Timeline for How is an $S^1$-equivariant elliptic cohomology theory affected as we continuously vary the underlying elliptic curve?
Current License: CC BY-SA 3.0
26 events
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| S Jan 31, 2018 at 11:33 | history | suggested | Helene Sigloch |
removed tag "rigid analytic geometry"
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| Jan 31, 2018 at 10:09 | review | Suggested edits | |||
| S Jan 31, 2018 at 11:33 | |||||
| Sep 12, 2015 at 10:40 | comment | added | მამუკა ჯიბლაძე | Just googled to find a paper with the title Semistable principal $G$-bundles in positive characteristic :D | |
| Sep 11, 2015 at 20:33 | comment | added | მამუკა ჯიბლაძე | If I am not mistaken, their $\mathcal X_G$ makes sense in positive characteristic too. True, their $G$ is a compact Lie group, but the basic examples come from algebraic groups definable in a characteristic-free way. I am not sure about the definition of semistability for principal bundles, though. | |
| Sep 11, 2015 at 20:12 | comment | added | Catherine Ray | @მამუკა ჯიბლაძე I am having a lot of difficulty understanding why their construction should work in characteristic p. Grojnowksi's definition relies on choosing an isomorphism $S^1 \times S^1 \simeq E$ (for a complex elliptic curve $E$) to define the fixpoint set of an $S^1$-space $X$ over each point of $E$. How does their construction differ from Grojnowksi's to avoid using a lattice? | |
| S Jul 15, 2015 at 6:27 | history | bounty ended | CommunityBot | ||
| S Jul 15, 2015 at 6:27 | history | notice removed | CommunityBot | ||
| Jul 9, 2015 at 21:52 | comment | added | Catherine Ray | Pardon my ignorance. What do you mean by "algebraically vary the curve"? My guess is that you are talking about an algebraic group action (rather than a continuous group action). If so, which algebraic group? | |
| Jul 8, 2015 at 15:31 | comment | added | Will Sawin | Doesn't this suggest that it is not quite right to view the elliptic curve as varying continuously, but rather algebraically? For instance the $p$-torsion in the theory should depend on the elliptic curve's reduction modulo $p$, which does not depend continuously on its geometry. | |
| Jul 8, 2015 at 2:19 | history | edited | Catherine Ray | CC BY-SA 3.0 |
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| S Jul 7, 2015 at 4:46 | history | bounty started | Matt Privman | ||
| S Jul 7, 2015 at 4:46 | history | notice added | Matt Privman | Improve details | |
| Jul 7, 2015 at 4:39 | comment | added | Catherine Ray | @WillSawin I believe so, but it’s entirely obscure to me how to get from the outline of an equivariant derived theory to actually utilizing it in this circumstance. Grojnowski's construction has a very different flavour than Landweber-exact theories. | |
| Jul 7, 2015 at 4:18 | history | edited | Catherine Ray | CC BY-SA 3.0 |
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| Jul 7, 2015 at 4:05 | history | edited | Catherine Ray | CC BY-SA 3.0 |
Returned to the original question.; edited title
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| Jul 3, 2015 at 5:38 | comment | added | მამუკა ჯიბლაძე | Does not the Ginzburg-Kapranov-Vasserot version work everywhere? | |
| Jul 3, 2015 at 5:03 | history | edited | Catherine Ray | CC BY-SA 3.0 |
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| Jul 3, 2015 at 4:57 | history | edited | Catherine Ray | CC BY-SA 3.0 |
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| Jul 3, 2015 at 4:51 | history | edited | Catherine Ray | CC BY-SA 3.0 |
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| Jul 3, 2015 at 4:44 | history | edited | Catherine Ray | CC BY-SA 3.0 |
Moved Author's note.
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| Jun 27, 2015 at 17:46 | history | edited | Catherine Ray | CC BY-SA 3.0 |
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| Jun 27, 2015 at 17:46 | comment | added | Catherine Ray | The case where $G = S^1$ collapses Grojnowski’s construction quite a bit. Here, $G = T, \mathfrak{g} = \mathfrak{t}$, and $Hom(T, S^1) = Hom(S^1, S^1) = \mathbb{Z}$. So $\text{Spec } E_T = \mathbb{Z} \otimes_{\mathbb{Z}} E$, which gives us $\text{Spec } E_T = E$. | |
| Jun 27, 2015 at 17:45 | comment | added | Catherine Ray | @grghxy You're right. It’s not possible for $\text{Spec } E_{S^1}^*(pt)$ to give us the elliptic curve in the usual way (as a ring of global functions), rather, it's my understanding that we get $\text{Spec } E_{S^1}^*(pt) = E \otimes Hom(S^1, T)$, where we are identifying $Hom(S^1, T)$ with the lattices on the dual of the Lie algebra (of E), $S^1$ is the multiplicative circle group, and $T$ is a compact torus. | |
| Jun 27, 2015 at 17:21 | comment | added | Will Sawin | Isn't this what the derived algebraic geometry theory of elliptic cohomology is for? I'm not an expert, but I am talking about this sort of thing: math.harvard.edu/~lurie/papers/survey.pdf | |
| Jun 27, 2015 at 15:13 | comment | added | grghxy | Why do you say "$E$ is disconnected" in positive characteristic? That is false. And at the start you write that an elliptic curve is Spec of a ring, which is also false. | |
| Jun 27, 2015 at 8:03 | history | asked | Catherine Ray | CC BY-SA 3.0 |