(C)\Xreg(R)).
Proof of Theorem 1.1. It is well-known that ρX =b(n+ 2)/2c ([1], Re- mark 1.6). Fix anyP ∈UX(R). Apply Lemma 2.5 withα+ 2β=b(n+ 2)/2c.
Then apply Lemma 2.5 and conclude the proof.
Proof of Theorem1.2. Lemma 2.2 gives the uniqueness of the setS(X, P).
Thus σ(S) =S. Hence P has a unique signature and its extended signature does not contain ∅. The remaining assertions are particular cases of Theo- rem 1.1, concluding the proof.
Lemma 2.7. Let X be a real smooth curve of genus1 such that X(R)6=
∅. Fix an integer d > 0 and L ∈ Picd(X)(R). Let |L|(R) denote the set of all σ-invariant elements of the linear system |L|. Fix any integer t such that 0≤t≤dandt≡d (mod 2). LetA(t)be the set of allS ∈ |L|(R) such thatS is reduced and it has exactly t real points. Then A(t)6=∅ and A(t) contains a non-empty open subset of the (d−1)-dimensional real projective space|L|(R).
Proof. Fix any setS ⊂X(C) such that](S) =d−1 and eitherS ⊂X(R) (case t = d) or S is the union of t real points and ((d−t)/2−1) pairs of complex conjugate points and one point Q∈X(C)\X(R) (caset6=d). The line bundle L(−S) has degree 1. Thus there is a unique Q0 ∈ X(C) such that L(−S) ∼= OX(Q0). Since L is real and S\ {Q} is σ-invariant, we have Q0 =σ(Q). First assume t < d. SinceQ∈X(C)\X(R), we have σ(Q)6=Q.
Since S\ {Q}isσ-invariant and reduced, we get that S∪σ(Q) is reduced and σ-invariant. By construction S∪ {σ(Q)} ∈ A(t). VaryingS as above we get the lemma when t 6= d. Now assume t = d. In this case there is a unique effective divisor S+Q0 ∈ |L| containingS. Since S is real, the uniqueness of this divisor implies that it is defined overR, i.e., Q0 ∈X(R). We only need to prove that S+Q0 is reduced for general S, i.e.,Q0 ∈/ S for general S. This is
true because the set of all unreduced elements of |L|(C) is an algebraic set of dimension ≤d−2 by Bertini’s theorem.
Proof of Theorem 1.3. Since deg(OX(1)) = 2k+ 2 is even, there are 4 non-isomorphic L ∈ Pick+1(X) such that L⊗2 ∼= OX(1). Call Li, 1 ≤ i ≤4, these line bundles. For eachP ∈UX\U0X there isi∈ {1,2,3,4}and Z∈ |Li| such that P ∈ hZi. Since dim(|Li|) = k and dim(hZi) = k, we see that UX \U0X is contained in a hypersurface of P2k+1(C). Hence U0X contains a non-empty Zariski open subset of UX. Since the embedding X ,→ P2k+1 is real, σ(U0X) =U0X.
(i) FixP ∈σk+1(X)\σk(X) and assume the existence of 3 distinct degree k+ 1 subschemesAi,i= 1,2,3, ofX(C) computingzX(P). Thus dim(hAii) = k for all i. First assume Ai ∩Aj = ∅ for all i 6= j. Since P ∈ hA1i ∩ hA2i, A ∪B is contained in a hyperplane of Pn. Since deg(A1 ∪A2) = 2k+ 2, A1+A2 ∈ |OX(1)|. Thus A2 ∈ |OX(1)(−A1)|. Using A3 instead of A2 we get A3 ∈ |OX(1)(−A1)|. Thus OX(A2) ∼= OX(A3). In the same way we get OX(1) ∼= OX(A3). Thus OX(1) ∼= OX(2A1). Thus P ∈ UX \ U0X. Now assume Ai ∩Aj 6= ∅ for some i < j, say if (i, j) = (1,2). Let B1,2 be the maximal divisor contained both in A1 and A2. Hence D := A1+A2−B12 is a subscheme of degree at most 2k+ 1 of the linearly normal curve X. By Riemann-Roch every effective divisor of X with degree at most 2k+ 1 is a linearly independent subscheme of Pn, contradicting Lemma 1.2. Hence part (a) is proved.
(ii) Fix any real line bundleM on X such that deg(M) =k+ 1 and set N :=OX(1)⊗M∗. Hence N is a degreek+ 1 real line bundle. Assume the existence of A ∈ |M| such that σ(A) = A, A is reduced and A has exactly t1 real connected components. Assume the existence of B ∈ |M| such that σ(B) = B, B is reduced and B has exactly t2 real connected components.
Assume A∩B =∅. Since any zero-dimensional subscheme of X with degree at most 2k+ 1 is linearly independent, dim(hAi) = dim(hBi) = k. Since A+B ∈ |OX(1)|,A+B is contained in a hyperplane. Since A∩B =∅, we saw in step (i) that hAi ∩ hBi is a unique point, QA,B. Since hAi and hBi are real, QA,B ∈ Pn(R). Assume for the moment zX(QA,B) = k+ 1. Any such QA,B has signatures (k+ 1;t1) and (k+ 1;t2) and no other extended signature, because part (a), i.e., step (i), gives the non-existence of Z ⊂X(C) such that deg(Z) =k+ 1, QA,B∈ hZiand Z /∈ {A, B}. Sinceσk(X)∩Pn(R) has Hausdorff dimension ≤ 2k, to conclude it is sufficient to find families of sets A, B as above such that the set of the associated points QA,B (without imposing the condition zX(QA,B) = k+ 1) cover a non-empty open subset of Pn(R). Start with any real line bundle M on X. Apply Lemma 2.7 with respect to the integer d:=k+ 1 first to the integer t1 and M and then with
respect to the integer t2 and the real line bundleOX(1)⊗M∗. The proof of Theorem 1.3 is over.
Remark2.8. LetX ⊂P4 be a real rational normal curve. There is a non- empty open subset Ω of UX(R) formed by points with real rank 4 ([6], case d = 4 of the Main Theorem). Recall that ρX = 3. Fix any P ∈ Ω and take any signature (3, a) of P. Since 3 is odd, a >0. Since P has not real rank3, a6= 3. Thus (3,{1}) is the only signature of P with respect to X.
Acknowledgement.The author would like to thank the referee for his/hers helpful remarks and suggestions.
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Received 26 April 2010 University of Trento
Department of Mathematics 38123 Trento, Italy ballico@yscience.unitn.it
