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7

Let $z_1, z_2, z_3, z_4 \in \Bbb P ^1 (\Bbb C) = \Bbb C \cup \{ \infty \}$, all of them distinct. The cross-ratio is essentially the only projective invariant of this system of points. More rigorously, let $$X = \{ (z_1, z_2, z_3, z_4) \in \big( \Bbb P ^1 (\Bbb C) \big) ^4 \mid z_i \ne z_j \; \forall i \ne j \} ;$$ then $F : X \to \Bbb C$ is invariant ...


2

The cross-ratio is the unique rational function of four complex numbers up to multiplication by a constant such that the cross-ratio of four distinct points is real if and only if the points are concyclic.


2

Suppose that $(x:y:z) \in \mathbf{Z}(XZ - Y^2)$. Then we know that $xz - y^2 = 0$, i.e. $$ y^2 = xz $$ Since $k$ is algebraically closed by hypothesis $\sqrt{x},\sqrt{z} \in k$, thus $$ (x:y:z) = (x : \sqrt{x} \sqrt{z} : z) \in V $$ so $\mathbf{Z}(XZ - Y^2) \subseteq V$ and we're done.


2

1) That is correct. 2) A few things. First, there is no intrinsic $\mathcal O_X(1)$, and once the surface is embedded, then clearly not every curve is a hyperplane section of the surface (this isn't even true in $\mathbb P^2$ for instance). More important, though, is that not every curve is even a candidate. For example, a single ruling on a smooth quadric ...


1

This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Dmitri below. What you want almost works, but with a little twist. The group $SO(3)$ is acting on $F_2$, so that there are two orbits that are $\mathbb CP^1$ and all other orbits are $SO(3)$. More precisely, $F_2$ can be seen as a compactification of ...


1

Just a note, lots of number theory of quadratic forms happens because all null vectors of your form, with rational integer entries, are integer multiples of $$ \left( u^2 , u v, v^2 \right) $$ with $\gcd(u,v)=1.$ I'm just sayin'


1

Hint Since (1) collinearity is preserved by affine transformations and (2) one can choose an affine transformation that maps $A, B, C$ as given resp. to, e.g., $(0, 0), (1, 0), (0, 1)$, we may as well assume that $A, B, C$ are those points. (To see the latter, note that we can apply the translation that maps $A$ to the origin and then apply the linear ...


1

For 2), let $\pi:X=\mathbb{P}(E)\to S$ be the natural map. We have a canonical surjection $\pi^*E\to\mathcal{O}_X(1)$ and thus a map by composition $\pi^*E'(-1)\to\mathcal{O}_X$. The image is the ideal sheaf defining $Y=\mathbb{P}(E")$.



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