Timeline for Does the spectrum of Morava E-theory depend only on height?
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 9, 2022 at 9:37 | vote | accept | Sofía Marlasca Aparicio | ||
| Jan 21, 2022 at 0:52 | answer | added | kiran | timeline score: 12 | |
| S Jan 12, 2022 at 18:06 | history | bounty ended | CommunityBot | ||
| S Jan 12, 2022 at 18:06 | history | notice removed | CommunityBot | ||
| Jan 6, 2022 at 20:16 | comment | added | Tyler Lawson | Even with the multiplicative formal group over $\Bbb F_q$, this is $H^1(\hat{\Bbb Z}, \Bbb Z_p^\times) \cong \Bbb Z_p^\times$, and under this identification the extension from $\Bbb F_q$ to $\Bbb F_{q^k}$ sends a classifying element $\pi$ to the element $\pi^k$. This means that two classifying elements $\pi$ and $\pi'$ become equal over a finite extension if and only if the ratio $\pi / \pi'$ is a root of unity. | |
| Jan 6, 2022 at 20:13 | comment | added | Tyler Lawson | @BenWieland Unfortunately there is not necessarily a finite extension where $f$ and $g$ become isomorphic. If $G$ is the automorphism group of $f$ over the algebraic closure, then "formal groups over $K$ whose extension to $\bar K$ is isomorphic to $f$" are parametrized by $H^1(Gal(\bar K/K), G)$, and extending from $K$ to $L$ comes from restriction along $Gal(\bar K/L) \to Gal(\bar K/K)$. | |
| Jan 5, 2022 at 19:57 | comment | added | Ben Wieland | You have a lot of structure because $E(K)$ is a ring. Choosing a basis for $W(L)/W(K)$ give you a map $E(K)^n\to E(L)$ which exhibits it as a free module. Composing with the projection you get a map $E(f,K)\to E(g,K)$ which is an isomorphism on $W(K)$, but you have to do something to check that it is nontrivial on the deformation parameters. | |
| Jan 5, 2022 at 4:31 | comment | added | Eric Peterson | @Ben The last step, descending from L to K, is a bit opaque to me—especially since I recently learned that there exist examples in finite spectra like X = C(7 nu) and Y = C(nu), which are distinct, with X v X and Y v Y nonetheless equivalent. Can you elaborate? | |
| Jan 5, 2022 at 2:40 | comment | added | Ben Wieland | Yes, the underlying additive spectra are the same. Choose a finite extension $L/K$ such that the formal groups become isomorphic over $L$. Then $E(f,L)=E(g,L)$. The key is that $E(f,L)$ is a wedge of copies of $E(f,K)$. Note that the homotopy groups of $E(f,L)$ are determined from $E(f,K)$ by tensoring from $W(K)$ to $W(L)$. | |
| S Jan 4, 2022 at 16:28 | history | bounty started | Sofía Marlasca Aparicio | ||
| S Jan 4, 2022 at 16:28 | history | notice added | Sofía Marlasca Aparicio | Improve details | |
| Jan 4, 2022 at 16:27 | comment | added | Sofía Marlasca Aparicio | @CharlesRezk Exactly! I guess fully faithful makes it quite unlikely that the spectrum doesn't change, but I would like to at least see an example | |
| Dec 30, 2021 at 17:31 | comment | added | Charles Rezk | These comments don't seem to address the question, which I believe is asking whether the underlying spectra of Morava E-theories are equivalent. | |
| Dec 29, 2021 at 7:26 | comment | added | Jack Davies | It very much depends on the formal group (law). A reference is a 2004 paper by Goerss--Hopkins called "Moduli spaces of commutative ring spectra" (sec. 7) where they originally prove what Maxime and Denis refer to. Another nice reference is Lurie's "Elliptic Cohomolgy II" (sec. 5) for an alternative (constructive) construction, and his chromatic homotopy theory lecture notes for a bit of background using the Landweber Exact Functor Theorem (lecture 21-2). In the latter, Lurie also discusses that the Bousfield class of Morava E-theory depends only on the height (lecture 23) | |
| Dec 28, 2021 at 20:13 | comment | added | Denis Nardin | @MaximeRamzi Not only it's functorial, it's even fully faithful (in E_oo-ring spectra) | |
| Dec 28, 2021 at 18:52 | comment | added | Maxime Ramzi | I don't know a reference, but I think the point is that if $k$ is algebraically closed, then all formal group laws of a given height are isomorphic, and so any two Morava $E$-theories over such a thing are isomorphic. But actually, without "a" this is a bit of an abuse of notation - in principle, it should be $E(k,\Gamma)$ for some formal group $\Gamma$ over $k$ , and the assignment $(k,\Gamma)\mapsto E(k,\Gamma)$ is functorial iirc | |
| Dec 28, 2021 at 17:37 | history | asked | Sofía Marlasca Aparicio | CC BY-SA 4.0 |