most recent 30 from http://mathoverflow.net 2013-05-25T20:53:15Z http://mathoverflow.net/feeds/question/39155 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/39155/geometric-explanation-of-an-orbit-space-integer-action-on-the-affine-line philip314 2010-09-17T22:01:35Z 2010-09-17T23:19:20Z

<p>Let $k$ be a field of char $0$ and let $\mathbb{Z}$ act on $\mathbb{A}^1_k$ by the action induced by $G\to\mathrm{Aut}_k(k[X]), n\mapsto X+n$. It is rather easy to show that the orbit space $\mathbb{A}^1_k/\mathbb{Z}$ is just $\mathrm{Spec}(k)$. At least to me this is surprising at the moment since dividing out the analogue action of $\mathbb{Z}$ on the topological spaces $\mathbb{R}$ resp. $\mathbb{C}$ gives the sphere $S^1$ (up to homotopy). </p> <p><b>Q.:</b> Is that simply a pathological example of a categorical quotient or is there a geometric explanation why this is the right orbit space?</p>

http://mathoverflow.net/questions/39155/geometric-explanation-of-an-orbit-space-integer-action-on-the-affine-line/39167#39167 Heinrich Hartmann 2010-09-17T23:19:20Z 2010-09-17T23:19:20Z

<p>Your observation $\mathbb{A}^1/\mathbb{Z}=Spec(k)$ refers to the fact that there are no non-constant, $\mathbb{Z}$-invariant <strong>polynomial</strong> functions. On the other hand there are plenty $\mathbb{Z}$-invariant <strong>continues</strong> functions on $\mathbb{A}^1(\mathbb{R})$, hence you get a non-trivial quoteint.</p> <p>Also note that $\mathbb{Z}$ is no group-scheme on the nose. Taking the group ring yields $k[\mathbb{Z}] = k[t,t^{-1}]$ which is a torus. So to get a group-scheme $G$ with $G(k)=\mathbb{Z}$ you need to consider an infinite union of copies of $Spec(k)$, which would be not finitely generated.</p> <p>There is no good way to take quotients of schemes by infinite, discrete groups in general. For example there is no way to produce curves as quoteints of the upper half plane $\mathbb{H}/\Gamma$, or abelian varietis as quotients $\mathbb{C}^n/ \Lambda$, in the realm of scheme theory.</p>