Timeline for Leopoldt's conjecture and cup-products
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 8, 2012 at 1:45 | answer | added | Filippo Alberto Edoardo | timeline score: 3 | |
| Feb 7, 2012 at 21:47 | comment | added | Rob Harron | Dear Joël, thanks, you're quite right. | |
| Feb 7, 2012 at 19:52 | comment | added | Joël | Dear Rob, one has $H^3(G_{K,S},M) = \prod_{v | \infty} H^3(K_v,M)$. This is Tate's "Duality theorem in Galois cohomology", ICM1962, Theorem 3.1(c). When $v$ is complex, the local $H^3$ is always 0. When $v$ is real, it is $H^3(Z/2Z,M)$, which, for a module with trivial action of the complex conjugation at $v$, is just $M[2]$. When $M={\bf Z}_2$ in particular, it is again $0$. | |
| Feb 7, 2012 at 17:49 | comment | added | Rob Harron | @Guillermo: Ah, yes. Thanks. @Joël: If $p=2$, $K$ has a real place, and $S$ contains the primes above 2 and this real place, then $\mathrm{cd}_2(G_{K,S})=\infty$, and, unless I'm missing something, $H^3(G_{K,S},\mathbf{Z}_2)\neq0$ (but finite). If $H^2(G_{K,S},\mathbf{Q}_2)=0$, I think $H^2(G_{K,S},\mathbf{Q}_2/\mathbf{Z}_2)\neq0$, though still finite. | |
| Feb 7, 2012 at 3:59 | comment | added | Joël | @Rob: Right. And this is also equivalent to $H^2(G_{K,S},\mathbb{Q}_p/\mathbb{Z}_p)=0$ when $p>2$ as I have said, and actually, even when $p=2$. Look at the long exact sequence of cohomology attached to the short exact sequence $0 \rightarrow \mathbb{Z}_p \rightarrow \mathbb{Q}_p \rightarrow \mathbb{Q}_p / \mathbb{Z}_p \rightarrow 0$. One gets a surjection $H^2(G_{K,S},\mathbb{Q}_p) \rightarrow H^2(G_{K,S},\mathbb{Q}_p/\mathbb{Z}_p) \rightarrow H^3(G_{K,S},\mathbb{Z}_p)$ but the latter is $0$ in any case, even when $p=2$. So my formulation and Henri's are equivalent. | |
| Feb 7, 2012 at 3:25 | comment | added | Guillermo Mantilla | @Rob: Yes. The point is that $H^2(G_{K,S},\mathbb{Q}_p/\mathbb{Z}_p)$ is the dual of $H_2(G_{K,S},\mathbb{Z}_p)$ and the $\mathbb{Z}_p$-Rank of the later is precisely the Leopoldt defect. | |
| Feb 7, 2012 at 0:59 | comment | added | Rob Harron | According to Neukirch–Schmidt–Wingberg Corollary 10.3.10, the $\mathbf{Z}_p$-rank of $H^2(G_{K,S},\mathbf{Z}_p)$ equals the Leopoldt defect. This rank is the $\mathbf{Q}_p$-dimension of $H^2(G_{K,S},\mathbf{Q}_p)$, so Joël's formulation is correct (at least away from $p=13$, right Joël?). Shouldn't this be the same as $H^2(G_{K,S},\mathbf{Q}_p/\mathbf{Z}_p)$ being finite? | |
| Feb 6, 2012 at 21:43 | comment | added | Joël | @Henri: Hum, I do think the statement I wrote is equivalent to Leopoldt's conjecture. Actually, your statement and mine are equivalent if $p>2$, or if $K$ is totally complex. This is easy to see since the the $p$-cohomological dimension of $G_{K,S}$ is at most 2 under these hypotheses (Cf e.g. Jannsen, On the l-adic cohomology of varieties over number field and its Galois cohomilogy, in Galois groups over Q, Lemma 1). For the somewhat anecdotical reminding cases, I am not sure. | |
| Feb 6, 2012 at 20:56 | comment | added | Henri Johnston | Shouldn't the $\mathbb{Q}_p$ be replaced by $\mathbb{Q}_p/\mathbb{Z}_p$ in the statement of Leopoldt's conjecture? So that the statement should be $H^{2}(G_{K,S},\mathbb{Q}_p/\mathbb{Z}_p)=0$? | |
| Feb 6, 2012 at 19:55 | history | edited | Joël | CC BY-SA 3.0 |
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| Feb 6, 2012 at 19:44 | history | asked | Joël | CC BY-SA 3.0 |