Jesko Hüttenhain
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 Jan 19 comment Equivalent definition of almost geometric quotient @Rise: I thought I had a more geometric proof using your definition, but it doesn't quite work. Maybe I will come up with something better. Jan 19 revised Equivalent definition of almost geometric quotient rolled back to a previous revision Jan 19 revised Equivalent definition of almost geometric quotient deleted 285 characters in body Jan 19 comment Equivalent definition of almost geometric quotient Any nonempty open subset of an irreducible variety is dense. Jan 19 comment Equivalent definition of almost geometric quotient @Rise: Well, notation would be lighter if $X$ was affine, and the problem is trivial if $X$ is irreducible. Is any of that the case? If not, then at some point I am sure you will have to use the fact that the quotient is good. What is your definition of a "good" quotient, if not via $G$-invariant sections? Jan 19 answered Equivalent definition of almost geometric quotient Jan 18 answered Proving that $A^{T}A = M$ for all symmetric complex matrices $M$. Jan 16 comment Equivalent definition of almost geometric quotient For (a): You have that $U_0=\pi^{-1}(U)$. Let $x\in U_0$ and $y\in \overline{Gx}$ (the closure in $X$). Then, by definition of the quotient, $\pi(y)=\pi(x)$ and therefore you have $y\in\pi^{-1}(\pi(x))\subseteq \pi^{-1}(U)$. Since $y$ was arbitrary, you have $\overline{Gx}\subseteq U_0$ and since $Gx$ was already closed in $U$, we have $\overline{Gx}=Gx$. I'm a bit stuck with (b) right now, and I have an appointment in 5 ... I'll look at it again soon. Jan 16 comment Equivalent definition of almost geometric quotient What is your base field? Is $G$ assumed connected? Dec 22 answered Closed orbits for reductive group actions Dec 21 revised Can every variety appear as singular locus? LaTeX Dec 20 comment How to describe the points of a quotient stack? The image of a $G$-equivariant map from $G$ is certainly always an orbit. Let $\alpha:G\to X$ be such a map, let $x:=\alpha(1)$ and then you have $\alpha(g)=\alpha(g\cdot 1)=g.\alpha(1)=g.x$ for all $g\in G$. Dec 17 comment Can every variety appear as singular locus? If $V$ is smooth, the answer is easily yes: Let $C$ be a curve with a unique singular point $p$. Then, the singular locus of $C\times V$ is equal to $\{p\}\times V\cong V$. It's not that easy when $V$ itself is singular, though. Dec 5 comment Restriction of closed immersion to closed subset is a closed immersion Since the inclusion $\iota: Z\to X$ is a closed immersion, it induces a surjection $\mathcal{O}_{X,\iota(z)}\twoheadrightarrow\mathcal{O}_{Z,z}$ for a point $z\in Z$. You furthermore know that $f$ induces a surjection $\mathcal O_{Y,f(z)}\twoheadrightarrow\mathcal O_{X,z}$ for any $z\in X$, in particular for $z=\iota(z)\in Z$. The composition of these two surjections is the local map induced by $f\circ\iota$, and it is a surjection as well. Does this answer the question? Dec 3 awarded Popular Question Nov 12 answered Why is the secant variety a variety? Nov 12 comment Why is the secant variety a variety? @kaiser: I think Hoot's question is relevant. Is a variety always irreducible by your definition? It is an unfortunate communication barrier in algebraic geometry that "variety" is not entirely unambiguous. Nov 12 comment Why is the secant variety a variety? It is obvious that this is a variety because you take the closure. By definition, the result is a Zariski closed subset of projective space. Nov 3 answered Categorical Quotient Oct 21 answered Does the character with the following properties exist?