22

There is, naturally, a huge history regarding a basic question like this. This particular problem was figured out between 1930 and the early 1960's. The main names are P.A. Smith, Conner, Floyd. Here is a math review to get you going:
MR0130929 (24 #A783) Reviewed
Kister, J. M.
Examples of periodic maps on Euclidean spaces without fixed points.
Bull. Amer....

21

### Do mutually dual finite vector spaces have the same orbit cardinalities under a linear group action?

Yes. In particular, it can happen that $V$ has non-zero fixed points, but $V^*$ doesn't.
For example, let $G$ be the symmetric group of degree 3 acting in the obvious way on the set $\{e_1,e_2,e_3\}$, and let $W$ be the corresponding permutation module over the field of 3 elements. Let $V$ be the submodule spanned by $e_1-e_2$ and $e_1-e_3$. Then $V$ has ...

20

Your question translates into the language of orbifolds as saying: what is known about spherical $n$-orbifolds with underlying space homeomorphic to $S^n$?
In $S^2$, the examples you give are all there are.
Orbifolds with the geometry of $S^3$ were enumerated by William Dunbar in his thesis. His published paper MR1118824 contains the enumeration of the ...

17

Proof of (1):
(a). Suppose $X$ and $Y$ are $G$-spaces, the action of $G$ on $X$ is free, and $X\to X/G$ is
a principal bundle, then the space of $G$-equivariant maps
$$
F(X,Y)^G
$$
is the same thing as the space of sections of the fibration $X\times_G Y \to X/G$.
(b). If $E\to B$ is a Hurewicz fibration, with $B$ connected and the fiber contractible, ...

17

If $G$ is compact, the inclusion $H(M^G) \to H(M)^G$ is an isomorphism. The inverse map is defined as follows: Take a class $\omega$ in $H(M)^G$ and lift it to a closed form $\alpha \in \Omega(M)$. Put $\beta = \int_{g \in G} g^{\ast} \alpha$, where the integral is with respect to Haar measure normalized to have volume $1$. Clearly, $\beta \in \Omega(M)^G$. ...

answered Mar 11 '18 at 13:57

David E Speyer

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16

There is no such universal constant $\epsilon > 0$. Work with the complex Hilbert space $L^2[0,1]$ (which of course is also a real Hilbert space). Fix $n \in \mathbb{N}$.
Let $\Gamma_0$ be the set of continuous piecewise linear increasing bijections from $[0,1]$ to itself. [1] It is a group with composition as product. [2] It acts by isometries of $L^2[0,...

14

I was referred to this question by Uri Bader. Thanks Uri.
Here are two counter examples.
Consider the set $K$ of all continuous strictly increasing functions $f$ from $[0,1]$ onto $[0,1]$. Let $T$ be the map from $K$ onto $K$ that sends $f$ to its inverse. Endowing $K$ with the $L_1$ norm, $T$ is an isometry (the $L_1$ distance between two functions is the ...

13

we can clearly assume that $f(z)=-z$ in the standard metric on $S^1\subset \mathbb C$ (as we can assume that $f$ is isometric with respect to some Riemannnian metric on $S^1$). Then $g(z)=z\cdot e^{i\alpha(z)}$ with $0<\alpha(z)<2\pi$. If $\alpha(z_0)<\pi$ then $\alpha(g(z_0))>\pi$ and there is a point $z_1$ with $\alpha(z_1)=\pi$ by the ...

answered Nov 25 '11 at 15:39

Vitali Kapovitch

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13

What you should take as a model is the homotopy quotient $EG \times_G BG$. From the homotopy sequence of the fibration $EG \times_G BG \to BG$ (projection on first factor), you get that
$EG \times_G BG$ is aspherical and a short exact sequence
$$
1 \to \pi_1 (BG) \to \pi_1 (EG \times_G BG) \to \pi_1 (BG) \to 1.
$$
Since the action of $G$ on $BG$ has a ...

13

For infinite-dimensional vector spaces this not possible. According to this paper of Brenner and Ringel, if $A$ is any principal ideal domain that is not a field or a complete valuation ring, then there is an $A$-module $V$ such that $V\cong V\oplus V\oplus V$ but $V\not\cong V\oplus V$. In particular, if $k$ is a field and $A=k[\mathbb{Z}]=k[x,x^{-1}]$, ...

13

Regarding your question about weighted projective spaces, a lot is known about them, see for instance [1] and [2].
In particular, any weighted projective space $\mathbb{P}(\mathcal Q)$ is irreducible, normal, Cohen-Macaulay and has at most cyclic quotient singularities (hence rational singularities), see [2], p. 122.
However, weighted projective spaces ...

answered Oct 9 '15 at 12:17

Francesco Polizzi

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12

I think I have completely answered the question in the following form:
Theorem. For a finite subgroup $\Gamma < O(n)$ the quotient space $S^{n-1}/\Gamma$ is homeomorphic to $S^{n-1}$ if and only if $\Gamma$ has the form
\begin{eqnarray*}
\Gamma = \Gamma_{ps} \times P_1 \times \ldots \times P_k
\end{eqnarray*}
for a pseudoreflection group $\Gamma_{ps}$ ...

12

Let the root system be $v_1$, …, $v_n$ with all elements normalized to be length $1$. So $\langle v_i, v_i \rangle =1$, we have $\langle v_i, v_j \rangle \leq - \cos (\pi/3) = -1/2$ for at least $n$ pairs $(i,j)$, and we have $\langle v_i, v_j \rangle \leq 0$ for all $i \neq j$.
But then
$$\left\langle \sum_{i=1}^n v_i, \sum_{i=1}^n v_i \right\rangle \leq ...

answered Aug 26 '14 at 19:33

David E Speyer

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12

The maps $\psi_k$ are all isomorphisms; this is a simple application of the transfer ("averaging") construction. See Theorem 2.4, Chapter III, of Bredon's book "Introduction to compact transformation groups".

12

There are very precise results on the repartition of orbits of $SL(2,\mathbb{Z})$ of irrational points in $\mathbb{R}^2$, such as this one.
In particular these orbits are dense, and this can be seen easily like this.
If $\theta$ is irrational, there are coprime $(a,b)$ such that $a+b\theta$ is less than $\epsilon$ for any $\epsilon>0$. Then there are $(...

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

Let me follow up on, and try to clarify some of what has been said. Given a group $G$, we want to count the number of $G$-orbits on $G \times G$, where the action is conjugation in each coordinate. The number of fixed points of $x \in G$ is thus $|C_G(x)|^2$, so by the "Burnside" lemma (due to Cauchy and Frobenius) the number $N$ of orbits is given by
$$
N = ...

11

Even without knowing an explicit set of generators, you can compute the Hilbert series with very little work as follows. In general, suppose a finite group $G$ acts on a vector space $V$ over a field $k$ of characteristic not divisible by $|G|$ via an action map $\rho : G \to \text{GL}(V)$. Then $G$ acts on the symmetric algebra $S(V^{\ast})$ (a coordinate-...

answered Feb 23 '16 at 8:26

Qiaochu Yuan

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11

The space $BG/G$, with the model you describe, has been studied, but is not completely understood. It's $\pi_1$ identifies with the abelianization, but there may be higher homotopy.
It identifies with a hocolim over the Quillen category with objects all subgroups and morphisms given with conjugation and inclusion.
I'm not sure the group $K$ will have a ...

11

Division by two, with your naturality condition, is equivalent to division by two in the category of representations of a group:
If $V$ and $W$ are two finite-dimensional representations of a group $G$ such that $V + V = W+ W$, then $V=W$.
As Eric said, this is all about stabilizers. Given no other conditions, there is a $GL(V) \times GL(W)$-equivariant ...

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.

10

There are various compact manifolds of negative curvature which are not homnotopy-equivalent to closed hyperbolic manifolds: Locally symmetric ones (complex hyperbolic, etc) as well as Gromov-Thurston and Mostow-Siu examples. Their $\pi_1$'s are Gromov-hyperbolic, boundary is a topological sphere. The examples exist in all dimensions $\ge 4$ (Gromov-Thurston)...

10

For a fixed point your guess is right and one doesn't need Luna's slice theorem to prove it: Let $T$ be the tangent space in $x$ and let $\mathfrak m_x\subset\mathbb C[X]$ be the maximal ideal. Then the canonical surjective linear map $\mathfrak m_x\to\mathfrak m_x/\mathfrak m_x^2\cong T^*$ has a $G$-equivariant section giving rise to a $G$-linear map $T^*\...

9

Presumably you meant to assume the schemes are finite type over $k$. To work naturally with orbit questions for such schemes one just has to bring in appropriate use of flatness to adapt intuition and experience from the traditional smooth setting over algebraically closed fields (e.g., one uses the robust theory of quotients modulo possibly non-smooth ...

9

This question is connected to an extreme case of the odd analogue of Glauberman´s $Z^{\ast}$-theorem. This theorem asserts that if a finite group $G$ has no non-identity normal subgroup of order coprime to the prime $p,$ and $u$ is an element of order $p$ of $G$ which commutes with none of its other $G$-conjugates, then $u \in Z(G).$ This theorem was proved (...

9

I think the answer to your question is negative. Consider for instance
$$M = \begin{pmatrix}
1 & 0 \\
0 & 1
\end{pmatrix}, \quad N = \begin{pmatrix}
0 & 1 \\
1 & 0
\end{pmatrix}$$
Assume that there exist
$$A = \begin{pmatrix}
a_{11} & a_{12} \\
0 & a_{22}
\end{pmatrix}, \quad B = \begin{pmatrix}
b_{11} & b_{12} \\
0 & b_{...

9

Let $G$ be a two element group; the free group on two generators $x,y$ with the action of $G$ interchanging them is a free $G$-group (on one generator). Its subgroup generated by $xy^{-1}$ is closed under the $G$-action but is not a free $G$-group.

answered Oct 26 '18 at 18:43

მამუკა ჯიბლაძე

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9

Note added on 26 Nov 2018: I have corrected my answer, which had a serious mistake.
For simplicity of notation, let $(x,y,z) = (x_1,x_2,x_3)$. The Hessian form associated to $f_0 = {x_1}^3+{x_2}^3+{x_3}^3+6x_1x_2x_3$ is
$$
H(f_0) = \frac{\partial^2f_0}{\partial x_i\partial x_j}\,\mathrm{d}x_i\circ\mathrm{d}x_j\,.
$$
The determinant of this Hessian form is ...

answered Nov 25 '18 at 18:40

Robert Bryant

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8

Reposting VA's wonderful answer (with a trivial correction), since someone else just asked me this question:
This is just to add 1% to Dmitri's 99% complete answer. Change the
coordinates to $w_0,\dots, w_{n-1}$ defined by the formula
$$ w_i = x_0 + \mu^i x_1 + \mu^{2i} x_2 + \dots, $$
where $\mu$ is a primitive $n$-th root of identity. Then ...

8

As already pointed out, whenever you have a compact group $G$ acting on a manifold $M$, this action can be made isometric by constructing a metric on $M$ via a standard averaging process. In particular, if you have a (effective) circle action on $M$, you can find a metric $g$ on $M$ such that this action is isometric. In other words, the isometry group of $(...

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