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If ${L}$ is a line bundle over a complex manifold, what does the square root line bundle $L^{\frac{1}{2}}$ mean? After some google, I got to know that there are certain conditions for the existence of square root line bundle. In particular,I've following questions :

  1. What is the square root of a line bundle, what are the conditions for its existence.

  2. More importantly, how to think of the square root line bundle. Intuitively, It seems that a square root bundle $L^{\frac{1}{2}}$ is a line bundle s.t. the tensor bundle obtained by taking the tensor product of $L^{\frac{1}{2}}$ with itself gives the line bundle $L$.(correct if I am wrong).

I am particularly interested in square root of the Canonical Line bundle over a Riemann Sphere and the relation of square root bundle to Spinors in QFT. Please provide some references to look at.

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5 Answers 5

up vote 23 down vote accepted

If $L$ is any line bundle over a compex manifold $X$, a square root of $L$ is a line bundle $M$ such that $M^{\otimes2}=L$. So your guess in part (2) is correct.

This square root (if it exists) is not unique in general, and two of them will differ by a $2$-torsion line bundle, that is a line bundle $\eta$ such that $\eta^{\otimes 2}$ is trivial. In particular, if $\textrm{Pic}(X)$ is torsion free, then there is at most one square root.

In some cases no square root exists. Some general results are:

  1. A line bundle of degree $0$ has always at least one square root. This because $\textrm{Pic}^0(X)$ is a complex torus, hence a divisible group (in fact, there are roots of any order).

  2. A line bundle over a Riemann surface of genus $g$ has a square root if and only if it has even degree. The number of different square roots equals in this case $2^{2g}$, the number of $2$-torsion points in $\textrm{Pic}^0(X) \cong \textrm{Jac}(X)$.

  3. If $L$ is effective, that is $H^0(X, L) \neq 0$, and $Z \subset X$ is the zero locus of a holomorphic section of $L$, then the existence of a square root of $L$ is equivalent to the existence of a double cover $Y \to X$ branched over $Z$. In particular, non-trivial square roots of the trivial bundle correspond to non-trivial unramified double covers of $X$.

The square root of the canonical bundle of the Riemann Sphere $S$ is unique, since $\textrm{Pic}(S)=\mathbb{Z}$, and it is isomorphic to $\mathcal{O}(-1)$, the dual of the hyperplane bundle (the unique line bundle of degree $1$, whose transition function is $z \to 1/z$).

A readable introduction to spinor bundles is provided in the book of MOORE "Lectures on Seiberg-Witten invariants".

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A comment which I am surprised has not been mentioned yet.

There is a short exact sequence of sheaves on X $$ 0 \to \{\pm 1\} \to {\cal O}_X^\times \to {\cal O}_X^\times \to 0 $$ where the second map sends a nonvanishing holomorphic function to its square. You get a long exact sequence of cohomology groups, which in the relevant range is $$ H^0(X, {\cal O}_X^\times) \to H^1(X, \{\pm 1\}) \to Pic(X) \stackrel{2}{\to} Pic(X) \to H^2(X, \{\pm 1\}) $$ where map on the Picard group sends a line bundle $\cal L$ to its square ${\cal L}^{\otimes 2}$. One therefore gets an obstruction to extracting a square root of the line bundle in the cohomology group $H^2(X, \{\pm 1\})$, and an obstruction to uniqueness which is the aforementioned 2-torsion line bundle.

Chasing the homological algebra explicitly, in Mike Skirvin's answer he mentions taking the square root of the transition functions $g_{\alpha \beta}$ defining the line bundle. Generally, if you do this it will not work out - you must make nonunique choices of $h_{\alpha \beta}$ with $h_{\alpha \beta}^2 = g_{\alpha \beta}$, and you may not get a cocycle. The coboundary $\delta h$ will be a cycle, but because you know its square is $\delta g \equiv 1$ you find that $\delta h$ is made up of locally constant functions taking only the values $\pm 1$. This is a representing cocycle for the obstruction, and asking whether it is a coboundary is equivalent to asking whether one can fix the signs on the $h_{\alpha \beta}$ to make them match up and give you a line bundle.

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Thanks for mentioning this. This is more or less an expanded version of David Speyer's final comment on my "answer". –  Mike Skirvin Nov 4 '10 at 14:10
    
You're right, he beat me to it. –  Tyler Lawson Nov 4 '10 at 14:20

Yes, the square root of a line bundle $L$ is a line bundle $L'$ such that $(L')^{\otimes 2} \simeq L.$

I don't know of any general criteria for detecting the existence of a square root. A couple basic observations:

  • You certainly need the degree of $L$ to be even for a square root to exist.

(second edit: This second bullet point is still incorrect. See David Speyer's comments and Tyler Lawson's answer for a correct formulation.)

  • (edited based on David Speyer's comment below) If $U_i$ form an open cover of your manifold, then the transition functions for $L$ are given by maps $g_{ij}:U_i \cap U_j \to GL_1$ satisfying the cocycle condtion. These of course are the same as elements $g_{ij} \in \mathcal{O}(U_i \cap U_j)^{\ast}.$ To have a square root, it would suffice for a square root of the $g_{ij}$ to exist in $\mathcal{O}(U_i \cap U_j)^{\ast},$ as these would form transition functions for the square root of $L.$ As David Speyer notes in the comments, the converse is not true.

In regards to your specific question regarding the canonical line bundle of $\mathbb{P}^1,$ this is easy. The canonical bundle is $\mathcal{O}(-2),$ and its square root is the tautological line bundle $\mathcal{O}(-1).$ Since isomorphism classes of line bundles on $\mathbb{P}^1$ are determined completely by their degree, a line bundle on $\mathbb{P}^1$ has a square root if and only if its degree is even.

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The second bullet point is untrue. Consider $\mathbb{P}^1$, with the open cover $U_1 = \mathbb{P}^1 \setminus \{ \infty \}$ and $U_2 = \mathbb{P}^1 \setminus \{ 0,1 \}$. Take $g_{12} = x(x-1)$. Then the line bundle is isomorphic to $\mathcal{O}(2)$, so it has a square root, but $x(x-1)$ does not have a square root in $U_{12}$. –  David Speyer Nov 3 '10 at 16:36
    
David, thanks for pointing out my error. It should be correct now. –  Mike Skirvin Nov 3 '10 at 16:42
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Hate to say this, but I don't think the correction is right either. It certainly isn't in the analytic topology: Take a cover so that all the $U_i \cap U_j$ are contractible, then any nonvanishing analytic function will have a square root. –  David Speyer Nov 3 '10 at 17:06
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That is, indeed, the problem. When you take the square root, you have to choose a sign, and it isn't clear you can always do so in a way that preserves the cocycle condition. The obstruction to doing so lives in $H^2(X, \mu_2)$, where $\mu_2$ is the sheaf of locally constant $\pm 1$ valued functions. Now, in the Zariski topology, on integral domains, locally constant sheaves have no cohomology. So, in the Zariski topology on an integral variety, your statement is true. But it isn't true in the analytic topology and I suspect it isn't true for general schemes. –  David Speyer Nov 4 '10 at 13:24
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You have to be a little bit more careful here - you can't locally pick square-roots of functions in the Zariski topology, and so the first step of the construction for varieties needs to take place in the etale topology or something similar. –  Tyler Lawson Nov 4 '10 at 14:25

Try http://en.wikipedia.org/wiki/Theta_characteristic . Since the product of line bundles is tensor product, "square root" is in the same sense. So your comment is correct.

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A line bundle is given by the corresponding set of transition functions. These functions have values in the set of nonzero complex numbers (or in nonzero reals when working with real differentiable manifolds) and hence you can take their square root. The resulting functions may again satisfy the axioms for transition functions if you take the square roots properly and if the topology of the underlying manifold admits it. I guess only orientability is needed in general and hence for complex manifold square roots always exists.

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It is not true that all holomorphic line bundles on complex manifolds have square roots. See Polizzi's answer. –  Andy Putman Nov 3 '10 at 17:06
    
And what about smooth line bundles? –  Vít Tuček Nov 3 '10 at 17:56
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If $L$ is a line bundle such that $L \otimes L$ is holomorphic, then $L$ is holomorphic, so a counterexample in the holomorphic category also gives a counterexample in the smooth category. –  Andy Putman Nov 3 '10 at 18:01
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No. Let $X$ be a complex manifold and $L$ be a holomorphic line bundle on $X$ that does not have a holomorphic square root. We then forget the complex structure, so $X$ is just a smooth manifold and $L$ is a smooth complex line bundle on $X$. By what I just said, $L$ (as a smooth line bundle) does not have a smooth square root. –  Andy Putman Nov 3 '10 at 19:12
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@r0b0t: Note that $c_1$ is additive w.r.t. tensor product. Note that $c_1$ is defined for $C^\infty$ complex vector bundles. Take anything with odd $c_1$. –  Kevin H. Lin Nov 4 '10 at 12:56

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