Projective line as a quotient by a torus - MathOverflow

Projective line as a quotient by a torus

2012-07-09T13:43:01Z

Let $k$ be a field, and let $T$ an $n$-dimensional split torus over $k$. Let $X$ be a $k$-scheme with algebraic $T$-action. Solve for X:

$$X / T \cong \mathbf{P}^1_k$$

(The quotient should be a categorical quotient in the category of $k$-schemes. A categorical quotient is a morphism $\pi : X \to X/T$ which is universal among morphisms which are constant on $T$-orbits.)

For example, $X = \mathbf{P}^1_k \times T$ with $T$ acting by right multiplication on the right factor is a trivial solution. More generally, I expect to find such $X$ among $(n+1)$-dimensional toric varieties.

Would someone explain how nasty this question actually is and what restrictions could be made:

on the field $k$ or
on the scheme $X$ or
on the quotient morphism $\pi : X \to X/T$ or
on the action morphism $\gamma : X \times T \to X$ or
on the category in which the quotient is taken or
on the "goodness" of the categorical quotient

in order to have a simple classification of the solutions? In other words, the solution to this problem will be itself a problem, a possibly modified version of the above admitting a simple solution. This might be a complex question (or not!), in which case a partial answer would still be interesting.

Answer by Jason Starr for Projective line as a quotient by a torus

2012-07-09T13:57:14Z

I recommend you read about "universal torsors" in the sense of Colliot-Th'el`ene and Sansuc. These are torsors for (not necessarily split) algebraic tori which are universal among all such torsors. A universal torsor does not always exist, but it does exist when there is a $k$-rational point (or even just a degree $1$, zero cycle). In the case of $\mathbb{P}^n_k$, a universal torsor is the following $\mathbb{G}_m$-torsor $U$, $$\mathbb{A}^{n+1}\setminus {0} \rightarrow \mathbb{P}^n.$$ For every algebraic torus $T$, for every $T$-torsor, $X\to \mathbb{P}^n$, there exists a unique group-scheme homomorphism $\rho:\mathbb{G}_m\to T$ such that $X$ is isomorphic, as a $T$-torsor over $\mathbb{P}^n$, to $(T\times U)/\Delta(\mathbb{G}_m)$, where $\Delta$ denotes the diagonal action of $\mathbb{G}_m$, via $\rho$ and the standard action on $U$. Thus, the torsors $X$ for your torus $T$ (up to isomorphism of $T$-torsor) are in one-to-one correspondence with the group-scheme homomorphisms $\rho:\mathbb{G}_m\to T$, i.e., the group of cocharacters of $T$.

Answer by Will Sawin for Projective line as a quotient by a torus

2012-07-09T18:08:09Z

$\mathbb P^1$ viewed as a toric variety has $3$ orbits. If we take a quotient of a $n+1$-dimensional toric variety by an $n$-dimensional subtorus of its torus, the number of orbits does not change, so we are looking at toric varieties with $3$ orbits. The fan of such a toric variety consists of the central point with two rays coming out. The subgroup you quotient by corresponds to a hyperplane in $\mathbb Z^{n+1}$. To map to the correct points of $\mathbb P^1$ in the quotient, the rays must lie on opposite sides of the hyperplane.

Thus, classifying the toric ways of viewing $\mathbb P^1$ as a quotient is equivalent to classifying pairs of slopes up to automorphism. If we hold the torus fixed then we hold the hyperplane fixed, so the only automorphisms left are shears that add a fixed integral slope to the slope of both rays, meaning that these are classified by $\mathbb Q^{2n}/\mathbb Z^n$.

If we require $T$ to act without stabilizer, then we are just classifying principal $T$-bundles, which are just classified $H^1(\mathbb P^1,\mathbb G_m^n)=H^1(\mathbb P^1, \mathbb G_m)^n=\mathbb Z^n$. We can construct these by taking line bundles, removing the zero section, and then taking the fiber product. Torically these correspond to pairs of rays with integral slopes.

For toric varieties with rational slopes, $T$ acts with stabilizer above $0$ and $\infty$.