Timeline for A basis-free formula for the determinant as a polynomial
Current License: CC BY-SA 4.0
Post Revisions
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| 17 hours ago | history | edited | Rodrigo de Azevedo | CC BY-SA 4.0 |
added 1 character in body
|
| yesterday | history | became hot network question | |||
| yesterday | comment | added | Tomasz Kania | Would Theorem 3.2 count? arxiv.org/pdf/2510.16501 There is a slight error in this version of the paper (but not the theorem); will update it soon on arxiv. | |
| yesterday | comment | added | Martin Gisser | @KentaSuzuki, OK it seems your answer helps. Just the proof is lacking :-) But I'm a bit distracted today. And I'm far away from any useable library. One reason why my question is tagged "reference request", so I know what to download. | |
| yesterday | comment | added | Kenta Suzuki | @MartinGisser Sorry yes you are right, that was a typo. The point is that it is well-defined upon symmetrizing. | |
| yesterday | comment | added | Martin Gisser | @KentaSuzuki, your first map is not well-defined. I would start with $f_1=...=f_n$ and then polarize. That's messy, and I guess there is a better way to get your 2nd map. It dawns that I have seen this somewhere. Thanks a lot for this formula, but it is not what I'm looking for. | |
| yesterday | answer | added | Kenta Suzuki | timeline score: 9 | |
| yesterday | comment | added | Deane Yang | @KentaSuzuki, nicely done. | |
| yesterday | comment | added | Martin Gisser | @Kenta, The trick is to take the $n$-th derivative at $0$ of a homogeneous function of degree $n$ on the vector space $End(V)$. It is a general remark on p.7 of Lang's last variant of his differential geometry book. It is not motivated by determinant function. | |
| yesterday | comment | added | paul garrett | @KentaSuzuki, oh, wow, yes, without looking, that is surely what it is! :) | |
| yesterday | comment | added | Kenta Suzuki | Without knowing what Lang's definition is, my guess would be: An element $f_1\otimes\cdots\otimes f_n\in End(V)^{\otimes n}$ produces an endomorphism of $\wedge^n(V)$ as $v_1\wedge\cdots\wedge v_n\mapsto f_1(v_1)\wedge\cdots\wedge f_n(v_n)$. We symmetrize this construction, so $f_1\cdots f_n\in End(V)^{\otimes n}$ would act as $v_1\wedge\cdots\wedge v_n\mapsto\frac1{n!}\sum_{\sigma\in S_n}f_1(v_{\sigma1})\wedge\cdots\wedge f_n(v_{\sigma n})$. Now, note that since $\wedge^nV$ is one-dimensional, any endomorphism is multiplication by a scalar. This produces a map $S^n(End(V))\to\mathbb K$. | |
| yesterday | comment | added | Neil Strickland | See mathoverflow.net/questions/507049 for discussion of what it might mean to give a "basis-free definition" | |
| yesterday | comment | added | Kenta Suzuki | Could you repeat what the calculus trick is? I couldn't find what you were referring to in Lang's book. | |
| yesterday | history | asked | Martin Gisser | CC BY-SA 4.0 |