Timeline for Is there a "geometric" intuition underlying the notion of normal varieties?
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
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| Dec 28, 2012 at 5:29 | answer | added | P Vanchinathan | timeline score: 17 | |
| Dec 27, 2012 at 15:33 | answer | added | David C | timeline score: 19 | |
| Dec 27, 2012 at 15:03 | answer | added | Francois Ziegler | timeline score: 10 | |
| Oct 15, 2012 at 2:47 | comment | added | roy smith | to Damien, as i should have said, if i understood it well, it seems your remark is exactly on target, i.e. that S2 is precisely equivalent to hartogs' principle by S2 ( by Prop.1.1, Th.3.8 Groth. Local Cohomology). moreover it seems that hartogs holds for functions iff it holds for sections of all reflexive, i.e. dual sheaves. finally, Karl is teaching us that the hartogs'principle does not hold for any non reflexive sheaves, at least on normal varieties. I hope I have this right, but i readily admit i am more geometric than algebraic in my orientation. | |
| Oct 14, 2012 at 3:05 | comment | added | roy smith | indeed thm 2.8 of those notes show every dual is reflexive (with some hypotheses)! | |
| Oct 14, 2012 at 2:44 | comment | added | roy smith | thanks Karl, this algebraic stuff always confuses me. my impression is S2 (i.e. Hartogs, by Prop.1.1, Th.3.8 Groth. Local Cohomology) holds for the dual of any coherent F iff it holds for O(X) (lemma 21, Schlessinger, Rigidity of quotient sings.), e.g. on any normal or any cohen macaulay variety. is this right? moreover on a normal variety it seems this is best possible, i.e. then every S2 sheaf F is reflexive (is this equivalent to being a dual?) and let me reference a beautiful looking source: personal.psu.edu/kes32/Notes/GeneralizedDivisors.pdf | |
| Oct 13, 2012 at 21:51 | comment | added | Karl Schwede | Actually, on a normal variety, or even a variety that is S2 + G1 (Gorenstein in codimension 1), a coherent sheaf is reflexive if and only if it is S2. See the paper of Hartshorne, Generalized divisors on Gorenstein schemes. | |
| Oct 12, 2012 at 20:11 | comment | added | roy smith | normal varieties were introduced by Zariski, in a paper in the Amer. Journal of Math, vol. 61, 1939,p.249ff? but announced in an earlier paper in 1937. | |
| Oct 12, 2012 at 20:04 | comment | added | roy smith | see staff.uni-bayreuth.de/~btm113/AGlect3.pdf more generally, on a cohen macaulay variety, hartogs principle holds for any "reflexive" (e.g. locally free, like O(X),) coherent sheaf. | |
| Oct 12, 2012 at 19:57 | answer | added | Karl Schwede | timeline score: 73 | |
| Oct 12, 2012 at 19:36 | comment | added | roy smith | exactly right Damien, as i understand it. to link with my comment note the inverse of the map identifying two points of a smooth surface is regular except at the singular point, which has codimension two. since it cannot be extended to a full inverse, the singularity obtained by identifying the two points is not normal. the geometric version says just the existence of the finite map identifying the two points, which is not an isomorphism, implies non normality. if you identify two points of a smooth curve, it is not normal because singular in codim one, and again the finite map exists. | |
| Oct 12, 2012 at 19:01 | comment | added | DamienC | This might be a silly comment but it seems to me that normality is a way to get an algebraic version Hartog's principle. | |
| Oct 12, 2012 at 16:57 | history | edited | aglearner | CC BY-SA 3.0 |
The wording is improved slightly
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| Oct 12, 2012 at 4:11 | comment | added | roy smith | plus the existence of normalization. | |
| Oct 12, 2012 at 4:09 | comment | added | roy smith | this is zariski's main theorem i guess. | |
| Oct 11, 2012 at 23:47 | comment | added | aglearner | Dear Roy this is an interesting formulation of been normal! Could you give a reference for this? Do I understand correctly, that this should follow rather directly form the definition of been normal, or this requires some work? | |
| Oct 11, 2012 at 23:19 | comment | added | roy smith | mumford emphasizes that normal implies locally unibranch. i like the criterion that a variety is normal iff any surjective finite birational map onto it is an isomorphism. | |
| Oct 11, 2012 at 22:23 | answer | added | Leo Alonso | timeline score: 26 | |
| Oct 11, 2012 at 19:22 | comment | added | Sam Gunningham | There is a notion of a normal topological pseudomanifold, which means that the link of every point is connected. A version of Zariski's Main Theorem says that normal complex algebraic varieties are also normal in this sense. I don't know if the converse is true in general, but I would hope that under some reasonable hypotheses it should be (e.g. local complete intersection?) See e.g. maths.ed.ac.uk/~aar/papers/gormac2.pdf page 118 | |
| Oct 11, 2012 at 18:54 | comment | added | Thomas Kahle | Maybe you want to check out Sandor Kovacs answer to this question. mathoverflow.net/questions/35736/… | |
| Oct 11, 2012 at 18:45 | comment | added | aglearner | John, thanks! I also see that there are further nice links from Dao's answer, will study them. I don't have any specific example for which to check normality (maybe this is my problem). Maybe it would be sufficient for me to know just how people think about normality. | |
| Oct 11, 2012 at 17:48 | comment | added | J.C. Ottem | I don't know how helpful this is for you purposes, but there is the Serre criterion for normality: That $X$ satisfies your condtion (that is, regular in codimension 1, usually written $R_1$) and the more technical condition $S_2$, which can be difficult to verify. See also Hailong Dao's answer to this question: mathoverflow.net/questions/60097/… | |
| Oct 11, 2012 at 17:29 | history | asked | aglearner | CC BY-SA 3.0 |