Timeline for Normal Varieties
Current License: CC BY-SA 2.5
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 7, 2010 at 15:17 | comment | added | Qfwfq | (...For example the unit ball in $\mathbb{C}^n$ may not be an "analytic variety" in your sense, I suppose; or you just meant "variety" in the usual algebraic sense). | |
| Sep 7, 2010 at 15:13 | comment | added | Qfwfq | (Dear BCnrd: ok, I kind of read your sentence out of context -missing the fact that the hypothesis "constructible" was understood- and was thinking of an open set in the "usual" topology -such as the unit ball in $\mathbb{C}^n$- which is not Zariski open...) | |
| Sep 1, 2010 at 20:51 | answer | added | Dmitri Panov | timeline score: 5 | |
| Aug 30, 2010 at 22:35 | comment | added | BCnrd | Dear unknown (google): I meant it the way I wrote it. In general a constructible subset of a finite type $\mathbf{C}$-scheme (such as a subvariety, which is locally closed) is open for the analytic topology if and only if it is open for the Zariski topology (see SGA1, Exp. XII somewhere early). The question is posed with openness for the analytic topology, and so I was just pointing out that this actually implies openness for the Zariski topology, as I then wanted to work with that point of view (i.e., Zariski-open subscheme, so complement is Zar-closed, hence an analytic set, etc.) | |
| Aug 30, 2010 at 14:17 | comment | added | Qfwfq | (I'm sure BCnrd really meant "analytic" instead of "Zariski", and vice versa, in the sentence "A subvariety that is open in the analytic topology is also open in the Zariski topology"...) | |
| Aug 29, 2010 at 16:36 | comment | added | BCnrd | Donu,I should have remembered argument in Keerthi's answer. Variant works in analytic case. Let $p:E \rightarrow X$ be covering space corresponding to left $\pi_1(X)$-set $S := \pi_1(X)/\pi_1(U)$, so $E$ is conn'd & has unique structure of normal analytic space making $p$ local analytic isom. Then $E|_U$ corresponds to $S$ as left $\pi_1(U)$-set. But $E|_U$ is complement of nowhere-dense analytic set in $E$, By Riemann Ext'n Thm for normal analytic spaces (apply to idempotents on $E|_U$) we see $E|_U$ is conn'd, so $S$ is transitive as left $\pi_1(U)$-set. Orbit of 1 is pt, so $S$ is pt.QED | |
| Aug 29, 2010 at 15:58 | history | edited | Robert Garbary | CC BY-SA 2.5 |
added 39 characters in body
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| Aug 29, 2010 at 15:54 | comment | added | Donu Arapura | BCnrd, if $X$ is a nodal rational curve, and $U$ is the smooth part, $\pi_1(U)=\mathbb{Z}$ maps to $0$, but $\pi_1(X)=\mathbb{Z}$. | |
| Aug 29, 2010 at 15:49 | comment | added | Donu Arapura | I'm assuming this is over $\mathbb{C}$, but it should be stated. Also a more descriptive title would help. | |
| Aug 29, 2010 at 15:49 | answer | added | Keerthi Madapusi | timeline score: 9 | |
| Aug 29, 2010 at 15:49 | comment | added | BCnrd | Is normality really needed? A subvariety that is open in the analytic topology is also open in the Zariski topology, and consequently has complement with "topological" codimension at least 2. But loops are 1-dimensional, so in $X$ they can be deformed to be entirely in $U$. That's a sketch of a proof, ignoring care needed to deal with singularities (in $X$ and $X-U$), so does a problem really arise when the singularities are worse than normal? | |
| Aug 29, 2010 at 15:47 | comment | added | Pete L. Clark | I guess you are working over $\mathbb{C}$? E.g., over $\mathbb{R}$ the inclusion $\mathbb{A}^1 \subset \mathbb{P}^1$ gives a counterexample. | |
| Aug 29, 2010 at 15:39 | comment | added | Donu Arapura | If someone else doesn't give an answer, I'll try to write one later. In outline: reduce it to a local statement (normality enters because the links are connected) then use van Kampen. | |
| Aug 29, 2010 at 15:25 | history | asked | Robert Garbary | CC BY-SA 2.5 |