# Tag Info

23

Chevalley's theorem (see Theorem 4.19 in Milne) is very close to this. It says Let $G$ be a linear algebraic group and let $H$ be a Zariski closed subgroup. Then there is a representation $V$ of $G$ and a one dimensional subspace $L$ of $V$ such that $H$ is the stabilizer of $\mathbb{P}(L)$ in the action of $G$ on $\mathbb{P}(V)$. So this is close to what ...

17

One example is provided by the generalized Jacobian. For a smooth projective curve $C$ and a divisor $D$ on $C$, the generalized Jacobian is defined to be the moduli space parameterizing pairs consisting of a line bundle of degree $0$ on $C$ together with a trivialization of that line bundle over $D$. This admits a map to the usual Jacobian, whose kernel is ...

14

No. Let $C \subset \mathbb{P}^2$ be a smooth sextic curve, let $X$ be the double covering of $\mathbb{P}^2$ ramified over $C$, and let $G = \mathbb{Z}/2$ acting on $X$ be the involution of the double covering. Then $X$ is a K3 surface, hence its canonical bundle is trivial. But $X^G = C$ is a curve of genus $10$, its canonical class is not trivial.

8

We can generalize Sasha's counterexample in all dimensions as follows. Let $f \colon X \to \mathbb{P}^n$ be a double cover branched over a smooth hypersurface $Y=Y_{2n+2}$ of degree $2n+2$. Then $$\omega_X = f^*(\omega_{\mathbb{P^n}} \otimes \mathcal{O}_{\mathbb{P}^n}(n+1)) = \mathcal{O}_X$$ is trivial. On the other hand, the double cover involution gives ...

5

It is equal. Since the ground field is of characteristic zero and $v$ is symmetric one may as well compute the stabilizer of the polynomial $f:=\sum_{i=1}^nx_i^m$. Assume $g\in\mathrm{GL}_n(\mathbb C)$ stabilizes $f$. Then it also stabilizes the Hessian $$\det\nolimits_{ij}(\partial_i\partial_jf)\in\mathbb C^*(x_1\cdots x_n)^{m-2}$$ up to a factor. Hence it ...

4

(Comment converted to answer per request:) The “surprising result” about simply connected homogeneous spaces in your (currently) last paragraph is Montgomery's Theorem (1950): generally (in your notation) if $G/H$ is compact and $H$ closed connected, then $K$ is transitive on $G/H$. The theorem is also discussed in Samelson (1952, p. 17).

4

For a non-projective example, you could take your favorite smooth variety $Y$ with nontrivial canonical bundle, and let $X$ be the total space of the cotangent bundle of $Y$. Then I believe $X$ will have trivial canonical bundle. You can let $\mathbb{G}_m$ act by scaling on the fibers of $X$. The $\mathbb{G}_m$-fixed points will be the original variety $Y$ (...

3

I think the answer to Question 1 is yes, since under your hypothesis the set-theoretical quotient is essentially the GIT quotient. This follows from the fact that $G$-invariant regular functions on $X$ separate orbits (when these are closed). There is an analytic proof of this fact. Let $V$ and $W$ be two disjoint closed orbits. Then 1 can be written as the ...

2

Let $X$ be a complex affine algebraic set with the analytic topology and let $G$ be a complex affine algebraic group acting rationally on $X$. $\newcommand\sslash{/\hspace{-0.2ex}/}$Let $X\sslash G=\operatorname{Spec}_\text{max}\mathbb{C}[X]^G$ be the affine GIT quotient of $X$ by $G$. Give $X\sslash G$ the analytic topology. So $X\sslash G$ is necessarily ...

2

The answer is true if $M$ is compact with finite fundamental group. A theorem of Lichnerowicz says that the isometries of a riemannian manifold $M$ is a Lie group $G$. If $G$ acts transitively on $M$, there exists a compact subgroup of $H$ of $G$ which acts transitively on $M$ (your previous question). A Lie compact group is a subgroup of $SO(n)$. https://...

1

Answer is yes. By theorem of Mostow mentioned in this question Homogeneous manifold deformation retracts onto compact submanifold source is "Covariant Fiberings of Klein spaces" Mostow 1955 If G and G' both have finitely many connected components (for example if they are algebraic groups) then G/G' is a vector bundle over K/K' where K and K' are ...

1

The answer of your question is true if the orbit $O_v$ is simply connected. Montgomery, D. Zippin, L. Topological transformations groups p. 226

1

Here is a geometric proof which shows that the fundamental group of a linear group orbit must be finite by abelian (confirming the most general form of the two answers given above). By theorem of Mostow mentioned in this question Homogeneous manifold deformation retracts onto compact submanifold source is "Covariant Fiberings of Klein spaces" ...

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