Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was wondering if anyone visiting would be up for solving the following interesting little exercise out of Fulton's Algebraic Topology: A First Course.

Let $G$ a group act on a set $Y$. Say that $G$ acts evenly on $Y$ if any point in $Y$ has a nbhd. $V$ such that $g \cdot V = \{g \cdot y : y \in V\}$ (the image of $V$ under the group action) and $h \cdot V$ are disjoint for any distinct elements $g$, $h$ in $G$. Prove that the following actions are even:

(1) $\mathbb{Z}$ acts on $\mathbb{R}$ by translation: $n \cdot r = r+n$.

(2) $\mathbb{Z}$ acts on the right half plane (RHP) in $\mathbb{R^2}$ so that the polar coordinate covering map $$p: \{(r, \theta) \in \mathbb{R^2}: r>0\} \rightarrow \mathbb{R^2}\setminus \{(0,0)\}$$ given by $p(r,\theta) = (r \cos \theta, r \sin \theta),$ is the quotient of the RHP by this $\mathbb{Z}$-action.

(3) The cyclic group $G = \mu_n$ of $n$th roots of unity in $\mathbb{C}$ acts by multiplication on $S^1$ [regarded as the unit circle in the complex plane].

share|cite|improve this question
up vote 4 down vote accepted

For (1): given $r\in\mathbb{R}$, let $V_r=(r-\frac{1}{2},r+\frac{1}{2})$. Then $$n\cdot V_r=(r+n-\tfrac{1}{2},r+n+\tfrac{1}{2}), \quad m\cdot V_r=(r+m-\tfrac{1}{2},r+m+\tfrac{1}{2})$$ intersect if and only if $r+m-\frac{1}{2}<r+n+\frac{1}{2}$ and $r+n-\frac{1}{2}<r+m+\frac{1}{2}$, i.e. $m-n<1$ and $n-m<1$, hence $|n-m|<1$. Since $n,m\in\mathbb{Z}$ this forces $n=m$.

I'm not sure I understand the statement of problem 2. Am I correct that the action is intended to be $n\cdot (r,\theta)=(r,\theta+2\pi n)$?

If so, then given $p=(r,\theta)\in\text{RHP}$, let $V_p=B_\delta(p)=\left\{(a,b)\in\text{RHP}\mid \sqrt{(b-\theta)^2+(a-r)^2} < \delta \right\}$ where $\delta=\min\{r,\pi\}$. For example, if $\pi\leq r$, then we get a picture like this:

enter image description here

the point $p=(r,\theta)$ is marked, and the circles represent $V_p$ and a few of its translates. If we had $r<\pi$, i.e. the point $p$ was less than $\pi$ away from the $y$-axis, we'd have to shrink the circles to have radius $r$ (or less) to ensure they don't leave our set, $\text{RHP}$.

For (3): Given $p=(\cos(\theta),\sin(\theta))\in S^1$, let $V_p=\{(\cos(\psi),\sin(\psi))\in S^1\mid \theta-\frac{2\pi}{n}<\psi<\theta+\frac{2\pi}{n}\}$. I'll let you show that $\zeta\cdot V_p$ and $\omega\cdot V_p$ are disjoint for any distinct $\zeta,\omega\in\mu_n$.

share|cite|improve this answer
Thanks for responding! I think the action you note for problem (2) is the right one (and I don't recall Fulton making the group action explicit when he discusses the polar coordinate map in Chapter 11, but I will check). Again thanks. – Vulcan Jan 31 '12 at 2:53
@Robert: No problem, glad to help! I've included a choice of $V_p$ for (2), assuming our interpretation of the problem is right. – Zev Chonoles Jan 31 '12 at 3:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.