Timeline for Are the trace relations among matrices generated by cyclic permutations?
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 8 at 16:44 | comment | added | Mariano Suárez-Álvarez | This sort of thing is discussed in Victor Ginzburg's Calabi-Yau algebras, section 2. | |
| Jan 5 at 19:52 | answer | added | Nikita Safonkin | timeline score: 2 | |
| May 7, 2021 at 18:50 | comment | added | Asvin | @JosephVanName I sent an email, please let me know if you didn't get it. | |
| May 7, 2021 at 16:31 | comment | added | Joseph Van Name | Yes. I would like to take a look at what you have written so far. You may send the draft to circcash9192020 AT protonmail DOT com. | |
| May 7, 2021 at 3:10 | comment | added | Asvin | @JosephVanName I am writing up a paper about it but if you share an email id, I can send you something. | |
| May 7, 2021 at 2:07 | comment | added | Joseph Van Name | @Asvin That sounds like an interesting infinite product or limit especially since I don't hear too much about non-commutative infinite products. Do you have a reference for such a product? I personally have found non-commutative infinite product formulae $$\frac{1}{1-(x_{1}+\dots+x_{n})}=\lim_{n\rightarrow\infty}p_{n}(x_{1},\dots,x_{n})\dots p_{0}(x_{1},\dots,x_{n})$$ from rank-into-rank embeddings in set theory. | |
| May 6, 2021 at 20:31 | comment | added | Asvin | @JosephVanName That is more or less the context that inspired this question! I had an infinite sequence of products of matrices (over a polynomial ring) that seemed to converge $\ell$-adically and to prove this, I worked with every term in the expanded polynomial and showed that the coefficients converge. | |
| May 6, 2021 at 17:43 | comment | added | Joseph Van Name | Using this formula, we get $0=\log(\text{Det}(1))=\sum_{k=0}^{\infty}\log(\text{Det}(p_{k}))$. Now, for the right hand side, we can replace each instance of $\text{Tr}(x_{a_{1}}\dots x_{a_{r}})$ with $f(a_{1},\dots,a_{r})$ for any cyclically invariant function $f$ where the sum converges. We now no longer have to worry about matrices. | |
| May 6, 2021 at 17:43 | comment | added | Joseph Van Name | This problem is useful for transforming an infinite product of non-commutative polynomials (or something similar) into something commutative and easier to work with. For example, suppose that $1=\lim_{n\rightarrow\infty}p_{n}\dots p_{0}$ where each $p_{i}$ is a non-commutative polynomial. Then you can take the determinant of both sides to obtain a commutative infinite product, and you can take the logarithm to turn this into an infinite sum. But $\log(\text{Det}(I-A))=\sum_{k=1}^{\infty}\frac{\text{Tr}(A^k)}{k}$. | |
| May 5, 2021 at 14:37 | comment | added | Joseph Van Name | I was able to show that if our field is the complex numbers, if $\text{Tr}(f(\phi(g_{1}),\dots,\phi(g_{n})))=0$ for each finite group G, irreducible representation $\phi:G\rightarrow U(r)$, and elements $g_{1},\dots,g_{n}\in G$, then $f$ is in the subspace spanned by the cyclic permutations. The proof uses the fact that the characters of irreducible representations form a basis for the class functions, so the proof is not to unexpected. | |
| May 2, 2021 at 5:17 | answer | added | Joseph Van Name | timeline score: 4 | |
| May 2, 2021 at 3:54 | vote | accept | Asvin | ||
| May 2, 2021 at 3:30 | history | became hot network question | |||
| May 1, 2021 at 21:32 | answer | added | Will Sawin | timeline score: 17 | |
| May 1, 2021 at 21:29 | comment | added | LSpice | Well, I can't establish it because it's not true, as the example of $f(X_1, X_2) = X_1^2 X_2 X_1 - X_1 X_2 X_1^2$ shows. So I guess I should say, I can reduce it just to handling $f$ of the form $f(X_1, \dotsc, X_n) = X_1 g(X_1, \dotsc, X_n)$, where $g$ itself is already in the desired span. | |
| May 1, 2021 at 21:17 | comment | added | LSpice | $f$ is a non-commutative polynomial, right? I can reduce the problem (re-phrased as @BenjaminSteinberg and ChristianRemling suggested, in terms of a linear span) to showing that, if $f = X_1 g$ for some other non-commutative polynomial $g$, then $f = 0$. But I can't seem to establish that yet. | |
| May 1, 2021 at 20:02 | comment | added | Benjamin Steinberg | I think you want just the linear span as @ChristianRemling suggests and then I think it should be generated by all differences of two words that differ by a cyclic permutation | |
| May 1, 2021 at 19:41 | comment | added | Christian Remling | Is there a specific reason why you want to consider the ideal generated by these $f$ (I'm asking because the $f$ with $\textrm{tr }f=0$ don't seem to form an ideal). | |
| May 1, 2021 at 19:30 | history | asked | Asvin | CC BY-SA 4.0 |