Timeline for What is a square root of a line bundle?
Current License: CC BY-SA 2.5
9 events
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| Nov 4, 2010 at 14:25 | comment | added | Tyler Lawson | 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. | |
| Nov 4, 2010 at 14:13 | history | edited | Mike Skirvin | CC BY-SA 2.5 |
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| Nov 4, 2010 at 13:24 | comment | added | David E Speyer | 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. | |
| Nov 3, 2010 at 18:48 | comment | added | Mike Skirvin | I see. Even though the question was about complex manifolds, I've been thinking about everything in terms of complex algebraic varieties. In this setting, the statement seems potentially correct, but maybe even then it's not. I'm obviously somewhat confused because transition functions for $L^2$ are given by the squares of the transition functions for $L.$ So is the problem that a square root of a transition function might not be a cocycle? | |
| Nov 3, 2010 at 17:06 | comment | added | David E Speyer | 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. | |
| Nov 3, 2010 at 16:42 | comment | added | Mike Skirvin | David, thanks for pointing out my error. It should be correct now. | |
| Nov 3, 2010 at 16:40 | history | edited | Mike Skirvin | CC BY-SA 2.5 |
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| Nov 3, 2010 at 16:36 | comment | added | David E Speyer | 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 $U_{12}$. | |
| Nov 3, 2010 at 16:29 | history | answered | Mike Skirvin | CC BY-SA 2.5 |