I'm trying to understand forms of algebraic groups. There is a theorem that says that if $L/K$ is a finite Galois extension, then the $K$-isomorphic $L/K$-forms of an algebraic group $X$ over $k$, are in bijection with the first cohomology set $H^1(Gal(L/K),Aut_L(X))$, where Aut$_L(X)$ denotes the group of $L$-defined automorphisms of $X$. I'm trying to work this out for some examples. For example, if $X=\mathbb{G}_m$, the multiplicative group, then for any field $L$, Aut$_L(\mathbb{G}_m) \cong \{\pm 1\}$. According to Springer's book Linear Algebraic Groups, this implies that the classes of $\mathbb{G}_m$ correspond to quadratic extensions of $K$ contained in $L$. Could someone explain how this works?

Here's what I understand from how the forms are constructed from classes of cocycles. Suppose we are given a cocycle $a \in Z^1(Gal(L/K),Aut_L(X))$. This is by definition a map $a:Gal(L/K) \rightarrow Aut_L(X)$ such that for all $s,t \in Gal(L/K)$, $a(st)=a(s)s\cdot a(t)$. Denote the cohomology class by [a]. If $\sigma \in Gal(L/K)$, denote the image of $\sigma$ by $a_\sigma$. Then another notation for $a$ is $\{a_\sigma\}$. We want to construct a form from $X$, in other words, some algebraic group $Y$ over $k$ such that there exists an $L$-defined isomorphism $X \rightarrow Y$.

We can identify Aut$_L(X)$ with the automorphism group of the Hopf algebra $\mathcal{O}[X_L]=L \otimes_K \mathcal{O}[X]$. Then we define an action of Gal$(L/K)$ on $\mathcal{O}[X_L]$ by $^\sigma a = a_\sigma \circ \sigma a$. Then consider the set $B=\{a \in \mathcal{O}[X_L]:^\sigma a = a\} \subset \mathcal{O}[X_L]$. Then $L \otimes_K B \cong \mathcal{O}[X_L]$ and $B$ is a Hopf algebra. $B$ is the coordinate ring of $L$-defined functions on some algebraic group, denoted $_a X$, which is said to be obtained from $X$ by twisting using $a$. It depends only on the class of $a$.


For the multiplicative group, here is how it works (at least if $K$ is not of characteristic 2, I don't want to take any risk). There are the so called non split tori of rank 1. They are defined as follow : choose $a\in K^*$ and define $T_a\subset\operatorname{SL}_2(k)$ to be the set of matrices of the form $\left( \begin{array}{cc} x & ay \\ y & x \end{array} \right)$ such that $x^2-ay^2=1$.

For example, if $K=\mathbb{R}$ and $a=-1$, you get the circle.

You can prove that, if $a$ is a square in $K$, then $T_a$ is isomorphic to $\mathbb{G}_m$, however if $a$ is not a square, you get a new group.

More generally, $T_a\simeq T_b$ if and only if $a/b$ is a square in $K$. So the set of groups of the form $T_a$ up to isomorphism is isomorphic to $K^*/{K^*}^2$. And this set is isomorphic to the set of quadratic extensions of $K$.

Of course, on $K[\sqrt{a}]$, $a$ becomes a square and so $T_a\otimes_K K[\sqrt{a}]\simeq\mathbb{G}_m$. In other words, $T_a$ is a form of $\mathbb{G}_m$ and becomes isomorphic to $\mathbb{G}_m$ in an extension $L$ if and only if $a$ is a square in $L$.

In fact, every form of $\mathbb{G}_m$ is isomorphic to $T_a$ for some $a\in K$. So the forms of $\mathbb{G}_m$ is in bijection with $K^*/{K^*}^2$ and with the set of quadratic extension of $K$. Those that are split in $L$ correspond to the classes $a$ such that $a$ is a square in $L$ and to the quadratic extensions contains in $L$.

Now, as you said, $\operatorname{Aut}_L(\mathbb{G}_m)=\{\pm1\}$ (with trivial $\operatorname{Gal}(L/K)$-action). So $$H^1(\operatorname{Gal}(L/K),\operatorname{Aut}_L(\mathbb{G}_m))=H^1(\operatorname{Gal}(L/K),\{\pm1\})=\operatorname{Hom}(\operatorname{Gal}(L/K),\{\pm1\}).$$ Thus the class of a form of $\mathbb{G}_m$ corresponds to a morphism $\varphi\in\operatorname{Hom}(\operatorname{Gal}(L/K),\{\pm1\})$. Now such a $\varphi$ is completely determined by $\ker\varphi$ which is a subgroup of $\operatorname{Gal}(L/K)$ and thus corresponds to an extension of $K$ in $L$ by the main theorem of Galois theory. It can be two different things :

  • either $\ker\varphi=\operatorname{Gal}(L/K)$ ($\varphi$ is trivial). The corresponding extension is of course $K$ itself. And the form of $\mathbb{G}_m$ is $\mathbb{G}_m$ itself.

  • or $\ker\varphi$ is a subgroup of index two in $\operatorname{Gal}(L/K)$. The corresponding extension is a quadratic extension of $K$ contain in $L$. It is of the form $K[\sqrt{a}]$. And the form of $\mathbb{G}_m$ is isomorphic to $T_a$.

  • 1
    $\begingroup$ I'd say with $\varphi({\scriptstyle\begin{pmatrix}u & av \\ u & v \end{pmatrix}}) = u+\sqrt{a}v$ we get an isomorphism from the matrix field $K({\scriptstyle\begin{pmatrix}x & ay \\ y & x \end{pmatrix}}), x \in K,y \in K^* $ to $K(\sqrt{a})$. So clearly $\{T_a, a \in K^*/ {K^*}^2\}$ is isomorphic to the set of quadratic extensions of $K$. $\endgroup$ – reuns Dec 29 '16 at 12:59
  • $\begingroup$ @user1952009 Yes good point. $\endgroup$ – Roland Dec 29 '16 at 15:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.