Timeline for Does the truth of any statement of real matrix algebra stabilize in sufficiently high dimensions?
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
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| Aug 2, 2010 at 6:18 | comment | added | Victor Protsak | Mariano: for cubes, consider reps of $\mathfrak{sl}_2^{\oplus 3}$ whose restriction on the diagonal $\mathfrak{sl}_2$ in each pair of factors contains a trivial submodule (pairs 12 and 23 are sufficient). By replacing "trivial mod" with "$m$-dimensional simple mod", you can get off the pure powers. Basically, you can first get dim $n_1\ldots n_k$ from the direct sum of $k$ copies of $\mathfrak{sl}_2$ and then specialize $n_i$ or the difference $n_i-n_j,$ etc, to a chosen natural number. (That only produces pols that split over $\mathbb{Z}$, I'm not sure how to tweak it to get the rest.) | |
| Aug 2, 2010 at 5:20 | comment | added | Mariano Suárez-Álvarez | How do you do the cubes, for example? | |
| Aug 2, 2010 at 5:12 | comment | added | Victor Protsak | By increasing the number of factors, we can get $f(\mathbb{N})$ for any monic polynomial $f$ with integer coefficients as the truth set. I can almost see how to get any diophantine set ($\iff$ recursively enumerable, by Matiyasevich) in this way. | |
| Aug 2, 2010 at 5:06 | comment | added | Ryan Reich | Damn, I missed it. Ironically, this is why I used the construction I did. | |
| Aug 2, 2010 at 4:59 | history | edited | Mariano Suárez-Álvarez | CC BY-SA 2.5 |
added 42 characters in body
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| Aug 2, 2010 at 4:48 | comment | added | Victor Protsak | Excellent, Mariano! I was nearly there. | |
| Aug 2, 2010 at 4:43 | history | answered | Mariano Suárez-Álvarez | CC BY-SA 2.5 |