Timeline for Massive cancellations
Current License: CC BY-SA 3.0
4 events
| when toggle format | what | by | license | comment | |
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| Nov 27, 2014 at 3:33 | comment | added | Will Sawin | @ScottAaronson The way I think of it is that there's an eigen-linear form that sends the $\mathbb Q$-vector space $K$ to $K \subseteq \mathbb C$ by a linear map. It's easy to see that having an eigen-linear form with eigenvalue $\lambda$ implies an eigenvector with eigenvalue $\lambda$. | |
| Nov 27, 2014 at 3:28 | comment | added | Scott Aaronson | A small observation. By combining this argument with Piotr Achinger's construction, we get an elementary, self-contained proof that there exist transcendental numbers that can be explicitly constructed: namely, the same numbers that Liouville showed to be transcendental in his original proof -- en.wikipedia.org/wiki/…. Maybe this "ultimately" amounts to the same thing as Liouville's proof, but I think I prefer it to the proof on Wikipedia. | |
| Nov 27, 2014 at 1:55 | comment | added | Scott Aaronson | Thanks so much, Will -- this clarified things enormously! There was one step that wasn't obvious to me: why does the minimum eigenvalue of the matrix give a lower bound on the actual real valuation of the element? But I think I now see it: let $\alpha_1,\ldots,\alpha_k$ be the generators of the extension field; then $(1,1/\alpha_1,\ldots,1/\alpha_k)$ is always an eigenvector of the matrix corresponding to an element $v\in\mathbb{Q}[\alpha_1,\ldots,\alpha_k]$, and its associated eigenvalue is always $|v|$ (by calculation). | |
| Nov 26, 2014 at 18:21 | history | answered | Will Sawin | CC BY-SA 3.0 |