Timeline for Families of Galois representations over disks
Current License: CC BY-SA 4.0
24 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 20, 2023 at 10:44 | comment | added | David Loeffler | Please, just stop commenting and start actually thinking! We've already established that (with your somewhat weird definitions) all the rings $\mathbb{Z}_p\langle \tfrac{x}{r}\rangle$ for $0 \le r < 1$ are all actually the same ring. So please don't waste your time and ours trying to microscopically analyse the maps in between them. | |
| Nov 20, 2023 at 6:33 | comment | added | David Loeffler | But you must have ignored a great deal of "too many comments" warnings to get this far. | |
| Nov 20, 2023 at 6:33 | comment | added | David Loeffler | Your maps $f_r$ and $g_r$ are never both well-defined (one is only defined if $r < 1$, the other if $r >1$, and I'm too lazy to work out which). | |
| Nov 17, 2023 at 17:13 | comment | added | anon | Yes, I might at some point but I have to first resolve something that seems like a contradiction. I know that these kinds of moduli problems are studied by Emerton and Gee, but I am just trying to find a way to write down simple examples of families of Galois representations, say over a Tate algebra with $r=1$. | |
| Nov 17, 2023 at 15:37 | comment | added | Simon Wadsley | I wonder if it is worth rewriting the question at this point to address all the points of clarification. | |
| Nov 17, 2023 at 14:05 | comment | added | anon | Yes. $\mathbb{F}_p[x]$ for $r\geq 1$ and $\mathbb{F}_p[[x]]$ for $r<1$. | |
| Nov 17, 2023 at 13:37 | comment | added | Simon Wadsley | I have another question. You allow $R=\mathbb{F}_p$. What is the Tate algebra in this case? Just a polynomial ring? | |
| Nov 17, 2023 at 9:28 | comment | added | anon | I think I should have just asked about the case that all the $r_i$ are equal to $1$. | |
| Nov 16, 2023 at 16:30 | comment | added | Simon Wadsley | That is because $G$ is compact, right? | |
| Nov 16, 2023 at 15:56 | comment | added | David Loeffler | Surely any representation into $GL_n(\mathbb{Q}_p[[X]])$ can be conjugated into $GL_n(\mathbb{Z}_p[[X]])$? | |
| Nov 16, 2023 at 15:27 | comment | added | anon | Yes, they are the same as rings. I will rethink my question! I consider $\mathbb{Z}_p[[x]]$ with the projective/inverse limit or adic topology and on the other hand the topology on $\mathbb{Z}_p<\frac{x}{r}>$ is described by a single norm, so I am not sure that I would consider these as equivalent objects. I have to check. Anyway, certainly $\mathbb{Q}_p[[x]]$ and $\mathbb{Q}_p<\frac{x}{r}>$ are very different and so I could have asked the same question about lifting $\mathbb{Q}_p[[x]]$ reps (up to conjugation) to $\mathbb{Q}_p<\frac{x}{r}>$ reps. | |
| Nov 16, 2023 at 14:57 | comment | added | David Loeffler | Hold on, if r < 1 then your ring Zp<X/r> is equal to Zp[[X]]! I think you need to rethink your question. | |
| Nov 16, 2023 at 10:14 | comment | added | anon | I should also have said that I am using the topology on $R<\frac{x}{r}>$ generated by the norm $|\sum a_i x^i| = \sup |a_i|r^i$. | |
| Nov 16, 2023 at 8:49 | comment | added | David Loeffler | This is not such a contrast to Mazur's setup as you seem to think, since $\mathbb{Z}_p[[X]]$ embeds in the ring you are calling $\mathbb{Z}_p\langle X / r\rangle$ for any $r < 1$. | |
| Nov 16, 2023 at 8:32 | comment | added | anon | Thanks everyone for the input. Sorry about the non-standard notation. I was thinking about lifts to $GL_n(A)$ where (in contrast to the setup in Mazur) $A$ is now a regular non-local ring, whose formal completion at some ideal is $\mathbb{Z}_p[[x]]$ or $\mathbb{Q}_p[[x]]$. | |
| Nov 16, 2023 at 6:23 | comment | added | David Loeffler | FWIW, the notation R<X/r> already has a meaning and it isn’t that. But the notational question is unimportant. I think @tkr has very correctly diagnosed the misconception behind the question. | |
| Nov 16, 2023 at 4:00 | comment | added | tkr | But $\mathbb{Z}_p[[x]]$ is really the ring of bounded-by-1 functions converging on the open unit disc. Any such series converges on $|x|_p<1$! So, perhaps your perception of the problem you are studying is not quite accurate. | |
| Nov 15, 2023 at 22:27 | comment | added | anon | The thing is that I am thinking of $\mathbb{Z}_p[[x]]$ as functions on something of radius zero, a formal scheme, not an open disk. | |
| Nov 15, 2023 at 21:26 | comment | added | Simon Wadsley | Ah. I see. So you really want to study the representations up to equivalence. | |
| Nov 15, 2023 at 21:13 | comment | added | anon | Maybe it would have been better to ask if the image of $G$ can be conjugated to lie in the image of $GL_n(R<\frac{x_1}{r_1}, \dots, \frac{x_k}{r_k}>)$ by a single element of $GL_n(R[[x_1, \dots, x_k]])$ | |
| Nov 15, 2023 at 21:04 | comment | added | anon | Hi Simon, yes you are right. I meant to ask if the image of $G$ ever lands in that subgroup. I was thinking about the context of Mazur's article where he was sometimes able to get universal formal families of reps thought of as reps into $GL_n(\mathbb{Z}_p[[x_1, \dots, x_k]])$ specializing to given reps into $GL_n(\mathbb{F_p})$. I am asking about getting deformations beyond the formal ones. | |
| Nov 15, 2023 at 20:55 | comment | added | Simon Wadsley | I don't understand the question either. Does 'is there a lift?' mean any more or less than is image of the homomorphism contained in the subring? | |
| Nov 15, 2023 at 20:27 | comment | added | anon | I should have said that $R<\frac{x}{r}> \subset R[[x]]$ is the subring of formal power series $\sum a_i x^i$ for which $|a_i| r^i$ converges to $0$ and this subring induces a continuous group homomorphism $GL_n(R<\frac{x}{r}>) \to GL_n(R[[x]])$. This map is used to write down what lifting means in the one variable case and similarly for many variables. | |
| Nov 15, 2023 at 20:08 | history | answered | David Loeffler | CC BY-SA 4.0 |