Timeline for Ultrafilters as a double dual
Current License: CC BY-SA 4.0
32 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 20, 2019 at 17:25 | comment | added | Łukasz Lew | Is the embedding an instance of Yoneda embedding? | |
| Mar 9, 2019 at 23:22 | comment | added | Asaf Karagila♦ | @YCor Ah, you're right. Yes. | |
| Mar 9, 2019 at 23:21 | comment | added | YCor | @AsafKaragila I mean the sum, i.e. the subspace generated by these subspaces... | |
| Mar 9, 2019 at 23:11 | comment | added | Asaf Karagila♦ | @YCor: Why does the sum isomorphic to a subspace at all? | |
| Mar 9, 2019 at 23:08 | comment | added | YCor | @AsafKaragila define $V_{00}$ as the sum of all subspaces of $V$ with trivial dual. What's clear is that $V_{00}$ has trivial dual, and that $V_{00}\subset V_0$, as $V_0$ is the intersection of all kernels of all linear forms on $V$. | |
| Mar 9, 2019 at 23:04 | comment | added | Asaf Karagila♦ | @YCor: Ah, that what you meant. Well, yes, it seems plausible. Because the kernel is exactly those vectors which are never mapped to anything non-zero. I'm not willing to sign off on the "maximal", but it looks like a reasonable conjecture. | |
| Mar 9, 2019 at 22:42 | comment | added | YCor | @AsafKaragila Let $V$ be a vector space over a field. Denote by $V_0$ the kernel of the canonical homomorphism $V\to V^{\ast\ast}$. Answering to მამუკა ჯიბლაძე who asked "can we prove anything [about $V_0$]?" you replied "That it's a subspace?" (I agree!) "I'd guess that it's the maximal subspace which admits a trivial dual." The point of my last comment was to say I'd be surprised that "for every field $V$, the space $V_0$ has a trivial dual" be a theorem of ZF. Do you really think it's true? (Of course it's true for many $V$, e.g. when $V=0$ or when $V=V_0$.) | |
| Mar 9, 2019 at 22:31 | comment | added | Asaf Karagila♦ | @YCor: I am not sure what you mean by that (I mean, syntactically, I can't parse your comment). But I never said always. | |
| Mar 9, 2019 at 21:03 | comment | added | YCor | @AsafKaragila I see no reason (in ZF) that the kernel $V_0$ of $V\to V^{\ast\ast}$ always has a trivial dual. | |
| Mar 8, 2019 at 17:23 | comment | added | Manuel Bärenz | I meant to say that I simply misread your sentence "the canonical embedding of $V$ into $V^{**}$ uses choice in a subtle way" as "the canonical map of $V$ into $V^{**}$ uses choice in a subtle way". | |
| Mar 8, 2019 at 17:08 | comment | added | Asaf Karagila♦ | @ManuelBärenz: If $V\neq\{0\}$ you'd expect $V^*\neq\{0\}$ as well. But that's not necessarily the case without the axiom of choice. As for you not caring about that, that was the first in the list of similarities mentioned above. So I think in the context of this question, we should at least care a little bit. | |
| Mar 8, 2019 at 17:06 | comment | added | Manuel Bärenz | @AsafKaragila, ah I didn't care about that. Yes, I imagine that's the hard part. What do you mean by "$V^*$ might be trivial even though $V$ isn't? | |
| Mar 8, 2019 at 17:04 | comment | added | Asaf Karagila♦ | @ManuelBärenz: How do you prove that this function is injective? (You want it to be an embedding, after all...) | |
| Mar 8, 2019 at 17:03 | comment | added | Manuel Bärenz | @AsafKaragila $i: V \to V^{**}, i(v) = \omega \mapsto \omega(v)$ looks pretty constructive to me. What am I missing? Or is there just no constructive proof that this is an injection? | |
| Mar 8, 2019 at 14:38 | vote | accept | Adam P. Goucher | ||
| Mar 8, 2019 at 12:55 | comment | added | მამუკა ჯიბლაძე | @AsafKaragila You win. As always. | |
| Mar 8, 2019 at 12:05 | comment | added | Asaf Karagila♦ | @მამუკაჯიბლაძე Yes, but $\varnothing$ is. And in the usual settings the kernel of the embedding is $\{0\}$. | |
| Mar 8, 2019 at 12:05 | comment | added | მამუკა ჯიბლაძე | @AsafKaragila Well surely $\{0\}$ is not linearly independent :P | |
| Mar 8, 2019 at 12:02 | comment | added | Asaf Karagila♦ | @მამუკაჯიბლაძე: Sure, but what about $\{0\}$ when you are in a "usual setting" being the kernel of such embedding? | |
| Mar 8, 2019 at 12:01 | comment | added | მამუკა ჯიბლაძე | @AsafKaragila Maybe something along the lines of "every linearly independent subset is contained in a larger one"? | |
| Mar 8, 2019 at 11:57 | comment | added | Asaf Karagila♦ | @მამუკაჯიბლაძე: That it's a subspace? I'd guess that it's the maximal subspace which admits a trivial dual. It can be a proper subspace, of course, since if $V$ admits a trivial dual, then $V\oplus F$, where $F$ is the field, admits functionals which are null on $V$ but non-null on $F$, so $(V\oplus F)^{**}\cong F$. | |
| Mar 8, 2019 at 11:55 | comment | added | მამუკა ჯიბლაძე | @AsafKaragila Interesting, never thought of it - can one prove anything about the kernel of $V\to V^{**}$ without choice? Could you recommend a text about those things? | |
| Mar 8, 2019 at 8:57 | comment | added | Asaf Karagila♦ | (My point above, is that the canonical embedding of $V$ into $V^{**}$ uses choice in a subtle way, whereas the canonical embedding of $X$ into $\beta X$ does not.) | |
| Mar 8, 2019 at 8:05 | answer | added | Qiaochu Yuan | timeline score: 14 | |
| Mar 7, 2019 at 17:57 | history | became hot network question | |||
| Mar 7, 2019 at 16:59 | answer | added | Nik Weaver | timeline score: 32 | |
| Mar 7, 2019 at 16:30 | comment | added | Asaf Karagila♦ | Well, one major difference is that without choice it is always the case that $X$ embeds into $\beta X$, it's just not provable that the embedding is not surjective; whereas $V^*$ might be trivial, let alone $V^{**}$, even though $V$ isn't. | |
| Mar 7, 2019 at 16:19 | history | edited | YCor | CC BY-SA 4.0 |
fixed typo
|
| Mar 7, 2019 at 16:17 | comment | added | YCor | Close to Todd's comment, I'd view $\beta X$ as $F(X)=\mathrm{hom}_{\mathrm{Bool}}(\mathrm{hom}_{\mathrm{Top}}(X,\mathbf{Z}/2\mathbf{Z}))$. In general, I guess that for a topological space $X$, the map $X\to F(X)$ is the initial object for the category of continuous maps from $X$ to compact Hausdorff totally disconnected topological spaces. A difference with taking biduals is that $F(F(X))=F(X)$ by Stone duality. | |
| Mar 7, 2019 at 15:43 | comment | added | Todd Trimble | Well, one of the very first things that comes to mind that's sort of in this vein is that $\beta X = \hom_{\text{Bool}}(\hom_{\text{Set}}(X, 2), 2)$. But if you want to pursue your analogy at a deeper level, try golem.ph.utexas.edu/category/2012/09/…, where both the ultrafilter monad and the double dualization monad are reckoned to be codensity monads induced by the full inclusions of finitary objects. | |
| Mar 7, 2019 at 15:37 | history | edited | Martin Sleziak |
edited tags
|
|
| Mar 7, 2019 at 15:35 | history | asked | Adam P. Goucher | CC BY-SA 4.0 |