Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $L$ be a line bundle over a compex manifold $X$, a square-root of $L$ is a line bundle $M$ such that $M^{\otimes2}=L$. My question is when the square-root of Line Bundle is unique?

share|improve this question
1  
Square roots are unique iff the trivial line bundle does not have a non-trivial square root. –  Mariano Suárez-Alvarez Nov 19 '13 at 18:37
    
When trivial line bundle does not have a non-trivial square root? can you give a rererrence about your statement? –  Hassan Jolany Nov 19 '13 at 18:44
add comment

2 Answers

up vote 4 down vote accepted

Let's start with a counter-example. Take a one dimensional complex torus $X=\mathbb C/\mathbb Z\oplus \mathbb Z\tau$. Take a point which is a $2$-torsion in the group structure coming from $\mathbb C$, for instance the point $(\frac 12, \frac 12 \tau)$. The line bundle that corresponds to this point (if you don't know what that is, just take its ideal sheaf in the structure sheaf) will be a "square-root" of the trivial bundle.

In general, there are plenty torsion bundles, that is, line bundles with a finite power which is trivial. There are two ways you can get those.

(For simplicity) let $X$ be a smooth projective manifold over $\mathbb C$. Consider the exponential sequence $$ 0 \to \mathbb Z_X \to \mathscr O_X \to \mathscr O_X^\times \to 0$$ and the long exact cohomology sequence it leads to: $$ 0 \to \mathbb Z \to \mathbb C \to \mathbb C^\times \overset 0 \longrightarrow H^1(X,\mathbb Z)\to H^1(X,\mathscr O_X)\to H^1(X,\mathscr O_X^\times)\overset{c_1}{\longrightarrow} H^2(X,\mathbb Z)$$ Now here one has that $H^1(X,\mathbb Z)\simeq \mathbb Z^{2g}$, $H^1(X,\mathscr O_X)\simeq \mathbb C^{g}$, and $H^1(X,\mathscr O_X^\times)\simeq \mathrm{Pic}(X)$ and the last map $H^1(X,\mathscr O_X^\times)\to H^2(X,\mathbb Z)$ is just $c_1$ as in Henry's answer.

The image of $H^1(X,\mathscr O_X)$ in $H^1(X,\mathscr O_X^\times)$ is usually denoted by $\mathrm{Pic}^\circ(X)$ and is isomorphic to $\mathbb C^g/\mathbb Z^{2g}$, a complex torus (actually if $X$ is projective, then this is an abelian variety, called the Picard variety). The line bundles parametrized by this are the topologically trivial line bundles, that is those whose $c_1$ is zero. This being a complex torus there will be plenty of elements that are of finite order and they all correspond to torsion line bundles. In particular, if $g\neq 0$ then there will always be non-trivial line bundles whose square is trivial. To see how these appear geometrically, consider the Albanese morphism which maps $X$ to the abelian variety $\mathrm{Alb}(X)=H^0(X,\Omega_X)/H_1(X,\mathbb Z)^*$ (called the Albanese variety of $X$) which is just the dual abelian variety of the Picard variety. Take a torsion line bundle there and pull it back to $X$.

The other way you can get a non-trivial $2$-torsion line bundle is if there is a $\mathbb Z_2$ sitting in $\mathrm{Pic}(X)/\mathrm{Pic}^\circ (X)$, the image of $c_1$. According to Andy's comment below, any torsion in $H^2(X,\mathbb Z)$ is in the image of $c_1$, so one does not need to worry about the containment. This is what happens in the case of Enriques surfaces as pointed out by Lev. For an Enriques surface $g=0$, $c_1$ is an isomorphism and $\mathrm{Pic}(X)\simeq \mathbb Z^{10}\oplus \mathbb Z_2$. The $2$-torsion line bundle is the canonical line bundle, the determinant of the (co)tangent bundle.

share|improve this answer
1  
I'm pretty sure that it's the case that if $X$ is quasi-projective, then all of the torsion in $H^2(X,\mathbb{Z})$ is in the image of $c_1$. Topologically, these line bundles are exactly those that can be made trivial when you pass to a finite cover (and thus are algebraic). I wrote up the details of this in Lemma 2.8 of my paper "The Picard group of the moduli space of curves with level structures". –  Andy Putman Nov 20 '13 at 5:31
1  
Andy, thanks for that comment. I actually wondered about that. What I meant was that it is not obviously in the image of $c_1$. I'll make a correction. –  Sándor Kovács Nov 20 '13 at 6:59
    
@AndyPutman, that's a very nice paper! ($\pi$'s are rendered as $\neq$s, for some reason, in the PDF file linked to from arxiv.org/abs/0908.0555) –  Mariano Suárez-Alvarez Nov 20 '13 at 17:24
    
@MarianoSuárez-Alvarez : Thanks! I really enjoyed writing it. I looked at the arxiv posting, and it appears fine to me. Does the version on my webpage math.rice.edu/~andyp/papers render correctly? –  Andy Putman Nov 20 '13 at 17:52
    
I checked the one in the arXiv from other computer and it worked fine there; the one in your webpage works fine in both places. The PDF that arXiv is generating is not embedding all fonts, so the pi gets resolved locally. I wonder why they do that, as that can break in quite dramatic ways :-| –  Mariano Suárez-Alvarez Nov 20 '13 at 17:58
show 2 more comments

Complex line bundles over a manifold are classified by their first Chern class; we have a bijection $$\{\text{isomorphism classes of complex line bundles on $X$}\} \leftrightarrow H^2(X; \Bbb Z),$$ $$L \mapsto c_1(L).$$ The first Chern class is additive with respect to tensor products, so we see that $$c_1(L^{\otimes 2}) = 2c_1(L). \tag{$\ast$}$$ Now if $K$ is a complex line bundle with $K^{\otimes 2}$ trivial, then $$c_1((L \otimes K)^{\otimes 2}) = c_1(L^{\otimes 2}) + c_1(K^{\otimes 2}) = c_1(L^{\otimes 2}),$$ so $(L \otimes K)^{\otimes 2} \cong L^{\otimes 2}$ and therefore $L \otimes K$ is also a square root of $L^{\otimes 2}$. In general, from $(\ast)$ we see that square roots of the trivial line bundle will correspond to $2$-torsion elements in $H^2(X; \Bbb Z)$. Hence square roots of line bundles are unique when $H^2(X; \Bbb Z)$ has no $2$-torsion.

share|improve this answer
    
Is there any nesessary condition to say that $H^2(X; \Bbb Z)$ has no 2-torsion? –  Hassan Jolany Nov 19 '13 at 19:05
6  
The term complex line bundle is bit ambiguous. I guess you are thinking of $C^\infty$ complex line bundle, but I suspect the OP might have thinking of holomorphic line bundle. Here the classification is not just in terms of the first Chern class. –  Donu Arapura Nov 19 '13 at 19:20
1  
there is no 2-torsion if Pic is discrete. I'm not sure about char = p, but in char = 0 I think what you are looking for is $H^1(X,O_X) = 0$ –  aginensky Nov 19 '13 at 19:49
    
Donu Arapura@ can you explain more? –  Hassan Jolany Nov 19 '13 at 20:03
3  
@aginensky This is not quite true. Enriques surfaces have a torsion line bundle (the canonical bundle) but $h^{1,0}=0$. –  Lev Borisov Nov 19 '13 at 22:54
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.