Let us work over the complex numbers for simplicity. Let $M$ be an affine algebraic monoid and $X$ an affine variety on which $M$ acts regularly, i.e. there is a morphism $\alpha: M\times X\to X$. Let $x\in X$. To avoid missing any important conditions, let us for simplicity also assume that $M=\mathbb C^{n\times n}$ is the monoid of square matrices. Is $M.x$ an affine variety?
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$\begingroup$ The orbit need not be affine even if $M$ is an algebraic group. $\endgroup$– nafCommented Sep 19, 2013 at 10:20
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$\begingroup$ I suppose I need it to be reductive. I edited the question. I am more interested in the "good" cases, though, so let's assume that $M$ is just the square matrices. $\endgroup$– Jesko HüttenhainCommented Sep 19, 2013 at 10:35
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$\begingroup$ A constructible subset is in general not a variety, in any reasonable sense. $\endgroup$– Laurent Moret-BaillyCommented Sep 19, 2013 at 12:12
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2$\begingroup$ Possibly the result Jesko Hüttenhain is thinking of is that an orbit of a reductive group is affine if and only if its stabilizer subgroup is also reductive from Matsushima, Nagoya Math. J. 18 153-164 (1961) $\endgroup$– Steven SamCommented Sep 19, 2013 at 15:26
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2$\begingroup$ If $G$ is reductive, and its Borel acts with an open orbit, then $G$ (and its Borel!) act with finitely many orbits. $\endgroup$– Allen KnutsonCommented Sep 20, 2013 at 3:55
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2 Answers
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The orbits of a monoid aren't even varieties. Let $M$ be $\mathbb{A}^2$ with the monoid structure $(x_1, y_1) \cdot (x_2, y_2) = (x_1 x_2, y_1 y_2)$. Let $M$ act on $\mathbb{A}^2$ by $(x,y) \cdot (t,u) = (xt, xyu)$. Then $M \cdot (1,1)$ is the image of the map $(x,y) \mapsto (x,xy)$, which isn't a variety at all. Note that the unit group is $\mathbb{G}_m^2$, which is reductive.
answered Sep 19, 2013 at 13:56
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Let $SL_2$ act on ${\mathbb A}^2$. This has two orbits, $\{\vec 0\}$ and its complement, and the latter is not affine.
answered Sep 19, 2013 at 12:34
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1$\begingroup$ Conveniently the monoid of 2x2 matrices has 2 orbits on A^2, {0} and A^2. $\endgroup$ Commented Sep 19, 2013 at 13:02