Tag Info

Hot answers tagged

8

This follows from Sard's theorem. If $G \times M \to M$ is smooth, then in particular $G \times \{pt\} \to M$ is smooth. If $\operatorname{dim} G < \operatorname{dim} M$, then Sard's theorem implies the image has measure zero, and in particular is not all of $M$. Compactness is inessential here.


4

It's a group. The author explains why for every $z$ there exists a group element mapping $i$ to $z$. Composition of one of these maps with the inverse of another shows that for every $z$ and $w$ there is a group element mapping $w$ to $z$.


3

The definition of isotropic (which I found here) ensures only the existence of some isometries satisfying the conditions. Since $x$ and $y$ are isometries, so is $xyx^{-1}$. There is no requirement that the set of isometries that you "use" to satisfy these conditions (which is not necessarily unique) have any specific topology or even be closed under ...


3

First suppose that $\Omega$ is finite, and so $G/K$ is finite, where $K$ is the kernel of the action, so we can assume that $K=1$ and $G$ is finite. For any $\alpha \in \Omega$, there is a nontrivial $p$-subgroup $Q(\alpha)$ of $G$ with the unique fixed point $\alpha$. Let $P(\alpha)$ be a Sylow $p$-subgroup of $G$ containing $Q(\alpha)$. Since the orbits ...


2

Here is a counterexample where $H$ can be taken as the full automorphism group of $K$. Necessarily, $K$ is nonabelian. Namely, let $$ K = \left\{ \begin{pmatrix} a & x \\ 0 & 1 \end{pmatrix} \mid a, x \in \mathbb{F}_{16}, a^3 = 1 \right\}. $$ ($K$ is a semidirect product of $C_3$ ...


2

You're absolutely right. Without using the fact that the full isometry group is a Lie group, I think it would be very difficult to show that there's a Lie group containing all the elements of the form $x^{-1}yx$ for $x\in G$ and $y\in H$. I think the Wikipedia definition pointed out by @MattSamuel has it right: There's no good reason to include the ...


1

Let $p$ be a point in $M$. Since $M$ is Hausdorff and $G$ is finite, there is an open set $U$ such that $p\in U$ and $gp\not\in U$ for all $g\in G\setminus\{1\}$. Since $M$ is regular, there is an open set $V$ such that $p\in V\subseteq\overline V\subseteq U$. Now the set $W=V\setminus\bigcup_{g\in G\setminus\{1\}}g(\overline V)$ is open, contains $p$. What ...


1

Since $|G|=32 = 2^5$, every element of $G$ must have a power of $2$ as its order. Consider the isotropy group $$G(x) = \{g\in G : g(x)= x\}$$ This is a subgroup of $G$, for if $g,h\in G(x)$ then $g(h(x))=g(x)=x$. If we let $G_x$ denote the orbit of $x\in X$ then we have $|G_x|\cdot |G(x)| = |G|$. See http://mathworld.wolfram.com/GroupOrbit.html and ...


1

This set is not necessarily finite or countable. For example, let $\mathbb{Z}/2\mathbb{Z}$ act on $\mathbb{R}^2$ by$$(x,y)\mapsto(-x,y).$$The set of points with non-trivial isotropy group is $\{x=0\}$. There is, however, something else that can be said. Given a compact Lie group action on a manifold, a fundamental theorem says that the set of points of a ...


1

The Chevalley–Shephard–Todd theorem implies that if a finite group acting $G$ linearly (and faithfully, w.l.o.g) on $A^n$ has smooth quotient $A^n/G$, then $G$ is generated by pseudoreflections and the quotient is $A^n$. If we know further that $G$ is cyclic, then there is a generator $g$ of $G$ which is a pseudoreflection. Up to linear change of ...


1

This question doesn't make sense to me. How can the action of a subgroup $H$ on another subgroup $K$ be any sort of transitive on the parent group $G$? You've not even defined it as an action on $G$. The best sense I can make of the question is easily seen to be false. Suppose $K$ is cyclic of order two and $H=\mathbb Z_p\times\mathbb Z_p$ for $p\geq 2$ ...



Only top voted, non community-wiki answers of a minimum length are eligible