Timeline for Probability of satisfying a word in a compact group
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 15, 2018 at 4:20 | vote | accept | Sean Eberhard | ||
| Oct 13, 2018 at 17:56 | comment | added | Uri Bader | Sean, yes. This is indeed due to connectedness. I posted an answer, for clarity. | |
| Oct 13, 2018 at 17:53 | answer | added | Uri Bader | timeline score: 20 | |
| Oct 13, 2018 at 16:33 | comment | added | Sean Eberhard | @UriBader Why does full dimension imply measure 1? I'm thinking of for example the solution space to $x^2=1$ in $O(2)$. Of course $O(2)$ is not connected, so it's not a counterexample. | |
| Oct 13, 2018 at 16:27 | comment | added | Uri Bader | the solution space for a $k$ letters word will be a closed subvariety of $G^k$, and as such it will have either measure 0 or 1. Morover, if $G$ is not commutative then it contains a free group hence this subvariety will be proper, thus of 0 meaure. | |
| Oct 13, 2018 at 16:19 | comment | added | Nate Eldredge | Particularly in Lie groups, there's lots of other measure-theoretic and potential-theoretic notions of "smallness" that one could ask about. If it does have measure zero, what's its Hausdorff dimension? Is it a polar set? Et cetera. | |
| Oct 13, 2018 at 16:02 | history | asked | Sean Eberhard | CC BY-SA 4.0 |