Timeline for Doing scheme theory with Hausdorff spaces
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 12, 2019 at 6:54 | comment | added | KConrad | You should have put the motivation for the question in the question so it was clearer where it came from. That professor's statement about "normal people" is simply ignorant. What would the professor think if a student who is used to metric spaces said "normal people don't have a need for topological spaces"? Or you could tell the professor that the word "normal" is really overused in math and we don't need yet another meaning for it. | |
| Apr 11, 2019 at 23:30 | answer | added | Daniel Loughran | timeline score: 1 | |
| Apr 11, 2019 at 21:23 | comment | added | Somatic Custard | In my opinion, the worst part about non-Hausdorff spaces is dealing with people fussing about whether "compact" includes the Hausdorff condition or not. That is, if you are willing to spend a little bit of time acclimating to them. Try starting with finite spaces that are (weakly) homotopy equivalent to CW-complexes, like the pseudocircle. What fun! | |
| Apr 11, 2019 at 21:18 | comment | added | Somatic Custard | Tim Campion's answer reminds me that you may want to check out Stone spaces of complete types. If my memory serves me, there is a situation remotely similar to the following: to a polynomial algebra one can associate a Stone space of complete types X (compact & Hausdorff) and the prime spectrum Y (compact and not Hausdorff), and a natural, continuous map X --> Y which is (almost?) a bijection. Oh, I see that Denis already mentioned and poo-pooed this idea. | |
| Apr 11, 2019 at 18:34 | vote | accept | CommunityBot | ||
| Apr 11, 2019 at 18:33 | comment | added | user137767 | @DenisNardin honestly, it was just a stupid argument with a friend (one professor said in an introductory topology course that "normal people don't have a need for non-Hausdorff spaces"). I wanted to figure out if algebraic geometers count as "normal people". | |
| Apr 11, 2019 at 18:27 | comment | added | Denis Nardin | @StepanBanach Probably the condition you want is "conservative" (or possibly even "monadic" depending on how strict you are on what a forgetful functor is). However it is unclear to me why what you are trying to do is a sensible thing. Spectral spaces arise naturally in a vast array of places, and there's really no reason to try to avoid them. | |
| Apr 11, 2019 at 18:11 | answer | added | Tim Campion | timeline score: 4 | |
| Apr 10, 2019 at 23:00 | comment | added | Achim Krause | The constant functor from schemes to Hausdorff spaces that sends everything to the empty space is full and has a right adjoint which is the constant functor that sends everything to Spec $\mathbb{Z}$. Similarly, the constant functor with value the one-point space is full and has as left adjoint the constant functor with value the empty scheme. So I don't think "full + adjoint" is anywhere close to what you want. | |
| Apr 10, 2019 at 19:49 | history | asked | user137767 | CC BY-SA 4.0 |