# Is a connected algebraic group of dimension 2 necessarily commutative?

There is a conclusion for a connected algebraic group $G$:

If dim$G \leq 2$, then $G$ is solvable.

I am wondering whether the stronger statement is true:

If dim$G \leq 2$, then $G$ is commutative.

If dim$G=0$, then $G=\{ e\}$. This is clearly commutative.

If dim$G=1$, then $G \cong G_a$ or $G_m$, thus is commutative.

If dim$G=2$, suppose that $x_1, x_2$ span the vector space containing $G$ over $K$. In $G$, the identity element $e =ax_1+bx_2$, $a,b \in K$ and are not both $0$. Then,

$x_1=ex_1=(ax_1+bx_2)x_1=ax_1^2+bx_2x_1$, (1)

$x_1=x_1e=x_1(ax_1+bx_2)=ax_1^2+bx_1x_2$, (2)

$x_2=ex_2=(ax_1+bx_2)x_2=ax_1x_2+bx_2^2$, (3)

$x_2=x_2e=x_2(ax_1+bx_2)=ax_2x_1+bx_2^2$. (4)

From (1) and (2), $b(x_1x_2-x_2x_1)=0$. Similarly, $a(x_1x_2-x_2x_1)=0$ from (3) and (4). So, if $x_1x_2 \neq x_2x_1$, then $a=b=0$, contradiction. And $G$ is commutative from the commutativity of $x_1$ and $x_2$.

Is this alright?

Thank you very much.

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What vector space containing $G$ over $K$? And I am pretty sure the affine group of transformations of the form $x \mapsto ax + b$ is a two-dimensional connected algebraic group. –  Qiaochu Yuan Aug 4 '11 at 4:51
@Qiaochu: I considered $\mathbb{A}^n$ as a vector space. But thanks to your example, I see I was wrong. When the production is defined to construct the group structure, the addition and the production do not always obey the distributive law. Thank you very much. –  ShinyaSakai Aug 5 '11 at 3:02
I don't understand what you mean by that. What I'm saying is that to say that $G$ has dimension $2$ does not imply that $G$ can be embedded into a $2$-dimensional vector space. –  Qiaochu Yuan Aug 5 '11 at 3:07
@Qiaochu: I know what you mean. I may be wrong. I am considering about affine algebraic varieties, and it is a closed subvariety of $\mathbb{A}^n$. Is it wrong to consider $\mathbb{A}^n$ as an $n-$dimensional vector space over the base field $K$? As in the case of affine transformations in your example, $f: x \mapsto ax+b$ $(a \neq 0)$, can be denoted as $(a,b)$. For any two affine transformations $(a_1,b_1)$ and $(a_2,b_2)$, if $(a_1,b_1)+(a_2+b_2): x \mapsto (a_1,b_1)x+(a_2,b_2)x$, then $(a_1,b_1)+(a_2+b_2)=(a_1+a_2,b_1+b_2)$. –  ShinyaSakai Aug 5 '11 at 3:23
Why do you think a variety of dimension $d$ is necessarily a subvariety of $\mathbb{A}^d$? And why do you think an algebraic group is necessarily affine? –  Qiaochu Yuan Aug 5 '11 at 3:29

## 1 Answer

Consider $H=T(2, K)$ and let be G the subset of H given by the condition $x_{11}=1$.

Then $G$ is a closed subgroup of H and you can verify its not commutative.

For example take the matrices $a=\begin{pmatrix}1&-1\\0&1\end{pmatrix}$ and $b=\begin{pmatrix}1&1\\0&-1\end{pmatrix}$.

Is also clear that $dim\ G=dim\ H -1=2$.

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