Timeline for Nonzero subdeterminants conjecture: has anybody seen this anywhere?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 6, 2018 at 15:42 | comment | added | chizhek | @darij grinberg - This is the mistake in your initial reasoning: if $1\leq i\leq n$ and $\mu'\in M'$, then $c'_{i,\mu'}=(\mu'_i+1)\cdot c_{\mu'+e_i}$. This mistake - the missing nonzero coefficient - is immaterial, provided your argument, with the lonely diagonal product at the top of the lexicographic hierarchy of products in the determinant, holds water. | |
| Jun 6, 2018 at 13:09 | comment | added | darij grinberg | Oops, yes, I forgot to mention the coefficients (though I thought of them, as they're the reason for the characteristic-$0$ assumption). Thanks for clearing it up. You never said "integer coefficients", I believe, so I figured I should answer as generally as I could. | |
| Jun 6, 2018 at 10:03 | comment | added | chizhek | @darij grinberg - This is a promising way to attack the problem, though your reasoning at the start is incorrect. Note that the characteristic is 0 since the ground ring is the polynomial ring in the variables $c_\mu$, $\mu\in M$, with integer coefficients. | |
| Jun 4, 2018 at 16:46 | comment | added | darij grinberg | ... namely as the result of multiplying through the main diagonal. All other monomials are smaller in the lexicographic order of monomials (where the variables $c_\mu$ themselves are ordered by the lexicographic order of the $\mu \in M$). So this monomial isn't cancelled by anything, and thus is nonzero. Note that I'm assuming the ground field to have characteristic $0$ here; otherwise I don't know. | |
| Jun 4, 2018 at 16:45 | comment | added | darij grinberg | Okay, let me try directly attacking your Conjecture. Consider the $n\times n$-submatrix of $\Gamma$ obtained by taking columns $\nu_1, \nu_2, \ldots, \nu_n$. WLOG assume that $\nu_1 > \nu_2 > \cdots > \nu_n$ in lexicographic order. We regard elements of $M$ and of $M^\prime$ as vectors in the $\mathbb{Z}$-module $\mathbb{Z}^n$, and we let $\left(e_1,e_2,\ldots,e_n\right)$ be the basis of this $\mathbb{Z}$-module. Then I believe that the monomial $c_{\nu_1 + e_1} c_{\nu_2 + e_2} \cdots c_{\nu_n + e_n}$ appears only once in the determinant, ... | |
| Jun 4, 2018 at 16:41 | comment | added | darij grinberg | In other words, you want to prove that the vectors $\left(c_{\nu+e_1}, c_{\nu+e_2}, \ldots, c_{\nu+e_n}\right)$ for any $n$ distinct values of $\nu \in M^{\prime}$ (where $e_1,e_2,\ldots,e_n$ is the standard basis of the $\mathbb{Z}$-module $\mathbb{Z}^n$, and we regard $\nu$ as a vector in this module) are linearly independent over the polynomial ring in the variables $c_\mu$. Right? | |
| Jun 4, 2018 at 12:14 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
added link
|
| Jun 4, 2018 at 12:12 | history | edited | Suvrit |
added two related tags
|
|
| Jun 4, 2018 at 9:31 | history | asked | chizhek | CC BY-SA 4.0 |