Bull. Sci. math. 134 (2010) 747–749www.elsevier.com/locate/bulsci

On line bundles over real algebraic curves

Indranil Biswas

School of Mathematics, Tata Institute of Fundamental Research, Homi Bhabha Road, Bombay 400005, India

Received 7 January 2010

Available online 27 January 2010

Abstract

Let X be a geometrically irreducible smooth projective curve defined over the real numbers. Let nX bethe number of connected components of the locus of real points of X. Let x1, . . . , x� be real points from �

distinct components, with � < nX . We prove that the divisor x1 + · · · + x� is rigid. We also give a verysimple proof of the Harnack’s inequality.© 2010 Elsevier Masson SAS. All rights reserved.

MSC: 14F05; 14P99

Keywords: Real algebraic curve; Line bundle; Harnack’s inequality

1. Introduction

We consider line bundles over a geometrically irreducible smooth projective curve X definedover R; see also [2,1]. We assume that X has real points; let nX be the number of connectedcomponents of the locus of real points.

Choose � of the nX components, where � < nX , and choose a real point of X from eachof these � components. Let x1, . . . , x� be the chosen points. We prove that the effective divisorx1 + · · · + x� is rigid, equivalently,

H 0(X, OX(x1 + · · · + x�)) = R

(Theorem 2.1).The method of proof also yields a proof of the Harnack’s inequality.

E-mail address: indranil@math.tifr.res.in.

0007-4497/$ – see front matter © 2010 Elsevier Masson SAS. All rights reserved.doi:10.1016/j.bulsci.2010.01.006

748 I. Biswas / Bull. Sci. math. 134 (2010) 747–749

2. Rigidity and vanishing conditions

Let X be a geometrically irreducible smooth projective curve defined over R. Let XC :=X×R C be the complex curve obtained by changing the base field. The Galois group Gal(C/R) =Z/2Z acts on XC as an antiholomorphic involution

τ : XC −→ XC (1)

of the Riemann surface XC. The real points of X are the points of XC fixed by τ . Each connectedcomponent of the locus of real points is diffeomorphic to the circle S1.

The involution τ of XC define an involution τ̃ of the divisors on XC by sending any divisor∑ni=1 mixi to

∑ni=1 miτ(xi). The divisors on X are identified with the divisors on XC fixed by τ̃ .

Let nX denote the number of connected components of the locus of real points of X; letC1, . . . ,CnX

be the connected components. So {Ci}nX

i=1 are also the connected componentsof (XC)τ .

For any divisor D = ∑ni=1 mixi on X, and for any connected component C of (XC)τ , define

NC(D) :=∑xi∈C

mi. (2)

Theorem 2.1. Take any integer 0 � � < nX . Then

H 0(X, OX(x1 + · · · + x�)) = R,

where xi ∈ Ci .

Proof. Let D = ∑ni=1 miyi be a divisor on X. Define

B(D) := {i ∈ [1, nX] ∣∣ NCi

(D) ≡ 1 mod 2}, (3)

where NCi(D) is defined in (2). Assume that

degree(D) < #B(D). (4)

We will show that

H 0(X, OX(D)) = 0. (5)

To prove this by contradiction, let σ be a nonzero section of OX(D). Let

D′ := divisor(σ ) =a∑

j=1

njzj (6)

be the divisor on X associated to the section σ . So, we have ni > 0. Also,

degree(D′) =

a∑j=1

nj = degree(D) =n∑

i=1

mi. (7)

It can be proved that for any i ∈ [1, nX],NCi

(D) ≡ NCi

(D′) mod 2. (8)

Indeed, D and D′ differ by the divisor of a rational function ϕ on the real curve X. The function ϕ

takes Ci to the real points of P1 . Both Ci and the locus of real points of P

1 are diffeomorphic to

R R

I. Biswas / Bull. Sci. math. 134 (2010) 747–749 749

the circle S1. Finally, note that for any real analytic map φ : S1 −→ S1, and for any two pointss1 and s2 of S1, the number of pre-images of s1 (with multiplicity) differ by an even integer fromthe number of pre-images of s2. Hence (8) holds.

Since (8) holds, for all i ∈ B(D), where B(D) is defined in (3), we have NCi(D′) �= 0. As all

nj in (6) are positive, this implies that

a∑j=1

nj � #B(D).

In view of (7), this contradicts the given condition in (4).Therefore, we conclude that (5) holds.Consider the divisor x1 + · · · + x� in the statement of the theorem; it being effective, we have

R ⊂ H 0(X, OX(x1 + · · · + x�)). The proof will be completed by showing that

dimH 0(X, OX(x1 + · · · + x�))� 1.

Recall that � < nX . Fix a real point y ∈ C�+1. Set

D := x1 + · · · + x� − y.

Therefore, B(D) = {1, . . . , �, � + 1}. So #B(D) > degree(D) = � − 1. Hence we have

H 0(X, OX(D)) = 0

(see (5)). This implies that the evaluation map

H 0(X, OX(x1 + · · · + x�)) −→ OX(x1 + · · · + x�)y

that sends any η to η(y) is injective. Therefore,

dimH 0(X, OX(x1 + · · · + x�))� 1.

This completes the proof of the theorem. �Let g be the genus of X. The Harnack’s inequality says that nX � g + 1. We will give a very

simple proof of this inequality.Assume that nX � g + 2. Fix points xi ∈ Ci , where 1 � i � g + 2. Consider the divisor

D = −x1 +g+2∑i=2

xi

on X. Since degree(D) = g, from the Riemann–Roch theorem we know that

H 0(X, OX(D)) �= 0.

But this contradicts (5). Hence we conclude that nX � g + 1.

References

[1] U.N. Bhosle, I. Biswas, Stable real algebraic vector bundles over a Klein bottle, Trans. Amer. Math. Soc. 360 (2008)4569–4595.

[2] C. Ciliberto, C. Pedrini, Real abelian varieties and real algebraic curves, in: Lectures in Real Geometry, Madrid,1994, in: de Gruyter Exp. Math., vol. 23, de Gruyter, Berlin, 1996, pp. 167–256.