Timeline for algebraic VS topological ergodicity
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Mar 3, 2012 at 14:09 | history | edited | Matthew Daws | CC BY-SA 3.0 |
Added answer to other bit of question
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| Mar 3, 2012 at 10:37 | comment | added | johnnyblade | Of course thank you Matthew, I misread the word 'translate'. Regards | |
| Mar 3, 2012 at 9:17 | comment | added | Matthew Daws | @johnnyblade: I "translated" $f$ so that $f(0)=0$, and then conclude $f=0$. So actually, the starting $f$ could be anything. It also occurs to me that the same idea works on $X=[0,1]$, just let $\phi(t)=t^2$. | |
| Mar 3, 2012 at 7:36 | vote | accept | johnnyblade | ||
| Mar 3, 2012 at 7:22 | comment | added | johnnyblade | Thank you Matthew, but there is still something that I don't understand. Following your argument we prove that every invariant function is the zero function, contradicting the fact that every *-automorphism of $C(X)$ leaves $I$ fixed, i.e. $I$ must be fixed by an automorphic action (which is equivalent to the topological one in the commutative case). Tell me if it's non-sense. Thank you again, Best regards | |
| Mar 3, 2012 at 7:12 | vote | accept | johnnyblade | ||
| Mar 3, 2012 at 7:36 | |||||
| Mar 2, 2012 at 21:50 | comment | added | Matthew Daws | I'm not quite sure what "locally closed orbit" means, so I'm not sure how this example interacts with pm's answer... | |
| Mar 2, 2012 at 21:49 | history | answered | Matthew Daws | CC BY-SA 3.0 |