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Feb
2
comment Quotient $G \to G/N$ induces quotient $H \to H/N$ by restriction?
@MatthiasKlupsch: By $H/\!\!/N$, I mean the GIT quotient. It is precisely $\operatorname{Spec}(k[H]^N)$, i.e. the variety with coordinate ring $k[H]^N$. This is well-defined only if $N$ is reductive because a Theorem by Hilbert asserts that $k[H]^N$ is finitely generated in this case. The image of a morphism between affine varieties is not closed in general. You are right, however, the main argument is $k[G]^N\to k[H]^N$ being surjective, not the image of $\varphi$ being closed.
Feb
1
comment Quotient $G \to G/N$ induces quotient $H \to H/N$ by restriction?
This is true when $N$ is reductive: Since $N\subseteq H$, you know that $H$ is an $N$-invariant, closed subset of $G$. Therefore, $\pi_G(H)$ is a closed subset of $H/\!\!/N$. Since the image of $\varphi$ is equal to $\pi_G(H)$, the image of $\varphi$ is closed. Hence, $\varphi$ is indeed an isomorphism onto its image. The fact that the restriction $k[G]^N\to k[H]^N$ is a surjection follows by applying the Reynolds operator. I have no counterexample for $N$ nonreductive, off the top of my hat, but it feels to me like there might be one.
Jan
25
comment Group isomorphisms don't preserve everything?
You would want that there is an isomorphism $\phi\colon G\to H$ with $\phi(N)=K$, i.e. your two isomorphisms should be compatible.
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
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
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
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
20
comment When does “closed immersion” pass to the quotient?
If $\dim(Y)=\dim(X)-q$ and $\dim(H)=\dim(G)-p$ then there will be examples where $\dim(Y/H)=\dim(X)-\dim(G)+p-q$ and $\dim(X/G)=\dim(X)-\dim(G)$ (and everything remains a variety). So, this feels to me like you'd need at least $q>p$ to make this work. However, even much more pathological things can happen.
Dec
19
comment When does “closed immersion” pass to the quotient?
Unlikely to be true if formulated this way. Choose $H=\{1\}$, then your morphism $\bar f\colon Y\to X/G$, dimension alert: Choose $Y=X$ and $G$ acting transitively on $X$, then you'd have $\bar f: X\to\{\ast\}$ which will in general not be injective.
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?
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.
Oct
5
comment Algebra of invariants finitely generated
Definitely true if $k$ is of characteristic zero. A reference would be 27.4 in the book Lie algebras and algebraic groups by Tauvel & Yu.
Oct
5
comment Prove of Sheafication of sheaf is a sheaf isomorphism
In other words, what is your definition of sheafification? Also, to be on the safe side, how do you define presheaf and sheaf? =)
Sep
24
comment Euler character of etale finite cover
@poorna: sounds like a tame situation, but I don't know. Euler characteristic is sensitive to removing a finite set of points, so I am unsure. I think it might be best to ask this as a separate question, I am sure someone will know better.
Sep
23
comment Euler character of etale finite cover
@poorna: Ok, I was a bit slow there. But assuming that $\pi$ is etale outside a finite set of points does not even guarantee that the morphism is finite, it could for instance be the blow-up of a singular surface. I suspect one might be able to construct counterexamples this way.
Sep
23
comment Euler character of etale finite cover
@poorna: We need Hirzebruch-Riemann Roch only for the Bundle $\mathcal O_{\tilde X}$ on the smooth variety $\tilde X$, which works. The Lemma requires the morphism $\pi:\tilde X\to X$ to be flat and finite, according to Fulton. I would assume that this condition can not be left out. Flatness is implicit if both $X$ and $\tilde X$ are regular. So, I think it works if you can show that $\pi$ is flat. I do not know off the top of my hat if your conditions imply that, though. Flatness is quite hard to grasp for me. If something comes to mind, I will post it.