12

For circle foliations of compact $3$-manifolds, this is essentially answered by a theorem of Epstein: every such foliation is a Seifert fibration. Most Seifert fibrations are finitely covered by a product (surface)x(circle), and in these cases one of course has a free circle action. One of the exceptional cases of Seifert fibrations not covered by a ...


11

Dennis Sullivan constructed a non-vanishing Lispschitz vector field on a closed $5$-manifold such that all orbits are periodic but they amazingly have unbounded lengths (!). An addendum by Nicholas Kuiper shows that the flow can be chosen smooth, and another by William Thurston gives another construction which is real analytic.


8

Here is a picture of a train track, which approximates a lamination: Another picture: The leaf space of this lamination (obtained by collapsing each leaf and complementary polygon to a point) is an $\mathbb{R}$-tree. As you contract the leaves to points, you might get a picture looking somewhat like this: If you fill in with the boundary circle, you get ...


8

Setup Let $G$ be a finite group acting on a smooth projective variety $X$, and let $$ \rho: G \times X \to X $$ be the action morphism. For any $g \in G$ let $\rho_g$ denote the composition $$ X \simeq \{g\} \times X \subset G \times X \xrightarrow{\rho} X $$ The fixed point locus $X^G \subset X$ is a closed subscheme (possibly empty): Edit ...


7

I've never stumbled across a standard one in all of the literature, though I've definitely appreciated a remark by Ken Brown in his group cohomology bible: "The hypothesis that $G_\sigma$ fixes $\sigma$ pointwise is not very restrictive in practice. In the case of a simplicial action, for example, it can always be achieved by passage to the barycentric ...


6

It's enough to pick a contractible manifold $M$ with two non-homotopic actions. For example, let us pick $M=\mathbb{R}^2$ with two actions of $S^1$, the trivial one and the one given by rotations. These two actions are not homotopic. In fact for every action $\rho$ of $S^1$ on $\mathbb{R}^2$ we can consider the map $\mathbb{R}^2\times S^1\to GL_2^+(\mathbb{R}...


6

The answer to the edited question is no. Take a double covering $\pi :X\rightarrow \mathbb{P}^2$ branched along a smooth quartic curve $C$ (so $X$ is a Del Pezzo surface), and $G=\langle \sigma \rangle$, where $\sigma $ is the involution such that $\pi \circ\sigma =\pi $. Then $\operatorname{rk} H^*(X)=10$, but $\operatorname{rk} H^*(X^G)=\operatorname{rk} ...


5

There are examples of free actions on $\mathbb{R}^2$ where every orbit is discrete and closed but the action is not properly discontinuous and the quotient is non-Hausdorff. The example is rather standard. I will use $G\cong {\mathbb Z}$. Let its generator act on the punctured plane $P:=\mathbb{R}^2 - \{(0,0)\}$ via $$ (x,y)\mapsto (2 x, 2^{-1} y). $$ This ...


4

This is an attempt to understand the very nice argument by Peter Mueller (completed by OP, Sean Eberhard, in the comments). I am not completely happy with it, and would like to encourage others to look for more conceptual explanations. Denote $\Omega:=P^1\mathbb{F}_q=\mathbb{F}_q\cup \infty$, so that $|\Omega|=q+1$ and the group $G=PSL(2,q)$ acts on $\Omega$...


4

This is not a complete solution of the problem, but only an outline of a possible strategy, along with some numerical evidence: Let $S$ be a subset of $G=\operatorname{PSL}(2,q)$. For $g\in G$ let $P_g$ be the permutation matrix of $g$, and let $J$ be the all-$1$-matrix of the same size. Then regularity of $S$ is equivalent to $\sum_{g\in S}P_g=J$. Thus the ...


4

Yes, it's called effectively free (see for instance here, page 3, Fels/Olver, On relative invariants).


4

There is a paper by Stefan Waner from 1980, I think it's called "Equivariant Classifying Spaces", in which he proves an equivariant version of Milnor's theorem. It might do what you want.


4

Most of my work is with actions of compact Lie groups, so forgive me if I make any errors below related to other groups. I'm not sure how general a classification you're looking for, but I'm going to assume a framework in which I can say something: Consider a group $G$ acting on the base space $S^1$ by rotations, so the action is given by a group ...


4

Your definition implies that $$ \tilde S = \bigcup_{p\in M} T_pM^{G_p}. $$ In particular, $\pi(\tilde S) = M$, and $\tilde S$ will be a submanifold of $TM$ iff the dimension of $T_pM^{G_p}$ is the same for all $p\in M$. $G$ being connected won't necessarily make this happen: Another counterexample is $S^1$ acting by rotation on $S^2$, fixing the north and ...


4

Let $G$ be the fundamental group of the surface of genus 2. The group $Out(G)$ is the mapping class group $Mod_2$. Take a pseudo-Anosov $\phi$ from $Mod_2$. Then $G$ acts on the asymptotic cone $T$ of $G$ by $$g\cdot (g_1,...,g_n,...)=(\phi(g)\cdot g_1,...,\phi^n(g)\cdot g_n,....).$$That action is free by isometries and $T$ is an $\mathbb{R}$-tree. One ...


3

I believe, the examples you gave are essentially the only ones. Indeed, let $\rho\colon G\to O(d)$ be a faithful irreducible “strongly angle-preserving” representation. Claim. The image of $\mathbb{R}G$ under (the natural linear extension of) $\rho$ is a division algebra. Proof. By linearity of the scalar product, for each $x\in\mathbb RG$ we have $$ \...


3

I am afraid that I would have no idea where to look for a reference for this statement, but here is a very rough sketch proof. I can fill in details if necessary. Let $A_1,\ldots,A_s$ be the orbits of $G$ on $[m]$ and $B_1,\ldots,B_t$ the orbits of $H$ on $[n]$. We can assume that each $|B_j| > 1$, since otherwise the action of $H$ would be disconnected. ...


2

For the obvious reflection action of $\mathbb{Z}/2\mathbb{Z}$ on $S^1$ this set $\tilde S $ has two singular points at the zero vectors tangent at points $p=(1,0)$ and $q=(-1,0)$. Then $\tilde S$ is $T(S^1\setminus \{p,q\}) \cup \{p_0,q_0\}$ where $p_0, q_0$ are zero vectors at $p,q$. This is a connected set and after removing $p_0,q_0$ we obtain a ...


2

$O(2n)$ acts transitively on the space $C$ of compatible complex structures, by sending a complex structure $J$ to $OJO^{-1}$, it's easy to check this is a compatible complex structure. To show transitivity use block-diagonalisation for real matrices to show it any complex structure can be conjugated to a standard one in block-diagonal form. Compatibility ...


1

$EG$ is also the universal free $G$-space, meaning that, if $X$ is a free $G$-space (let's assume of the $G$-homotopy type of a $G$-CW complex), there is, up to $G$-homotopy, a unique $G$-map $X\to EG$. Taking quotients, you get a map $X/G \to BG$ which induces the action of $H^\ast(BG)$ on $H^\ast(X)$ with any coefficients. Edited to add: I really should ...


1

Assuming $G$ is a finite group: If $G_c$ is the stabilizer of $c$, then $$ |Gc| = |G|/|G_c| $$ and $$ |Hc| = |H|/|H\cap G_c| = |HG_c|/|G_c| $$ so $$ \frac{|Gc|}{|Hc|} = \frac{|G|}{|HG_c|}. $$ EDIT: For the infinite case asked for in the comment, we can do this: By choosing representatives $gc$ for each element of the orbit of $c$, the map $(gc,s) \mapsto gs$...


1

There is actually a simple answer. Pick X to be two $\mathbb{C}P^1$'s intersecting transversally, and a $\mathbb{Z}/2$ action that swaps two spheres. Then $\text{rk } H^*(X)=3$ whereas $\text{rk } H^*(X^G) = 1.$


1

[Too long for a comment. Not a definitive answer, but maybe as close as we're likely to get.] I'm assuming here, since you talk about $X_n$ being potentially exponentially large, that we may think of $X_n$ as being a subset of binary strings whose length is at most polynomial in $n$, and that the action of each generator of $G$ is efficiently computable in ...


1

Is it true that if $G$ has full spectrum with respect to a series $\sum_{n=1}^\infty a_n$ then $G$ has full spectrum also with respect to any other conditionally convergent series $\sum_{n=1}^\infty b_n$? If I understand the definitions correctly, clearly no. Let $G$ be the symmetric group of $2\mathbb{N}$ and consider conditionally convergent series ...


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