Timeline for Diagonalization over valuation rings
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 27, 2022 at 17:25 | comment | added | David Loeffler | The intro of this paper by Ribet link.springer.com/content/pdf/10.1007/BF01393341.pdf is one reference for the modular forms angle. | |
| Mar 27, 2022 at 9:57 | comment | added | FPV | @DavidLoeffler Really nice answer. Do you know any reference for this? | |
| Mar 24, 2022 at 21:35 | comment | added | David Loeffler | If $L$ has distinct eigenvalues modulo $\mathfrak{m}$, then it is diagonalisable over $\mathcal{R}$. By the way, people who work with modular forms care a lot about this kind of diagonalization problem (with $L$ being a Hecke operator) -- a very considerable chunk of Wiles' proof of Fermat's last theorem is about understanding the sizes of congruence ideals, which are precisely there to measure the failure of diagonalisability of operators on $\mathbf{Z}_p$-modules. | |
| Mar 24, 2022 at 12:08 | comment | added | Uriya First | Aside to my earlier comment: Choosing the $U_\alpha$ in advance such that $U_\alpha={\cal R}^n\cap KU_\alpha$ and $\bigoplus_{\alpha} U_\alpha\neq {\cal R}^n$, and then taking the $\alpha$-s to be in $1+{\frak m}^n$ for $n$ sufficiently large (to guarantee that $L({\cal R}^n)\subseteq {\cal R}^n$) gives rise to non-diagonal $L$ such tha $L_{R/\frak m}$ is diagonalizable. This works for any discrete valuation ring $\cal R$. | |
| Mar 24, 2022 at 12:00 | comment | added | Uriya First | Assuming $L_K:K^n\to K^n$ is diagonalizable, $K^n$ would factor as a sum of eigenspaces $K^n=\bigoplus_\alpha V_\alpha$ with $L_Kv=\alpha v$ for all $v\in V_\alpha$. This decomposition is unique. If $L$ were diagonalizable then you would have a similar factorization of $\cal R$-modules ${\cal R}^n=\bigoplus_{\alpha}U_\alpha$, and the uniqueness forces $V_\alpha=KU_\alpha$, equiv. $U_\alpha=V_\alpha \cap {\cal R}^n$. It follows that $L$ is diagonalizable iff ${\cal R}^n=\bigoplus_\alpha (V_\alpha\cap\cal R^n)$. | |
| Mar 24, 2022 at 11:15 | comment | added | Johannes Hahn | It is necessary for the eigenvalues to be in $R$ and for the reduction $\overline{L} : (R/\mathfrak{m})^n \to (R/\mathfrak{m})^n$ to be diagonalisable. Also the projections onto the eigenspaces projections would need to lift. Therefore I would guess that at some point Hensel's Lemma will need to be invoked. But I'm not sure if a Henselian ring is really all that's needed to turn the necessary condition into a sufficient one. It's certainly sufficient to separate the eigenvalues that are not $\mathfrak{m}$-close to each other. But what happens to close eigenvalues, I do not know. | |
| Mar 24, 2022 at 10:45 | history | edited | FPV | CC BY-SA 4.0 |
added 1 character in body; edited tags
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| Mar 24, 2022 at 10:38 | comment | added | Z. M | Do you have any good criterion when the valuation ring is $\mathbb Z_p$? | |
| Mar 23, 2022 at 22:34 | history | asked | FPV | CC BY-SA 4.0 |