Can the Hirzebruch Surface $F_2:=\mathbb{P}(\mathcal{O}\oplus \mathcal{O}(2))$ be obtained by some GIT quotient of $\mathbb{P}^4$ (or $\mathbb{C}^4$)?
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2$\begingroup$ GIT quotient can mean many things. But at least there are no surjective morphisms from $\mathbb{P}^4$ to $F_2$. $\endgroup$– MohanCommented Jul 27, 2017 at 15:05
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1$\begingroup$ In Audin's book she explains how to construct a projective toric variety, whose polytope has $n$ facets (4 in your case), as a GIT quotient of $\mathbb C^n$. Essentially it amounts to repeated use of symplectic cutting. $\endgroup$– Allen KnutsonCommented Aug 25, 2017 at 12:56
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$\begingroup$ @Allen Knutson which book is this? $\endgroup$– user100841Commented Aug 25, 2017 at 16:47
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$\begingroup$ Michele Audin, The Topology of Torus Actions on Symplectic Manifolds. $\endgroup$– Allen KnutsonCommented Sep 4, 2017 at 13:35
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1 Answer
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Depending on the interpretation of your question, the answer is Yes.
In fact the Hirzebruch surface $\mathbb{F}_n =\mathbb{P}(\mathcal{O}\oplus \mathcal{O}(n))$ is the quotient of $X = \mathbb{A}^2 \setminus \{0\} \times \mathbb{A}^2 \setminus \{0\}$ with respect to the group action $$ \mathbb{G}_m^2 \times X \to X, \quad (\lambda,\mu) \cdot (s,t;x,y) = (\lambda s, \lambda t; \mu x, \lambda^{-n} \mu y).$$
From a highbrow perspective, such a realisation exists as $\mathbb{F}_n$ is toric. (Every toric variety is a quotient of an open subset of an affine space by the action of some multiplicative group, by the theory of Cox rings).
answered Jul 27, 2017 at 15:21
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1$\begingroup$ As a comment: this is quite useful, as all automorphisms of $\mathbb{F}_n$ are in fact automorphisms of $X$ homogeneous for the action of $\mathbb{G}_m^2$. This allows then two write these globally. $\endgroup$ Commented Jul 27, 2017 at 18:36
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$\begingroup$ this is very useful ...but i wanted to realise as a quotient $\mathbb{A}^4$ by $\mathbb{C}\times \mathbb{C}^*$...this should be something modified GIT $\endgroup$– user100841Commented Aug 26, 2017 at 8:13