Example of linear algebraic group which has unique rational point

Question

Sorry for my bad English.

For a field $k$, I want to find examples of linear algebraic groups over $k$, which has a unique rational point (it is just unit element $e$).

Of course trivial group is one such example, so I want to know non-trivial such groups especially over $k=\mathbb{R}$.

Answer 1

Lukas Heger's example is good, and I want to point out that it is, in some sense, the only type of example.

Fact(cf. [1, Theorem 17.93]: Let $G$ be a connected linear algebraic group over a characteristic $0$ field $k$. Then, $G(k)$ is Zariski dense in $G$.

From this, we see that if $G$ is a connected linear algebraic group with $G(\mathbb{R})=\{e\}$ then

$$G=\overline{G(\mathbb{R})}=\overline{\{e\}}=\{e\},$$

and so $G$ is trivial. Thus, we see that if $G$ is a linear algebraic group with $G(\mathbb{R})=\{e\}$ then $G^\circ$ is trivial. Thus, $G$ is a finite group scheme and, as we are in characteristic $0$, this implies that $G$ is a finite etale group scheme.

Such finite etale group schemes are classified by a pair $(M,\sigma)$ where $M$ is a finite abelian group and $\sigma$ is an order two element of $\mathrm{Aut}(M)$. For instance, in Lukas Heger's example the pair corresponds to $(\mathbb{Z}/3\mathbb{Z},[-1])$ (where $[-1]$ means multiplication by $-1$).

Evidently for such a pair $(M,\sigma)$ the corresponding finite etale group $\mathbb{R}$-scheme $G$ satisfies $G(\mathbb{R})=M^\sigma$ (the fixed points of $\sigma$). Thus, it is easy to produce many examples of the phenomena you are looking for. They are precisely classified by pairs $(M,\sigma)$ where $M$ is a finite abelian group and $\sigma$ is an order $2$ automorphism of $M$ such that $M^\sigma=\{e\}$.

References:

[1] Milne, J.S., 2017. Algebraic groups: the theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.

Answer 2

For $k=\Bbb R$, we can consider $\mu_3=\mathrm{Spec}(\Bbb R[x]/(x^3-1))$. This affine (and hence linear) algebraic group has the property $\mu_3(A)=\{a \in A\mid a^3=1\}$ for any $\Bbb R$-algebra $A$. Thus $\mu_3(\Bbb R)=\{1\}$.