Timeline for Action of a Lie group with finitely many orbits
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 29, 2014 at 15:19 | comment | added | Misha Verbitsky | I apologize for misreading the question | |
| May 29, 2014 at 11:25 | comment | added | asv | Agree with Yves Cornulier. | |
| May 28, 2014 at 16:07 | comment | added | YCor | I don't think it answers the question, because the closure of the orbit $O$ of $(p,q,r)$ ($p$ times 1, $q$ times $-1$, $r$ times 0) consists of the union of the orbits $(p',q',r')$ with $p'\le p$ and $q'\le q$, thus $\bar{O}-O$ is the union of the orbits $(p',q',r')$ with $p'\le p$, $q'\le q$, $p'+q'<p+q$. Write $d=p+q$. While $O$ has dimension $f(d)=n^2-d(n-d)-(d-n)^2-d(d-1)/2$, $\bar{O}-O$ has dimension $f(d-1)$ , and $f(d)-f(d-1)=n-d+1$ for all $1\le d\le n$. Thus $\bar{O}-O$ has dimension smaller than the dimension of $O$. | |
| May 28, 2014 at 14:18 | history | answered | Misha Verbitsky | CC BY-SA 3.0 |