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Does anyone know an example of a rational ruled surface $X=\mathbb{P}(\mathcal{O}\oplus\mathcal{O}(-e))$ for $e\ge 0$ which admits a transitive algebraic group action? except the trivial case $\mathbb{P}^1\times\mathbb{P}^1$.

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I am pretty sure that $\mathbb{P}^1 \times \mathbb{P}^1$ is the only case, unfortunately though I don't have a reference or proof off hand. Note that the other rational ruled surfaces still admit certain "large" group actions, for example they are all toric. –  Daniel Loughran Feb 28 '12 at 6:59
Surely an automorphism has to preserve the negative section, no? So a point on that curve can not be moved away from it. –  quim Feb 28 '12 at 7:09
(So yes, P1xP1 is the only case) –  quim Feb 28 '12 at 7:10
Thanks are to all of you. Intersection theory is really a good way to approach such kind of question. –  Thunder Feb 29 '12 at 5:25

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up vote 5 down vote accepted

A rational ruled surface with $e>0$ has a unique irreducible curve with negative self-intersection, so any automorphism has to fix that. Therefore it cannot have a transitive automorphism group. (Actually it also has to fix the ruling, because it has to fix the cone of curves and the negative curve and the fiber of the ruling are the generators, but of course that in itself would allow for a transitive action as the fibers cover the entire surface).

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However, I think these are essentially the only restrictions, ie, for the full group of automorphisms of F_e there are exactly two orbits, the negative curve and the complement. Is this right? I don't have a reference at hand. –  quim Feb 28 '12 at 9:45

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