Timeline for Radon-Nikodym derivative in a compact Hausdorff space
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| S May 28, 2023 at 2:02 | history | bounty ended | CommunityBot | ||
| S May 28, 2023 at 2:02 | history | notice removed | CommunityBot | ||
| May 26, 2023 at 22:23 | comment | added | Sanae Kochiya | @JulianHölz If by "left-amenable", you mean $\mu(gA) = \mu(A)$ for all $g\in G$, I suppose it would be easier to start from searching for a $\mu$ such that "$\mu(g\cdot)\sim\mu(\cdot)$ for all $g\in G$. If I let $\delta_x$ be the Dirac measure defined on $x\in X$, then $g\delta_x = \delta_{gx}$. Since $G$ is assumed to be countable, the sum of $\delta_{gx}$ (after multiplying coefficients to make the infinite sum equal to $1$ will be a regular but very atomic probability measure. Since this example exists even when $X$ is just locally compact, I prefer to search for a non-atomic one. | |
| May 26, 2023 at 21:22 | comment | added | Julian Hölz | Concering your third question: I think the property you are looking for is called "(left/right) amenable" (as in amenable groups). Although I am not so sure that what you get as an invariant measure must be non-atomic. Examples of groups that is not amenable are free groups. | |
| May 23, 2023 at 17:09 | vote | accept | Sanae Kochiya | ||
| May 23, 2023 at 13:19 | answer | added | Thomas Lehéricy | timeline score: 5 | |
| S May 20, 2023 at 0:48 | history | bounty started | Sanae Kochiya | ||
| S May 20, 2023 at 0:48 | history | notice added | Sanae Kochiya | Draw attention | |
| May 20, 2023 at 0:32 | history | edited | Sanae Kochiya | CC BY-SA 4.0 |
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| May 20, 2023 at 0:27 | comment | added | Sanae Kochiya | I posted an attempt to the third answer but then was blocked by a few more questions. Please feel free to add extra conditions that you believe is necessary for answering any of the questions above. For instance, should there exist a necessary condition for the space $X$ to have a non-trivial homeomorphism group, please feel free to add it to your post, and it would be also great to see an answer when $X$ is a general topologically-homogeneous space. | |
| May 14, 2023 at 21:08 | comment | added | Taras Banakh | Manifolds modeled on various model spaces (like $\mathbb R^n$ or Hilbert cube) usually are topologically homogeneous, which means that for any points $x,y$ there exists a homeomorphism $h$ such that $h(x)=y$. | |
| May 14, 2023 at 20:26 | history | edited | Sanae Kochiya | CC BY-SA 4.0 |
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| May 14, 2023 at 20:24 | comment | added | Sanae Kochiya | About your second comment, I decided to only consider the case where $G$ is countable, so that it would be hard for $G$ to be compact, and hard for an infinite space $X$ to have $\operatorname{Homeo}(X)$ countable. | |
| May 14, 2023 at 20:19 | comment | added | Sanae Kochiya | Thank you for your comments. About your first comment, could you specify the condition for $\operatorname{Homeo}(X)$ to be non-trivial? I should have mentioned the space have infinitely many points, the topology is non-discrete and so on, but I suppose you mean other conditions. | |
| May 14, 2023 at 6:07 | comment | added | Taras Banakh | Concerning your question 3, then for any compact subgroup $G$ of the homeomorphism group of $X$ there exists the unique Haar measure on $G$ and its image under the action should be a probability measure on $X$, which is unvariant with respect to the action. If $G$ is non-compact, for example, if $G$ is the whole homeomorphism group of the (say) Cantor set, then the action has no invariant measure even in your weak sense. | |
| May 14, 2023 at 5:57 | comment | added | Taras Banakh | Your compact Hausdorff space can have no homeomorphisms at all (there are many such spaces among continua), in which case your question 2 has a negative answer. So you should restrict the class of considered compact Hausdorff spaces to obtain something interesting. | |
| May 13, 2023 at 16:58 | history | asked | Sanae Kochiya | CC BY-SA 4.0 |