TITLE: What is a square root of a line bundle?
QUESTION [25 upvotes]: 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 :

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

REPLY [25 votes]: 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.