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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
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Jan
19
revised Equivalent definition of almost geometric quotient
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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?