Timeline for How does one understand geometric CFT in terms of modularity?
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| Jun 15, 2020 at 7:27 | history | edited | CommunityBot |
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| Jul 9, 2012 at 0:22 | history | bounty ended | Makhalan Duff | ||
| Jul 6, 2012 at 22:15 | comment | added | John Pardon | What do you mean by "modularity", and what does it have to do with your question? It seems to be purely a question about geometric class field theory. The only thing that comes to mind for what you mean is modularity of elliptic curves, but that seems to be entirely unrelated. | |
| Jul 2, 2012 at 8:15 | comment | added | Dror Speiser | For a general number field it is not necessarily true that $K^\times \backslash I_K / D_K\prod_v O_v^\times$ is trivial. As you may know, the maximal unramified extension of a number field, called its Hilbert class field, has galois group over the field isomorphic to the class group. I don't remember if there is a small play on the infinite places, but in any case $K^\times\backslash I_K / D_K\prod_v O_v^\times$ is either the class group, or something very close to it (maybe the strict class group). This is all standard CFT, and appears in any CFT book. I highly recommend Milne's online notes. | |
| Jul 2, 2012 at 2:19 | comment | added | Makhalan Duff | $\mathbb{I}_{\mathbb{Q}}/\mathbb{Q}^{\times}$ by $D_{\mathbb{Q}}$, which is the product of the infinite places, and by $\prod \mathbb{Z}_p^{\times}$ (to account for not allow ramification anywhere). But I wonder where you're using something special about $\mathbb{Q}$, because this isn't true for all number fields, is it? | |
| Jul 2, 2012 at 2:19 | comment | added | Makhalan Duff | @Dror: that's very helpful. Let me ask you a few questions. 1. Is $D_K$ always the product over all infinite places? 2. Let $L$ be the maximal abelian extension of $K$ that is unramified over a specific set of primes $S$ of $O_K$. Are you saying that $L^{\times}N_{L/K}(\mathbb{I}_L)$ is equal to the product $\prod_{\mathfrak{p} \not \in S} O_{\mathfrak{p}} ^{\times}$? I guess what I'm trying to ask is: it seems that you're saying that the "reason" that there are no abelian unramified extensions of $\mathbb{Q}$ is that you get $Gal(\mathbb{Q}^{un,ab}/\mathbb{Q})$ by quotienting | |
| Jul 2, 2012 at 1:09 | comment | added | Dror Speiser | In general, to get maximal abelian extensions with general controlled ramification, you divide by a similar product over the places of the field. The galois group will be isomorphic to the double cosets, and these groups are finite. Two primes split in the same manner if they give rise to the same double coset. Finally, there is an interpretation of the double cosets as a ray class group, much like a generalised jacobian for a curve (Picard group corresponds to maximal unramified abelian extension). So primes split equivalently if they have the same class in the ray class group. | |
| Jul 2, 2012 at 0:55 | comment | added | Dror Speiser | It is not that $D_K$ is the product of completions. $D_K$ is some product over the finitely many infinite places. For example, for the rationals, there is an isomorphism $\mathbb{Q}^\times \backslash I_\mathbb{Q} \cong \mathbb{R}_+^\times \times \prod_p \mathbb{Z}_p^\times$, and the connected component of 1 is $\mathbb{R}_+^\times \times 1$. So $Gal(\mathbb{Q}^{ab}/\mathbb{Q})\cong \prod_p \mathbb{Z}_p^\times$. To get the unramified abelian extensions of $\mathbb{Q}$, we further divide by $\prod_p \mathbb{Z}_p^\times$, to get $\mathbb{Q}^{ab,un}=\mathbb{Q}$. | |
| Jul 1, 2012 at 23:50 | history | bounty started | Makhalan Duff | ||
| Jul 1, 2012 at 23:48 | history | edited | Makhalan Duff | CC BY-SA 3.0 |
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| Jun 30, 2012 at 20:04 | comment | added | Dror Speiser | @Makhalan: It is not true that $K^\times \backslash I_K / \prod_\nu O_\nu^\times$ is isomorphic (after taking profinite completions) to $Gal(K^{ab}/K)$, or even $Gal(K^{ab,un}/K)$. Something that is true: $I_K/K^\times$ is a topological group. Call its connected component $D_K$. Then there is an isomorphism $K^\times \backslash I_K/ D_K \cong Gal(K^{ab}/K)$. For example, the group $\mathbb{Q}^\times \backslash I_\mathbb{Q} / \prod_p \mathbb{Z}_p^\times$ is connected, and dividing by itself we can recover the classical fact that the rational field has no unramified extensions. | |
| Jun 30, 2012 at 16:03 | comment | added | Makhalan Duff | $K^{\times}\backslash \mathbb{I}_K/\prod_{v\not \in S} O^{\times}_v$ is isomorphic to $Gal(K^{ab}/K)$. What is the correct way to think about this? | |
| Jun 30, 2012 at 16:02 | comment | added | Makhalan Duff | @Felipe: sorry for asking so many questions, but I'm confused again. Serre's book would be easier to read with a little motivation, so hopefully you'll indulge me. My perception was that if $S$ is a finite set of places, then $K^{\times}\backslash \mathbb{I}_K/\prod_{v\not \in S} O^{\times}_v$ would have a profinite completion which is isomorphic to $Gal(L/K)$ where $L$ is the maximal abelian cover of $K$ that is ramified only in $S$. But it seems that what you're saying is that if S is the set of infinite places, then | |
| Jun 30, 2012 at 13:14 | comment | added | Felipe Voloch | The product $\prod_v O_v^{\times}$ does not include the archimedian places in the number field case. In the function field case, if you exclude a finite set of places you get a bigger group. Read Serre "Groupes algebriques et corps de classes" | |
| Jun 30, 2012 at 13:09 | history | edited | Makhalan Duff | CC BY-SA 3.0 |
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| Jun 30, 2012 at 13:06 | comment | added | Makhalan Duff | I'm sorry, my last comment was silly. You're right that $\pi_1^{ab}(C)\cong Gal(K^{ab,un}/K)$. My question is, why is the analogous statement for $K$ a number field that $K^{\times}\backslash \mathbb{I}_K/\prod_v O_v^{\times}$ is isomorphic (after taking profinite completions) to $Gal(K^{ab}/K)$ rather than $Gal(K^{ab,un}/K)$? Or am I wrong about that? For example $\mathbb{Q}^{\times}\backslash \mathbb{I}_{\mathbb{Q}}/\prod_p \mathbb{Z}_p^{\times}$ is far from trivial, isn't it? I guess I'm just finding it hard to draw the analogy. | |
| Jun 30, 2012 at 2:07 | comment | added | Felipe Voloch | The fundamental group only captures unramified covers, by definition. $\pi_1({\mathbb{P}}^1)$ is trivial. | |
| Jun 30, 2012 at 1:24 | comment | added | Makhalan Duff | In what sense? For example it is not true that every abelian cover of $\mathbb{P}^1_{\mathbb{F}_p}$ is unramified! (E.g., $y^2=x$ will define an abelian cover ramified over $0$ and $\infty$.) I feel like I'm missing the crucial point you're trying to get across... Can you tell me what I'm missing? | |
| Jun 30, 2012 at 0:55 | comment | added | Felipe Voloch | $Gal(K^{ab,un}/K) = \pi_1^{ab}(C)$ because $C$ is projective. | |
| Jun 30, 2012 at 0:37 | history | edited | Makhalan Duff | CC BY-SA 3.0 |
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| Jun 29, 2012 at 22:56 | comment | added | Makhalan Duff | @Felipe: I am confused by your statement. The analogous theorem to the theorem I cited in the number theory case describes $Gal(K^{ab}/K)$ in terms of (the profinite completion of) a double quotient of the ideles of $K$. If the theorem I cited is only unramified class field theory, wouldn't it describe $Gal(K^{ab,un}/K)$ instead? | |
| Jun 29, 2012 at 22:25 | comment | added | Felipe Voloch | That theorem you cite is only unramified class field theory. To get the full picture you need to allow ramification too. | |
| Jun 29, 2012 at 21:44 | history | edited | Makhalan Duff | CC BY-SA 3.0 |
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| Jun 29, 2012 at 21:38 | history | asked | Makhalan Duff | CC BY-SA 3.0 |