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Let $X$ be the analytification of a variety $V$ such that $X$ is homeomorphic to $\mathbb R^2$.

What can we say about $X$? Can $V$ be singular?

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  • $\begingroup$ Yes, of course: the analytification of $\{(z,w)\in \mathbb{C}^2 : z^2 = w^3\}$ is homeomorphic to $\mathbb{R}^2$. $\endgroup$
    – Jason Starr
    Commented May 9, 2014 at 17:15
  • $\begingroup$ @JasonStarr Thank you. If we ask $V$ to be smooth, then $X$ is isomorphic to $\mathbb A^1$ I think. Is that true? $\endgroup$
    – Ste3an
    Commented May 9, 2014 at 17:29
  • $\begingroup$ Yes to the follow up question, because $\mathbb{A}^1$ can be characterized as the unique nonsingular simply connected curve which isn't compact. $\endgroup$
    – Donu Arapura
    Commented May 9, 2014 at 19:34

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