Posted by Jason Polak on 29. March 2016 · Write a comment · Categories: math

Let $V$ be a variety over a field $k$ with separable closure $\bar k$. A $\bar k/k$-form of $V$, or simply form, is a variety $W$ such that $W_{\bar k} = V_{\bar k}$, where $\bar k$ is the separable closure of $k$. If $V$ is quasiprojective then then forms of $V$ are in bijection with the Galois cohomology set $H^1(k,\mathrm{Aut}(V))$. Here, the Galois action on $\mathrm{Aut}(V)$ is given by $\sigma\circ f = (1\otimes \sigma^{-1})f(1\otimes\sigma)$.

Let’s illustrate this with an example: the multiplicative group $\G_m$. Here, we are looking for forms that are also algebraic groups. Are there any nontrivial ones? Yes indeed! Let $E/k$ be a quadratic extension with $\langle\sigma\rangle = \mathrm{Gal}(E/k)$. Then the algebraic $k$-group $F$ defined for each $k$-algebra $R$ by

$$F(R) = (\mathrm{Res}^1_{E/k}\G_m)(R) = \{ x \in E\otimes_k R : x\sigma(x) = 1\}$$

is a form of $\G_m$, because $F_E\cong \G_{m,E}$. It’s also not isomorphic to $\G_m$ over $k$ (why? though we’ll see one reason why later). In any case, we’ve found two kinds of one-dimensional tori: $\G_m$, and $F(R)$ for a quadratic extension $E/k$. Are there any other ones? In this post we’ll classify all one-dimensional tori.

The Cohomology Group

To figure out the answer we need to compute $H^1(k,\mathrm{Aut}(\G_m))$. First, what are the automorphisms of $\G_m$? It’s not difficult to figure out that the homomorphisms correspond the $k$-algebra homomorphisms $k[t,t^{-1}]\to k[t,t^{-1}]$ given by $t\mapsto t^n$; therefore, the automorphisms correspond to $t\mapsto t^{\pm 1}$. So, $\mathrm{Aut}(\G_m)\cong\Z/2$. Furthermore, since the automorphisms don’t involve coefficients and won’t for any field extension, the Galois action on $\Z/2$ is trivial. (By what we’ve said in this paragraph we see that $\mathrm{Aut}(\G_m^n)\cong\GL_n(\Z)$ and the action of the Galois group is trivial on this group — though, figuring out all the tori here gets much more complicated!)

So, since the action of $\Gamma = \mathrm{Gal}(\bar k/k)$ is trivial, we have to determine the continuous homomorphisms $\Gamma\to \Z/2$; if such a homomorphism is nontrivial then it surjects so all of them are determined by quadratic extensions. Therefore, all the nontrivial forms of $\G_m$ split over a quadratic extension. Let $E/k$ be such an extension with Galois group $G$ and $\sigma$ the nontrivial Galois automorphism.

Calculating the Form

There is an action of $G$ on $\G_m\times_k\mathrm{Spec}(E) = \G_{m,E}$ that is twisted by the cocycle $a_\sigma$ defined by the homomorphism $G\to\Z/2$. The nontrivial element $\sigma$ acts by $a_\sigma(1\otimes\sigma)$. The nontrivial element of $\Z/2$ corresponds to inversion. Hence, the action of $\sigma$ on $\G_{m,E}$ is given by $x\mapsto \sigma(x)^{-1}$.

The theory of Galois descent then tells us that the map $X\to X\otimes_k E$ is an equivalence of categories between quasiprojective $k$-schemes and quasiprojective $E$-schemes together with a $\Gamma$-action, compatible with the $\Gamma$-action on $E$. This equivalence allows us to figure out the functor of points of the resulting form: it is the algebraic $k$-group that is defined on a $k$-algebra $R$ by
F(R) = \{ x\in \G_m(R\otimes_k E)^\times : x\sigma(x) = 1\}.
The condition $x\overline{x} = 1$ is just the $\Gamma$-equivariance condition imposed upon the morphisms in the category of quasiprojective $E$-schemes. Notice too how the theory of forms has provided a proof of the simple proposition that $F\not\cong\G_m$ as $k$-schemes – they can’t be because they correspond to distinct elements in the cohomology group $H^1(k,\Z/2)$.

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