Let G be a connected unipotent k-group acting on a k-variety V transitively and k-morphically. Is it possible that V(k) is empty?

  • $\begingroup$ What is $V(k)$? It usually denotes $k$-rational points. $\endgroup$ – Jesko Hüttenhain Jan 22 '18 at 23:40
  • $\begingroup$ yes...V(k) is k-rational points of variety V. $\endgroup$ – user371913 Jan 24 '18 at 8:40

This might be a somewhat silly example: Take $G$ to be any unipotent group and consider the $\Bbb R$-variety $V:=\operatorname{Spec}(\Bbb C)$, which clearly satisfies $V(\Bbb R)=\emptyset$. If we let $G$ act on $V$ trivially, it nevertheless acts transitively and $\Bbb R$-morphically. The fundamental issue is visible however: Given a $K$-variety $V$ with transitive $K$-morphic $G$-action, we may consider $V$ as a $k$-variety for a proper subfield $k\subsetneq K$. Since $V(k)=\emptyset$ (there can be no $k$-morphism $K\to k$), this is also an example because the action of $G$ remains transitive and is also $k$-morphic because it is $K$-morphic.

So for a more sophisticated example, take the $\Bbb R$-variety $V=\operatorname{Spec}(\Bbb C[x])=\Bbb A_{\Bbb C}^1$ and $G=\Bbb G_a$, i.e. $G=V$ acting on itself by addition.

  • $\begingroup$ In last part V(R) is non empty. $\endgroup$ – user371913 Jan 25 '18 at 16:11
  • $\begingroup$ An $\Bbb R$-rational point of $V$ is a morphism $\operatorname{Spec}(\Bbb R)\to\operatorname{Spec}(\Bbb C[x])$, which corresponds to a ring homomorphism $\Bbb C[x]\to\Bbb R$. Such a thing does not exist, because it would restrict to a ring homomorphism $\Bbb C\to\Bbb R$. $\endgroup$ – Jesko Hüttenhain Jan 26 '18 at 8:01

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