Timeline for Exterior powers and choice
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 22, 2019 at 17:37 | vote | accept | Phil-W | ||
| May 21, 2019 at 18:52 | comment | added | KConrad | Thanks for asking this question. I added a link to this page in a footnote to the proof of Corollary 5.9. | |
| May 21, 2019 at 11:22 | answer | added | Jeremy Rickard | timeline score: 5 | |
| May 21, 2019 at 10:56 | comment | added | Jeremy Rickard | I'll write an answer with more details. | |
| May 21, 2019 at 10:54 | comment | added | Phil-W | I'm confused, is it the same $x$ as before (kernel element of $\Lambda^k \varphi$) ? In my mind $U$ was precisely the finite-dimensional subspace spanned by the finitely elements of $V$ involved in the injectivity proof of the exterior powers. That's why I don't see how we can "replace" the infinite-dimensional $V$ by $U$ without first proving the injectivity of the exterior power of the inclusion of $U$ in $V$. Maybe I'm not seeing something trivial here. | |
| May 21, 2019 at 10:33 | comment | added | Jeremy Rickard | But if $V$ is also finite dimensional then you can do this without choice by choosing a basis of $U$ and extending to a basis of $V$. And you can assume that $V$ is finite dimensional, because if $\Lambda^kU\to\Lambda^kV$ has a non-zero element $x$ in the kernel, you can replace $V$ by the finite dimensional subspace spanned by the image of $\Lambda^k\varphi$ and the finitely many elements of $V$ involved in the proof that $\Lambda^k\varphi(x)=0$. | |
| May 21, 2019 at 10:26 | comment | added | Phil-W | But to apply such reduction, don't we need to know already that $\Lambda^k U$ is injected in $\Lambda^k V$ when $U$ is a finite dimensional subspace of $V$? Isn't that circular ? | |
| May 21, 2019 at 10:05 | comment | added | Jeremy Rickard | An element $x$ in the kernel of $\Lambda^k\varphi$ can be written in terms of finitely many $v\in V$. As well as the $\varphi(v)$, a proof that $\varphi(x)=0$ using the relations in the exterior power only uses finitely many other elements $w\in W$. Doesn't this reduce (1) to a question about finite dimensional spaces? | |
| May 21, 2019 at 4:50 | comment | added | YCor | Note that (2) is a particular case of (1) with $(V,W)$ replaced with $(F^k,V)$ and $F$ the underlying field. | |
| May 21, 2019 at 1:32 | comment | added | Gabe Goldberg | I am pretty sure the statements (1) and (2) can be checked to be $\Pi_1$ sentences (in the Lévy hierarchy). The point is that they hold if and only if they hold in all transitive models of a sufficiently large fragment of ZF. By Schoenfield absoluteness, since they are provable in ZFC, they are provable in ZF. (The relevant form of Schoenfield absoluteness is (2) of this question: mathoverflow.net/questions/269682/…) | |
| May 20, 2019 at 19:53 | comment | added | Phil-W | I tried to apply the finite-dimensional reduction but since we are not dealing with the same type of map (in this case we are dealing with exterior powers, and in the tensor case, it was not about tensor powers), I didn't see how to transpose the arguments (at least it is not trivial to me how to do this, but I might be wrong), and it didn't lead to anything. | |
| May 20, 2019 at 19:23 | comment | added | YCor | Jeremy Rickard gave an efficient way to deal with such questions in your previous question you link at (mathoverflow.net/questions/325037). What have you tried? does his approach fail here and why? | |
| May 20, 2019 at 18:08 | history | asked | Phil-W | CC BY-SA 4.0 |