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I'm sure this is a silly question, but suppose we have a proper holomorphic fibration $f : X \to B$ of complex manifolds where the fibers of $f$ are complex tori. Each torus $X_b = f^{-1}(b)$ has a group structure, and thus an identity element $0_b$. Does the function $b \mapsto 0_b$ give a holomorphic section $B \to X$?

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    I have no intuition on complex manifolds, so this may be entirely silly: I don't understand your heuristic at all. I mean, of course you can equip each fiber with a group structure, but how is this group structure determined purely in terms of $f$? In other words, I don't understand how $b \mapsto 0_b$ is supposed to be defined in the first place.2011-07-20
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    As an example of what I have in mind: consider the unit tangent bundle on a $2$-dimensional Riemannian manifold. This certainly is an $S^1$-bundle, but how do you equip the fibers with a group structure here? There must be some magic going on in the complex setting.2011-07-20
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    No, no magic, you're quite right, I should have thought about this more before asking. One can also fabricate examples of such fibrations where no holomorphic section exists (so one can't get a "holomorphic" variation of group structures), by taking a 2-dimensional torus that doesn't split as a product and quotienting by a sub-torus.2011-07-20
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    That makes sense, yes. Thank you for clarifying.2011-07-20
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    The problem is that the group structure of the complex torus is not canonical. You need to specify beforehand a point on the torus (which is going to be the identity of the group). So, if this zero-point is chosen "randomly", the zero section does not need to be even continuous.2011-07-20
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    Thank you all for your comments. @David Speyer, itsn't that a torus in the sense of toric geometry? I get that an elliptic curve and $\mathbb C^* / \mathbb Z$ are the same thing, but there seems to be a leap there somewhere.2011-07-20
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    Oh, my bad. I didn't read carefully enough and thought that was the kind of torus you wanted. I'll delete my comment.2011-07-20

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