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answered factors in polynomial rings with field coefficients
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
29
revised Cardinality of the conjugate class of $\sigma$
LaTeX
Jan
29
revised Closed form for a binomial identity another solution
LaTeX .. mostly.
Jan
29
answered $P$ is a Sylow subgroup of $G$ then $P$ is normal in $G$.
Jan
29
answered Inverse of the element in the multiplicative group
Jan
25
answered Group isomorphisms don't preserve everything?
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
awarded  Cleanup
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